Transmission Lines Lecture Notes

Renato OrtaLecture Notes on Transmission Line Theory November 2012 DEPARTMENT OF ELECTRONICS AND TELECOMMUNICATIONS POLITECNICO DI TORINO Contents Contents 1 1 Transmission line equations and their solution 4 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 1.2 Electromagnetism background . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Circuit model of a transmission line . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.4 Lossless lines. Wave equations and their solutions . . . . . . . . . . . . . . . . . . . 11 1.5 Review of Fourier transforms and phasors . . . . . . . . . . . . . . . . . . . . . . . 14 1.6 Transmission line equations in the frequency domain . . . . . . . . . . . . . . . . . 16 1.7 Propagation of the electric state and geometrical interpretations . . . . . . . . . . 21 1.8 Solution of transmission line equations by the matrix technique . . . . . . . . . . . 23 2 Parameters of common transmission lines 27 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.2 Coaxial cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2.3 Two-wire line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29 2.4 Wire on a metal plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 2.5 Shielded two-wire line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.6 Stripline . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31 2.7 Microstrip . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3 Lossless transmission line circuits 38 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.2 Definition of local impedance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 3.3 Reflection coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.4 Energy balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46 3.5 Line voltage, current and impedance diagrams . . . . . . . . . . . . . . . . . . . . . 47 3.6 The Smith Chart . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.7 Analysis of simple circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 1 CONTENTS 4 Energy dissipation in transmission lines 61 4.1 Dielectric losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 4.2 Conductor losses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 4.3 Loss parameters of some transmission lines . . . . . . . . . . . . . . . . . . . . . . 68 4.3.1 Coaxial cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 4.3.2 Two-wire line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 5 Lossy transmission line circuits 72 5.1 Solution of transmission line equations . . . . . . . . . . . . . . . . . . . . . . . . . 72 5.2 Computation of the power flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79 5.3 Frequency dependence of phase constant and characteristic impedance . . . . . . . 80 6 Matching circuits 84 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 6.2 Types of impedance matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 6.3 Impedance matching devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87 6.3.1 L cells with lumped reactive elements . . . . . . . . . . . . . . . . . . . . . 87 6.3.2 Resistive matching pad . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 6.3.3 Single stub matching network . . . . . . . . . . . . . . . . . . . . . . . . . . 91 6.3.4 Double stub matching network . . . . . . . . . . . . . . . . . . . . . . . . . 96 6.3.5 λ/4 matching networks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 7 The Scattering matrix 101 7.1 Lumped circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 7.2 Distributed parameter circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 7.3 Relationship between [S] and [Z] or [Y ] . . . . . . . . . . . . . . . . . . . . . . . . 104 7.4 Computation of the power dissipated in a device . . . . . . . . . . . . . . . . . . . 105 7.5 Properties of the scattering matrix [S] of a device . . . . . . . . . . . . . . . . . . . 106 7.6 Change of reference impedances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 7.7 Change of reference planes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 7.8 Cascade connection of structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 7.9 Scattering matrix of some devices . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.9.1 Ideal attenuator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.9.2 Isolator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.9.3 Circulator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 7.9.4 Ideal directional coupler . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 7.10 Examples of analysis of structures described by S matrices 2 . . . . . . . . . . . . . 114 . 140 Bibliography 141 3 . . . . . . .10. . . . . . . 127 8. . . . . . . . . . . . .2 Interconnection of two two-ports by means of a length of transmission line . . . . . . . . . . . . . . . . . . 115 7. . 132 8. .3 Change of reference impedance for a one-port load . . .1 General solution of transmission line equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1 Cascade connection of a two-port and a load . . . . . . . . . . . . . . . . . .3 Real interconnections . . . . . . . . . . . . . . . 131 8. . . . . 131 8. . . .2 The group velocity . . . . . . . . . . . . . . . . 118 8 Time domain analysis of transmission lines 122 8. . .5 Mismatched ideal transmission lines . . . . . . . . . . . . . . . 130 8. . . . . . . . . . . .5. . . .CONTENTS 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4 Digital communication . . . . . . . . . .2 Mismatched ideal lines . . . . . . . . . . . . . . .11 Transmission matrix . . . . . 117 7. . .5. . . . . .10. . . . .10. . . . . .1 Introduction . 122 8. . . . . . . . . . . . 123 8. . . . . . . . . . . 116 7. . . . . . . . . . . . . . . .3 Distortions . . . . . . . . . . . . . . . . . . . . . . .5. . . . . . but those of the first type are characterized by the fact that their fundamental propagation mode is TEM (transverse electromagnetic) . is related to the TEM mode only. This implies that they can be used also at very low frequency .1 are waveguides.even at dc . as in a broadcasting system or in a cell phone network. In conclusion. 1. Hollow metal pipes.1.since they consist of two conductors. 1. which depends on their transverse size. whereas the others are more appropriately called metal or dielectric waveguides. In this text we will deal only with structures consisting of two metal conductors. Twisted pairs and coaxial cables are used for cabling a building but coaxial cables can also be used for intercontinental communications. In telecommunications this behavior can be useful when the user position is not known in advance. The various line types are used for different applications in specific frequency ranges.irrespective of their size. has a natural tendency to spread in the whole space at a speed close to 300. irrespective (at least with a good approximation) of the bends that the line undergoes because of installation needs. Striplines and microstrips are used only inside devices.1 Introduction Electromagnetic energy. all the structures of Fig. such as coaxial cables. as in the case of optical fibers.Chapter 1 Transmission line equations and their solution 1. In other applications. are used to deliver large amounts of microwave power over short to moderate distance.or quasi-TEM in the case of microstrips . These can be defined transmission lines in strict sense.000 Km/s. a one dimensional propagation phenomenon takes place on a transmission line. transmission lines are waveguides whose behaviour. instead. in general. have a lowest frequency of operation. a transmission line is a system of metal conductors and/or dielectric insulating media that is capable of “guiding” the energy transfer between a generator and a load. 4 . In the most general terms. once generated in one place. More rigorously. and their lengths never exceeds some centimeters. at sufficiently low frequency. From this point of view. Waveguides can also be made of dielectric materials only. microstrips and striplines. known as waveguides. such as amplifiers or filters. some examples of which are shown in Fig. electromagnetic energy must be transferred from one place to the other along a well defined path without any spreading at all: an example is the cabling of a building. Waveguides. There are many types of transmission lines. t) magnetic induction Wb/m2 J (r. that describe the link 5 . Examples of transmission lines: (a) coaxial cable.t)) are specified by Maxwell equations. E(r.t) electric induction C/m2 B(r. 2012) n3 n2 n1 b a c d e Figure 1. (d) microstrip . (e) stripline.t) = × H(r.t) magnetic field A/m D(r.Renato Orta .t) and the magnetic induction B(r. (b) two wire line.t) + J c (r.t) electric field V/m H(r.t). The relationships between these fields and the sources (described by the current density J (r.t). the electric displacement (or electric induction) D(r.t) (conduction) current density [A/m2 ] These equations must be supplemented with the constitutive relations.t) + J (r. 1.t) = − ∂ B(r.t) current density (source) A/m2 J c (r. they are completely described by four vector fields: the electric field E(r.1) ∂ D(r.t) ∂t (1.t) ∂t A general reference for electromagnetism is [1].t). the magnetic field H(r. Let us review the meaning of the symbols and the relevant measurement units. from a quantitative point of view. (c) optical fiber.Transmission Line Theory (Nov.1. that are written in MKSA units as × E(r.2 Electromagnetism background The physical phenomena that take place in a transmission line belong to the realm of electromagnetism and hence. t) where µ = = µ0 µr 0 r and µr . It is known from Mathematics that a field with “arbitrary” time dependence can be represented as a summation of sinusoidal fields with frequencies contained in a certain band (Fourier theorem). the time harmonic (sinusoidal) regime with frequency f is very important. All non ferromagnetic materials have values of µr very close to 1. isotropic. the ratio between the magnitudes of the electric and magnetic fields is called wave impedance and has the value Z0 = µ0 ≈ 120π ≈ 377 Ω 0 In the case of linear.t) (1. Moreover.t): J c (r. the presence of an electric field E(r. In this case λ0 denotes the minimum wavelength.t) = γ E(r. and µ0 magnetic permeability.Transmission Line Theory (Nov.Renato Orta . i. Even if an electromagnetic field can have an arbitrary time dependance.t) (1. the constitutive relations (1. essentially three regimes can be identified: 6 . non dispersive dielectrics. The simplest case is that of free space in which B(r. The size L of the structures with which the electromagnetic field interacts must always be compared with wavelength. measured in S/m.t) = where 0.t) = E(r. the one that corresponds to the maximum frequency. both from a theoretical and from an application point of view. which is a sort of characteristic length of the field spatial structure. electromagnetic waves are characterized by a spatial period λ0 = c/f .2) D(r.t) dielectric permittivity. have the values µ0 = 0 = 4π · 10−7 H/m 1 1 ≈ · 10−9 2 µ0 c 36π F/m where the speed of light in free space c has the value c = 2.99792458 · 108 m/s. 2012) between fields and inductions. in the case of a plane wave.t) where γ is the conductivity of the dielectric.e.t) = µ H(r.t) = µ0 H(r.3) D(r. In these conditions.t) gives rise to a conduction current density J c (r. The ratio L/λ0 is defined electrical length of the structure and is a pure number. Depending on the value of L/λ0 . r (pure numbers) are the relative permittivity and permeabilities. 0 E(r.2) are substituted by B(r. When the dielectric contains free charges. called wavelength. t) and i(z. that are applicable only to lumped parameter 7 . 1. For this reason one says that a lumped parameter circuit operates in quasi-static regime. with L/λ0 ∼ 1. with a transverse cross section that is independent of the longitudinal coordinate z. these quantities are defined uniquely only in static conditions. 2012) • quasi-static regime. mirrors. but they are commonly used also in the frequency band for which the electrical size of the network is very small.3 Circuit model of a transmission line Consider a length of uniform transmission line.. capacitors. analyzed in this text • the optical regime. i. but for teaching convenience we will proceed in circuit terms.e.e. their transverse size is small with respect to wavelength but their length can be very large. an electromagnetic system can be considered lumped provided the propagation delay is negligible with respect to the period of the oscillations.g. Typically. but also in the space surrounding the conductors of a line.. Indeed L L1 τ L = = = λ0 c/f cT T where T is the period of an oscillation with frequency f = 1/T and τ is the time that an electromagnetic wave requires to go from one end of the network to the other. 1. Consider now one of the transmission lines shown in Fig.1. Hence. The equations that determine the dynamics of a transmission line could be obtained directly from Maxwell equations. etc. 1. hence its behavior cannot be predicted by Kirchhoff laws.) The solution technique of electromagnetic problems and even their modeling is different depending on the regime of operation. In Fig. by generalizing the properties of lumped parameters networks.2a a coaxial cable is shown as an example. its schematic and conventional representation in the form of two parallel “wires” in which a current flows and between which a potential difference exists. Lumped parameter circuit theory deals with the dynamics of systems made of elements of negligible electrical size. typical of the usual optical components studied by classical optics (lenses. A circuit containing transmission lines is often called “distributed parameter circuit”to underline the fact that electromagnetic energy is not only stored in specific components. 1.2b shows its symbol. The state variables employed in the model are the potential difference vrs (t) between two nodes Pr and Ps of a network and the electric current irs (t) that flows in the branch defined by the same two nodes. The state variables of such a system are then v(z. with L/λ0 1.Renato Orta . typical of distributed parameter circuits.Transmission Line Theory (Nov. Then. with L/λ0 1.e. i. a transmission line is characterized by inductance and capacitance per unit length. a transmission line can be long with respect to wavelength. while a lumped parameter circuit is modeled as point like. This condition can be reformulated in terms of transit time. As previously remarked. inductors. It is evident that all two conductor transmission lines have the same circuit symbol shown in Fig. at DC. a transmission line is a one dimensional system.2b. typical of lumped parameter circuits • the resonance regime . e. As a consequence. Fig.t). Rigorously. 1. i. in which voltage and currents depend on time and on a longitudinal coordinate that will always be indicated with z. 2012) (a) Figure 1.t) (b) Figure 1. the equivalent circuit contains a series resistance with value R∆z.3. Analogously. a) Element ∆z of a coaxial cable.t) /∆z 5∆z &∆z v(z. measured in H/m is the inductance per unit length of the line. This field gives rise to a linked flux through the rectangle shown in Fig. so that we will be able to derive a set of partial differential equations that can be solved in closed form. To this end we will let ∆z go to zero. The proportionality constant that relates the charge on 8 .t) *∆z v(z+∆z. (b) (a) Length of coaxial cable and (b) its symbolic representation circuits. The surface used to define L is shown dashed.Transmission Line Theory (Nov. power is dissipated in the metal conductors because of their limited conductivity: hence. Hence. Moreover. 1. Fig.3 shows an element of the line with its equivalent circuit. 1. However we can subdivide the line in a large number of sufficiently short elements ∆z λ. b) Equivalent circuit of the element we use physical arguments. We start by observing that the current flowing in the conductors produces a magnetic field with force lines surrounding the conductors.t) (a) i(z+∆z. L. This is actually the modeling technique used in some circuit simulators. We will instead follow a different route because we are interested in an analytical solution of the problem.3. as a consequence of the potential difference maintained between the inner and outer conductors. the inductance of the element that we can write as L∆z because the surface of the rectangle is clearly proportional to ∆z. but for the other transmission lines one can proceed similarly . To obtain the equivalent circuit ∆z i(z.2. we make reference to the coaxial cable.Renato Orta . where R is the resistance per unit length of the line. a charge is induced on them. by definition. derive a lumped equivalent circuit for each of them and then analyze the resulting structure by the usual methods of circuit theory. The proportionality factor relating the flux to the current is. expressed in Ω/m. 2012) the ∆z element to the potential difference is. The incremental ratios in the left hand side become partial derivatives with respect to z and.t) ∀t ≥ 0 where e(t) is a given causal function.5)   ∂ ∂   − i(z. coupled.0) i0 (z) 0≤z≤L = 9 .6) RL i(L. both the load impedance and the internal impedance of the generator have been assumed real. 1.4. the initial condition that specify the initial state of the reactive components (only of the line. the capacitance of the element.t) + C v(z. where G is the conductance per unit length of the line.t) + L ∂ i(z.t) − v(z + ∆z. by definition. 1. From a circuit point of view.Renato Orta . It is clear that the boundary conditions to be associated to (1.t). a line connects a generator to a load. for simplicity.t) (1. 1.t) − i(z + ∆z. where. in this case) is v(z. such as those of Fig.t)    ∂t     i(z.t) = G v(z.t) + L ∆z ∂ i(z. which is responsible of a current flowing from one conductor to the other through the insulator. as sketched in Fig. Moreover.4) ∂ G ∆z v(z + ∆z.t) ∂t = Next divide both sides by ∆z and take the limit for ∆z → 0.3b:   v(z. Figure 1. partial differential equations. Heaviside 1880):   − ∂ v(z. Usually.0) = v0 (z) 0≤z≤L i(z. we obtain the transmission line equations (Telegrapher’s equations.5. This is the simplest circuit comprising a transmission line.4.t) = In z = L v(L.t) ∀t ≥ 0 (1. measured in S/m.t) = v(0.t)   ∂z  ∂t (1.5) are a system of first order. that we write C∆z. measured in F/m.t) ∂z ∂t It is to be remarked that any other disposition of the circuit elements. where C is the capacitance per unit length of the line. Equations (1. Alternative equivalent circuits of an element of transmission line. this phenomenon is accounted for by the conductance G∆z. that must completed with boundary and initial conditions. noting the continuity of v(z.Transmission Line Theory (Nov. Since ∆z λ Kirchhoff laws can be applied to the circuit of Fig.5) are: In z = 0 e(t) − Rg i(0.t) + C ∆z v(z + ∆z. leads exactly to the same differential equations.t) = R i(z.t) = R ∆z i(z. the dielectric between the conductors has a non zero conductivity. Finally. 6. In problems of electromagnetic compatibility one studies the effect of a wave that impinges on the transmission 10 . i(L. For example. i.t) + i(L.t) L C vC(L.6. at t = 0 the line is at rest and. but is formulated as an ordinary differential equation of the type D( d d ) v(L.t) dt dt (1. the first is nonhomogeneous. if the load network is that of Fig. we can say that the system is excited via the boundary condition in z = 0. without forcing term. not always is a transmission line excited only at its ends. consisting of a series connection of a resistor R. eq.7) takes the form: d d d2 1 v(L. Concerning the boundary conditions (1. Load network with reactive components. the second is homogeneous.t) R v(L.5. L Fundamental circuit comprising a generator and a load connected by a transmission line.t) dt dt dt C The initial conditions to be specified are vc (0) and i(0). 2012) Rg + e(t) RL 0 Figure 1.Renato Orta . (1.t) = R i(L.6). the boundary condition is not of algebraic type. where v0 (z) e i0 (z) are known (real) functions. In the case the line is initially at rest.e. D and N are two formal polynomials in the operator d/dt. 1. Typically.5) is a system of homogeneous equations.7) to be completed with the initial conditions for the reactive components of the load network. hence.t) Figure 1.t) = N ( ) i(L. a capacitor C and an inductor L. In the applications. In the case the load network contains reactive elements.Transmission Line Theory (Nov. v0 (z) ≡ 0 e i0 (z) ≡ 0 0 ≤ z ≤ L We observe that (1. which express the voltage across the capacitor and the current in the inductor at the time t = 0.t) + L 2 i(L. o /∆z 5∆z ] v ++ o i &∆z ] *∆z Figure 1.t)+ i (z.t) + C v(z. (1. Wave equations and their solutions A transmission line is called ideal when the ohmic losses in the conductors and in the insulators can be neglected.t) describe source terms and therefore are to be considered as known. 2012) line: the phenomenon is not a point-like excitation and can be modeled by means of a set of voltage and current generators “distributed” along the the line with a density per unit length ◦ ◦ v (z.t) ∂z = R i(z. without sources. Other common names are free evolutions.Renato Orta . Eq.t). The line equations. (1.4 Lossless lines. called forward wave and backward wave. We will find that the general solution is the linear combination of two normal modes of the system.7 and correspondingly eq.7.t) ∂t (1. proper evolutions.8) ◦ ∂ G v(z. Differentiate the first equation with respect to z and the 11 .5) become   − ∂ v(z.8) define a non-homogeneous problem.t) =   ∂z      ◦ ∂ − i(z. In this case the equivalent circuit of a line element has the form shown in Fig.t) ∂t ◦ The functions v (z. 1. It is well known that the general solution of a linear non-homogeneous differential equation is given by the sum of a particular solution of the non-homogeneous equation and the general solution of the associated homogeneous equation.t) alone. We are going to focus first on the homogeneous equation.t)+ v (z. become in this case   ∂v + L ∂i    ∂z ∂t =   ∂i ∂v   +C ∂z ∂t = 0 (1.t) and i (z. 1. Equivalent circuit of a line element ∆z when distributed generators are present on the transmission line.t) + L ∂ ◦ i(z.9) 0 From this system of first order partial differential equations we can obtain a single second order equation for the voltage v(z.t) e i (z. resonant solutions. since they contain a forcing term.Transmission Line Theory (Nov. in order to obtain the current i(z. Recall in fact that on a transmission line.t) obeys a wave equation identical to (1.Renato Orta . Define the new independent variables ξ = z − vph t.10) can be solved by a change of variable technique. 2 t= 1 (η − ξ).Transmission Line Theory (Nov.9) with respect to t and the second with respect to z. i(z. Obviously one of the two (1. 2vph Now rewrite the wave equation in the new variables. We need the chain rule of multivariable calculus.t) and we obtain ∂2v ∂2v − LC 2 = 0 ∂z 2 ∂t This equation is known as wave equation (in one dimension) because its solutions (obtained by √ d’Alembert in 1747) are waves propagating along the line with speed ±vph = ±1/ LC. To obtain it.0) = v0 (z). differentiate the first of (1. 2012) second with respect to t:  2 2  ∂ v +L ∂ i   ∂z 2  ∂z ∂t = 0     ∂2i ∂2v +C 2 = 0 ∂t ∂z ∂t The two mixed derivatives are equal under the usual regularity conditions for i(z. Observe that also the current i(z.t).0) = i0 (z) (1. voltage and current are inextricably linked. with the initial conditions v(z. The wave equation for an infinitely long ideal transmission line. the wave equation in the new variables becomes ∂2v =0 ∂ξ∂η that is ∂ ∂η ∂v ∂ξ 12 =0 .9 ) must be associated to (1. η = z + vph t The old variables are expressed in terms of the new ones as z= 1 (ξ + η).10). ∂v ∂v ∂ξ ∂v ∂η ∂v ∂v = + = + ∂z ∂ξ ∂z ∂η ∂z ∂ξ ∂η ∂v ∂v ∂ξ ∂v ∂η = + = −vph ∂t ∂ξ ∂t ∂η ∂t and also ∂2v ∂ = ∂z 2 ∂ξ ∂ ∂2v = vph ∂t2 ∂η ∂v ∂v + ∂ξ ∂η ∂v ∂v − ∂η ∂ξ + vph − ∂ ∂η ∂ ∂ξ ∂v ∂v + ∂ξ ∂η ∂v ∂v − ∂η ∂ξ ∂v ∂v − ∂ξ ∂η = ∂2v ∂2v ∂2v +2 + 2 ∂ξ 2 ∂ξ∂η ∂η 2 vph = vph ∂2v ∂2v ∂2v −2 + 2 2 ∂η ∂ξ∂η ∂ξ Using these two last expressions.10). an arbitrary function ξ. C/L is called characteristic admittance of the line and is measured in In conclusion. we get v(ξ. We have introduced the symbol f1 (ξ) to denote the integral of the arbitrary function f (ξ). Rewrite the previous equation as v(ξ.Renato Orta . the general solution of the transmission line equations can be written as v(z.t) = − − 1 vph v + (ξ)dξ + v − (z + vph t)dt 1 vph v − (η)dη = Y∞ {v + (z − vph t) − v − (z + vph t)} were the quantity Y∞ = Siemens. yield v0 (z) = v + (z) + v − (z) i0 (z) = Y∞ v + (z) − Y∞ v − (z). Returning to the original variables. consider (1. we must obtain the functions v + (ξ) and v − (η) in such a way that the initial conditions (1.t) = v + (z − vph t) + v − (z + vph t) i(z. (1. we get v(z. ∂z ∂v = v + (z − vph t) + v − (z + vph t) ∂z and 1 L 1 = − L v + (z − vph t)dt + i(z.η) = f1 (ξ) + f2 (η) This is the general solution of the wave equation.t) = v + (z − vph t) + v − (z + vph t) (1.12). (1.t) = − From (1.η) = f (ξ)dξ + f2 (η) where f2 is an arbitrary function of η.11) we compute 1 L ∂ v(z.t)dt. eq. 2012) whose solution is ∂v = f (ξ) ∂ξ where f is a constant with respect to η. To derive the expression of the current.10) are satisfied. Now.e.Transmission Line Theory (Nov. S.9) from which ∂i 1 ∂v =− ∂t L ∂z that is i(z. 13 .11) where the more appropriate symbols v + e v − have been introduced in place of f1 e f2 .12) = To complete the solution of the initial value problem. written for t = 0. i. By integrating the previous equation.t) Y∞ v + (z − vph t) − Y∞ v − (z + vph t). or spectrum. 2 2 Y∞ 1 [v0 (z − vph t) + v0 (z + vph t)] + [i0 (z − vph t) − i0 (z + vph t)] . based on the use of Fourier transforms. It is possible also to employ another method. Note that the electric state on the line depends on z e t only through the combinations t − z/vph e t + z/vph : this is the only constraint enforced by the wave equation. i(z.Transmission Line Theory (Nov. 2 v − (z) = 1 [v0 (z) − Z∞ i0 (z)].Renato Orta . The arbitrariness is removed when a particular solution is constructed. The solution for t > 0 is obtained by substituting the argument z with z − vph t in v + and z + vph t in v − . obtained for the first time by d’Alembert. ∞ | f (t) | dt < ∞ −∞ the spectral representation exists: f (t) = 1 2π ∞ F (ω) ejωt dω (1. 1. defined by ∞ F (ω) = f (t) e−jωt dt = F{f (t)} −∞ 14 (1.t) = v(z.14) .t) = 1 1 [v0 (z − vph t) + Z∞ i0 (z − vph t)] + [v0 (z + vph t) − Z∞ i0 (z + vph t)] . as it follows from (1. we find v + (z) = 1 [v0 (z) + Z∞ i0 (z)]. The solution method just presented is the classical one. these equations can be rewritten i(z. 2 In this way the functions v + e v − are determined. 2 2 one can immediately verify that these expression satisfy the initial conditions. i. 2 2 Alternatively.t) = Z∞ 1 [v0 (z − vph t) + v0 (z + vph t)] + [i0 (z − vph t) − i0 (z + vph t)] .12): v(z.t) = Recall that the general solution of an ordinary differential equation contains arbitrary constants. which satisfies initial/boundary conditions. 2 2 Y∞ Y∞ [v0 (z − vph t) + Z∞ i0 (z − vph t)] + [v0 (z + vph t) − Z∞ i0 (z + vph t)] . of f (t). This is the only possible one in the case of finite length lossy lines and will be presented now after a short review of phasors and Fourier transforms. whereas a partial differential equation contains arbitrary functions.e.13) −∞ where F (ω) is the Fourier transform. 2012) Solving by sum and difference.5 Review of Fourier transforms and phasors It is known that for every absolutely integrable function of time f (t). Renato Orta - Transmission Line Theory (Nov. 2012) The meaning of (1.13) is that the function f (t) can be represented as a (continuous) sum of sinusoidal functions, each one with (infinitesimal) amplitude F (ω) dω. This representation underlines the importance of sinusoidal functions in the analysis of linear systems. A very useful property of Fourier transforms is the following: df dt F = jω F{f (t)} = jω F (ω) (1.15) In other words, there is a one-to-one correspondence between the derivative operator in time domain and the multiplication by jω in the frequency domain. Even if the Fourier transform is defined for complex time functions, provided they satisfy (1.13), the physical quantities such as voltage and current are real functions. This implies that the following relation holds: F (−ω) = F ∗ (ω) (1.16) i.e. the spectrum of a real function is complex hermitian; the part of spectrum corresponding to the negative frequencies does not add information to that associated with the positive frequencies. In the applications, very often signals are sinusoidal (i.e. harmonic), that is of the type f (t) = F0 cos(ω0 t + φ) (1.17) Let us compute the spectrum of this signal by means of (1.14); by Euler’s formula ∞ F (ω) = F0 cos(ω0 t + φ) e−jωt dt = −∞ = = ∞ F0 2 ej(ω0 t+φ) e−jωt dt + −∞ F0 2 ∞ e−j(ω0 t+φ) e−jωt dt = −∞ πF0 ejφ δ(ω − ω0 ) + πF0 e−jφ δ(ω + ω0 ) (1.18) This spectrum consists of two “lines” (Dirac δ functions) at the frequencies ±ω0 , so that the signal (1.17) is also called monochromatic. F(ω) -ω0 Figure 1.8. ω0 ω Spectrum of a sinusoidal signal. Let us now proceed in the opposite direction and derive the time domain signal from its spectrum (1.18) through the inverse transform formula (1.13): f (t) = = F0 2 ejφ ∞ δ(ω − ω0 ) ejωt dω + e−jφ ∞ δ(ω + ω0 ) ejωt dω = −∞ −∞ F0 jφ jω0 t e e + e−jφ ejω0 t = 2 = Re F0 ejφ ejω0 t (1.19) 15 Renato Orta - Transmission Line Theory (Nov. 2012) The quantity F = F0 exp(jφ) is generally called phasor of the harmonic signal f (t) and coincides, apart from the factor π, with the coefficient of the Dirac δ function with support in ω = ω0 . Moreover, eq. (1.19) can be defined as the inverse transform formula for phasors. Observe further that, calling Ph the one-to-one correspondence that associates a time-harmonic signal to its phasor, F = Ph{f (t)} the following property holds Ph df dt = jω0 F This equation is formally identical to (1.15); Note, however, that ω denotes a generic angular frequency, whereas ω0 is the specific angular frequency of the harmonic signal under consideration. Because of the very close connection between phasors and Fourier transforms, we can say that any equation in the ω domain can be interpreted both as an equation between transforms and as an equation between phasors and this justifies the use of the same symbol F for the two concepts. It is important to remember, however, that phasors and transforms have different physical dimensions: • phasors have the same dimensions as the corresponding time harmonic quantity • transforms are spectral densities. For example, the phasor of a voltage is measured in V, whereas its transform is measured in V/Hz. This is obvious if we consuider eq. (1.18) and note the well known property ∞ δ(ω) dω = 1 −∞ which implies that the Dirac function δ(ω) has dimensions Hz−1 . 1.6 Transmission line equations in the frequency domain Let us apply now these concepts to the ideal transmission line equations, that we rewrite here for convenience:   ∂v + L ∂i = 0    ∂z ∂t   ∂i ∂v   +C = 0 ∂z ∂t Take the Fourier transforms of both sides, observing that z is to be considered as a parameter in this operation:   − d V (z,ω) = jω L I(z,ω)   dz  (1.20)   d   − I(z,ω) = jω C V (z,ω) dz where V (z,ω) = F{v(z,t)} and I(z,ω) = F{i(z,t)} are the Fourier transform of voltage and current. Note that the transmission line equations have become ordinary differential equations 16 Renato Orta - Transmission Line Theory (Nov. 2012) in the spectral domain. Moreover, the spectral components of voltage and current at different frequencies are uncoupled, as it is obvious since transmission lines are a linear time-invariant (LTI) system. Proceeding in a similar way on the wave equation (1.10), we obtain d2 V (z,ω) + k 2 V (z,ω) = 0 dz 2 and d2 I(z,ω) + k 2 I(z,ω) = 0 dz 2 √ where the quantity k = ω LC, with the dimensions of the inverse of a length, has been introduced. These equations can be called Helmholtz equations in one dimension. Their counterpart in two or three dimensions are very important for the study of waveguides and resonators. These equations have constant coefficients (because of the assumed uniformity of the transmission line) and their general solution is a linear combination of two independent solutions. As such one could choose sin kz and cos kz but exp(+jkz) and exp(−jkz) have a nicer interpretation. Hence, we can write V (z,ω) = V0+ (ω) e−jkz + V0− (ω) e+jkz (1.21) I(z,ω) = + − I0 (ω) e−jkz + I0 (ω) e+jkz ± where V0± (ω) and I0 (ω) are arbitrary constants with respect to z (but dependent on ω, of course, which is a parameter). We must remember, however, that the transmission line equations are a 2×2 first order system (see eq. (1.9)) and hence, its solution contains only two arbitrary constants. ± Then, between V0± (ω) and I0 (ω) two relations must exist, which we can find by obtaining I(z,ω) from the first of (1.20) by substituting (1.21): 1 jωL I(z) = − dV dz = (1.22) 1 jkV0+ e−jkz − jkV0− (ω) e+jkz jωL = Note that √ k ω LC C 1 = = = Y∞ = ωL ωL L Z∞ where we have introduced the characteristic admittance and characteristic impedance of the line. The characteristic impedance is denoted by the symbol Z∞ since it coincides with the input impedance of a semi-infinite line, as it will be shown in section 5.1. Eq. (1.22) can be rewritten as I(z,ω) = Y∞ V0+ (ω) e−jkz − Y∞ V0− (ω) e+jkz From the comparison between this equation and the second one of (1.21), it follows + I0 (ω) = Y∞ V0+ (ω) − e I0 (ω) = −Y∞ V0− (ω) which are the desired relations. In conclusion, the general solution of transmission line equations in the spectral domain are V (z,ω) = V0+ (ω) e−jkz + V0− (ω) e+jkz I(z,ω) Y∞ V0+ (ω) e−jkz − Y∞ V0− (ω) e+jkz (1.23) = 17 Note that also the first term of the expression of the current describes a forward wave: in particular. 1.9. The two curves are obviously periodic and we can define two periods: • the temporal period T = 2π/ω0 is the time interval during which the wave phase changes of 2π radians (note that ω0 is the time rate of change of the wave phase) 18 . so that the propagation mode is rigorously TEM.Renato Orta .t) (1. it can be shown that c vph = √ r and. It is to be remarked that when the dielectric is homogeneous.25) = Y∞ v + (z. LC = r c2 Consider now the plots of Fig. By enforcing its differential to be zero d (ω0 t − k0 z + arg(V0+ )) = ω0 dt − k0 dz = 0 we find the condition that must be satisfied: dz ω0 ω0 1 = = √ =√ = vph dt k0 ω0 LC LC Hence we say that the first term of (1.t) i(z.Transmission Line Theory (Nov. It is clear that the value of the cosine function is constant if the argument is constant. 1.t) = = − Y∞ | V0+ | cos(ω0 t − k0 z + arg(V0+ )) + Y∞ | V0− | cos(ω0 t + k0 z + arg(V0− )) (1.26) √ where k0 = ω0 LC. The propagation velocity of a wave (phase velocity) can be defined as the velocity an observer must have in order to see the wave phase unchanging. in which only one spectral component at ω0 is present. We can use the inverse transform rule of phasors f (t) = Re{F ejω0 t } (1.t) + v − (z. The first (a) shows the time evolution of the forward voltage in a specific point of the line z = z0 .t) = = | V0+ | cos(ω0 t − k0 z + arg(V0+ )) + + | V0− | cos(ω0 t + k0 z + arg(V0− )) v(z.24) so that we obtain: = v + (z. so that the signals are monochromatic. as a consequence.t).10. Consider the first term of the expression of v(z. called wave. sketched in Fig. 2012) To understand fully the meaning of these two equations. The second (b) shows the distribution of the forward voltage on the line at a specific time instant t = t0 . Consider first the simplest case.t) − Y∞ v − (z.25) represents a forward wave because it moves with positive √ phase velocity equal top 1/ LC. It is a function of z and of t. the current is proportional to the voltage via the characteristic admittance. it is necessary to transform them back to time domain. plotted in Fig.Renato Orta . • the spatial period or wavelength λ = 2π/k0 is the distance over which the wave phase changes by 2π radians (note that k0 is the space rate of change of the wave phase) From this definition and from that of k0 we find at once fλ = ω0 2π ω0 1 = vph = =√ 2π k0 k0 LC and also T vph = λ: in other words. with similar argument as above. 1. that it describes a backward wave. Consider now the second term of the expression of the voltage (1. (b) a backward wave and (c) a stationary wave on a short circuited transmission line. 2012) Figure 1. a wave moves over the distance of a wavelength during the time interval of a temporal period.25). Tree dimensional representation of (a) a forward wave.9 the straight lines z = vph t are clearly recognizable as the direction of the wave crests . We find immediately. In the spacetime plot of Fig.Transmission Line Theory (Nov. moving with negative phase velocity 1 ω0 = −√ vph = − k0 LC 19 . 1.9b.9. with | V − |=| V + | can be rewritten in factorized form: 1 1 v(z. the wave crests are aligned on the straight lines z = −vph t. In conclusion.10.t). t ) V0+ + v ( z. Each wave is made of voltage and current that. is improper since a wave is always travelling at the phase speed. we find again the result of Section 1. 2012) 2 v ( z0 . even at intuitive level. This definition. in a certain sense. It is important to observe that the two waves are absolutely identical since the transmission line is uniform and hence is reflection symmetric. even if ordinarily used. In any case. The proportionality between voltage and current of the same wave (called impedance relationship) + I0 (ω) = Y∞ V0+ (ω) − e I0 (ω) = −Y∞ V0− (ω) is only apparently different in the two cases. Forward and backward waves on the line are the two normal modes of the system. the current is proportional to the voltage via the factor −Y∞ . The minus sign in the impedance relation for the backward wave arises because the positive current convention of the forward wave is used also for the backward one. (1. When on a transmission line both the forward and the backward wave are present with the same amplitude.28) 2 2 i.9a e b suggest. Actually. we say that a (strictly) stationary wave is present. the name given to the phenomenon is related to the fact that eq. this 20 .Renato Orta . an idea of movement. 1.e. (a) Time evolution of the forward wave in a fixed point of the line and (b) distribution of the forward voltage on the line at a specific time instant. what is referred to by the term stationary wave is the interference pattern of two waves. t 0 ) 1 V0+ 0 -1 0 λ 2 T + 1 0 5 10 (a) 15 -1 0 t 5 10 (b) 15 z Figure 1.27) 2 2 and 1 1 i(z.Transmission Line Theory (Nov.4: the general solution of the transmission line equations is expressed as linear combination of two waves. Also in this case.25). 1. Fig.9c shows a spacetime plot of v(z. whereas they are in general coupled by the boundary conditions (generator and load) if the line has finite length. moving in the opposite direction. Moreover.t) = 2Y∞ | V0+ | sin[ω0 t + (arg(V0+ ) + arg(V0− ))] · sin[k0 z − (arg(V0+ ) − arg(V0− ))] (1. a forward one propagating in the direction of increasing z and a backward one. as a product of a function of z and of a function of t. are the two sides of a same coin. They are independent (uncoupled) if the line is of infinite length. Whereas Figs.t) = 2 | V0+ | cos[ω0 t + (arg(V0+ ) + arg(V0− ))] · cos[k0 z − (arg(V0+ ) − arg(V0− ))] (1. Renato Orta .7 Propagation of the electric state and geometrical interpretations We have obtained the general solution of the transmission line equations in the form V (z) = V0+ e−jkz + V0− e+jkz I(z) Y∞ V0+ e−jkz − Y∞ V0− e+jkz = (1. in particular.29) in vector form: V (z) I(z) = V0+ 1 Y∞ e−jkz + V0− 21 1 −Y∞ e+jkz (1. Equations (1. Further considerations will be made in Section 3. In the light of these considerations. V (z) = V0 cos kz − jZ∞ I0 sin kz I(z) I0 cos kz − jY∞ V0 sin kz (1.e.. It is useful to describe the propagation phenomenon on the transmission line in geometric terms. also in z = 0: V (0) = V0+ + V0− = V0 I(0) = Y∞ V0+ − Y∞ V0− = I0 (1.30) from which V0+ e V0− can be obtained: V0+ = 1 2 (V0 + Z∞ I0 ) V0− = 1 2 (V0 − Z∞ I0 ) (1.33) = This form of the solution is called stationary wave type solution whereas eq.32) + I0 ) e−jkz − 1 (Y∞ V0 − I0 ) e+jkz 2 i.29) hold in any point z and. Suppose then that the electric state of the line is given at z = 0.31) Substituting these relations into (1. i. V (0) = V0 and I(0) = I0 are given: we want to find the state V (z). we solve the initial value problem associated to eq. we can rewrite (1.34) .20). I(z) in an arbitrary point z. (1. (1.Transmission Line Theory (Nov. 2012) plot is clearly characteristic of a stationary phenomenon.29) is called travelling wave type solution.29) we find V (z) = 1 2 (V0 I(z) 1 2 (Y∞ V0 = + Z∞ I0 ) e−jkz + 1 (V0 − Z∞ I0 ) e+jkz 2 (1. Since voltage and current in a point of the line define the system state.5. 1. we can introduce a two dimensional complex state space (isomorphic to C2 ) each point of which correspond to a possible operation condition of the transmission line.29) where the two arbitrary constants V0+ e V0− appear.e. The state is a function of z and the corresponding point moves on a trajectory in the state space. In order to understand better the meaning of these equations. via Euler’s formula. Forward and backward voltages are then interpreted as excitation coefficients of these waves. Assume for instance that the backward wave is not excited in the point z = 0: it will be absent on the whole transmission line. four real dimensions would be necessary for this type of plot. respectively. in the state space we can describe the excitation of the line with reference to the “natural basis” V e I or to the vectors ψ1 e ψ2 . Geometric representation of the electric state of a transmission line. (1. It is convenient to rewrite also eq.11. This matrix is known as transition matrix in the context of dynamical systems (in which the state variables are real and the independent variable is time) but coincides with the chain matrix (ABCD) of the transmission line length.0)] which relates the state in a generic point z to that in the origin z = 0. 2012) In other words. 1 −Y∞ ψ2 = (1. Obviously the two basis states are the forward and backward waves discussed before.Transmission Line Theory (Nov. 1.11.35) with complex coefficients V0+ e−jkz e V0− e+jkz . the state in a generic point z is obtained as a linear combination of two basis states ψ1 = 1 Y∞ .0)] where we have introduced the matrix [T (z.Renato Orta .36) [T (z. In algebraic terms this state vector is 22 .38) In geometric terms. Assuming for simplicity of drawing that in a point of the line voltage and current are real. As in the cartesian plane of analytic geometry different reference systems can be used.36) we find immediately V (z) I(z) = V0+ 1 Y∞ e−jkz (1. In the general case. Indeed. V ψ1 ψ2 I Figure 1. we can say that in the propagation the state vector remains parallel to itself since it is only multiplied by the scalar exp{−jkz}. in the origin V0 I0 = V0+ 1 Y∞ (1. The basis of the two vectors ψ1 e ψ2 has peculiar properties with respect to all the other bases that could be introduced in the state space.33) in vector form: V (z) I(z) = cos kz −jY∞ sin kz −jZ∞ sin kz cos kz V0 I0 (1. the situation is that sketched in Fig.37) By means of (1. viewed as a two-port device. 1.ω) = exp −jωA(z − z0 ) · ψ 0 (1. if we wish that on a transmission line only one of the basis states is excited.ω)|  z=z0 = jω A · ψ(z. For comparison.8 Solution of transmission line equations by the matrix technique In the previous sections we have found the solution of transmission line equations from the second order equation.40) where the exponential of the matrix is defined by the series expansion: exp −jωA(z − z0 ) = I − jωA(z − z0 ) − where I is the identity matrix.Renato Orta . Otherwise.0)]. which has the advantage that the geometrical interpretation of forward and backward waves as modes of the system is almost automatic. Suppose we know voltage and current in the point z0 of the line and we want to compute the corresponding values in an arbitrary point z. with eigenvalue exp{−jkz}. with a more abstract technique. 2! (1. it is not identically zero on the line (apart for the trivial case of a non excited line).39) = ψ0 = V0 I0 where we use a double underline to denote matrices and A= 0 C L 0 It is well known that the solution of this problem can be written in the form ψ(z. In other words.Transmission Line Theory (Nov. Conversely. Consider again the transmission line equations in the spectral domain   − d V (z. In this section we obtain the same result directly from the first order system. A completely analogous property holds for the backward wave (vettore ψ2 ). 2012) eigenvector of the transition matrix [T (z.31). with coefficients given by (1. notice that if the total voltage is zero in a point. we want to solve the initial value problem   − d ψ(z. whose components in the natural basis are total voltage and current. Hence these equations describe the change of basis. Note that ψ1 . . e ψ2 are not orthogonal (if Z∞ = 1Ω).ω)   dz     ψ(z.ω) (1.ω) The system can be rewritten as a single differential equation for the state vector ψ(z).41) . . it is necessary that V0 /I0 = ±Z∞ .ω)   dz      − d I(z. both modes are excited.ω) dz = jω C V (z. 23 1 2 2 ω A (z − z0 )2 + .ω) = jω L I(z. Renato Orta - Transmission Line Theory (Nov. 2012) It is simple to verify that (1.40) satisfies (1.39). Indeed, by differentiating (1.41) term by term, (which is allowed by the fact that the series converges uniformly for all matrices A and all (complex) z) we find d − exp −jωA(z − z0 ) = jωA exp −jωA(z − z0 ) dz so that d d − exp −jωA(z − z0 ) · ψ 0 = − exp −jωA(z − z0 ) · ψ 0 dz dz = jωA exp −jωA(z − z0 ) · ψ 0 = jωA · ψ The matrix exponential can be computed directly by eqs. (1.41) and (1.40). Note first that A2n = 0 C and A2n+1 = 0 C 2n L 0 2n+1 L 0 √ = ( LC)2n I √ = ( LC)2n 0 C L 0 Hence the series (1.41) reduces to exp −jωA(z − z0 ) 1 √ 1 √ (ω LC(z − z0 ))2 + (ω LC(z − z0 ))4 + . . .] I+ 2! 4! = [1 − − j[ω(z − z0 ) − + √ 1 (ω(z − z0 ))5 ( LC)4 + . . .] A 5! √ 1 (ω(z − z0 ))3 ( LC)2 + 3! We modify slightly the previous equation as follows 1 √ 1 √ = [1 − (ω LC(z − z0 ))2 + (ω LC(z − z0 ))4 + . . .] I+ exp −jωA(z − z0 ) 2! 4! − √ √ 1 1 j √ [ω(z − z0 ) LC − (ω(z − z0 ))3 ( LC)3 + 3! LC + √ 1 (ω(z − z0 ))5 ( LC)5 + . . .] A 5! In the first square parenthesis we recognize the Taylor expansion of cos k(z − z0 ) and in the second one the expansion of sin k(z − z0 ). Moreover   L 0 1 0 Z∞ C  √ = A= Y∞ 0 C LC 0 L so that, in conclusion, exp −jωA(z − z0 ) = cos k(z − z0 )I − j sin k(z − z0 ) 24 0 Y∞ Z∞ 0 Renato Orta - Transmission Line Theory (Nov. 2012) i.e. cos k(z − z0 ) −jZ∞ sin k(z − z0 ) −jY∞ sin k(z − z0 ) cos k(z − z0 ) exp −jωA(z − z0 ) = Even if we are now in the position to obtain the solution of the initial value problem (1.39), we will use instead a different method that allows a more fruitful physical interpretation. Indeed, it is known that a function of a (diagonalizable) matrix is easily computed in the basis of its eigenvectors, because in this basis the matrix is diagonal. Hence we compute first the eigenvectors of A, by solving 0 L 1 0 u1 −λ =0 C 0 0 1 u2 We find immediately  √   λ1 = LC  λ= [u1 ] = √   λ2 = − LC  [u2 ] = 1 C/L 1 − C/L The eigenvectors have an arbitrary norm, since they are solutions of a homogeneous problem; we have chosen to set to one their first component (i.e. the “voltage” component). Notice that they coincide with the basis states of (1.35). Define the modal matrix M , whose columns are the two eigenvectors :  M = 1 C L 1 −  C  L The matrix M , together with the eigenvalue diagonal matrix, satisfies 0 C L 0 M =M λ1 0 0 λ2 . (1.42) It can be shown that if f (x) is an analytic function, then 0 C f L 0 M =M f (λ1 ) 0 0 f (λ2 ) from which, by left multiplication by M −1 , 0 C f L 0 f (λ1 ) 0 0 f (λ2 ) =M M −1 . Applying this property to the exponential of the matrix in (1.40), we obtain: V (z,ω) I(z,ω) =M exp{−jk(z − z0 )} 0 0 exp{+jk(z − z0 )} where Td = M −1 exp{−jk(z − z0 )} 0 0 exp{+jk(z − z0 )} 25 V (z0 ,ω) I(z0 ,ω) (1.43) (1.44) Renato Orta - Transmission Line Theory (Nov. 2012) √ is the evolution matrix in the modal basis and k = ω LC. The inverse of M is  [M ]−1 = 1 1  2 1  L  C  L  − C so that (1.43) is rewritten as V (z,ω) I(z,ω) cos k(z − z0 ) −jY∞ sin k(z − z0 ) = −jZ∞ sin k(z − z0 ) cos k(z − z0 ) V (z0 ,ω) I(z0 ,ω) (1.45) [T (z,z0 )] This equation is identical to (1.36), apart from the fact that the initial point is in z = z0 instead of the origin. Eq. (1.45) is the final result of the computation, but (1.43) is fundamental for the interpretation, because it makes explicit the change of basis, from the natural basis V , I to the modal basis of forward and backward waves. Fig. 1.12 shows pictorially the method described.         basis V0+ V0− M Td     modal −1         M V ( z) I ( z) V0+ ( z ) V0− ( z )     I0     basis V0     natural evolution initial point z0 Figure 1.12. final point Method of solution of transmission line equations 26 z 2.l.l. we show only the expressions of the inductance and capacitance p. L e C versus the ratio of the conductor diameters.).1 shows the field lines of the electric and magnetic fields of the TEM mode.u. 2. separated by a dielectric (see Fig.1 Introduction In chapter 1 we have obtained the transmission line equations on the basis of a phenomenological model that contains four primary parameters: L (inductance per unit length. 2. are often made of braided small diameter copper wires. Fig. d d r c vf = √ . .2 shows a plot of Z∞ .l. Fig.2 Coaxial cable The coaxial cable is a transmission line consisting of two coaxial cylindrical conductors.u. the line parameters are given by: C= 2π 0 r .l. The parameters related to the losses will be shown in chapter 4. the fundamental one of this structure viewed as a waveguide.u.).). p.u. The expressions that yield these parameters as a function of the geometry of the structure require the solution of Maxwell equations for the various cases.1).l. 2. here shown as homogeneous. In particular.2) (2.u.1) (2.3) r where the logarithms are natural (basis e). log(D/d) Z∞ = L= µ0 1 log 0 r 2π µ0 log 2π D d D 60 D ≈ √ log( ). The two conductors. (2. R (resistance p. We can observe that 27 . G (conductance p. C (capacitance p.Chapter 2 Parameters of common transmission lines 2. If r denotes the relative permittivity of the insulator. In this chapter we limit ourselves to a list of equations for a number of common structures: the reader can consult the books in the bibliography for further details .). a point is reached in which higher order modes start to propagate. The field lines of the electric field are shown by solid lines.1.z) = V (z) 1 log(D/d) ρ 28 . Figure 2. If the operation frequency increases.5) 2 The electric field in the cable is radial and its magnitude is given by E(ρ. consistently with the fact that the TEM mode has zero cutoff frequency. (2. those of the magnetic field by dashed lines. 2012) d D Figure 2.4) fmax = π(D + d) The corresponding minimum wavelength is π λmin = (D + d). Coaxial cable. Parameters of the coaxial cable vs. the electric field configuration is that of a cylindrical capacitor. (2.the geometrical dimensions.2.ϕ. The maximum frequency for which the coaxial cable is single mode is approximately 2vf .Renato Orta .Transmission Line Theory (Nov. Renato Orta - Transmission Line Theory (Nov. 2012) where V (z) is the voltage. Hence the maximum electric field, not to be exceeded in order to avoid sparks, is on the surface of the inner conductor and has the value Emax = V (z) 1 log(D/d) d Example Compute the parameters of a cable, with inner conductor diameter d =1.6 mm, outer conductor diameter D = 5.8 mm, r = 2.3. √ Applying the previous formulas we get L = 0.26 µH/m, C = 99.35 pF/m, Z∞ = 50.92 Ω, vf /c = 1/ r = 65.9%, fmax = 17.0 GHz. The normalized maximum electric field is Emax = 485.3V/m if the voltage V is 1V. It is to be remarked that the coaxial cable is an unbalanced line, which means that the return conductor is connected to ground. Hence the voltage of the inner conductor is referred to ground. 2.3 Two-wire line The two-wire line consists of two parallel cylindrical conductors. This structure has a true TEM mode only if the dielectric that surrounds the conductors is homogeneous and the formulas reported hereinafter refer to this case. In practice, of course, the conductors are embedded in a thin insulating support structure, which causes the fundamental mode to be only approximately TEM. The parameters of the two-wire transmission line, whose geometry is shown in Fig. 2.3 are: C= π 0 r , cosh−1 (D/d) Z∞ = 1 π µ0 0 r µ0 cosh−1 (D/d), π L= cosh−1 D d 120 ≈ √ cosh−1 r D d (2.6) , (2.7) c vf = √ . (2.8) r It may be useful to recall that cosh−1 x = log(1 + x2 − 1) ≈ log(2x), se x 1. d D Figure 2.3. Two-wire transmission line. The field lines of the electric field are shown solid, those of the magnetic field dashed. 29 (2.9) Renato Orta - Transmission Line Theory (Nov. 2012) Example Compute the parameters of a two-wire line, in which the wires have a diameter of 1.5 mm and a separation of 5.0 mm and are located in air. we find that C = 14.84 pF/m, L = 750 nH/m, Z∞ = 224.71, vf = c. It is to be remarked that the TEM fields are non negligible up to large distance from the line itself, so that the two-wire line is never isolated from the other nearby conductors, which entails problems of electromagnetic compatibility. On the contrary, in a coaxial cable with sufficiently good outer conductor, the operation of the line is completely shielded from external interference. For this reason, the two-wire line is always used in a balanced configuration, i.e. the two wires have opposite potentials with respect to ground. 2.4 Wire on a metal plane This line consists of a single wire running parallel to a grounded metal plate, see Fig. 2.4a. If the metal plate were infinite, this line would be perfectly equivalent to a two-wire line, because of the image theorem (Fig. 2.4b). When the ground plane is finite, the equivalence is only approximate, but if its size is much larger than the distance h between the wire and the plane, the errors are negligible. d d h D= 2h (a) Figure 2.4. (b) (a) Wire on a metal plane and (b) equivalent two-wire transmission line. The parameters of the two-wire line are: C= π 0 r , cosh−1 (2h/d) Z∞ = 1 π µ0 0 r µ0 cosh−1 (2h/d), π L= cosh−1 2h d 120 ≈ √ cosh−1 r c vf = √ . 2h d , (2.10) (2.11) (2.12) r Example Consider a wire with diameter d = 3.2 mm in air, placed at an height h = 5.74 cm on a ground plane. We find C = 6.51 pF/m, L = 1.71 µH/m and Z∞ = 512.4 Ω. 30 Renato Orta - Transmission Line Theory (Nov. 2012) 2.5 Shielded two-wire line To avoid the electromagnetic compatibility problems of the two-wire line, the structure of Fig. 2.5 can be used. Note that this is a three conductor line (two plus a grounded one). In this case there are two TEM 2h Electric field D Figure 2.5. Magnetic field d Shielded two-wire line and field configuration of the symmetric (balanced) TEM mode. modes, a symmetric (balanced) one where the potentials of the two inner conductors are symmetric with respect to that of the outer one, connected to ground, and an asymmetric (unbalanced) one, with different parameters. The parameters for the symmetric mode can be computed from the following equations: C= log π 0 r 2h(D2 − h2 ) d(D2 + h2 ) Z∞ = , 1 π L= µ0 log 0 r vf = 2h(D2 − h2 ) d(D2 + h2 ) c 2h(D2 − h2 ) d(D2 + h2 ) µ0 log π , . , (2.13) (2.14) (2.15) r Example Consider a shielded two-wire line with diameter of the outer conductor D = 100 mm, inner conductors with diameter d = 15 mm e spacing 2h = 50 mm. Using the previous formulas we get: C= 25.77 pF, L = 0.43 µH, Z∞ = 129.39 Ω. 2.6 Stripline The stripline consists of a metallic strip placed between two grounded metal planes (Fig. 2.6). This is clearly an unbalanced structure, which is used only inside components and devices. Since the two planes have the same potential, this is a two conductor line and the fundamental mode is TEM. The relevant parameters cannot be expressed in terms of elementary functions. We report below an approximate expression for the characteristic impedance, which is valid in the case the strip thickness is negligible: 30π Z∞ ≈ √ r b weff + 0.441b 31 (2.16) 266 cm. Fig.Transmission Line Theory (Nov.20) r Z∞ Example Design a stripline with characteristic impedance Z∞ = 50 Ω.Renato Orta . For the design activity.19) = 0. (2. we can use the following equations.6.441 (2. in which the dimensions of the structure are known. √ Since Z∞ r = 74. whose lower face is covered with a metal ground plane.2.6 − x if b r Z∞ > 120 Ω where 30π x= √ − 0.1065 cm−1 vf c/ r c and λ= 2π = 2. in which the dimensions are to be determined in order for the line to have a desired characteristic impedance. 2012) w b Figure 2. r = 2. x if √ √ r ∞ (2. Hence w = 0.2 Ω (< 120 Ω) we compute x = 0.16) and (2. w 0. the 32 .0212 cm.17): √ w Z < 120 Ω.20) and this is already the value of w/b. 2.85 − 0.32 cm.17) (2.35. then the propagation constant is computed from √ 2πf r ω ω k= = √ = = 3.830 by means of (2. as shown in Fig. separation between the ground planes b = 0. Stripline geometry . as for all TEM structures is given by c vf = √ . Since the transverse cross section is not homogeneous.7 Microstrip A microstrip consists of a conducting strip deposited on a dielectric layer. obtained by inversion of (2. where the equivalent strip width weff is computed from w weff = − b b 0 if w/b > 0.35 − b 2 if w/b < 0.8.7 shows plots of the characteristic impedance of a stripline where the strip thickness f is non negligible.247 ns.35 The phase velocity.18) r The previous equations are appropriate in an analysis problem. Find then the value of the propagation constant and the wavelength at the frequency f = 10 GHz and the delay τ = l/vf introduced by line lentgth l = 5 cm. 2. 2. k τ = √ l l = vf c r = 0. Characteristic impedance of a stripline vs. In practice. fundamental mode is not rigorously TEM. In an analysis problem. (2. we compute first an equivalent dielectric constant eff . in which the dimensions of the line are known. only approximate formulas are available for the characteristic impedance. but exploiting this effective permittivity c vf = √ eff 33 (2.7.Renato Orta . the longitudinal field components are very small with respect to the transverse ones and the so called “quasi-TEM approximation” is used. w εr h Figure 2.8. its dimensions. which is a weighted average of the permittivities of air and of the substrate: eff = r +1 2 1 1 + 12h/w 1+ .Transmission Line Theory (Nov.21) The phase speed is computed as always.22) . Even in this case. Ground conductor Microstrip geometry. 2012) Figure 2. 985 and C = 2. from (2.Renato Orta . Moreover. which takes into account the frequency dispersion of eff due to the longitudinal field components.Transmission Line Theory (Nov.18◦ /cm. three auxiliary quantities are computed: A= Z∞ 60 r +1 + 2 B= r r −1 +1 377π √ 2Z∞ 0.44 eff h h if w < 1. Then the propagation constant is given by √ 2πf eff k= = 71.87 rad/m = 41. we can use the approximate formula (Getzinger. 2. 2.11 (2.39 − 0. which introduces a phase shift of 90◦ at the frequency f = 2. h if w >1 h (2.19 cm.28) eff = r − 2 1 + (f 2 /fp ) G where eff (0) is the zero frequency value given by (2.393 + 0. From this w = 0.159. Note that the effective permittivity eff given by (2.25) r C = log(B − 1) + 0. which is in the domain of vality of the equation and hence it is acceptable. First of all. eff = 1.081.61 (2. 1973) r − eff (0) (2. h (2.27) we get w/h = 3.21) and the other parameters are fp = Z∞0 /(2µ0 h) 34 (2.391 cm results.24) r (2.21) the effective dielectric constant is computed.9 and Fig.23 + 0. We compute A = 1.056.2.88. for various values of r of the substrate.5 GHz. instead. 2.26) r Next  8eA    2A  e −2  if w =  2 h  r −1   B − 1 − log(2B − 1) + C  π 2 r w < 2.10 show the plots of eff versus w/h in the two ranges of wide and narrow strip. B = 7. as it is to be expected in the case of a TEM mode. c If the phase shift must be kl = π/2. Next.23) where natural logarithms are used.667 log + 1. From the second. Fig. The substrate thickness is 1/20 and r = 2. 2012) and the characteristic impedance is given by   √ log 8h + w  60   w 4h eff  Z∞ =  120π    √ w w  + 1. it is not acceptable. we obtain l = 2. we get w/h = 3. 2.27) w if >2 h Example Compute the width w and length l of a microstrip with characteristic impedance Z∞ = 50 Ω. For design these formulas are not convenient and the following are used instead.12 show the analogous plots of the characteristic impedance Z∞ . from the first of (2.11 and Fig.29) . Fig. If we desire a more accurate model.125.21) does not depend on frequency. Since this result is greater than 2. 31) and where Z∞0 is the zero frequency characteristic impedance (in Ω).6 + 0. (2.23) with this value of eff (f ).9. Effective permittivity eff versus microstrip dimensions (wide strip approximation).Transmission Line Theory (Nov. 2012) or fp (GHz) = 0.398Z∞0 /h(mm) (2. Figure 2. 35 .30) G = 0. The characteristic impedance at the operating frequency is then computed by (2.009Z∞0 .Renato Orta . Renato Orta . 2012) Figure 2. Characteristic impedance Z∞ versus microstrip dimensions (wide strip approximation). 36 . Effective permittivity eff versus microstrip dimensions (narrow strip approximation). Figure 2.Transmission Line Theory (Nov.10.11. Renato Orta .Transmission Line Theory (Nov.12. Characteristic impedance Z∞ versus microstrip dimensions (narrow strip approximation). 37 . 2012) Figure 2. we indicate how a shorted transmission line of suitable length can be used to realize capacitors.1. where ordinary lumped parameter components are not available. we can start to study some simple circuits. 3. The relationship between these two quantities is displayed in graphic form by means of a famous plot. whose IL I(z) ZL = VL VL IL Z ( z) = (a) Figure 3. which can be considered the trademark of microwave circuits. The fundamental concepts we are going to introduce are the local impedance on a line and the reflection coefficient.2 Definition of local impedance In the analysis of lumped parameter circuits a fundamental quantity is the impedance of an element. defined as the ratio between the phasors of the voltage at the terminals and that of the ingoing current.1 Introduction In Chapter 1 we have obtained the general solution of the transmission line equations.Chapter 3 Lossless transmission line circuits 3. inductors or resonators that can work at high frequencies. called Smith chart. Finally. V ( z) I ( z) (b) V(z) ZL 0 z (a) Impedance of a one-port circuit element and (b) local impedance on a transmission line. value depends on the longitudinal coordinate z: Z(z) = 38 V (z) I(z) (3. Next we discuss the power flow on the transmission line. In the case of a transmission line terminated with a load ZL we can define a local impedance Z(z).1) . With this result in our hands. ζ0 x rmin 1 rmax r Figure 3.4).3) 1 − jY∞ ZL tan kz It is convenient to introduce the normalized impedance ζ(z) = Z(z)/Z∞ . and the previous equation becomes ZL − jZ∞ tan kz Z(z) = (3. This curve is shown in Fig.2) Note that the origin has been placed on the load. The intersections with the real axis. Example 1 Shorted piece of lossless transmission line of length l. as shown in Fig.5) Zing = jXing = jZ∞ tan kl Note that this input reactance is purely imaginary. rmax e rmin have the property rmax rmin = 1 Consider now some particularly important examples. 39 .3a. which is completed when the variable z increases by λ/2. This equation defines a curve in the complex plane ζ with z as parameter.Transmission Line Theory (Nov. 3. We have ζL = 0 ζ(z) = −j tan kz (3.Renato Orta .4) 1 − jζL tan kz Obviously this formula allows the computation of the input impedance of a transmission line length loaded by the normalized impedance ζL . 3. due to the periodicity of the tangent function. as it is to be expected in the case of a lossless circuit of finite size. Representation in the complex plane of the normalized impedance ζ = r+jx of the curve ζ(z) defined by (3.2. Then the local impedance Z(0) = V0 /I0 coincides with the load impedance ZL .2 and it can be shown to be a circumference. Its transformation law is easily deduced from the previous equation: ζL − j tan kz ζ(z) = (3. so that V0 and I0 are the load voltage and current. 2012) Substitute in this equation the expressions (1.33) of voltage and current on the line: = V0 cos kz − jZ∞ I0 sin kz I0 cos kz − jY∞ V0 sin kz = Z(z) V0 − jZ∞ I0 tan kz I0 − jY∞ V0 tan kz (3. It is clearly a closed curve. of that frequency for which the line is half wavelength long.e.3. (b) 2π (a) Shorted transmission line and (b) corresponding input reactance. 40 . Note also that Xing (ω) is an ever increasing function of frequency.75 1 1. Example 2 Length of lossless transmission line terminated with an open circuit. i. the input reactance Xing (ω) has a behavior similar to that of the reactance of a series LC resonator.3b. We have ζL → ∞ ζ(z) = j cot kz Zing = jXing = −jZ∞ cot kl The behavior is analogous to that of the shorted line. 3. apart from a kl = π/2 translation of the plot. Analogously. Recalling that k = ω/vf . Typical of the distributed parameter circuits is that Xing (ω) is a periodic meromorphic function. the plot of this function is still given by Fig. we can obtain any input reactance. the input impedance of a lumped parameter circuit is a rational function. the line behaves as a shunt resonator.25 0. can always be written as the ratio of two polynomials. say l0 . we note that the input reactance is a function of frequency: Xing = Z∞ tan ωl0 vf and. 2012) X ing Z∞ 10 5 0 -5 Zing -10 0 0. the input impedance is that of an open circuit.e.5 0. Suppose now to fix a certain value of the line length. We can observe that in the neighborhood of f0 = vf /(2l0 ). lumped or distributed (Foster theorem). in the neighborhood of f0 = vf /(4l0 ). obviously. If the line is λ/4 long. either inductive or capacitive.Transmission Line Theory (Nov. We observe that the input impedance is a periodic function of kl with period π. On the contrary. for which the line is λ/4 long. Example 3 Length of lossless transmission line terminated with a reactive load.Renato Orta . If we choose the line length conveniently.25 kl (a) Figure 3. i. as typical of all lossless circuits. 375 0. terminated with the characteristic impedance Z∞ .875 (b) kl 1. Example 4 Length of lossless transmission line. (a) Open circuited length of lossless transmission line and (b) corresponding input reactance. 2012) X ing 10 5 Z∞ 0 -5 Zing -10 0 0.Transmission Line Theory (Nov. X ing 10 Z∞ 5 xL 0 XL -5 Zing -10 0 (a) 0.Renato Orta . 41 .4.5.25 kl (b) 2π Figure 3.25 2π Figure 3.75 (a) 1 1. (a) Length of lossless transmission line closed on a reactive load and (b) corresponding input reactance. We find ZL Z∞ xL − tan kz ζ(z) = j 1 + xL tan kz It is useful to set xL = tan φL because the previous equation becomes ζL = jxL = ζ(z) = j tan φL − tan kz = j tan(φL − kz) 1 + tan φL tan kz from which we get Zing = jXing = jZ∞ tan(kl + φL ) We see that changing the load produces a rigid displacement of the input reactance plots.5 0.25 0. 25 kl (a) (b) 2π Figure 3.Transmission Line Theory (Nov. (a) Transmission line terminated with the characteristic impedance and (b) corresponding input resistance (Xing = 0). If we evaluate the limit of ζ(z) for z → −λ/4 by de l’Hospital rule we find ζing = 1 . discussed in Section 6.3.6. the argument of the tangent in (3. terminated with a generic impedance ZL .75 1 1.25 0.5 1 Z∞ 0. λ/4-length of lossless transmission line.5. 2012) We find ζL = 1 ζ(z) = 1 Zing = Z∞ The line is said to be matched and this is the only case in which the input impedance does not depend on the line length. terminated with a generic impedance ZL .5 Zing 0 0 0.7. If l = λ/4.Renato Orta .4) is π/2 and we are in presence of an undetermined form. ZL Zing λ 4 Figure 3. 42 . Example 5 l = λ/4 length of lossless transmission line.5 0. ζL Zing = 2 Z∞ ZL (3. Ring 2 Z ∞ 1.6) This length of transmission line behaves as a normalized impedance inverter and is commonly employed to realize impedance transformers. another procedure is more convenient. connected by a transmission line.8. We find immediately VA = ZA Vg ZA + Zg IA = Vg ZA + Zg Voltage and current in all points.zA )] IA I(z) where [T (z. The results of Examples 1 and 2 can be used as a basis for a measurement technique of the parameters Z∞ e k of a length l of line. A B Circuit consisting of a generator and a load. Zg ZA Lumped equivalent circuit.Renato Orta .9. + Vg Figure 3.9.7 VA V (z) = [T (z. ZA . We can now perform the complete analysis of a simple circuit. Example 6 Analysis of a complete circuit.zA )] = cos k(z − zA ) −jY∞ sin k(z − zA ) −jZ∞ sin k(z − zA ) cos k(z − zA ) We will see that. can be computed by the (ABCD) chain matrix. computed in Section 1. hence it is ZA = ZL + jZ∞ tan kl 1 + jY∞ ZL tan kl So we are left with the lumped parameter circuit of Fig. connected by a transmission line. consisting of a generator and a load. Recall that the input impedance Zsc of this piece. hence also on the load. 3.Transmission Line Theory (Nov. in practice. Compute the impedance seen by the generator. is Zsc = jZ∞ tan kl 43 . when it is shorted. 2012) + Vg Zg ZL ZA Figure 3. Example 7 Measurement of the parameters Z∞ and k of a length of transmission line. This is also the input impedance of a piece of transmission loaded by ZL . total voltages and currents are not the most convenient quantities for the description of the electric state on a transmission line.10. If the line is shorter than λ/4. 3. since these are the basis states of the system. If instead the load impedance is arbitrary. 2012) whereas Zoc given by Zoc = −jZ∞ cot kl is the corresponding input impedance of the length of line when it is open.3 Reflection coefficients In lumped circuits the state variables are voltages and currents and circuit elements are characterized by their impedance (or admittance) that plays the role of transfer function.7. The natural state variables are instead the amplitudes of forward and backward waves. 3.Transmission Line Theory (Nov. Consider a transmission line with characteristic impedance Z∞ and phase constant k. satisfy the boundary condition: V inc (0) + V ref (0) = ZL (I inc (0) + I ref (0)) 44 . n = 0. as depicted in Fig. sum of the forward and backward components. z Scattering description of a load.Renato Orta . As discussed in section 1. Its value can be determined only if we know an estimate of the wavelength on the line. These equations can easily be solved with respect to Z∞ e k in the form √ Z∞ = Zcc Zca k= 1 arctg l − Zcc + nπ Zca The presence in this formula of the integer n is related to the fact that the tangent function is periodic with period π.10: V+ V− ZL 0 Figure 3. Hence we must define the behavior of a generic load in the basis of forward and backward waves. Obviously V inc and I inc satisfy this relation only if ZL = Z∞ : in this case the forward wave alone is capable of satisfying the boundary condition. V inc (z) = V0+ e−jkz I inc (z) = Y∞ V0+ e−jkz Saying that the line in z = 0 is loaded by ZL is equivalent to saying that in this point voltage and current are related by V (0) = ZL I(0). excited by a generator that produces a forward wave incident on the load. loaded by the impedance ZL . necessarily on the load a backward (reflected) wave must be generated V ref (z) = V0− e+jkz I ref (z) = −Y∞ V0− e+jkz with a suitable amplitude V0− in such a way that the total voltage and current. it is a fractional bilinear transformation. whereas its phase is proportional to z. it is to be expected that their transformation law is simple.7) Γ= ζ +1 y+1 1 − VΓ 1 + IΓ I Γ=− ζ −1 y−1 = . In other words. ζ +1 y+1 y= 1− 1+ V Γ 1 + IΓ = VΓ 1 − IΓ with y = 1/ζ. which are the basis states of the line. 45 . The local voltage reflection coefficient in a point z is defined as the ratio of the backward and forward voltages in that point: V − e+jkz V − (z) V Γ(z) = + = 0+ −jkz = V Γ0 e+j2kz (3. We prove now that this is the case. 2012) that is V0+ + V0− = ZL Y∞ (V0+ − V0− ) From this the unknown amplitude V0− is immediately deduced as V0− = ZL Y∞ − 1 + V ZL Y∞ + 1 0 The proportionality coefficient that relates the backward voltage to the forward one is called voltage reflection coefficient − ZL Y∞ − 1 ζL − 1 ZL − Z∞ def V V ΓL = V Γ0 = 0+ = = = ZL Y∞ + 1 ζL + 1 ZL + Z∞ V0 This voltage reflection coefficient is the transfer function of the circuit element when forward and backward + − voltages are used as state variables. V Γ(z) moves on a circumference with center in the origin of the complex plane. We have seen that the transformation law of the local impedance on a transmission line is fairly complicated. ζ= = (3. All these relations are fractional bilinear transformations of the general type w= az + b cz + d This class of complex variable mappings are well known and have a number of properties that will be discussed later on. for which k = ω LC is real the magnitude of the reflection coefficient is independent of z. Since the reflection coefficients are defined with reference to the forward and backward waves. the same circuit element can be characterized either: • by the impedance ZL or the admittance YL (with ZL = 1/YL ) • or by the voltage V Γ or current I Γ reflection coefficient (with V Γ = − I Γ) The equations that relate the reflection coefficients to the corresponding normalized impedances and admittances are y−1 1 + VΓ 1 − IΓ ζ −1 V =− .Renato Orta .Transmission Line Theory (Nov.8) V (z) V0 e √ In the case of a lossless transmission line. By the way. Obviously also the forward and backward currents I0 e I0 could be used as state variables: this choice would lead to the definition of a current reflection coefficient: I ΓL = I Γ0 = def − I0 −Y∞ V0− = − V Γ0 + = I0 Y∞ V0+ Hence. 3.Renato Orta .23). which is said to be a “matched” load for the line. in the same point we can define an ingoing active power P (z) 1 P (z) = Re{V (z) I ∗ (z)} 2 as well known from circuit theory. Conversely. Hence the two waves are power-orthogonal (i. power uncoupled).4 Energy balance In the study of lumped circuits. It is useful to express this power in terms of the amplitudes of the forward and backward waves. 2012) 3. due to the absence of the reflected wave. The net power coincides with the incident one.11). Indeed.Transmission Line Theory (Nov. is absorbed by the part of the circuit lying to the right of z (note the current sign convention) and is interpreted as the power flowing in the line in the point z. Recalling (1. the reflected power is equal to the incident one. if ZL = jXL V Γ= jXL − Z∞ jXL + Z∞ and numerator and denominator have the same magnitude. loaded with the impedance ZL (see Fig.e.e. the net power flowing is the same in every point of the line. the line is matched. we have = 1 Re{[V + (z) 2 = ∗ 1 Re{Y∞ [|V + (z)|2 2 ∗ − |V − (z)|2 ] + Y∞ (V − (z)V +∗ (z) − V + (z)V −∗ (z))} = = P (z) ∗ + V − (z)] Y∞ [V +∗ (z) − V −∗ (z)]} = ∗ 1 Re{Y∞ [|V + (z)|2 2 ∗ − |V − (z)|2 ] − j2Y∞ Im{V + (z)V −∗ (z)}} For an ideal line Y∞ is real and then P (z) = |V + |2 1 1 Y∞ |V + |2 − Y∞ |V − |2 = (1 − | V Γ|2 ) 2 2 2Z∞ We can make the following remarks: • Since | V Γ| = constant on an ideal line. in harmonic regime. We can also say that the net power is the difference between the incident and the reflected power. Consider an ideal transmission line. in these conditions. If in the point z voltage and current are V (z) e I(z). We extend them to the realm of distributed circuits containing transmission lines. ZL = Z∞ ). and consequently the net flowing power is zero. This is obviously related to the fact that an ideal line is lossless . 0 z Transmission line terminated with a generic load impedance. since these variables yield the most natural description of the system. if positive. the whole incident power is absorbed by the load. ZL z’ Figure 3. This power. 46 . This condition takes place when the load is a pure reactance. the energy considerations play an important role.11. • If | V Γ| = 1. one says that. Hence the power absorbed by the load impedance ZL is PL = P (0) = P (z). • In a lossless line the net active power flowing in a point is the difference between the active powers flows associated to the forward and backward waves. • If V Γ = 0 (i. Renato Orta . expressed in dB. Both of them are very often used in practice (see Table 3.12. The quantity 1 − | V Γ|2 is called power transmission coefficient.13. 2012) • For a passive load. RL = 0 dB for a reactive load and RL → ∞ dB for a matched load. Ideal transmission line terminated with a generic load impedance. will always use the one for voltage also in the current expression.Transmission Line Theory (Nov.12. Hence.1). The analytic expression of |V (z)| is then 47 . current and impedance diagrams Consider a loaded transmission line. as we will prove in Section 3. Our goal now is to obtain plots of the magnitude and phase of voltage. 3. Return loss e Standing Wave Ratio (VSWR). defined as RL = −10 log10 |Γ|2 It yields the ratio (in dB) between the reflected power (which is “lost”. 3. recall that Γ(z) = Γ0 exp{+j2kz} (see Fig. as shown in Fig. current and impedance on the line. As for the second.6. because it is equal to the ratio of the power absorbed by the load and the incident power. from the point of view of the load) and the incident one. Let us start with the magnitude plot. the voltage reflection coefficient will be written Γ(z) without superscripts. A quantity frequently used in practice to characterize a load is the return loss RL. shown in Fig. 3. the reflected power is smaller or equal to the incident one.5 Line voltage. The same coefficient. This condition is equivalent to Re{ZL } ≥ 0. which will be introduced in the next section. Voltage and current on the line can be expressed in the following way in terms of forward and backward waves: V V (z) = V + (z) + V − (z) = V + (z)(1 + I(z) = I + (z) + I − (z) = Y∞ V + (z)(1 − Γ(z)) V Γ(z)) Since the reflection coefficients for voltage and current are just opposite one of the other. Since there is no ambiguity. is called reflection loss. The magnitude of voltage and current is given by |V (z)| = |V + (z)| |1 + Γ(z)| |I(z)| = |Y∞ V + (z)| |1 − Γ(z)| The first factor |V + (z)| is constant on an ideal line. 3. This shape is easily explained. express the mismatch of the load with respect to the line in different but equivalent manners. hence | V Γ| ≤ 1.14). for simplicity we ZL z v L Figure 3. Transmission Line Theory (Nov.5 -1 -0.5 -1 -0. local impedance on a transmission line loaded by ZL = (1 + j)Z∞ . current. = |V + | |1 + |Γ0 | exp(j(arg(Γ0 ) + 2kz))| = = |V + | 1 + |Γ0 |2 + 2|Γ0 | cos(arg(Γ0 ) + 2kz) 48 1/2 . 2012) 2 V ( z) V+ 1 -1.5 kz (2π ) 0 0 2 I ( z) I+ 1 -1. Γ 1 + ( z) (z ) Γ 1 -1 1 − ( z) Γ Γ |V (z)| Γ Figure 3.13.14. Plot of the magnitude of voltage. − (z ) Plot of the local reflection coefficient in the complex plane.Renato Orta .5 kz (2π ) 0 0 Figure 3.5 kz (2π ) 0 0 3 Z (z) Z∞ 2 1 -1.5 -1 -0. 2012) Note that the curve is only apparently sinusoidal.0316 ∞ 5. The ratio between the maximum and minimum voltage magnitude is called VSWR (Voltage Standing Wave Ratio) S= 1 + |Γ| Vmax = Vmin 1 − |Γ| Clearly.1778 0. we find that the circumference of Fig.0653 0 3. Table 3. Correspondence between values of return loss.5623 0. 3.8480 3.6508 0. It is evident from the figure that |V (z)| and |I(z)| reach the maximum and minimum value when Γ(z) is real. the ratio between the maximum and minimum current magnitude is also S.5697 1.0043 We have seen in Fig.Renato Orta . The VSWR is normally used in practice to specify the mismatch of a load with respect to a reference resistance (Z∞ ).4). VSWR. 3. the VSWR is always greater than 1.Transmission Line Theory (Nov.4325 1.4575 0.1 0. and moreover: |1 + Γ(z)|max = 1 + |Γ| |1 + Γ(z)|min = 1 − |Γ| In Fig. 3. Return loss. VSWR and reflection loss Return Loss (dB) |Γ| VSWR Reflection Loss (dB) 0 3 5 10 15 20 30 1 0.7079 0. Since the magnitude of Γ(z) of a passive load is always comprised between 0 (matched load) and 1 (reactive load) (see Section 3. Table 3.2222 1. Hence. Hence.2 that the normalized local impedance ζ(z) moves on a circumference in the complex ζ plane. the magnitude of the impedance is an oscillating function and the maxima and minima are reached when ζ(z) is real and their value is Rmax = Vmax Imin = |V + |(1 + |Γ|) |Y∞ V + |(1 − |Γ|) = Z∞ S Rmin = Vmin Imax = |V + |(1 − |Γ|) |Y∞ V + |(1 + |Γ|) = Z∞ S On the basis of these results.3162 0.0436 0. magnitude of the reflection coefficient.0206 1.1395 0.2 has center in ζc and radius R given by 1 + |Γ|2 1 1 S+ = ζc = 2 S 1 − |Γ|2 and R= 1 2 S− 1 S 49 = 2|Γ| 1 − |Γ|2 .1 yields examples of correspondences.9249 1.1. reflection loss and magnitude of the reflection coefficient express the mismatch in equivalent manners.14 the vectors 1 + Γ e 1 − Γ are shown. as well as measurement instruments such as the Network Analyzer.5 kz (2π ) -1 0 Figure 3. Nowadays.15).5 -1 -0.e.6 The Smith Chart The Smith Chart is a graphical tool of great importance for the solution of transmission line problems. 2012) 10 + arg(I(z)/I0 ) + arg(V(z)/V0) -1. has a periodic behavior. With a little algebra it is possible to find the expressions of the phase of voltage.5 -1 -0.5 kz (2π ) 5 0 0 1 arg Z (z) Z∞ 0 -1. its usefulness is no longer that of providing the numerical solution of a problem. display the results on a Smith Chart.Renato Orta . The normalized impedance. i. all modern codes for the Computer Aided Design (CAD) of distributed parameter circuits. when the load becomes purely reactive. 50 .Transmission Line Theory (Nov. Plot of the phase of voltage.e. i.15. 3. but that of helping to set up a geometrical picture of the phenomena taking place on a transmission line. since computers are widespread. that tend to resemble a staircase when |Γ0 | → 1. It is clear that the circumference degenerates in the imaginary axis if |Γ| → 1. as it is shown in Fig. Hence. the load becomes reactive. as already shown by (3. current and local impedance on a transmission line loaded by ZL = (1 + j)Z∞ .4). current and normalized impedance: |Γ0 | sin(2kz + arg Γ0 ) arg V (z) = arg V0+ − kz + arctan 1 + |Γ0 | cos(2kz + arg Γ0 ) + arg I(z) = arg I0 − kz − arctan arg ζ(z) = + |Γ0 | sin(2kz + arg Γ0 ) 1 − |Γ0 | cos(2kz + arg Γ0 ) |Γ0 | sin(2kz + arg Γ0 ) 1 + |Γ0 | cos(2kz + arg Γ0 ) |Γ0 | sin(2kz + arg Γ0 ) arctan 1 − |Γ0 | cos(2kz + arg Γ0 ) arctan The phases of voltage and current are decreasing functions for increasing z. 3. (3.17). is shown in Fig. because the straight lines r = constant and x = constant are orthogonal in the ζ plane and the mapping (3. An example of Smith chart. on which suitable coordinate curves are displayed. shown in Section 3.9) has the following property: if the variable ζ moves on a circumference in its complex plane. V Γ = | V Γ| exp{j arg( V Γ)}.16).9) is analytic in the whole complex plane.18. we can place it on the chart by viewing the set of lines as constant resistance and constant reactance circles: in this way V ΓA is automatically defined. (r=const. 3.) are mapped onto the circumferences with equation V Γr − All of them pass through the point centers on the real axis (Fig. For this to be true without exceptions we must regard straight lines as (degenerate) circumferences of infinite radius. the left half plane Re{ζ} < 0. also the corresponding V Γ values are located on a circumference. if we know the normalized impedance ζA . corresponding to passive loads. • The two sets of circumferences meet always at right angle (except at V Γ = 1). on which constant conductance and constant susceptance curves are drawn.Transmission Line Theory (Nov. the Smith chart can be considered equally well as: • The complex V Γ plane. I Γ = − V Γ. is mapped onto the unit radius circle. | V Γ| > 1. 3. | V Γ| ≤ 1. the Smith chart consists of a portion of the complex V Γ plane. Hence. the complex number V Γ is always given in polar form. 2012) Mathematically. If we recall the relation between the two types of reflection coefficients. it is clear that we can exploit the Smith chart to compute the admittance corresponding to a given impedance and viceversa. We have seen (Eq. The opposite 51 . The Smith chart is equipped with scales to measure magnitude and phase of V Γ. In this way the transformation ζ → V Γ and its inverse are geometrically straightforward. 3. on those of the second the imaginary part of the impedance is constant (constant reactance curves). It can be shown [2] that the bilinear fractional transformation (3. • the vertical lines of the ζ plane. With some algebraic manipulation it can be shown that • The right half plane Re{ζ} ≥ 0. Because of the form of the evolution law of the reflection coefficient on a line. it is based on the two relations. parallel to the imaginary axis. but have their centers on a vertical line.e. I • The complex Γ plane. we can draw two sets of curves in the complex V Γ plane: on the curves of the first set the real part of the impedance is constant (constant resistance curves). is mapped onto the region external to the unit circle. Both of them are complex variables and in order to provide a graphical picture of the previous equations. equipped with all the necessary scales. ζ +1 ζ= 1+ 1− V Γ VΓ where ζ = Z/Z∞ = r + jx is the normalized impedance and V Γ = V Γr + j V Γi = | V Γ| exp{j arg( V Γ)} is the voltage reflection coefficient.3: V Γ= ζ −1 . passing through the singular point V Γ = 1. V r 1+r 2 + 1 1+r V 2 Γi = 2 Γ = 1. which is a singular point of the mapping. (x=cost.7)) that the relation between V Γ and ζ is formally the same as that between I Γ and y. apart from V Γ = 1). i. on which constant resistance and constant reactance curves are drawn.Renato Orta .) are mapped onto the the circumferences with equation V Γr − 1 2 + V Γi − 1 x 2 = 1 x 2 Also these circumferences pass through the singular point. Indeed. (Fig. In particular. and have the • the horizontal lines of the ζ plane. 2012) ζ PLANE Γ PLANE 1 Imag( Γ ) Imag( ζ ) 5 0 −5 Figure 3. The reflection coefficient at point A is given by V ΓA = V ΓB exp(−j2klAB ) = 52 V ΓB exp(−j 4π lAB ) λ .5 0 −0.5 −1 −5 0 2 Real( ζ ) 4 −1 0 Real( Γ) Figure 3. which is solved with the same simplicity is the following. labeled now with resistance and reactance values. limited point with respect to the center is I ΓA and. solves the problem.5 0 −0. viewed this time as constant conductance and constant susceptance circles. we obtain the desired value of yA . 3. Fig.16. 3. Place ζB on the Smith chart. so that V ΓB is determined. compute the normalized impedance ζB at point B.Transmission Line Theory (Nov. 0 2 Real( ζ ) 0.5 −1 −1 4 0 Real( Γ ) 1 Constant resistance lines in the ζ plane and their image in the ζ PLANE V Γ plane. Fig. already solved in 3. This property is clearly very useful when we have to analyze transmission line circuits containing series and parallel loads. Suppose we must find in the V ΓA plane the set of impedances with conductance greater than one.Renato Orta . by reading its coordinates with respect to the set of lines.20a shows hatched the region of the I ΓA plane where g ≥ 1. Now let us see how the use of the Smith chart simplifies the analysis of the circuit of Fig. Γ PLANE 5 Imag( Γ ) Imag( ζ ) 1 0 0.17. to read the coordinates of the points.20b displays the symmetric region with respect to the origin. Constant reactance lines in the ζ plane and their image in the to the region inside the unit circle. Using the standard curves. 3.2.21. V 1 Γ plane. A more complex problem. From the load impedance ZL and the line characteristic impedance Z∞ . Renato Orta .9) . with center in the origin. passing through lAB arg( V ΓA ) = arg( V ΓB ) − 4π λ 53 V ΓB and with a phase (3. Hence V An example of Smith chart that can be used for analysis and design purposes ΓA is on the circumference. 2012) Figure 3.18.Transmission Line Theory (Nov. 20. 2012) V A I A Figure 3. Computation of impedances and admittances.19. it is straightforward to draw the plots of magnitude and phase of voltage and current. Regions of the Smith chart: (a) loads with conductance g ≥ 1. (a) (b) Figure 3. Find then the (total) voltage at the line input VA = The voltage at point z is given by ZA Vg Zg + ZA V (z) = V + (z)(1 + where + V + (z) = VA e−jk(z+lAB ) = and V Γ(z) = V V Γ(z)) VA e−jk(z+lAB ) 1 + V ΓA ΓB e+j2kz assuming that the origin has been chosen in B. as explained 54 . In conclusion V (z) = Vg ZA e−jklAB −jkz (e + Zg + ZA 1 + V ΓA V ΓB e+jkz ) and e−jklAB −jkz ZA (e − Zg + ZA 1 + V ΓA To write the expression of the current. taking into account that it is just necessary to study the behavior of 1 ± V Γ(z). After identifying V ΓA it is enough to read the coordinates of this point on the basis of the constant resistance and constant reactance circles to obtain ζA and therefrom ZA = ζA Z∞ . so that the coordinate of A is z = −lAB .Transmission Line Theory (Nov. we have used I(z) = Y∞ Vg V ΓB e+jkz ) + + IA = Y∞ VA and I(z) = I + (z)(1 + I Γ(z))) = I + (z)(1 − V Γ(z))) From a graphical point of view. Notice that the phase values in this equation must be expressed in radians. (b) impedances of the loads with g ≥ 1.Renato Orta . 9). Note that the “generator” in the label has nothing to do with the one present in the circuit.25 shift is related to the fact that the origin of these scales is on the negative real axis. in Section 3. 3.Renato Orta . (a) Complete circuit. Appropriate scales are provided on the chart to simplify these operations. transmission line and load and (b) Smith chart solution.10) The rotation sense on the chart is clockwise.Transmission Line Theory (Nov. the second a counterclockwise one. in the range [0.2]. This quantity is called Transmission coefficient for a reason explained in the next section. wavelengths toward load. see Fig.9) becomes l λ TG = eqA l λ TG + eqB lAB λ (3. In this way eq. concentric with the first. (3.(3. is to be read on the scale with the label Transmission coefficient E or I. In particular. the magnitude of 1 ± V Γ(z). The outer one has the label Wavelengths toward generator and displays the quantity TG l λ def = 0. The second scale. but is the driving point impedance generator that one imagines to connect in the point of interest of a circuit to define the relevant impedance.9) (remember that the phase of complex numbers increases counterclockwise).25 − eqB arg( V ΓB ) 4π A second one.22. is labeled Wavelengths toward load and displays the quantity TL l λ def = 0. the first is a clockwise scale. Moreover.5.21. even if the symbol suggests an interpretation as equivalent electrical length. consisting of generator.14 and Fig. Two scales drawn on the periphery of the chart simplify the evaluation of eq. The phase is read by means of an angular scale with the label Angle of Transmission Coefficient in degrees. and with the ticks pointing toward the point V Γ = −1. 2012) + Vg ζB Zg ZL ZA A rm λ ζA B − l AB l AB z 0 (a) (b) Figure 3. has values that increase counterclockwise and is useful when the input impedance is known and the load impedance value is desired: l λ TL = eqB l λ 55 TL + eqA lAB λ . The presence of the 0. 3. as specified by the sign of the exponent in (3.25 + eqB arg( V ΓB ) 4π both of them being measures of the phase of the reflection coefficient. 3.Transmission Line Theory (Nov.22. (3. measured on the Smith Chart. A similar interpretation eqB holds for (l/λ)T L . It is useful to clarify the reasons for using the symbol (l/λ)T G as a measure of arg( V Γ). Moreover. In this way.21 eq the intersection of the circle | V Γ| = | V ΓB | with the negative real axis is the point labeled rm .10) has the appearance of a sum of homogeneous quantities.5). terminated with a resistor of value RL = Z∞ rm .(3.9). V ΓB can be viewed as the input reflection coefficient of a line with electrical length (l/λ)T G . Magnitude and phase of the transmission coefficient 1+ V ΓB . because in this point the normalized local impedance ζ(z) has the minimum real part and zero imaginary part (see Section 3. Clearly. more than eq eq. 2012) ζB 1+ ΓB arg(1+V ΓB ) V Figure 3. from the picture. eq. only sums are carried out. 56 . In Fig.Renato Orta . so that V V ΓA− VΓ A+ ΓA+ = 0. As for the forward voltage. by recalling the general formula V (z) = V + (z)(1 + + VA− (1 + that is V V + ΓA− ) = VA+ (1 + + VA+ + VA− V 1+ 1+ = V Γ(z)). A+ Cascade connection of two lines with different characteristic impedance. 2012) 3. Apply Kirchhoff laws at the node A: 57 . The normalized impedance is instead discontinuous (ζA− = ζA+ since Z∞1 = Z∞2 ).Renato Orta .7 Analysis of simple circuits Sometimes two transmission lines with different characteristics are cascaded or lumped loads are connected in series or in shunt with the transmission line. hence the different size of the “conductors” is just a graphical convention to denote lines with different characteristic impedance. Then + + VA+ = VA+ = VA− (1 + V ΓA− ) Defining a Transmission Coefficient def TV = in this case is TV = 1 + For this reason. we find ΓA+ ) ΓA− VΓ A+ Also the ratio of the forward currents is obtained immediately: + IA+ + IA− = + Y∞2 VA+ + Y∞1 VA− Y∞2 1 + Y∞1 1 + = Suppose that the second line is matched.23. Let us see how the analysis is carried out in such cases. Shunt connection of a lumped load Consider now the case of of a line with the lumped load Yp connected in shunt at A. The very circuit scheme adopted implies that both the voltage and the current are continuous at point A: IA− = IA+ VA− = VA+ Dividing both sides of the first equations by IA− = IA+ yields the continuity of the local impedance ZA− = ZA+ . the quantity 1 + + VA+ + VA− V V ΓA− Γ is always called Transmission coefficient. on the Smith chart. Notice that the picture uses the symbols of the transmission lines: Z ∞1 Z∞2 - Figure 3.Transmission Line Theory (Nov. but it has nothing to do with the actual geometry of the lines. Cascade connection of transmission lines Consider first the cascade connection of two lines with different characteristic impedance. A+ Series connection of a lumped load on a transmission line.Transmission Line Theory (Nov.Renato Orta . as in the previous case. we can obtain. Kirchhoff law at node A yield Vs Zs AFigure 3. 2012) Ip Yp A Figure 3. it is convenient to work on the current.25. the relation between the forward voltages + VA+ 1 + V ΓA− + = 1 + V ΓA+ VA− and therefrom the corresponding one for the forward currents + IA+ + IA− = Y∞2 1 + Y∞1 1 + V ΓA− VΓ A+ Series connection of a lumped load Consider now the case of a lumped load Zs connected in series on a transmission line at A. VA− = VA+ IA− = IA+ + Ip YA− = YA+ + Yp Exploiting the continuity of the total voltage at A. Shunt connection of a lumped load on a transmission line.24. VA− = VA+ + Vs IA− = IA+ ZA− = ZA+ + Zs To find the link between forward and backward waves. which is continuous: + IA+ 1 + I ΓA− 1 − V ΓA− = = + 1 + I ΓA+ 1 − V ΓA+ IA− As for the voltage + VA+ + VA− = Z∞2 1 − Z∞1 1 − 58 V ΓA− VΓ A+ . since they have been applied to lumped elements. Zs could be the input impedance of a distributed circuit. In other words. and then apply the usual lumped circuit theory. Y .26.Renato Orta . See also Chapter 7 for a review of these matrices. Shunt connection of a distributed load on a transmission line.Transmission Line Theory (Nov. Open circuit impedance matrix [Z] Defining equations V1 V2 = Z11 Z21 Z12 Z22 I1 I2 We get Z = −jZ∞ cot kl csc kl csc kl cot kl Note that the current I2 is assumed to be positive when it enters into the port. 3. positioned at right angle with respect to the main line. characterized via its matrices Z. it may be convenient to represent a transmission line length as a two-port device. We derive now the expression of these matrices for a length l of transmission line with characteristic impedance Z∞ and propagation constant k. as in Fig. A+ Series connection of a distributed load on a transmission line. Yp could be the input admittance of a distributed circuit positioned at right angle with respect to the main line. 3. as shown in Fig. Zs AFigure 3.11) . Likewise.26.27.27. It is interesting to note that the loads in the circuits above are lumped in the z direction but not necessarily in others. We will see examples of such circuits in Chapter 6 on impedance matching. Transmission line length as a two-port device Two analyze more complex cases. Yp A Figure 3. or ABCD. 2012) Note that in these cases the use of Kirchhoff laws is completely justified. 59 (3. Also useful are the T and Π equivalent circuits. also in this case. as it is typical of distributed parameter circuits. 60 . Moreover.Transmission Line Theory (Nov.Renato Orta . shown in Figure 3. YP 1 = YP 2 = jY∞ tan Z T1 YP12 Z T2 Z T12 YP1 (a) Figure 3.12) Note that. Chain matrix ABCD Defining equations V1 I1 A C = −B −D V2 I2 We get ABCD = cos kl jY∞ sin kl jZ∞ sin kl cos kl (3. YP2 (b) (a) T equivalent circuit and (b) Π equivalent circuit of a transmission line length. This is the reason of the minus signs in the defining equations. the current I2 is assumed to be positive when it goes out of the port.13) Note that. YP 12 = −jY∞ csc kl 2 Note that all the matrix elements are periodic functions.28. 2 ZT 12 = −jZ∞ csc kl kl . the current I2 is assumed to be positive when it enters into the port.28. differently from before. 2012) Short circuit admittance matrix [Y ] Defining equations I1 I2 Y11 Y21 = Y12 Y22 V1 V2 We get − cot kl csc kl Y = jY∞ csc kl − cot kl (3. obviously. Y = Z −1 . The values of the elements are ZT 1 = ZT 2 = jZ∞ tan kl . 1) applies. and cross section S. which is in phase with the applied electric field because γd in (4. is summed with the displacement current. The detailed study of these phenomena requires the solution of Maxwell’s equations in the structures of interest. Indeed. This attenuation has two origins: one is the energy loss in the dielectrics. for each point of which (4. which have very high but not infinite conductivity.1) where E is the applied electric field. From this point of view. the other is the energy loss in the conductors. caused by the applied field. In accordance with the circuit point of view. In every real dielectric there are electrons that are not strictly bound to atoms and are set in motion by an applied electric field: in this way an electric current is produced. Assuming that the current density is constant in the cross section and the electric field is constant along the wire. It is useful to note that this equation is the microscopic form of Ohm’s law. adopted in these notes.1 Dielectric losses The phenomenon of energy dissipation in insulators is the simplest to describe. the material is characterized by a conductivity γd .Chapter 4 Energy dissipation in transmission lines Wave propagation in real world transmission lines is always affected by attenuation.1) is real. Jc is the resulting current density per unit surface and the subscript “c” reminds us that this is not an independent source but a conduction current. defined by Jc = γd E (4. consider a metal wire of length L. Usually the conduction current. Substituting we get γd S V = GV I= L which we recognize as the macroscopic form of Ohm’s law. 4. which is in quadrature. we limit ourselves to a qualitative discussion of the subject. which have a small but not negligible conductivity. measured in S/m. A much more detailed treatment can be found in [3]. we can write |Jc | = I S |E| = V L where I is the current in the wire and V the potential difference between the two wire ends. so that a complex dielectric 61 . Generally.ω) ω jω εE (r. with a density per unit surface of the cross-section that decays exponentially toward the inside of it. This phenomenon has two consequences: • energy is dissipated in the metal because the electric field and the conduction current are in phase. it can be shown that the current is no longer confined to the conductor surface but is distributed also inside the metal. becomes × H (r.1).ω) ˜ jω ε − j (4. Indeed. Indeed ε is always understood to be complex unless ˜ specific indications are given. so that ε is assumed to be pure imaginary. In these conditions.Renato Orta . whose real part is ˜ the usual dielectric constant and whose imaginary part is related to the conductivity. the displacement current is negligible with respect to the conduction current. on the very surface of the conductors there is an electric current strictly related to the electromagnetic field. 2012) constant is introduced. recall the second Maxwell’s equation × H (r. the loss angle δ. for frequencies up to the millimeter wave range. This procedure will be briefly illustrated in the next section. according to (4.ω) + γd E (r. has a magnitude equal to that of the tangential magnetic field in the points of the dielectric facing the conductor. We have seen in Chapter 1 that dielectric losses are accounted for in circuit form by means of the conductance per unit length G. In a transmission line in which the conductors can be assumed perfect. Actually. Also the magnetic field penetrates the metal.3) In very straightforward way we have introduced a complex equivalent permittivity ε. for a good conductor the loss angle δ → π/2. as well as that of all line parameters. 4.2) which. The computation of this quantity. see Fig.ω) + Je (r. 4.t) + γd E (r.3. and γd are functions of frequency.4) Then the relationship between loss angle and conductivity is tgδ = γd ωε0 εr (4. The formulas that allow the computation of G for some examples of lines are reported in Section 4.5) Obviously. From the knowledge of the fields it is possible to derive the values of the line parameters. Observe that if the frequency behavior of ε is of ˜ interest.1.ω) + Je (r.t) = ε ∂E (r. requires the solution of Maxwell’s equations for the structure under consideration. with a similar exponential decay. Its direction is orthogonal to that of the magnetic field. whose density per unit length Jσ . measured along the boundary of the conductor cross-section. where conductor losses are analyzed. we must keep in mind that both εr . the electromagnetic field is different from zero only in the insulators. If now we imagine that the metal conductivity is very large but finite.2 Conductor losses The complex dielectric permittivity can describe also a good conductor. 62 .Transmission Line Theory (Nov. in the spectral domain. in conductors such as copper.ω) = = = jωεE (r.t) + J e ∂t (4. Finally it is to be noted that the symbol ε has been introduced only for clarity. starting from the geometry and the physical parameters of the materials. It is a surface current.ω) + Je (r.ω) γd E (r. is introduced: it is defined as the argument of the complex number ε: ˜ ε = ε − jε = ε0 εr ˜ 1−j γd ωε0 εr = ε0 εr (1 − jtgδ) (4. 2012) Jσ ds Figure 4. Fig. so that the y-variations of fields and current can be neglected. only on z e da x.7) δ= ωµγ We see that if h/δ → 0 the current is uniformly distributed in the conductor.2.1) shows the data for some common conductors. adjacent to the interface between the metal and the insulator. if h/δ is large. since we want to obtain the line parameters per unit length. 63 .2. to be summed to the external one.6) w sinh (T h) where T = (1 + j) /δ and I/w is the total current (per unit length along the y direction) flowing in the conductor.ω) versus the normalized depth x/h and the parameter h/δ. Assume w/h >> 1. has the direction ˆ and is given by z x I cosh T h h − 1 Jz (x) = T (4. A case that lends itself to a simple analysis is that of a planar transmission line. the current flows essentially in a thin film. This behavior is analyzed in greater detail below.1.Transmission Line Theory (Nov. Here we focus on the x dependance. In these conditions. which accounts for the flux in the insulator. so that they depend z x w y h d Figure 4. which justifies the name of the phenomenon. • the magnetic field in the conductor produces an induction flux described by means of an internal inductance. Planar transmission line. Perfect conductor and surface current on it. shown in Fig. The skin depth δ is inversely proportional to the square root of frequency and of metal conductivity. On the contrary.Renato Orta . Jz (x) decays exponentially with decay rate δ. since the skin depth δ can be shown to be related to frequency by 2 (4. 4. This corresponds to showing the frequency dependance.3 shows a 3D plot of the current density per unit surface Jz (x. Table (4. 4. It can be shown that the current density per unit surface in the left conductor. Its density Jσ is the current that flows through the line element ds. In the right conductor the current flows in the opposite direction. 3 ×10 7 0.78 3.19 2.081 1.93 2.26 4.870×10 1.4 1.19 4.2 7 0.65 3.4 0. Table 4.90 2.3 1. Characteristics of some good conductors Skin Depth Material Aluminum 1 1 δf 2 [m Hz 2 ] γ [S/m] 50 Hz [cm] 1 kHz [mm] 1 MHz [mm] 3 GHz [µm] 3.38 0. The electric field Ez (x = 0) is found from (4. depth x and frequency (through the parameter h/δ.1.014 0.0117 2.15 2.7 0.50 50. (I/w).1 0.86 ×10 Starting from the expressions of the current density and of the electric field it is possible to compute the metal surface impedance.12 7 0.3 5.171 3. defined as the ratio between the electric field Ez at the interface x = 0 and the current density per unit length along y.3.06 2.50 ×10 7 0.59 29 Nickel Gold Brass Copper Tin Zinc 1.014 0.6) and 64 .075 1.59 22.085 1.126 1.59 ×10 7 0.Renato Orta .085 1.98 0.14 0.117 1.6 0.075 1.171 2.15 ×10 Chromium 3.0 ×105 1.066 1.066 0.8 ×107 0.064 0.80 ×10 7 0.Transmission Line Theory (Nov.41 0.6 7 0. which is small at low frequency).03 0.064 1.081 1.41 5.70 0. Plot of the current density Jz vs.2 Silver 6.54 ×107 0.5 Graphite 1.126 2. 2012) Figure 4. since the various elements of the conductor are in parallel.2 0.01 1. 1.9) becomes Z = R + jωLi ≈ 65 2Rs (1 + j) w (4.6) simplifies and becomes an exponential.995 0 1 0. This surface resistance depends on frequency and is measured in Ω.8) γ γ w Hence the surface impedance.Renato Orta .5. It is to be noted that this impedance per unit length coincides with the series impedance of the equivalent circuit of an element ∆z of transmission line (see Fig.8 1 x/h x/h Figure 4. if the frequency is very high. In particular. (4.9) where we have introduced the parameter Rs Rs = 1 = γδ ωµ 2γ (4.Transmission Line Theory (Nov. At the end of this chapter we will see that traditionally its numerical value is expressed in “Ω per square”. which actually coincides with R{Z} only if h δ.4). Plot of Jz (x) for h/δ = 10.6) is valid for all frequencies. which is much smaller than in the left figure. The expression (4.6 0. the factor 2 in Eq. Finally.9) is proportional to the internal inductance. The imaginary part of Z in (4.005 5 1 0 0 0.025 15 J z (x ) J z (x ) I (wh ) 10 I (wh ) 1. Figure 4.2 0. The normalization quantity I/(wh) is the average current density. Jz (x) ≈ I (1 + j) x I T exp (−T x) = exp − (1 + j) w w δ δ (4. The expression of the conduction current density (4. the impedance per unit length along z has the value Z/w. associated to the magnetic flux inside the conductor. If the conductor has width w. is Ez (x = 0) = Z T coth (T h) = γ = R + jωLi = 2 = h h 1+j coth (1 + j) = 2Rs (1 + j) coth (1 + j) 2 γδ δ δ (4.02 1. the skin depth δ is small and h/δ 1. 4.10) called surface resistance.4 0.015 1. per unit length in the z direction and per unit width in the y direction.12) . Plot of Jz (x) for h/δ = 0.9) takes into account the presence of two identical conductors. (Ω/ ).1): 1 T I Jz (x = 0) = coth (T h) (4.5.4 0.3).4.6 0.11) (see also Fig. related to the magnetic flux in the dielectric between the conductors. 2012) (4.8 0. 1. apart from the contribution of the external inductance. so that the conductor behaves as if it had infinite thickness. while (4. Note the range on the vertical axis. the resistance per unit length of each conductor is given by (4.15) γwh for each conductor.5 2 2. dashed line: imaginary part. Fig.13) If we substitute this into (4.5 1.5 1 1 0. The normalization impedance is the surface resistance Rs in (a) and the dc resistance Rdc in (b).6b shows a similar plot.5 1 1. i.5 0. Normalized series impedance of the planar line.12) Rs 1 R= = (4.16) we can derive the following interpretation of the skin depth δ: at high frequency.5 Zs / Rs 3 1. 2012) Conversely.6. In this low frequency condition the resistance per unit length has the value 1 R= (4. normalized to the surface resistance Rs versus the normalized thickness h/δ. with the direct current resistance Rdc . Since wh is the conductor cross-section area. the resistance per unit length is the same as that a direct current would feel flowing with uniform density in layer with thickness δ. when the frequency is sufficiently low.15) and (4.5 3 Normalized thickness h /δ (b) Figure 4.5 2 2. h/δ 1. as is to be expected. Fig. at high frequency. 4. In these conditions the current flows with almost uniform density in the whole conductor cross-section (see Fig.16) w γδw By comparing (4. 4. but the impedance is normalized to the dc resistance Rdc = 1/ (γwh).7) of δ has been used and the factor 2 refers always to the presence of two identical conductors that contribute to the result. Solid line: real part.9) by recalling the small argument expansion of the hyperbolic cotangent coth (z) ∼ = 1 z + .9).Transmission Line Theory (Nov. We note that the normalized resistance becomes very 2. this result coincides.5 0 0. Actually.Renato Orta .e. z 3 (4. large at low frequency.5 1 1.9). given by (4. the absolute resistance tends to the finite value Rdc (as it is evident from 66 .5 3 Normalized thickness h /δ (a) 0.6a shows a plot of the impedance per unit length.5) and the impedance per unit length can be obtained from (4. 4. On the contrary. when the conductor thickness is much larger than δ. we get ∼ = 1 h 2 (1 + j) δ + (1 + j) = w γδ h (1 + j) 3 δ = Z 2 1 2h +j 2 γhw 3γ w =2 1 ωµh 1 +j γhw 3 w (4.5 2 2 Zs / Rdc 3 2.14) = R + jωLi where the expression (4. 18) Le = w whereas the dc internal inductance is 1µ h (4. this equivalent inductance depends on frequency.19) Li0 = 3w and even smaller if the frequency increases.(4. The frequency on the horizontal axis is normalized to the demarcation frequency fd .7. we find that Rs = Rdc f fd Fig. 67 .17) By exploiting the previous equations. 4.9 shows a plot of the internal inductance. Since δ depends on ω. Since in general d h. the internal inductance is negligible with respect to the external one. Fig.0 10 2. versus the normalized thickness h/δ. Eq. 4. has been chosen to be that for which δ = h: fd = 1 πµγh2 (4. but the asymptotes relative to the low and high frequency behavior are added.5 2 2 δ/h γhR s 3 1.7b shows a plot of the skin depth vs. They cross at h/δ = 1. (b) Normalized skin depth δ/h. the normalized frequency.5 1 1 0.0 7. the external inductance is given by µd (4.Renato Orta . this condition determines the demarcation frequency fd that separates the low and high frequency regimes. as shown in Fig. 4.7a.5 0. Indeed.5 2. This figure shows a plot of the normalized surface resistance Rs h = Rs γh = Rdc δ versus the normalized frequency f /fd . 4. 4.5 5.5 Normalized frequency (a) 0 0.5 Normalized frequency (b) 10. 4.Transmission Line Theory (Nov.14) shows that it approaches zero as ω → 0.8 shows again the plot of Fig.5 0 0.5 1. 2012) 3 2. Fig. Figure 4. (a) Normalized surface resistance Rs γh = h/δ.6b. normalized to the dc value.5 5.6b) whereas the surface resistance Rs goes to zero. also called demarcation frequency. As far as the series reactance is concerned.0 2. where the normalization frequency fd . Since δ is a function of frequency. Fig.0 7. Note that the internal inductance is always small with respect to the external one. 5 0 0 0.5 2 2.Transmission Line Theory (Nov.5 1 0.3.5 1 1. 4. The asymptotic behaviors are also plotted and define the demarcation frequency.1 Loss parameters of some transmission lines Coaxial cable Dielectric losses G= 2πγd D log d 68 (4.4 0.5 Normalized thickness h /δ 3 Figure 4.Renato Orta .2 0 0 2 4 6 Normalized thickness h/δ 8 10 Figure 4.8. 2012) 3 R / Rdc 2. normalized to the dc value. 1 0. normalized to Rdc .8 0. Internal inductance Li /Li0 of a planar line.6 0. Real part of the series impedance per unit length.9.3 4.5 2 1.20) . versus the normalized thickness h/δ. Metal losses • Low frequency Resistance per unit length: R= 1 1 + 2 d2 De − D 2 1 π 4 (4. 69 .(4. in the case of a planar line. D = j2 √ 2δs 2δs and J0 . The same interpretation was already given in connection with Eq. When the skin effect is well developed. The equivalent width of the conductor is 1/πd for the inner conductor and 1/πD for the outer one: these quantities are obviously the circumferences of the conductors. 3 d = j2 √ • High frequency R + jωLi = Rs 1+j π 1 1 + d D (4. K1 are modified bessel functions.22) • Medium frequency 3 R + jωLi = Rs j 2 √ 2J0 d πdJ1 d 1 + Rs j 2 √ 2K0 D πDK1 D (4.24) This formula has a simple interpretation. the series impedance is the same as the one we would have if the whole current flew with uniform density in a layer one skin depth thick.Renato Orta . Coaxial cable.10.23) where 1 d D . 2012) D d De Figure 4. J1 are Bessel functions of first kind and K0 .21) γc Internal inductance: µ Li = 1− 4π D De 2 −2 2 D De − 1 − 2 log D De (4.16).Transmission Line Theory (Nov. δ J w w Figure 4. 2012) 4. • High frequency R + jωLi = 2Rs 1+j πd (4. Prismatic conductor with length and width w and thickness δ.25) D d −1 d D Figure 4. refer to Fig.27) πdJ1 d where 3 d = j2 √ d 2δs e J0 .11.26) • Medium frequency (if D >> d) 3 R + jωLi = Rs j 2 √ 2J0 d (4.3.12 where a prismatic conductor with length and width w and thickness δ is considered. in order to understand why the surface resistance Rs is measured in Ω/ .Transmission Line Theory (Nov. 4.Renato Orta .12.28) Finally. Metal losses • Low frequency R= 2 πd2 γc Li = µ 4π (4. 70 .2 Two-wire line Dielectric losses πγd G= cosh (4. J1 are bessel functions of first kind. Two-wire line. every square. has the same resistance.Renato Orta . 71 . with arbitrary side. the surface resistance has the value R= l w 1 = = γS γδw γδ which is independent from the sides of the square. 2012) We said that the surface impedance is the same we would have for a uniform current flow in a layer of thickness δ. In such conditions. Hence.Transmission Line Theory (Nov. i.t) = G v(z.ω)   dz =    − d I(z.ω) We could now repeat step by step the analysis carried out for the ideal lines.2) take the form   − d V (z.ω) (5. to understand the role played by the parameters R (conductor resistance per unit length) and G (dielectric conductance per unit length).Chapter 5 Lossy transmission line circuits 5.t) + C ∂ v(z.ω) formally identical to that of ideal lines. we get the real transmission line equations in the spectral domain. but it is simpler to resort to the trick of introducing a complex inductance and capacitance per unit length Lc = L+ R R =L−j jω ω Cc = C+ G G =C−j jω ω in such a way that Eq.t) ∂t ∂t Fourier transforming both sides. The relevant equations are   − ∂ v(z.t)   ∂t ∂t (5.ω)    − d I(z. It is just enough to take the solution of the ideal case and obtain its “analytic continuation” from the real values L e C to the complex ones (Lc and Cc ).1)    − ∂ i(z.e. that is   − d V (z.t) + L ∂ i(z. Note that the 72 .ω) dz = jω Cc V (z.2) (G + jω C) V (z.(5. we turn back to the complete equations.t) = R i(z. without losses.1 Solution of transmission line equations After explaining in detail the analysis technique of circuits containing ideal transmission lines.ω)   dz = jω Lc I(z.ω) dz = (R + jω L) I(z. and hence has a phase between −π and 0: −π < arg(k2 ) ≤ 0 73 .Transmission Line Theory (Nov. shown in Fig.ω) = V0+ (ω) e−jkz + V0− (ω) e+jkz I(z. in the equivalent circuit of an elementary length ∆z.4) is given by the product of two factors with phase between −π/2 and 0. As for the propagation constant. Hence.G. substituting the expressions of the complex inductance and capacitance.L.3) when k and Y∞ are complex. the case of practical interest is that in which Lc e Cc have very small imaginary parts.3) holds for any value of R. respectively.1.1. 5.ω) = Y∞ V0+ (ω) e−jkz − Y∞ V0− (ω) e+jkz (5. C ∆z G ∆z Time constants of the RL and RC groups of an elementary length of transmission line equivalent inductance and capacitance Lc e Cc can be written Lc = L 1−j 1 ωτs Cc = C 1−j 1 ωτp where τs = L/R and τp = C/G can be viewed as the time constants of the series RL group and of the parallel RC group.Renato Orta .5) and. the quantity below the square root sign in (5. k = ω L−j R ω C−j G ω √ = ω LC 1−j 1 ωτs 1−j 1 ωτp (5.3) where the (complex) propagation constant is √ k = ω Lc Cc (5. Obviously the time constants τs . 2012) τs τp R∆z R L∆z ∆ Figure 5.4) and the (complex) characteristic admittance is Y∞ = 1 = Z∞ Lc Cc −1 (5. Let us analyze now the properties of (5. the expressions of the voltage and current on a lossy line are given by V (z.6) Y∞ = 1 = Z∞ G C −jω L−jR ω = 1 1 − j ωτp C L 1 1 − j ωτs Observe that even if the solution (5. τp go to infinity for an ideal transmission line.C. 13). in which we had chosen k = ω LC. To ascertain whether k belongs to the fourth or to the second √ quadrant. we can take the limit for R.2. 5. instead of the propagation constant k = β − jα. Moreover.2. one introduces γ = jk = α + jβ in terms of which the general expression of the line voltage.(3. as shown in Fig.11) . Sometimes. become hyperbolic functions of γl. Re 2 Y∞ complex plane (left) and Y∞ complex plane (right) Computing the square root yields the values ±k = ±(β − jα) with β ≥ 0 and α ≥ 0. Re{k} > 0. Im Im 2 ∞ Y∞ Y Y∞2 Y∞ Re Figure 5. we choose Y∞ with positive real part. see Fig. Hence. i. As for the characteristic admittance.3.3.5) belongs to the right halfplane: for continuity with the case of a lossless line. This means also that Im{k} < 0: this choice agrees also with the fact that the forward wave must attenuate for increasing z.ABCD of a line length. the elements of the matrices Z. In these notes we will always use the phase constant k. we recall that Y∞ is the input admittance of a semi-infinite line: since it is a passive load.G → 0. given in (3.Transmission Line Theory (Nov. instead of circular trigonometric of kl.Renato Orta . 74 . by continuity. in the lossy case k belongs to the fourth quadrant.Y . for example. the real part of its admittance must be positive. 2012) Im Im k2 k −k Re Re k2 k k2 complex plane (left) and k complex plane (right) Figure 5. which is the ideal line case.e. is V (z) = V0+ e−γz + V0− eγz Moreover. 5. the radicand in (5. This choice is natural when transients are studied and the line equations are solved by the Laplace transform technique instead of the Fourier transform. we compute the time evolution of voltage and current relative to the first term of (5. where the real and imaginary parts of the complex propagation constant have been introduced.3). 75 . From the analysis of (5. which represented a forward wave in the case of ideal lines.8) and (5.3) represents a wave traveling in the direction of increasing z with phase velocity vf = ω0 ω0 = β0 Re{k0 } Note that vf depends on the value of ω0 . • The wave amplitude has an exponential decrease vs.Transmission Line Theory (Nov.9) we can conclude that: • The first term of (5. z. Fig. since Re{k0 } is a nonlinear function of frequency. The inverse of the imaginary part of the propagation constant (α0 ) is the distance over which the amplitude undergoes a decrease of the factor 1/e = 0. 2012) v + ( z0 . z at time t = t0 . which refers to a loss-less line.3) as phasor equations. for which the following inverse transform formula holds: v + (z.10. Time evolution of the forward voltage wave at point z = z0 To understand better the meaning of the solution of lossy transmission line equations.7) = | V0+ | cos(ω0 t − β0 z + arg(V0+ )) e −α0 z and. Note that it is identical to the plot of Fig.8) assuming k0 = k(ω0 ) = β0 − jα0 .4 shows a plot of the time evolution of the forward voltage wave in a specific point z = z0 . 1. Fig.5 shows instead a plot of the same wave vs.ω) ejω0 t } In this way we obtain the expression of the forward wave in the form v + (z.t) = + Re{| V0+ | ej arg(V0 ) e−j(β0 −jα0 )z ejω0 t } (5. 5.36788 = 8. for the forward current: i+ (z.68589 dB of voltage or current (see Fig.t) + = Re{| Y∞ | ej arg(Y∞ ) | V0+ | ej arg(V0 ) e−j(β0 −jα0 )z ejω0 t } = | Y∞ || V0+ | cos(ω0 t − β0 z + arg(Y∞ ) + arg(V0+ )) e−α0 z (5. We interpret (5. t ) V 0+ T t Figure 5. 5. as it can be expected.Renato Orta .t) = Re{V + (z. because of the power dissipation taking place on the lossy line.4.6). 5. Renato Orta . t0 ) V 0+ λ z Figure 5. we have V + (z) | V + (z) | def def = 20 log10 = 20 log10 e−α0 z = −α0 z 20 log10 e = −α0dB z V + (0) dB | V + (0) | 76 .ω ) V0+ e −1 z 1 α0 Figure 5. • The measurements units of β0 and α0 are β0 → rad/m α0 → Np/m or dB/m As for α0 .5.Transmission Line Theory (Nov. 2012) v + ( z. Forward voltage wave vs z at time t = t0 V + ( z . since V + (z) | V + (z) | def = ln = ln e−α0 z = −α0 z + (0) V | V + (0) | Np it is natural to express α0 in Np/m. If we express the voltage ratio in dB. defined as usual as the spatial period of the wave. is given by λ= 2π 2π = β0 Re{k0 } • The current is proportional to the voltage. Note also that Y∞ depends on frequency. z-plot of the amplitude of the forward voltage wave on a lossy line • The wavelength.6. but shows the phase shift arg(Y∞ ) with respect to it. We can observe that the crests are parallel to the straight lines z = ±vf t where the upper sign refers to the forward wave and the lower to the backward one. + Vg Plot of the backward voltage wave vs z Zg ZL A Figure 5. in which the natural evolution of the phenomenon takes place. The time domain expressions of the voltage and current backward wave are v − (z. Actually.t) = − | Y∞ || V0− | cos(ω0 t + β0 z + arg(Y∞ ) + arg(V0− )) eα0 z The plot of the backward voltage wave vs.68589. The same considerations can be carried out for the second term of (5. It is in this direction. which represents a backward wave identical to the forward one (apart. 5. Fig. in this reference the amplitude decays as exp{−ω0 α0 t/β0 }. 77 .8 the forward wave is created in A by the generator.7. 2012) v − (z . of course. from the propagation direction).8. z = −vf t = −ω0 t/β0 . because of the reflection symmetry of the transmission line. The presence in these expressions of an exponential that increases with z seems to contradict the dissipative character of the lossy line.t) = | V0− | cos(ω0 t + β0 z + arg(V0− )) eα0 z i− (z.7.3). in Fig. 5. t 0 ) V0− z Figure 5. B Lossy transmission line loaded with a generic impedance Hence.Transmission Line Theory (Nov. z that the amplitude of the backward wave reduces. whereas the backward wave is excited in B at the load position and then it propagates in the backward direction −ˆ. z at the time t = t0 is shown in Fig. the conversion factor for the attenuation constant is 20 log10 e = 8.Renato Orta . 5. The same conclusion can be reached by introducing the reference in which the backward wave is at rest.9 shows the space-time plots of the forward and backward voltage waves. 78 . This fact justifies the use of the symbol Z∞ . This result has also an intuitive explanation.10.9. v(z. 2012) 5 v(z.2 (a). For this reason.Transmission Line Theory (Nov. 1 0 0 −5 −1 2 2 1 z/λ 1 0 0 1 z/λ 3 2 0 t/T Figure 5. so that V Γ traces a logarithmic spiral in the complex plane. with the origin as pole. Actually the generator power is only partially delivered to the load: the rest is dissipated in the line. the transformation law of V Γ(z) = V Γ(0) ej2kz = V V Γ becomes Γ(0) ej2βz e2αz We can observe that if we move from the load toward the line input. it is useful to introduce the notion of reflection coefficient: V Γ(z) = V − (z) V + (z) In the case of a lossy transmission line. Plot of the curve V Γ(z) for α/β = 0. Indeed. 0 1 2 3 t/T Space-time plots of the forward and backward voltage waves (a ) (c ) (b ) Figure 5. t) progr.10. there is also an appreciable contribution of the backward wave. 0 (c) In Chapter 3 we have seen that in the analysis of circuits containing transmission lines. as shown in Fig. created in B by the load mismatch. apart from the forward wave. 5.1 (b). we can say that the input impedance of a semi-infinite real (lossy) line coincides with its characteristic impedance. the backward wave in A is negligible and the line appears to be matched.Renato Orta . the fact that the input impedance of a line is different from Z∞ means that in A. 0.t) regr. both the phase and the amplitude of the reflection coefficient decrease. originally produced by the generator. If the product of the attenuation constant times the line length is very large (α0 l → ∞). There is also a physical explanation: in each point of the line an active power flow exists even if the load is lossless.Transmission Line Theory (Nov. PB is also the power delivered to the load ZL .11. because it is algebraically impossible to write the voltage on the line as the product of a function of t times a function of z. If this is small (αl << 1) then the effects of losses can be neglected altogether. is dissipated in the line length comprised between the point under consideration and the load. connected to a reactive load. hence we must always consider the quantity αl.9) G 1− | V Γ(z) |2 + 2B Im{ V Γ(z)} Where Y∞ = G + jB is the characteristic admittance. in the points A and B: PB = PA 1 G 2 1 G 2 + | VB |2 (1− | + | VA |2 (1− | V Γ B |2 ) 1− | = e−2αl V Γ |2 ) 1− | A 79 V Γ B |2 2 A | VΓ (5. This power.2 Computation of the power flow The general formula that allows the computation of the power flow in each point of any transmission line has been derived in Chapter 3 and is reported her for sake of convenience: = 1 1 Re{V (z) I ∗ (z)} = G | V + (z) |2 − | V − (z) |2 + 2B Im{V +∗ (z)V − (z)} = 2 2 = P (z) 1 | V + (z) |2 2 (5.10) twice. 5. a purely stationary wave is established on it and the net power flux is zero. obviously. We can ask ourselves if also on a lossy transmission line. rigorously.11. A line is defined to be a low-loss line if B << G. we note that it is always multiplied times the line length. since e−αl 1 P (z) = When this condition is not satisfied losses must be accounted for and both | V + (z) | and | functions of z. 2012) + Vg Zg ZL A Figure 5.10) 2 As for the propagation constant.11) . so that also the power flow changes from point to point of the line.(5. The power flow in this case can be computed by the formula that. and the characteristic admittance can be taken as real. The ratio PB /PA is found readily by applying Eq. The answer is no. 5. V Γ(z) | are Apply now this formula to the circuit of Fig. Denote by PA (PB ) the net power flowing in A(B). obviously. holds only in the case of ideal lossless lines: 1 | V + (z) |2 G 1− | V Γ(z) |2 (5. B Length of lossy transmission line terminated with an arbitrary load impedance We have seen in Chapter 3 that when an ideal line is connected to a reactive load. a purely stationary wave can be formed.Renato Orta . which is defined nominal attenuation.11) is expressed in dB: PB PA = −αdB l + (1− | V ΓB |2 )dB − (1− | V ΓA |2 )dB dB The Smith chart is provided with a scale that allows the fast evaluation the attenuation increase due to the line mismatch. Considering the equations (5.3 Frequency dependence of phase constant and characteristic impedance The expressions (5. 2012) where we have used the equation + + | VB |=| VA | e−αl Moreover | V ΓA |=| V ΓB | e−2αl If the load impedance ZL coincides with the characteristic impedance of the line (matched line).7). where ωτs By recalling that if | x | 1. R. since the fraction that describes the additional attenuation due to the mismatch is always less than 1. where ωτs 1 and ωτp R ωL • a low frequency range. For the high frequency range we find k = √ ω LC 1−j √ 1 ω LC 1 − j 2 80 R ωL 1−j R G + ωL ωC G ωC . V ΓB = 0 and the ratio PB /PA equals the factor exp(−2αl). In this section we analyze it. we recognize that we can define two frequency ranges: • a high frequency range. by assuming that the primary constants L. G do not depend on frequency.Renato Orta . as it would be rigorously true only in the case of a lossless line. The amount of power dissipated in the line length AB is readily found by taking into account the energy conservation: Pdiss = PA 1 − PB PA Often the expression (5. i.e.Transmission Line Theory (Nov. This amounts to neglecting the dielectric dispersion and the frequency dependence of the skin effect (see Chapter 4). If the load impedance is arbitrary (mismatched line) the ratio PB /PA reduces. G ωC 1 1 and ωτp 1 1 1: 1+ 1 (1 + x)− 2 1 x 2 1− 1 (1 + x) 2 1 x 2 it is possible to obtain simple approximate expressions valid for each range. 5. Finally. if αl 1 the line has negligible losses and PB /PA ∼ 1.7) show clearly that both the phase constant and the characteristic admittance have a frequency dependence. C. (5.12) = G + jωC R + jωL √ −j RG 1+j Y∞ from which we get k = √ 1 −j RG 1 + j 2 ωL R 1+j ωL G ωL ωL + R G From this equation the expressions of β and α follow by inspection β (ω) = ω 2 α (ω) = √ RG C R +L G G R (5. which usually holds true in practice. always in the high frequency range. 2012) and hence β √ ω LC α 1 2 C +G L R L C As for the characteristic admittance. but the slope of the two straight lines is different. A simple computation shows that the low frequency slope is larger than the high frequency one if τp > τs .Renato Orta . we find Y∞ G 1 − j ωC C L = R 1 − j ωL C 1 1+j L 2 G R − ωL ωC from which C L G B C 1 L 2 j G R − ωL ωC We see that the terms R/ωL and G/ωC. To obtain the expressions for the low frequency range it is convenient to first rewrite (5. from Eq.13) Note that β (ω) is linear at both high and low frequency. As for the low frequency approximation of the characteristic admittance.13) we find Y∞ = G R 1 + j ωC G 1 + j ωL R ω G 1+j R 2 81 C L − G R . are summed in the expression of α but subtracted in that of Z∞ .Transmission Line Theory (Nov.7) as follows: k = −(R + jωL)(G + jωC) (5. beside being small by hypothesis. but the two constant values are different. since R L = Z∞ = G C Note that if the losses are so large that the imaginary part of the characteristic admittance B cannot be neglected. as it will be discussed in Chapter 8.5 mH/m G = 0.12 shows plots of β (ω).3 µS/m C = 5 nF/m The time constants have the values τp = 0. it can be shown that the line inductance per unit length is increased by the quantity L/d. the curve β (ω) becomes a straight line. where L is the inductor value and d their spacing.Renato Orta . We note that the imaginary part of Y∞ is maximum when the real part has the maximum slope. from Eq.7) must be used. which is a causal time function. α (ω). related only to the fact that Z∞ (ω) can be considered to be a transfer function. Since in practice the line parameters do not fulfill this condition. Z∞ is frequency independent. Indeed. This is a general property (Kramers Kr¨nig o relations). The other plots are instead of semi-log type. It is important to note that when τp = τs = τ . one can load the line with periodically spaced series inductors. the forward and backwards waves are no longer power orthogonal This implies that the amplitude of the reflection coefficient has no energy interpretation. 82 . while α (ω) is a constant. The imaginary part instead tends to zero in both regimes. B (ω) for a realistic transmission line with the following values of the primary constants: R = 25 Ω/m L = 2. If the spacing is much smaller than the wavelength. If the Heaviside condition is fulfilled. 2012) and hence the real and imaginary parts of the characteristic admittance are G R G B j ω 2 C L − G R G R We notice that both at low and high frequency the real part of the characteristic admittance is essentially frequency independent. In the intermediate frequency range no approximation is possible and the general expressions (5. Fig.7) it follows √ 1 k = ω LC 1 − j ωτ (5.Transmission Line Theory (Nov.0167s τs = 10−4 s The plot of β(ω) is of log-log type. (5. 5. so that the different slopes in the two frequency ranges is represented as a vertical translation.14) from which β (ω) = α (ω) = √ ω LC √ RG The condition τs = τp is called Heaviside condition and is very important since it guarantees distortion free propagation. that is the Fourier transform of the impulse response. G (ω). 01 0. B (ω) for a realistic transmission line. 2012) β =Re(k) ( rad / Km ) α =−Im(k) ( Np / Km ) 0. G (ω). Plots of β (ω).5 5 4 1 3 2 0.Renato Orta . α (ω).5 1 0 0 10 0 τ−1 105 s FREQUENCY ( Hz ) τ−1 0 10 p τ−1 τ−1 105 s FREQUENCY ( Hz ) p Figure 5.005 0 10 0 τ−1 τ−1 105 p s FREQUENCY ( Hz ) Re(Y∞) ( S ) −3 x 10 0 10 −1 τs 105 τ−1 p FREQUENCY ( Hz ) Im(Y∞) (S) −4 x 10 1. The values of the primary constants are specified in the text 83 .12.Transmission Line Theory (Nov.015 0 10 0. capable of eliminating the presence of reflected waves on the line. one is matching to the line. connected to a generator and a load.1.Chapter 6 Matching circuits 6.1. We have already analyzed this circuit in Section 3. 6. since the line is lossless: PB = PA = 1 1 ∗ Re{VA IA } = |IA |2 Re{Zin } = 2 2 84 .7. As for the latter. not specific of distributed parameter circuits. When a transmission line must be connected to a load with an impedance different from the characteristic impedance of the line. The power delivered to the load coincides with that absorbed by the input impedance Zin = ZA . The other type of matching. to find voltages and + Vg Zg ZL ZA Figure 6. i. Actually there are two types of matching. A B Circuit consisting of a lossless transmission line. the notion of impedance matching. currents in every point of the line. which can be realized either in lumped or distributed form.1 Introduction In this chapter we address a subject with great practical importance in the field of distributed parameter circuits. where a real generator and an arbitrary load are connected by a transmission line with negligible losses.e. it is necessary to introduce a matching device.2 Types of impedance matching Consider the circuit of Fig. has the property of allowing a generator to deliver its available power. several solutions will be described. 6. the other is matching to the generator. These two objectives can be reached by means of lossless impedance transformers. the power delivered to the load is maximum when the load is matched. so that Zin = Z∞ . 2012) = |Vg |2 1 Re{Zin } 2 |Zg + Zin |2 (6.2) we recognize the importance of a VSWR as close to one as possible. Indeed.6.Transmission Line Theory (Nov.2) Depending on the values of the internal impedance of the generator ZG and of the load impedance.1 the generator is fixed but the value of the input impedance Zin = ZA can be changed at will.1) recalling that che Re{Zin } = (Zin + Zin )/2: PB = ∗ |Vg |2 Zin + Zin ∗ + Z∗ ) 4 (Zg + Zin )(Zg in and differentiate with respect to Zin : ∗ |Vg |2 ∂PB Zin + Zg − Zin − Zin 1 = ∗ + Z∗ 2 ∂Zin 4 Zg (Zg + Zin ) in = ∗ Zg − Zin |Vg |2 1 ∗ ∗ 4 Zg + Zin (Zg + Zin )2 85 . since for every transmission line there is maximum voltage that must not be exceeded in order to avoid sparks that would destroy the line.Renato Orta . This remark is important in high power applications. We can ask what is the optimum value of Zin that allows the maximum power to ∗ be extracted from the generator. as well as that in A. B) Generator matching Suppose that in the circuit of Fig. Rewrite (6. the maximum line voltage Vmax has the minimum value when the load is matched to the line. The power delivered to the load is PB = 1 Z∞ |Vg |2 . fixing the active power delivered to the load PB . the standing wave diagram is flat (VSWR = 1) since only the forward wave is present on the line. express this power first in terms of the forward voltage PB = + 1 |VB |2 1 − | V ΓB |2 . the reflection coefficient in B is zero. two different cases can be considered: A) Matching of the load to the line If ZL = Z∞ . + Eliminating |VB | between the two equations we find: PB = 2 1 Vmax 1 . 2 |Z∞ + Zg |2 Observe that. 2 Z∞ The maximum voltage on the line is + Vmax = |VB |(1 + | V ΓB |). In this situation.1) The standing wave ratio (VSWR) on the line is given by S= 1 + | V ΓB | Vmax = Vmin 1 − | V ΓB | The power absorbed by the load can also be expressed in terms of the maximum voltage on the line. defined as uniformity matching. we can say that fixing the maximum voltage on the line. Alternatively. From (6. 2 Z∞ S (6. the locus reduces to the point V Γ∗ . as it will be discussed in Chapter 8. Moreover. 1 − (1 − m) (a2 + b2 ) mb . − b). 86 .2): x2 + y 2 − 2αx − 2βy + γ = 0 where α = ma . The optimum operating condition for the circuit of Fig.e. the Kurokawa reflection coefficient is coincident with the ordinary one. The power delivered in this case is the available power of the generator and has the value 1 |Vg |2 Pav = 2 4Rg where Rg = Re{Zg } is the internal resistance of the generator. The radius is √ 1 − m(1 − a2 − b2 ) r= . the voltage on the line is the minimum for that value of active power flow. Finally. a condition defined as conjugate matching. it is recognized as the usual reflection coefficient of the equivalent impedance Zeq = Zin + jXg with respect to Rg and hence can be determined graphically by means of the Smith chart. In other words. 6.e. i. in these conditions the generator delivers the maximum power. the delivered power is zero. lying on the segment joining the point coordinates (a.β). the line attenuation would be the minimum one and would be coincident with nominal one. It is interesting to note that the power delivered by a generator in an arbitrary load condition can be written as PB = Pav (1 − | k Γin |2 ) where k Γin is a generalized reflection coefficient of the impedance Zin with respect to the internal generator impedance. introduced by Kurokawa: k Γin = ∗ Zin − Zg Zin + jXg − Rg = Zin + Zg Zin + jXg + Rg Note that when Zg is real. whereas it is a different concept when Zg is complex. 1 − (1 − m) (a2 + b2 ) β = − γ = m − 1 + a2 + b 2 .Transmission Line Theory (Nov. i. to the origin. the locus is the unit circumference with center in the origin of the plane V Γin : this result is obvious. energy matching and line matching are independent. Indeed. with In particular. the VSWR can be greater than one. because of the line matching. When m = 1. the generator delivers its available power. is the circumference with equation (see Fig. 1 − (1 − m) (a2 + b2 ) V ∗ Γg .1 is that both the load and the generator are matched to the line. Zg − Z∞ Γg = = a + jb. 6. since it is related to | V ΓB |. In the rest of this chapter we will show how to design impedance transformers that allow the matching condition to be reached. 2012) ∗ This derivative is zero for Zin = Zg . If we now set V V Γin = x + jy. It can be readily checked that it corresponds to a maximum.Renato Orta . However. the line matching condition is essential to minimize distortions. since the corresponding Zin is a pure reactance. If the losses were not negligible. g Note that when the energy matching condition holds. Zg + Z∞ it can be proved that the locus in the plane V Γin of the points for which it is PB /Pav = m with m a constant. 1 − (1 − m) (a2 + b2 ) The center is in the point with coordinates (α. when m = 0. 6. Scheme of impedance transformer.2. 6.Renato Orta . Impedance matching devices First of all. 2012) m=0 m=0.3. First we address the simplest case of single frequency matching.8 m=1 Figure 6.3.3.5(−1 + j). Zin is the charac- ZL Z ing Figure 6. We have seen in the previous section that for several reasons it is useful to be able to design impedance transformers that perform as indicated in Fig. i. There are various solutions to this problem. teristic impedance Z∞ of the feeding line. We will discuss • “L” cells with lumped reactive • Single stub cells • Double and triple stub cells • λ/4 impedance transformers 6. all consisting of ideally lossless networks.4) For certain combinations of Zin and ZL both can be used. If the network contains more than two independent elements. In the case of matching to the line. one on the real and one on the imaginary part of the input impedance.Transmission Line Theory (Nov. at several frequencies or on a frequency band. multiple matching conditions can be enforced. with V Γg = 0.2 m=0. since two conditions must be enforced.e. In the case of conjugate matching.6.3 PB /Pav = m loci on the Smith chart. we observe that a matching network must be formed by at least two components.1 L cells with lumped reactive elements Two structures are possible (see Fig.6 m=0. Zin is the complex conjugate of the generator internal impedance.4 m=0. for 87 . The condition to be enforced at the input terminals is 1 Rin + jXin = jX + 1 jB + RL +jXL This is a complex equation in the two real unknowns B and X.3). The two possible schemes of L matching networks.4b. we must use the configuration of Fig. others only one of them. Consider the first configuration. we find = jX + RL + jXL = 1 − BXL + jBRL = jX + (RL + jXL )[(1 − BXL − jBRL ] = (1 − BXL )2 + (BRL )2 = jX + j + RL (1 − BXL ) − BRL XL = (1 − BXL )2 + (BRL )2 = RL + 2 2 B 2 (RL + XL ) − 2BXL + 1 + RHS j 2 XL (1 − BXL ) − BRL + (1 − BXL )2 + (BRL )2 X+ 2 2 XL − B(RL + XL ) 2 2 B 2 (RL + XL ) − 2BXL + 1 Enforcing the equality of left and right hand side yields Rin Xin = RL 2 2 B 2 (RL + XL ) − 2BXL + 1 = 2 2 XL − B(RL + XL ) X+ 2 2 2 B (RL + XL ) − 2BXL + 1 (6. In this case the condition to 88 . After some algebra.Renato Orta .4. 2012) jX jX ZL jB jB Z in ZL Z in (a) (b) Figure 6.Transmission Line Theory (Nov. which can be solved by separating real and imaginary part of the right hand side (RHS).4) The corresponding X values are found from the second of (6.3) From the first we obtain a quadratic equation in B: 2 2 (RL + XL )B 2 − 2XL B + (1 − RL )=0 Rin with solutions B= 1 2 2 RL + XL XL ± RL Rin 2 2 RL + XL − Rin RL (6. 6. Obviously the square root must be real: if this condition is not satisfied. Renato Orta .Transmission Line Theory (Nov. 287). for frequencies in the microwave range. this region is the part of the right half-plane lying outside of the circle with radius RL /2 and center in the point (RL .4) we see that the radicand is positive. and matching 89 . the formulas will simplify because Xin = 0. This means that with present day technology this matching technique can be used up to some GHz (see [4] p. It is interesting to note that the problem can also be solved graphically by means of the Smith chart. The susceptance B and the reactance X can be realized by lumped elements (inductors and capacitors) if the frequency is low enough. if X = −XL ± 2 2 Rin + Xin − RL Rin Rin ≤ Rc = 2 2 RL + XL RL Geometrically.2 Resistive matching pad The matching network described in the previous section has the advantage of being lossless (in the case of ideal components) and the disadvantage of being narrowband. From eq. as in Fig.6). 6. and hence the circuit of Fig. From eq. as discussed in section 3. Suppose that ZL is specified.6.2. 2012) enforce is 1 1 = jB + Rin + jXin jX + (RL + jXL ) from which RL − j(X + XL ) Rin − jXin 2 2 = jB + 2 Rin + Xin RL + (X + XL )2 By equating real and imaginary parts of the two sides we get the two equations Rin 2 2 Rin + Xin − Xin 2 2 Rin + Xin = RL 2 RL + (X + XL )2 = B− From the first we get X: (X + XL )2 = X + XL 2 RL + (X + XL )2 (6. in the plane Rin . the radicand is positive for 2 2 Rin + Xin − RL Rin ≥ 0 Geometrically. In some cases. In this way we have solved the matching problem in the most general case. Fig. the radicand of the square root must be positive.0). Obviously.6) Rin Finally.4a can be used. This is the case when two matched transmission lines of different characteristic impedances have to be connected. We see that matching is possible only with the circuit of type a for Zin inside the circle and only with the circuit of type b for Zin to the right of the vertical line. this region is the strip comprised between the imaginary axis and the vertical line Rin = Rc . from the second of (6. it is so important to have a wideband behavior that a certain amount of losses is accepted. 6. consider the circuit of Fig.3. It is interesting to ascertain for which combinations of load and input impedance each form of the L circuit can be used.5 shows these regions. have a purely reactive input impedance.6.5) RL 2 2 2 (Rin + Xin ) − RL Rin from which RL (6. terminated in short circuit or open circuit which. Also in this case.5) we get B. 6.4b. both circuits can be used. B e X can be realized with transmission line lengths. in the case of line matching. The upper limit can be identified as the frequency for which the component size is of the order of λ/10. Alternatively. Xin . Next.(6. For other values of desired input impedance. (6. 6. only network b. we get R1 = Z∞1 − 2 2 (Z∞1 − Z∞2 )R2 − Z∞1 Z∞2 = 0 and finally    R1 = Z∞1    1−    R2 = Z∞2   Z ∞1 1− R1 R2 Figure 6. The two equations to be enforced are Z∞1 = R1 + R2 //Z∞2 (6.Transmission Line Theory (Nov. 2012) y Ŷŝ ď Ă Z Z Đ Ŷŝ Figure 6. for Zin to the right of the vertical line.e. Realizability of L matching networks. is required in the two directions. After a bit of algebra.5. For Zin in the circle.6. only network a can be used. from left to right and from right to left. Z∞2 Z∞1 Z∞2 Z∞1 −1 Z∞2 Resistive matching pad 90 . i.7) Z∞2 = R2 //(R1 + Z∞1 ) that is   Z∞1 = R1 + R2 Z∞2   R2 + Z∞2   Z∞2 = R2 (R1 + Z∞1 )  R2 + (R1 + Z∞1 ) Solve the first with respect to R1 : R2 Z∞2 R2 + Z∞2 and substitute in the second.Renato Orta . 27 → −5. The power transmission coefficient is then P out Z∞1 = P in Z∞2 1− Z∞2 Z∞1 2 Z∞1 − Z∞2 = Z∞1 −1 Z∞2 2 = Z∞1 + Z∞2 Z∞1 −1 Z∞2 −2 From the last expression it is evident that the transmission coefficient is always less than 1.3 Single stub matching network The matching networks of this type consist essentially of a a transmission line length and a reactance.8). Z∞ jbs A+ ZL B Matching network with shunt stub: matching to the line. called “stub”. 2012) Obviously. the active power leaving the device is P out = Y∞2 |V out |2 /2 and the two voltages are related by R2 //Z∞2 V in R1 V out = V in = R2 //Z∞2 = V in 1 − R1 + R2 //Z∞2 Z∞1 Z∞1 where the first of (6. 6.6. As a numerical example.7. the series branch of the matching device must be connected to the line with larger characteristic impedance.30 Ω   R2 = 86. called “minimum loss matching pad”. 6. We know that the locus on the admittance Smith chart of yA+ when lAB is changed is a circumference with center in the origin and radius equal to |ΓL |.7) has been used. be yA− = 1 when it is terminated with yL . This reactance is realized by another length of transmission line. The matching network is an impedance transformer: its normalized input admittance must Z∞ - Figure 6. consider the connection between a 75 Ω line to a 50 Ω one. In other words.Renato Orta .6. to match a load to the feeding transmission line. Both points refer to values of yA+ that have the required real part. one gets    R1 = 43.Transmission Line Theory (Nov. the configuration of Fig.7).72dB P in The configuration of Fig. (Fig. that can be connected in shunt or in series to the line itself.6 can be used only if Z∞1 > Z∞2 . Assume also that the matching network employs transmission lines with the same characteristic impedance as the feeding line. Likewise. terminated with an open or short circuit.60 Ω  P out    = 0. is not the only possible one: sometimes T and π configurations are used. 6.6.3. Using the previous expressions. Suppose that a shunt stub matching network is to be designed. In order to compute the attenuation introduced by this device. as it is to be expected. The net active power absorbed is P in = Y∞1 |V in |2 /2 since there is no reflection. so that the resistance values turn out to be real. From these points the 91 . assume that a voltage wave of amplitude V in is incident from the left. This circumference intersects the constant conductance circumference g = Re{yA+ } = 1 in the points I1 e I2 (see Fig. In this case the length of AB becomes 0. Since (l/λ)T G + = 0. 6.67 and equivalent electrical length (TG) (l/λ)T G = 0. having a shunt short circuited stub. but its bandwidth would be smaller. corresponding to the complex conjugate of the generator internal impedance. The procedure described above to design a line matching network can be generalized to solve the problem of designing a conjugate matching network.Transmission Line Theory (Nov.1843. 92 .8.0326)λ = 0. Using again the Smith chart. | I ΓL | = 0. Of course there is also the solution bs = 1. The matching network structure is the same as before: • a transmission line length that allows the desired real part of the input admittance to be obtained. Let us make reference to a shunt stub.0326 (see Fig. 2012) yL I1 y=1 zL Figure 6.Renato Orta . the length of AB has the value (0. In this case the arrival point on the Smith chart is not the origin but a generic point.84.10. read on the chart. 6. 6. In this case we would have employed an impedance Smith chart: no other change would be required. still short circuited. The length of the line AB is deduced from the angle between I ΓL and I1 (or I2 ). 6. The data are: ZL = 125 − 125j Ω e Z∞ = 50 Ω. The input impedance of the matching network would still be Z∞ at the design frequency.84j and then bs = −1.283λ and that of the stub.2 + 0. the design procedure would be only slightly modified.1843 − 0. I2 Smith chart relative to the design of the matching network of Fig. The rotation takes place on the unit circle from the short circuit point y → ∞ to the point ys = 0 + jbs . A similar remark holds for the stub. yL = 0. There is indeed a general rule: the bandwidth of a device is inversely related to its electrical length. We find ζL = ZL /Z∞ = 2.421λ.079λ.84j. The relevant Smith charts are shown in Fig.25)λ = 0.9 and Fig. From the intersection of the constant | I Γ| circle eqB with the Re{y} = 1 circle we read yA+ = 1 + 1.9). is ls = 0. Because bs < 0 we say that the stub is inductive.329 − 0. the stub length is readily found as soon as its termination (short or open circuit) has been chosen. The stub length is then ls = (0.10 The line AB could be lengthened by any multiple of λ/2.5j and. 6.7 origin of the Smith chart ( I Γ = 0) is reached by selecting the susceptance bs : bs = −Im{yA+ }. Example 1 Design a line matching network. If the stub were to be connected in series to the main line.1517λ. eqA The length of the stub is found from the Smith chart of Fig.84 (capacitive stub) corresponding to yA+ = 1 − 1.2j.5 − 2. 0326 yL l yA- AB λ ζL yA Figure 6. Indeed. now.7 only the points of the region Rd can be reached. 6. so that it has the desired value.9. 2012) AB λ l 0. In fact. it is evident from Fig. 6.Transmission Line Theory (Nov. + Smith chart relative to the design of the matching network of Example 1 • a shunt susceptance that modifies the imaginary part of the input admittance. The problem. with a matching network of the type shown in Fig. is that the first step is not always successful.1843 yA + 0. the real part of the input admittance is always comprised between gm and gM defined by the intersection of the |Γ| =constant 93 . for a given load admittance YL .11 that.Renato Orta . when the line length AB is changed. so that the line matching is always possible. 2012) yA + 0. Hence the region Rd is the part of plane internal to the circle Re{y} = gm and external to the circle Re{y} = gM . gM 1 + |ΓL | S where S is the load VSWR. it is simple to recognize that whatever the value of yL .10. Incidentally.329 Figure 6. We find readily gm = 1 − |ΓL | 1 1 = = . Smith chart relative to the design of the stub for the matching network of Example 1 circle through yL and the real axis. we can still use a stub matching 94 .25 s l λ bs 0. the origin belongs always to Rd .Renato Orta . ∗ When the admittance yg to be reached lies outside of the region Rd .Transmission Line Theory (Nov. 12. bs = 2. in this case.14. ∗ We compute ζg = 2 + 4j and ζL = 1 + j. being that of Fig.5 − 2j. The domain Rr is the annular region between the concentric circles with radii |Γ| = 1 and |Γ| = |gL − 1|/|gL + 1|.e.13).1 − 0. If we choose the other intersection yB − = 0. We see easily that the constant | I Γ| circle through yL and the constant conductance ∗ circle through yg have no intersections.5 − 0. 6.344λ.5 (capacitive stub). Indeed.5j. We see that the union of Rr and Rd equals the entire Smith chart.5 (capacitive stub). network. The data are: ZL = 75 + 75j Ω. The points of the region Rd represent the input admittances of a matching network of the type of Fig. for example. ZL jbs B “Reversed L” stub matching network. the values of yA− that can be obtained Z∞ Z∞ . provided the structure is reversed. 6. from yL are those belonging to the region Rr (see Fig. i.7.Transmission Line Theory (Nov. 6.Renato Orta . yB − = 0. loaded by yL . so that the solution for certain values of yA− can be obtained with both types of networks. 6. The intersection between ∗ the constant | I Γ| circle through yg with the constant conductance circle through yL defines two points. Zg = 150 − 300j Ω and Z∞ = 75 Ω. The length of AB is 0.2j and yL = 0.190λ. we obtain bs = −1. 6.147λ and ls = 0.12. hence every matching problem can be solved by a stub network (either straight or reversed L). The relevant Smith chart is shown in Fig. then the corresponding admittances are read on the Smith chart: ∗ yg = 0.+ A Figure 6. Example 2 Design a conjugate matching network with an open circuit shunt stub.11.289λ and that of the stub is ls = 0. and the length of AB becomes 0. as shown in Fig. of which. 95 . 2012) y L gM gm Rd Figure 6. Let us try a reversed L configuration.7 a “straight L ”. Moreover the intersection of Rr e Rd is not empty.5 + 2j. a “reversed L matching network” is used. hence only the reversed L configuration is possible. but are not guaranteed to exist. 2.17). 3. having found b2 . they are also called Π networks. 4. 2. this circle intersects the constant conductance circle through yL in the points I3 and I4 .Renato Orta . These points define yB − . Because of their form. even if the diagrams on the Smith chart are different in the two cases. In this case the solution are at most two. 96 . sometimes double stub networks are used. (see Fig.Transmission Line Theory (Nov. 6. 6. It is clear that the stub susceptances and their separation can be chosen in an infinite number of different ways. from which b2 = Im{yB − } − Im{yL } is found. obtain yA+ and therefrom b1 = Im{yA− } − Im{yA+ }. The points belonging to the region Rr represent input admittances of a stub matching network loaded by yL . rotate the whole circle by d/λ toward the generator: the locus of the corresponding yA+ is obtained. draw the constant conductance circle through yA− : this is the locus of all possible yA+ as the stub susceptance in A is changed. for example by the microstrip technology 6. The design can be carried out in two different ways. having fixed the value of b1 . Sometimes the distance AB is fixed a priori.15 in the case of a shunt stubs. this circle intersects the constant conductance circle through yA− in the points I1 and I2 . 4. starting from the load or from the generator. The reason for which only examples of shunt stubs have been discussed is that this type of connection is more common. shown in Fig.16) the procedure is the following: 1. 2012) yL gL Rr Figure 6. 6. rotate the whole circle by d/λ toward the load: the locus of the corresponding yB − is obtained. draw the constant conductance circle through yL : this is the locus of all possible yB − as the stub susceptance in B is changed.13. If we start from the load.4 Double stub matching network Even if the straight or reversed L matching networks can solve any practical problem. obtain yB − and then b2 = Im{yB − } − Im{yL }. is the following: 1. but starting form the generator (see Fig. 3. The procedure for the design of the same matching network. because it is easier to realize. Obviously.3. the values of b1 e b2 turn out to be the same. From either of these points move to yA− by means of the stub with susceptance b1 = Im{yA− } − Im{yA+ }. 18).Renato Orta .Transmission Line Theory (Nov. Smith chart relative to the design of the conjugate matching network (reversed L) discussed in Example 2. in fact. the solution for certain load and input admittances is not guaranteed to exist. If we want to design such a matching network. we can use the method described above. 2012) 0. The detailed 97 .321 λ l Figure 6. provided convenient stub lengths are selected.14. where the stub lengths are changed by means of sliding pistons. the desired matching can always be obtained. If the distance between the stubs is fixed a priori.468 AB 0. In this case. This limitation is not present in the case of a triple stub matching network. even if the relative distances are fixed a priori (see Fig. 6. This device can be useful in the laboratory: indeed.179 l y B− ζL AB λ ζ g∗ ∗ yg yL y B− 0. some implementation exist. 6)): ζA = from which ZA = 98 1 . The normalized input impedance is the inverse of the normalized load impedance (see Eq. The wavelength is to be evaluated at the design frequency. can be used only to match real impedances and consists of a λ/4 length of transmission line of suitable characteristic impedance. Double stub matching network yi yL I1 I2 d/λ Figure 6. From these nontrivial examples we can appreciate the power of the Smith cart as a design tool.Renato Orta .16. 6. The scheme is shown in Fig.(3. 3. procedure is the following: 1.5 λ/4 matching networks This type of matching network. 2012) A B ZL Figure 6. 2. draw the constant conductance circles through yL and yA− . They represent the loci of yB − and yB + . ζL 2 Z∞ .15. They meet in two points and the value of b2 is given by b2 = Im{yB − } − Im{yB + }. Design of a double stub matching network.3. rotate the first toward the generator by d2 /λ and the second toward the load by d1 /λ.Transmission Line Theory (Nov. having defined yB − and yB + . b1 e b3 . therefrom. respectively. starting form the load. obtain yA+ and yC − and. RL .19. 6. in its simplest form. the characteristic impedance of the line must be the geometric mean of the two resistances to be matched. the characteristic impedance of the central line BC is √ Z∞m = RB RC 99 .18. as shown in Fig. starting form the generator. Their purpose is that of transforming the complex impedances into pure resistances. inserted between two line lengths (of arbitrary characteristic impedance Z∞ ).Renato Orta .21. we can use a matching network consisting of a λ/4 line length. as indicated in Fig. 6. 6.17. jb2 jb3 Triple stub matching network.20. A B C ZL jb1 Figure 6. By enforcing the condition that the input impedance ZA coincides with the desired input resistance Ri . In conclusion. Design of a double stub matching network. In the case the two impedances to be matched are complex.Transmission Line Theory (Nov. we find √ Z ∞ = RL Ri . 2012) d/λ yi yL I 1 I2 Figure 6. If these resistances are called RB = Z∞ rB − and RC = Z∞ rC + . Renato Orta . 100 .20. B λ/4 matching network between real impedances.19. λ 4 Z1 A Figure 6. ζ L rB− rC+ ζin Figure 6. d2 Z∞ d1 Z in RL Z2 B C ZL D Structure of the λ/4 matching network for complex impedances. Smith chart for the design of the λ/4 matching network for complex impedances.Transmission Line Theory (Nov.21. 2012) Z∞ Ri A Figure 6. 1 Lumped circuits The simplest two-lead circuit element is characterized by its impedance ZL . 7. the admittance YL ). say N .1. Often a couple of leads of a device is called a “port”: hence. so that the impedance does not depend on the excitation (I) but only on frequency. to which we will limit our attention. in the case of distributed parameter circuits a change of basis is highly convenient: we will introduce the so called power waves.e.1). The relation (7. in the case of linear networks. defined as the ratio between the voltage V at its leads and the absorbed current I. As discussed at length in the previous Chapters. i. As known in circuit theory. First we review the matrix characterization of multiport devices based on the use of total voltage and total current as state variables. I1 I2 V2 V1 Figure 7. the two voltages V1 and V2 depend on both I1 and I2 : V1 = Z11 I1 + Z12 I2 V2 = Z21 I1 + Z22 I2 (7. Suppose this element is linear. these concepts can be generalized to the case of devices with several ports. (or its inverse. This description is appropriate to the case of lumped networks.1) where Zij are only functions of frequency. We start this presentation by focusing our attention to the important case of two-port devices (see Fig.Chapter 7 The Scattering matrix In this Chapter we develop a convenient formalism to describe distributed parameter circuits containing multiport devices. such a circuit element is also called a one-port device.7. Two-port device with the definitions of voltage and current at the ports In general.1) can be written in matrix form: [V ] = [Z][I] 101 . a normalized form of forward and backward waves. must be open circuited. except for the j-th.Renato Orta . an amplifier is non reciprocal. Another useful matrix characterization of two-port devices is that based on the equations V1 V1 I1 AV2 − BI2 I1 or = = CV2 − DI2 = A B V2 C D −I2 The relevant matrix is called ABCD matrix and is a kind of transmission matrix.k=j In other words. the [Y ] and [Z] matrices of lossless devices are pure imaginary. In the N = 2 case. the linear dependence between currents and voltages is expressed in the form: I1 = Y11 V1 + Y12 V2 I2 = Y21 V1 + Y22 V2 that is. the other elements are trans-impedances. in this case [Z] is a N × N complex matrix. The total power dissipated in the device is the sum of the powers entering through the various ports: N Pdiss = 1 1 1 ∗ ∗ ∗ ∗ {V1 I1 + V2 I2 + . In fact it relates the electric state at the input to that at the output of the device. Then Pdiss = 0 = 1 1 {[V ]T [I]∗ } = {[V ]T [Y ]∗ [V ]∗ } 2 2 Due to the arbitrariness of [V ]. ferrite devices). [I] = [Y ][V ] From the comparison with (??) we get [Y ] = [Z]−1 . + VN IN } = { Vi Ii } = {[V ]T [I]∗ } 2 2 i=1 2 If the device is lossless. The diagonal elements are input impedances. at which the exciting current is applied. [I] = [I1 I2 ]T are column vectors and the 2 × 2 matrix [Z] is called open circuit impedance matrix. derived from (7.Transmission Line Theory (Nov. transmission lines is always reciprocal.. also for an N -port structure we can introduce a short circuit admittance matrix [Y ]. It can be shown that the ABCD matrix of a reciprocal device has unit determinant. Recall that a circuit made of resistors. It can be shown [4] that the matrices [Z] and [Y ] of reciprocal devices are symmetrical. On the contrary. as well as devices containing a magnetic material maintained in a static magnetic field (e. in matrix form. The advantage of the matrix notation is that (??) can describe also a N -port structure.1): Zij = Vi Ij Ik =0. inductors. it follows {[Y ]∗ } = 0 Hence. The name given to the matrix [Y ] comes from the definition of its elements Ii Yij = Vj Vk =0.k=j An important role in circuit theory is played by reciprocal and lossless networks. this dissipated power is zero for any excitation. Note that the matrix [Z(ω)] can be interpreted as a set of transfer functions between the applied currents (inputs) and the voltages at all ports (outputs).. 102 .g. The name is justified by the definition of its elements. 2012) where [V ] = [V1 V2 ]T . all ports. capacitors. As in the case of a one-port device we can introduce the admittance YL = 1/ZL . a one-port device connected to the line is more conveniently described in terms of a reflection coefficient rather than in terms of impedance or admittance.7. b= √ Y∞ V − the previous equation is rewritten as: 1 2 1 2 |a| − |b| 2 2 so a and b are directly related to the power flow. with respect to which the reflection coefficient is computed.. however. which is arbitrary. a single two-port is obtained.2. 7. To explain this name. In order to define power waves independently from the orientation of the z axis. In this case. if the two two-port devices are connected in cascade. In the case of a two-port device. Consequently. the short circuit admittance matrix of the resulting circuit is the sum of the ones of the two sub-blocks.2) i. with characteristic impedance Z∞i and propagation constant ki . which is generally known as scattering matrix or S matrix. On this line we define power waves: √ √ ai = Yri Vi+ . bi = Yri Vi− where the z axis points always into port i.Transmission Line Theory (Nov. the amplitudes of the forward and backward waves on the line are specified as power waves amplitudes a e b. To each port we assign a reference impedance Zri that can be interpreted as the characteristic impedance of a transmission line connected to the port. Introduce 103 .N . recall that in Chapter 3 we have shown that the net active power flowing on a line with real characteristic impedance is given by: Pt = If we set a= √ 1 1 Y∞ |V + |2 − Y∞ |V − |2 2 2 Y∞ V + . by introducing a matrix reflection coefficient. If the two two-port devices are connected in parallel. In this Chapter we are going to generalize this concept to N -port networks. the scattered waves on the various lines depend in general on the incident waves at all ports. (see Fig. Note also that the line characteristic impedance plays the role of a reference impedance. it is useful to describe the signal a as a wave incident on the load and b as a wave scattered from the load. Obviously.2 Distributed parameter circuits Suppose now that each port i of an N -port device is connected to a transmission line. labeled with the subscript i = 1. they are related by b(z) = Γ(z)a(z) where Γ(z) is the local reflection coefficient. whose open circuit impedance matrix is the sum of the ones of the two sub-blocks. 2012) The various matrices are useful for the characterization of the connection of multi-port devices. Finally. if two two-port devices are connected in series. the z dependence of the signals a and b is the same as that of V + and V − : a(z) = a(0)e−jkz Pt = b(z) = b(0)e+jkz Moreover.e.Renato Orta .. in order to guarantee that ai is actually incident on the device. Let us generalize these concepts to the case of a device with N ports.2) we have: b1 = S11 a1 + S12 a2 b2 = S21 a1 + S22 a2 (7. We have seen in Chapter 1 that the electric state of a transmission line is specified in the more natural (and simplest) way by giving the amplitudes of the forward and backward waves V + and V − . the ABCD matrix of the resulting structure is the product of the ones of the sub-blocks. instead of voltages or currents. For example. Transmission Line Theory (Nov. The terms on the main diagonal (i = j) of [S] are the usual reflection coefficients at port i when all the others are terminated with the relevant reference impedances.3) where [S] is a 2 × 2 complex matrix. Obviously. [S] being a N × N matrix.k=j from which it is evident the character of generalized reflection coefficients of the elements of the S matrix.Renato Orta . it is in practice the only one to be employed in the microwave field. it is easier to construct a wide band matched load rather than an open circuit.2. for example in a waveguide • the power waves a e b can be measured directly by means of an instrument called Network Analyzer. 2012) a1 a2 Zr2 Z r1 b2 b1 Figure 7. from a theoretical point of view. which generalizes the scalar one (eq.5) . There are several reasons. completely equivalent to that in terms of the matrices [Z] or [Y ] (apart from the singular cases). 7.7)). In general a wide band characterization of the devices is of interest and. the form of (7. In this way the access line is matched and only an outgoing wave is present on it.2) we find that the elements are defined as: bi Sij = (7.(3.3) is valid also in the case of a N -port network.2) can be rewritten in matrix form: [b] = [S][a] (7. Two-port device and definition of the relevant power waves the column vectors [a] = [a1 a2 ]T and [b1 b2 ]T . so that Eq.(7. The condition ak = 0 at port k is obtained by terminating the access transmission line.4) aj ak =0. called Scattering Matrix.3 Relationship between [S] and [Z] or [Y ] In (7. in practice. Let us start from the device characterization in terms of [Z] matrix: [V ] = [Z][I] Express now [V ] and [I] in terms of power amplitudes [a] e [b]: [V ] = [V + ] + [V − ] = [Zr ]1/2 ([a] + [b]) [I] = Y∞ [V + ] − [V − ] = [Yr ]1/2 ([a] − [b]) 104 (7. with characteristic impedance Zrk . From (7. among which: • voltage and current are not always well defined quantities. with a load impedance numerically equal to Zrk itself. The terms out of the main diagonal are usually called transmission coefficients from port j to port i. There is also a noteworthy relation between the [S] matrix and the open circuit impedance matrix [Z].4) we have given an explicit definition of the [S] matrix elements. Even if the characterization of a device by means of its scattering matrix [S] is. that is the reference load for the [Z] matrix. Remember that √ [Zr ]1/2 is the diagonal matrix having the values Zri on the main diagonal.7) The inverse relation of (7. characterized by its scattering matrix [S]. [b]T ∗ = [a]T ∗ [S]T ∗ hence 1 T∗ [a] ([1] − [S]T ∗ [S])[a] 2 It is straightforward to realize that this equation.Transmission Line Theory (Nov.5): [Zr ]1/2 ([a] + [b]) = [Z][Yr ]1/2 ([a] − [b]) Expanding the products and factoring [Zr ]1/2 at the left of both sides. for a one-port device (N = 1) reduces to Pd = Pd = 1 |V + |2 (1 − |Γ|2 ) 2 Z∞ 105 (7. 2012) where [Zr ] is the diagonal matrix constructed with the reference impedances of all ports. The power dissipated in it is the sum of the net active powers flowing into all ports: Pd = 1 2 N (|ai |2 − |bi |2 ) = i=1 1 ([a]T ∗ [a] − [b]T ∗ [b]) 2 Now recall that [b] = [S][a].8) .6) is [Z] = [Z∞ ]1/2 {[1] + [S]}{[1] − [S]}−1 [Z∞ ]1/2 In a similar way we can obtain the following relations between the scattering matrix and the short circuit admittance matrix: [S] = {[1] − [y]}{[1] + [y]}−1 where the normalized admittance matrix is defined as: [y] = [Zr ]1/2 [Y ][Zr ]1/2 The inverse relation is : [Y ] = [Z∞ ]−1/2 {[1] − [S]}{[1] + [S]}−1 [Z∞ ]−1/2 7.4 Computation of the power dissipated in a device Consider a N -port device.6) where [1] is the identity matrix of size N and [ζ] is the normalized open circuit impedance matrix of the device: [ζ] = [Zr ]−1/2 [Z][Zr ]−1/2 (7. Substitute into (7.Renato Orta . we find [Zr ]1/2 [Yr ]1/2 [Z][Yr ]1/2 + [1] [b] = [Zr ]1/2 [Yr ]1/2 [Z][Yr ]1/2 − [1] [a] Canceling the common factor [Zr ]1/2 we get [S] = {[ζ] − [1]}{[ζ] + [1]}−1 (7. with respect to a given set [Zro ] of reference impedances.Transmission Line Theory (Nov. by requiring that Pd = 0 for arbitrary excitation [a]. Its reflection coefficient Γn with respect to Zrn is derived as: Γn = Z − Zrn Z + Zrn with Z = Zro 1 + Γo 1 − Γo Substituting and with a little of algebra. which has reflection coefficient Γo with respect to Zro . 2012) 7. Suppose that we want to compute the scattering matrix [Sn ] of the same device. The proof can be found in [5].Renato Orta . with respect to a new set of reference impedances [Zrn ]. • A passive device has a scattering matrix [S] such that all the eigenvalues of [S]T ∗ [S] have a magnitude less than (or at most equal to) 1.8). • an active device has a scattering matrix [S] such that at least one eigenvalue of [S]T ∗ [S] has magnitude greater than 1. by considering first the case of a one-port device. • A lossless device has unitary scattering matrix: [S]T ∗ [S] = [1]. Let us address the problem gradually. [6] 7. It is to be remarked that the eigenvalues of [S]T ∗ [S] are the squares of the singular values of [S]. Zro − Zrn Γno = Zro + Zrn 106 .6 Change of reference impedances Let [So ] be the scattering matrix of a device.5 Properties of the scattering matrix [S] of a device • A reciprocal device has a symmetric [S] matrix: [S] = [S]T . we find: 1 + Γo − Zrn (Zro − Zrn ) + Γo (Zro + Zrn ) Γno + Γo 1 − Γo Γn = = = 1 + Γo (Zro + Zrn ) + Γo (Zro − Zrn ) 1 + Γno Γo + Zrn Zro 1 − Γo Zro where we have induced the reflection coefficient of the old reference impedance with respect to to the new one. In the case N = 2 the previous equation yields: |S11 |2 + |S21 |2 = 1 2 2 = 1 2 2 = 1 2 2 = 1 + ∗ S21 S22 = 0 + ∗ S12 S22 = 0 |S12 | + |S22 | |S11 | + |S12 | |S21 | + |S22 | ∗ S11 S12 ∗ S11 S21 These relations have a geometrical interpretation: the rows and the columns of the unitary matrix [S] form an ortho-normal basis in the N dimensional complex linear vector space CN . This follows immediately from (7. The complex amplitudes of these waves are function of the longitudinal coordinate.7).Transmission Line Theory (Nov. we can express the desired scattering matrix as [ζn ] − [1] [Sn ] = (7. exp{±jkN lN } } Denote with [S] the scattering matrix of the new structure obtained by the translation of the reference planes. b0i .N ) are indicated by a0i and b0i . Hence. [b].3). a given S matrix refers always to a specific choice of reference planes.. Moreover. so that the previous equations can be written in matrix form : [a] = exp{+j[k]l} [a0 ] (7. The amplitudes of incident and reflected waves at port i (i = 1. . = [Zrn ]−1/2 [Z][Zrn ]−1/2 = [Zrn ]−1/2 [Zro ]1/2 [ζo ][Zro ]1/2 [Zrn ]−1/2 = = [ζn ] [R][ζo ][R] where we have introduced the diagonal matrix [R] = [Zrn ]−1/2 [Zro ]1/2 . [a0 ]. hence they commute.10) [b] = exp{−j[k]l} [b0 ] where exp{±j[k]l} is the diagonal matrix exp(±j[k]l) = diag{ exp{±jk1 l1 } . Introduce now the column vectors [a]. from (7. respectively. bi = b0i e−jki li where ki is the propagation constant on the line connected to port i. . .6). Notice that these matrices are diagonal.7 Change of reference planes The scattering matrix [S] of a device describes its input/output characteristic on the basis of the incident and scattered waves on the transmission lines attached to each port.Renato Orta . Express now [ζo ] in terms of [So ] and substitute in (7. with components ai . [b0 ]. We find = [1] + [So ] [1] + [So ] [R] − [1] [R] − [R]−1 [1] − [So ] [1] − [So ] = = [1] + [So ] [1] + [So ] [R] [R] + [1] [R] + [R]−1 [1] − [So ] [1] − [So ] = ([R] − [R]−1 ) + ([R] + [R]−1 )[So ] [S]no + [So ] = ([R] + [R]−1 ) + ([R] − [R]−1 )[So ] [1] + [S]no [So ] [R] [Sn ] where we have introduced the diagonal scattering matrix of the old reference impedances with respect to the new ones [R] − [R]−1 [R]2 − 1 [S]no = = [R] + [R]−1 [R]2 + 1 7.9).7. bi . Consider a N -port device with scattering matrix S0 (see Fig. If [S0 ] is the scattering matrix of the original structure.2. given by ai = a0i ejki li . hence they commute. From (7. bi .. 2012) We can now address the case of an N -port device along the same lines. we have: [b0 ] = [S0 ][a0 ] 107 . . The same amplitudes at a distance li (> 0) away form the device are denoted by ai . We want now to examine the transformation of the [S] matrix induced by a change of these planes. a0i .9) [ζn ] + [1] Notice that the equation can be written in this form because both ([ζn ]−[1]) and ([ζn ]+[1])−1 are functions of the same matrix [ζn ]. 11) Make this equation explicit. separating the ones referring to the ports that are going to be connected from the others.8 Cascade connection of structures Very often it is convenient to view a complex system as made out of interconnected simpler blocks. 2012) ai a0i bi b0i port i a0j li port j aj b0j bj [S0] lj [S] Figure 7. which must be connected through the K ports. [S12 ]  . This operation implies a partition in blocks of the scattering matrices of the two structures. For the elements on the main diagonal Sii = S0ii e−2jki li We recognize at once the analogy with the transformation rule of reflection coefficients (3. since these matrix elements are indeed reflection coefficients at port i (when all the other ports are terminated with the respective characteristic impedances). the first with N + K ports.4.Renato Orta . the problem arises of computing the scattering matrix of the complete structure. Change of reference planes Substitute (7. i=j because the incident and scattered waves propagate on different transmission lines. With reference to Fig.12)   . it is convenient to use a matrix formulation. [a1 ] }N   ···  =  ··· ···  ···  (7. For the other elements the previous relation becomes Sij = S0ij e−j(ki li +kj lj ) . 7. 22 108 . consider two structures. K{ [b ] [a ] }K . The first is written as   . 7.8). There should be no surprise. so that the complete structure has N + M ports.3.10): [b] = [exp{−jkl}][S0 ][exp{−jkl}][a] In other words [S] = [exp{−jkl}][S0 ][exp{−jkl}] (7. Assuming that their scattering matrices are known.     N{ [b1 ] [S11 ] .Transmission Line Theory (Nov. the second with K + N ports. [S ] [S21 ] . In defining the power waves at the various ports. This condition can always be obtained by suitable exchanges of rows and columns.4. [S22 ] .Renato Orta .Transmission Line Theory (Nov. Connection of two structures where the blocks [Sij ] have the dimensions [S11 ] → N ×N [S12 ] → N ×K [S21 ] → K ×N [S22 ] → K ×K Implicit in this partition is the assumption that the ports to be connected are the last K ones. Indeed: • the exchange of bi with bj requires the exchange of the rows i and j in [S ] • the exchange of ai with aj requires the exchange of the columns i e j in [S ] Likewise.13) where the blocks [Sij ] have the dimensions [S11 ] → K ×K [S12 ] → K ×M [S21 ] → M ×K [S22 ] → M ×M In this case. the equations that define the connection are [a ] = [b ] [a ] = [b ] 109 (7. matrix [S ] in the following way  [a ] }K ···  [a2 ] }M (7. Suppose also that the ports to be connected have the same reference impedances.14) . 2012) 1 [a1] 2 [a’] [b’] S’(N+K)(N+K) [b1] [a’’] [b’’] K ports K ports N ports [a2] [b2] S’’(K+M)(K+M) M ports S(N+M)(N+M) Figure 7. it has been assumed the the ports to be connected are the first K ones. the second structure is characterized by the scattering   . M{ [b2 ] [S21 ] . [S12 ]  .   ···  =  ··· ···    . In these conditions.    K{ [b ] [S11 ] . the structure is accessible from the external world through N ports on the “1” side and M ports on the “2” side and. obviously partitioned in the following way   . it is necessary to eliminate the variables [a ]. it is described by a scattering matrix [S] with dimension (M + N ) × (M + N ).19) The curly parenthesis that multiplies [S12 ][a2 ] can be simplified. The steps to be performed are the following: 1. The expression to be rewritten is: [X] ([1] − [Y ][X])−1 [Y ] + [1] = factor [X] to the left and [Y ] to the right: = [X] ([1] − [Y ][X])−1 + [X]−1 [Y ]−1 [Y ] = factor ([1] − [Y ][X])−1 to the left: = [X] ([1] − [Y ][X])−1 [1] + ([1] − [Y ][X]) [X]−1 [Y ]−1 [Y ] = factor [X]−1 [Y ]−1 to the right: = [X] ([1] − [Y ][X])−1 {[Y ][X] + [1] − [Y ][X]} [X]−1 [Y ]−1 [Y ] = 110 .Renato Orta .14). the first of which becomes [a ] = [S11 ][b ] + [S12 ][a2 ] (7.16) and (7. [S ] [S21 ] .16) we find the expression of [a ] as a function of [a1 ] and [a2 ]: [a ] = + [S11 ] ([1] − [S22 ][S11 ]) −1 [S11 ] ([1] − [S22 ][S11 ]) [S21 ][a1 ]+ −1 [S22 ] + [1] [S12 ][a2 ] (7.     N{ [b1 ] [S11 ] .17). [S12 ]  . The second of (7.12) and (7.12) is Eliminating [a ] between (7. yields [b ] = [S21 ][a1 ] + [S22 ] [S11 ][b ] + [S12 ][a2 ] from which we find the expression of [b ] as a function of [a1 ] e [a2 ] [b ] = [1] − [S22 ][S11 ] −1 [S21 ][a1 ] + [S22 ][S12 ][a2 ] (7.18) into (7. Set [X] = [S11 ] and [Y ] = [S22 ] for a simpler reading. Substitute (7.13). Substituting (7. [b ] from (7.16) [b ] = [S21 ][a1 ] + [S22 ][a ] (7. 2012) After the connection. [b ]. via (7.15)   . 22 where the blocks [Sij ] have the dimensions [S11 ] → N ×N [S12 ] → N ×M [S21 ] → M ×N [S22 ] → M ×M In order to determine the resulting [S] matrix.13).14) into (7.18) 3.17) 2. M{ [b2 ] [a2 ] }M .Transmission Line Theory (Nov. hence. [a ]. [a1 ] }N   ···  =  ··· ···  ···  (7. we get the expressions of [S11 ] e [S12 ]. Now substitute in the second of (7. to obtain [b1 ] as a function of [a1 ] e [a2 ]: [b1 ] = + [S11 ] + [S12 ][S11 ] ([1] − [S22 ][S11 ]) [S12 ] ([1] − [S11 ][S22 ]) −1 −1 [S21 ] [a1 ]+ [S12 ][a2 ] From the comparison of this equation with (7. . is characterized by [S11 ] alone. so that it behaves as a K-port load. Sostitute (7. . this happens only for complex frequency values.12). the latter being given by (7. as given by (7.Renato Orta . 111 . In this case M = 0 and the whole structure. having only N access ports. If the Taylor expansion is used to compute the inverse matrix ([1] − [A])−1 = [1] + [A] + [A]2 + . (7.Transmission Line Theory (Nov. we collect the expressions of the four blocks: [S11 ] [S12 ] = = [S11 ] + [S12 ] [S11 ] ([1] − [S22 ] [S11 ]) [S12 ] ([1] − [S11 ] [S22 ]) −1 [S21 ] [S12 ] −1 −1 [S21 ] [S21 ] = [S21 ] ([1] − [S22 ] [S11 ]) [S22 ] = [S22 ] + [S21 ] ([1] − [S22 ] [S11 ]) −1 (7.20) 4. + [A]n + .15).15) we derive the expressions of the remaining elements [S21 ] e [S22 ]. If the constituents structures are passive. hence non invertible. The computation of the comprehensive scattering matrix requires the inversion of a matrix with dimension equal to the number of ports K that are connected.20) in the first of (7. Another particular case is the one in which all the ports of the second structure are connected. . located in the upper half-plane for stability reasons. when the comprehensive structure that originates from the connection is resonant.18). 2012) Simplify: = [X] ([1] − [Y ][X])−1 [X]−1 = rewrite as the inverse of a matrix: −1 = [X] ([1] − [Y ][X])−1 [X]−1 = {[1] − [X][Y ]}−1 In conclusion. and resuming the usual notation.19) is rewritten as [a ] = + [S11 ] ([1] − [S22 ][S11 ]) ([1] − [S11 ][S22 ]) −1 −1 [S21 ][a1 ]+ [S12 ][a2 ] (7. For reference sake. so that the eigenvalues of [A] have amplitude less than 1. which is convergent if the structures are passive.13) [a ] = [b ].18): [b2 ] = [S21 ] ([1] − [S22 ][S11 ]) + −1 [S21 ] ([1] − [S22 ][S11 ]) [S21 ][a1 ]+ −1 [S22 ][S12 ] + [S22 ] [a2 ] Comparing this equation with the second of (7. 5. it is useful to know the value of [b ]. This matrix is singular. all the sub-matrices are scalars. . we write it here again for ease of reference: −1 [b ] = [1] − [S22 ][S11 ] [S21 ][a1 ] + [S22 ][S12 ][a2 ] Note that if the structures of interest are two-ports to be connected through one of their ports. we obtain the characterization of the comprehensive structure in terms of the multiple reflection series.21) [S22 ] [S12 ] In some applications. without any power coming out of port 1. instead of two leads. we obtain a two-port structure that behaves as an isolator. 7. lossless three-port structure is necessarily non reciprocal and is a circulator. Indeed. the behavior is due to a magnetized ferrite.9. that allows the undisturbed passage of a signal from port 1 to 2. 7.Renato Orta . an incident signal at port 1 goes to port 2 without being influenced by the load on port 3. 7. Note that if port 3 is terminated with a matched load. It can be shown that a matched. In the drawings we will use the common convention of using one line as the symbol of a port.Transmission Line Theory (Nov. An isolator contains a magnetic material 2 1 Figure 7. because of the non reciprocity of the device.5. Symbol of an ideal isolator. a signal incident on port 2 is directed to port 3. that produces an attenuation AdB on the signals through it. whose symbol is shown in Fig. see Fig. maintained in a static magnetic field. 7. but prevents it in the opposite direction. The phase shift is related to the physical size of the device.9 Scattering matrix of some devices In this section we present the scattering matrix of some devices of common use in the microwave technology. The power incident on port 2 is completely dissipated in the device. 2012) 7. 7. On the contrary.6.5. is a nonreciprocal device. Its scattering matrix is then: [S] = 0 Ae−jϕ Ae−jϕ 0 where A = 10−AdB /20 .2 Isolator An ideal isolator. 7. (ferrite). where it is dissipated in the matched load. as done previously.9.1 Ideal attenuator An ideal attenuator is a reciprocal device. whose symbol is shown in Fig. Also in this case. 112 .7. Its S matrix is then:   0 0 e−jϕ1 .9. [S] =  e−jϕ2 0 0 −jϕ3 0 e 0 This matrix is clearly non symmetrical. The S matrix of the device is: [S] = 0 e−jϕ 0 0 The non-reciprocity is evident from the fact that S12 = S21 . matched at both ports. It is matched at all ports and the power propagating in the direction of the arrow suffers no attenuation. matched at both ports. while the one flowing in the opposite direction is completely dissipated.3 Circulator An ideal circulator is a nonreciprocal device. 7. Indeed.7.9.Transmission Line Theory (Nov.8. Symbol of an ideal three port circulator. is a matched. its S matrix can be written: √   1 − k2 jk 0 √ 0   1 − k2 0 0 . S31 of practical devices is not small enough. Ideal directional coupler An ideal directional coupler. 2 Isolator constructed by terminating port 3 of a circulator with a matched load. 3 1 Figure 7. It is clear that in practice this configuration works only if the transmission and reception frequencies are different and a bandpass filter is inserted between the circulator and the receiver. √ jk [S] =   1 − k2  jk 0 0 √ 0 jk 1 − k2 0 The quantity C = −20 log10 k = −20 log10 |S31 | 113 .Renato Orta .9.4 Circulator used as a diplexer. whose symbol is reported in Fig. 7. when in a communication system both the transmitter and the receiver are connected to the same antenna. With a suitable choice of the reference planes. TX 1 2 3 RX Figure 7.8. reciprocal four port structure. 2012) 1 3 2 Figure 7. A circulator can be used to realize a “diplexer”.6. as it is shown in Fig. 7. in which the two ports on each side are uncoupled. From the ratio (in amplitude and phase) of the signals coming out of ports 3 and 4.10. 10 dB-.10. 3 Symbol of directional coupler.10 1 2 Measurement of the reflection coefficient of a load by means of a directional coupler. an instrument capable of measuring directly the scattering parameters of a device. A directional coupler is the heart of the Network Analyzer. The isolation I is related to the same concept. Other parameters used to characterize a real directional coupler are the directivity D: D = −20 log10 |S41 | |S31 | and the isolation I: I = −20 log10 |S14 |. 7.Renato Orta . An ideal coupler has infinite isolation and directivity. by specifying how port 4 is isolated from 1.Transmission Line Theory (Nov. 1 2 1− k 2 jk 1 − k2 4 Figure 7. The coupling C denotes the fraction of the incident power at port 1 that is transferred to port 3. 2012) is called coupling coefficient: there are 3 dB-couplers (called hybrids). 114 . 20 dB-couplers. jk L 1 − k 2V1+ 4 3 1 − k2 jkV1+ jk L + 1 V Figure 7. etc.9. The directivity measures the ability of the directional coupler to discriminate the incident waves at port 1 from those incident at port 2. The concept of the measurement is illustrated in Fig. 7. hence the following relation holds I =D+C (in dB). it is possible to derive the value of the reflection coefficient. Examples of analysis of structures described by S matrices In this section we illustrate the use of the scattering matrix for the analysis of simple networks. 22) where ΓL is computed with reference to Zr2 and Γin is referred to Zr1 . by the scalar Γin .Transmission Line Theory (Nov.Renato Orta . 2012) 7.e. Alternatively. the third one stipulates that port 2 is loaded. whose port 2 is loaded with an impedance Observe finally that also the relation between ΓL and Γin is a bilinear fractional transformation. it can also be useful to derive it directly. 7.1 Cascade connection of a two-port and a load We want to compute the reflection coefficient at the input of a two-port whose port 2 is terminated with a load. 1 − S22 ΓL S11 + (7. substitute a2 in the first two equations b1 = S11 a1 + S12 ΓL b2 b2 = S21 a1 + S22 ΓL b2 From the second we get the important relation b2 = S21 a1 1 − S22 ΓL Substitute this b2 into the first of (??) and get b1 = S11 a1 + S21 S12 ΓL a1 1 − S22 ΓL From this. i. that can be found by using equations (7.(7. 115 . Two-port network. S' ΓL Γi Figure 7.21): Γin = = = S12 S21 ΓL = 1 − S22 ΓL S11 − S11 S22 ΓL + S12 S21 ΓL = 1 − S22 ΓL S11 − det [S] ΓL .10.22) is immediately derived. It can be remarked that when ΓL = 0.11. The desired result Γin is obtained by eliminating a2 and b2 . Eq. as shown in Fig. Γin = S11 .11. To this end. The comprehensive structure is a one-port load and is characterized by a 1 × 1 scattering matrix. The equations of the structure are:   b1   b2    a2 = S11 a1 + S12 a2 = S21 a1 + S22 a2 = Γ L b2 The first two equations describe the general operation of the two-port. as it is obvious from the definition of scattering parameters. Transmission Line Theory (Nov. but if the devices are not resonating or very large. Suppose that the S matrices of the two two-ports do not depend on frequency. on θ. In practice this is not strictly true. B Interconnection of two two-ports by means of a length l of transmission line.7. In these conditions it is simple to obtain a plot of the amplitude of the total transmission coefficient S21 (θ).21) to cascade the two structures. by means of (7.e.Renato Orta .13a. their frequency dependence is much weaker than that of the exponential. i. several devices can be modeled in this way. 7. Moreover. symmetrical with respect to the real axis with center in C= 1 1 − |S22 |2 |S11 |2 R= |S22 | |S11 | 1 − |S22 |2 |S11 |2 and radius 116 . In particular. [2]. for example a one-resonator filter or a Fabry-Perot interferometer. D−1 (θ) traces a circle.23) becomes D(θ) = 1 − S11 S22 e−j2(θ−(ϕ11 +ϕ22 )/2 The plot of D(θ) in the complex plane is clearly a circle with center in 1 + j0) and radius |S11 | |S22 |. In practice. in which two two-ports are connected by means of a length of transmission line. The S matrix of the comprehensive structure can be obtained in two steps: l S’ S’’ _ S’’ A S Figure 7. Setting S11 = S11 ejϕ11 S22 = S22 ejϕ22 the denominator of (7. 2012) 7. • shift the reference plane of the right two-port from B to A.10. We obtain S12 S11 e−2jθ S21 1 − S22 S11 e−2jθ S11 = S11 + S21 = S21 S21 e−jθ 1 − S22 S11 e−2jθ (7.2 Interconnection of two two-ports by means of a length of transmission line Consider the structure of Fig. it is known that the operation of inversion transforms circles into circles. as shown in Fig.12.23) where θ = kl = ωl/vf is the electrical length of the line.12.11) ¯ [S ] = e−jθ 0 0 1 S e−jθ 0 0 1 = S11 e−2jθ S21 e−jθ S12 e−jθ S22 • use (7. θmax θ Plot of |S21 (θ)|.Transmission Line Theory (Nov.3 Change of reference impedance for a one-port load Consider a load characterized by means of its reflection coefficient with respect to the reference impedance Zr1 . We can make reference to Fig. the structure would behave as pass-band filter.13. 7. Hence the plot of |S21 (θ)| is the oscillating curve shown in Fig.10. 7.7. |S21| |S21|max |S21|min θmin Figure 7. 2012) ℑm ℑm ℜe 1 1-|S22'||S11''| ℜe 1+|S22'||S11''| Figure 7.14 where the minimum and maximum values of the transmission coefficient are |S21 |MIN = |S21 | |S21 | 1 + |S22 | |S11 | |S21 |MAX = |S21 | |S21 | 1 − |S22 | |S11 | and their position is 1 (ϕ11 + ϕ22 + (2m + 1) π) 2 1 θMAX = (ϕ11 + ϕ22 + 2mπ) 2 It is also clear that from the shape of the curve |S21 (θ)| (for example obtained through a measurement) it is possible to infer the characteristics of the discontinuities present on the line and their separation. (1+|S22'||S11''|)-1 (1-|S22'||S11''|)-1 Plot of D(θ) (left) and of D−1 (θ) (right).14. 117 .Renato Orta . θMIN = If the discontinuities on the line had a very high reflection coefficient.15. We want to compute the reflection coefficient of the same load with respect to another reference impedance Zr2 . By means of (7. we introduce a single z axis for the two sides and define forward and backward waves in accordance with it.Transmission Line Theory (Nov. The transmission matrix relates the electrical states at the two sides of the structure in the power wave basis. we can obtain a direct link between ΓA− and ΓL .16. which can be viewed as generalizations of two-port structures. the junction between the two lines.21) we find ΓA− = ΓF + 1 − Γ2 ΓL ΓF + Γ L F = 1 + Γ F ΓL 1 + Γ F ΓL which is the desired equation.Renato Orta . 118 . 7.15. has the scattering matrix  S = Zr2 Zr1 ΓF Zr2 Zr1 (1 − ΓF ) where ΓF is the reflectin coefficient ΓF =  (1 + ΓF )  −ΓF Zr1 − Zr2 Zr1 + Zr2 This is called Fresnel reflection coefficient because of its analogy with the reflection coefficient defined in optics at the interface between two semi-infinite media. The first. this can also be obtained by applying (7.21) to analyze the two cascaded discontinuities. as required by the reciprocity of the structure. hence its scattering matrix S reduces to the reflection coefficient ΓL . 7.11 Transmission matrix Structures in which we can identify two “sides” with the same number N of ports. Referring to Fig. 2012) - Figure 7. However. are often characterized by means of a transmission matrix. ΓL Zr1 Zr2 A+ A Change of reference impedance. We obtain the result by passing through the impedance: ZA+ = Zr1 ΓA− = 1 + ΓL 1 − ΓL ZA+ − Zr2 ZA+ + Zr2 Substituting the first equation into the second one. we can verify that they are equal. Note that S12 and S21 are only apparently different: if the expressions of ΓF are substituted. The second discontinuity is a one-port load. Transmission Line Theory (Nov. .16. . 2012) The characterization of the device by means of the transmission matrix [T ] is then the following:   [c+ ] 1  [T11 ]  ···  =  ···  [c− ] 1 [T21 ]  .17. Generalized two-port structure: definition of forward and backward waves on the two sides of it. A characteristic of the transmission matrix is that its elements cannot be defined directly by circuit type equations but must be obtained by other groups of parameters such as the scattering ones via algebraic relations.Renato Orta .16. [T’’] [T’] [T] Figure 7. 7.17 the transmission matrix of the comprehensive structure is found by means of [T ] = [T ][T ] The convenience of the transmission matrix is related to the simplicity of this composition law. 7. with transmission matrices [T1 ] and [T2 ] are connected via N ports as shown in Fig. .   +  [T12 ]  [c2 ] ···  ···   [c− ] 2 [T22 ] where all blocks are square with dimension N × N . . side 1 side 2 1 1 + 1 [c ] + [ c2 ] [c1− ] − [ c2 ] N N z Figure 7. 119 . Cascade connection of two generalized two-ports. When two structures of the type of Fig. . The comprehensive structure is perfectly defined by its scattering matrix [S]. the transformation formulas are: [T11 ] = [S21 ]−1 [T12 ] = −[S21 ]−1 [S22 ] [T21 ] = [S11 ][S21 ]−1 [T22 ] = [S12 ] − [S11 ][S21 ]−1 [S22 ] (7. such a structure cannot be represented by a transmission matrix.Renato Orta . a notoriously non invertible operation.21). we can obtain [c+ ] from the second of (7. Observe that the dimensions of the various factors are [S21 ] → K ×N [S21 ] → ([1] − [S22 ][S11 ]) −1 N ×K → K ×K hence the dimension of [S21 ] is N × N . which allow the computation of [S] from [T ]: [S11 ] = [T21 ][T11 ]−1 [S12 ] = [T22 ] − [T21 ][T11 ]−1 [T12 ] [S21 ] = [T11 ]−1 [S22 ] = −[T11 ]−1 [T12 ] 120 .25). An example is the structure of Fig.27) Obviously.26) we can identify the expressions of [T11 ] and [T12 ].24) into this [c− ] = [S11 ][c+ ] + [S12 ][c− ] 1 1 2 [c+ ] = [S21 ][c+ ] + [S22 ][c− ] 2 1 2 (7.18 if K < N . However.  ···  =  ···  . [S21 ] represents an operator that maps vectors belonging to the complex vector space C N into vectors of C N through C K .24) matrix [S] is    [S12 ]  [a1 ] ···  ···   [a2 ] [S22 ] Substitute (7. Indeed. .26) into the first of (7. If K < N . 2 Since is expressed as a function of and Substituting (7.Transmission Line Theory (Nov. so that the transmission matrix [T ] does not exist. From the linear algebra point of view. it can be verified that the submatrix [S21 ] is not invertible. consider the expression of the block [S21 ] given by (7. we obtain (7. However. 2012) To derive the relation between the [T ] and [S] matrices of the same device. [b2 ] [S21 ] .27). [b1 ]  [S11 ] .25) Since the blocks [Sij ] are square. (7. hence it has zero determinant if K < N . we are in presence of a projection. this matrix has rank K at most. [c− ] = [S11 ][S21 ]−1 [c+ ] + [S12 ] − [S11 ][S21 ]−1 [S22 ] [c− ] 1 2 2 from which the expressions of the other elements of the [T ] matrix can be obtained. 7.   . For completeness we list also the inverse relations of(7. observe first that the forward and backward power waves are related to the incident and scattered ones by c+ = [a1 ] 1 c+ = [b2 ] 2 c− = [b1 ] 1 c− = [a2 ] 2 The characterization of the device via the scattering  . In conclusion.25) 1 [c+ ] = [S21 ]−1 [c+ ] − [S21 ]−1 [S22 ][c− ] 1 2 2 [c+ ] 1 [c+ ] 2 [c− ]. if the submatrix [S21 ] of a structure is not invertible. If K < N the transmission matrix of the comprehensive structure is not defined.Transmission Line Theory (Nov. 2012) N ports K ports N ports Figure 7.Renato Orta .18. Structure with 2N ports that arises from the “back-to-back” connection of two substructures with N + K ports each. 121 . Also the load impedance ZL and the internal impedance of the generator are generic complex functions of frequency.2. as shown in Fig. because these parameters do not depend on time. we can compute directly the circuit response to time harmonic signals. because the line parameters k and Z∞ . as shown in Fig. the input-output relation can be expressed in time domain as a convolution product +∞ vB (t) = tv (t − τ ) e (τ ) dτ (8.2) where the impulse response tv (t) is related to the transfer function by a Fourier transform: tv (t) = 1 2π +∞ −∞ 122 TV (ω) ejωt dω (8. Take the case of a simple circuit. the load voltage vB (t) is the output. as well as the impedances Zg and ZL are independent of the voltages and currents in the circuit. in general complex functions of frequency.1) −∞ or in the frequency domain as an algebraic product as VB (ω) = TV (ω) E (ω) (8. In this chapter we consider instead signals with arbitrary time dependence. • time-invariant. We want to compute the load voltage vB (t) that is produced by a generator with open circuit voltage e (t). It is well known that for linear time invariant systems (LTI). 8. The generator waveform e (t) is the system input. consisting of a generator and a load.1. in the language of system theory. By this technique. This problem can be conveniently described. Suppose that the line is characterized by a phase constant k (ω) and characteristic impedance Z∞ (ω).1 Introduction In the previous chapters we have discussed in detail the frequency domain analysis of transmission line circuits. connected by a transmission line. 8. The system is: • linear.Chapter 8 Time domain analysis of transmission lines 8.3) . • vB (t) is obtained by summing all these signals.4) −∞ Obviously. 2012) Zg e(t) + ZL A Figure 8. the integral (8.4): • the input is decomposed into a linear combination of harmonic signals ejωt e (t) = 1 2π +∞ E (ω) ejωt dω (8. in the sense that they travel through the system unchanged. 8. the transfer function TV (ω) is obtained without difficulty by the methods described in the previous chapters. here we wish to focus on the role of the transmission line.1 Eq.5) It is useful to note the following interpretation of (8. • since the system is LTI. such as the Fast Fourier Transform (FFT). 8.2 The group velocity In general. with amplitude VB (ω) dω = TV (ω) E (ω) dω. the output signal is still harmonic.1.Renato Orta . connected by a transmission line AB tv(t) TV(ω) e(t) Figure 8.4) must be evaluated by numerical techniques.Transmission Line Theory (Nov. We find TV (ω) = ZA (ω) 1 e−jk(ω)l (1 + ΓB (ω)) ZA (ω) + Zg (ω) 1 + ΓA (ω) (8.2) yields the solution to our problem in the form vB (t) = 1 2π VB (ω) ejωt dω = 1 2π +∞ TV (ω) E (ω) ejωt dω (8. [7]. which is the transfer function evaluated at the frequency ω. vB(t) Linear system interpretation of the circuit of Fig. However. This property explains the usefulness of the Fourier transform technique in the analysis of LTI systems. (8. B Circuit consisting of a generator and a load. Time-harmonic signals ejωt are “eigensignals” of LTI systems. except for the multiplication by a complex number. hence we simplify the model 123 .2.6) −∞ each with amplitude E (ω) dω. 8. since vB (t) is real. = Observe that vB (t) is computed from (8. we have vB (t) = 1 2π 0 VB ω ejω t dω + −∞ 1 2π +∞ VB (ω) ejωt dω (8. Note that the envelope is slowly varying in comparison with the carrier of Fig. However. by assuming that the generator is ideal (Zg = 0) and the line is matched (ZL = Z∞ ).9) 2 2 where M (ω) is the Fourier transform of m (t). we assume that the signal e (t) is not too different from a time-harmonic signal. we find +∞ +∞ 1 1 ∗ vB (t) = VB (ω) e−jωt dω + VB (ω) ejωt dω = 2π 0 2π 0 124 .1. Letting now ω = −ω and using (8. we choose e (t) = m (t) cos ω0 t (8. is amplitude modulated and. These functions are plotted in Fig. where ωc is the highest frequency present in the spectrum of m(t).8) where m (t) (envelope) is a slowly varying signal that is almost constant in a period T = 2π/ω of the cosine (carrier). ∗ VB (−ω) = VB (ω) (8.e.7) Moreover. i.11) 0 where the integration variable in the first integral has been called ω .3. by decomposing the integration domain into two parts.Transmission Line Theory (Nov. so that the transfer function becomes TV (ω) = e−jk(ω)l (8. Example of an amplitude modulated signal. Requiring that the envelope is slowly varying with respect to the carrier is equivalent to assuming ωc ω0 . We find E (ω) = {m (t) cos ω0 t} = 1 2π {m (t)} ∗ {cos ω0 t} = 1 M (ω) ∗ π {δ (ω − ω0 ) + δ (ω + ω0 )} = 2π 1 1 = M (ω − ω0 ) + M (ω + ω0 ) = (8. is quasi-monochromatic. 2012) e(t) t Figure 8. In particular. Hence the signal e(t) is indeed quasi-monochromatic. in these conditions.4) as an integral over both negative and positive frequencies. This signal. 8.4. of the type shown in Fig.10).Renato Orta . Indeed. 8. its spectrum is hermitian.10) and the spectral integral can be limited to the positive omega half-axis.3. as it can be ascertained by computing its spectrum. (8.15) α (ω) α (ω0 ) This truncation. i. Since ωc e (t) is almost time-harmonic 1 2π = 2Re +∞ VB (ω) ejωt dω ω0 . without changing the value of the integral. Substitute (8. Suppose we truncate the previous expansion at the second term for the real part and at the first term for the imaginary one.13) vB (t) Re ej(ω0 t−β(ω0 )l) e−α(ω0 )l 1 2π +∞ M (ω − ω0 ) e−jβ (ω0 )(ω−ω0 )l j(ω−ω0 )t e dω (8. the propagation constant is complex.4. (8.15) into (8. so that we can recognize the inverse Fourier transform of M (Ω) evaluated in t − β (ω0 ) l: vB (t) Re{ej(ω0 t−β(ω0 )l) e−α(ω0 )l m t − β (ω0 ) l } = 125 .12) 0 This transformation is equivalent to introducing the analytic signal associated to vB (t). whose spectrum is zero for ω < 0 and 2VB (ω) for ω > 0. apparently asymmetrical. we can think of substituting the function k (ω) with its Taylor expansion around ω = ω0 dk 1 d2 k k (ω) = k (ω0 ) + (ω − ω0 ) + (ω − ω0 )2 + . k (ω) = β (ω) − jα (ω). the previous equation can be rewritten vB (t) Re ej(ω0 t−β(ω0 )l) e−α(ω0 )l 1 2π +∞ M (Ω) e+j (t−β (ω0 )l)Ω dΩ (8.13) 0 Taking into account that the signal is quasi-monochromatic. Letting Ω = ω − ω0 .Renato Orta .16) 0 where actually the integral receives contribution only in the support of M (ω − ω0 ). Also the spectrum of m (t) is shown. assume β (ω) β (ω0 ) + β (ω0 ) (ω − ω0 ) (8. that is the support of its spectrum is a small neighborhood of ω0 . which is by hypothesis a small neighborhood of ω = ω0 . Spectrum of e (t).e.Transmission Line Theory (Nov. is justified by the actual behavior of k(ω) in the usual cases... At this point the problem is reduced to the evaluation of vB (t) = Re +∞ 1 2π M (ω − ω0 ) e−jk(ω)l ejωt dω (8. 2012) |Ε(ω)| Μ(ω) −ω0 −ωc ωc ω0 Figure 8.14) dω ω0 2 dω 2 ω0 In general.17) −∞ where the lower limit has been shifted to −∞. eq.23) Note that the concept of group velocity is the most physically important of the two.20) β(ω0 ) These quantities get their names from the fact that the signal e (t) consists of a “group” of frequencies. the transfer function has constant magnitude and linear phase. everything happens as if the envelope m (t) moved at the group velocity and the carrier with the phase velocity. it is the propagation velocity of information and of energy on the line. If we recall the definitions of phase velocity vph (ω0 ) = ω0 β (ω0 ) (8.Transmission Line Theory (Nov. we observe that (8. the information is associated to the envelope.24). On the basis of the previous definitions. As a consequence of the constructive and destructive interference phenomena that here take place. The concepts of phase and group velocity lend themselves to a geometrical interpretation.(8. so that the propagation can be defined as distortion free.Renato Orta .19) with the consequent definition of the group velocity vg (ω0 ) vg (ω0 ) = 1 dω = dβ dβ dω ω0 (8. as required by the theory of relativity. It turns out to be always smaller than the speed of light in vacuum. 8. 126 . at least in the band of the signal that is propagating. this is only an interpretation of (8. Indeed. Note.24) dω where H (ω) is the transfer function of the device. In the applications.18) The quantity β (ω0 ) l has the dimensions of a time and is called group delay τg (ω0 ) τg (ω0 ) = β (ω0 ) l = l vg (ω0 ) (8. We see also that.19) is in agreement with this definition. in the limit in which (8.22). Obviously. The higher order terms in the expansion (8. since envelope and carrier are not signals with an independent existence. that the concept of group delay can be defined both for lumped and distributed devices. Recalling (8. finally.5. they cannot be neglected when the bandwidth of e (t) is not small. each of which appears in the point B weighted by the transfer function.14) are responsible for the distortions. 2012) = m t − β (ω0 ) l e−α(ω0 )l cos (ω0 t − β (ω0 ) l) (8. a group delay appears in the cases in which the device can store energy. Obviously a resistor network has a real transfer function and τg = 0 according to (8. The signal e (t) is not monochromatic but consists of a “packet” of harmonic components. a transmission line does not introduce distortions if. The general definition is in fact d τg = − arg (H (ω)) (8. Hence. Consider a dispersion curve as the one sketched in Fig. In general terms. we can write vph (ω0 ) = tan ϕph vg (ω0 ) = tan ϕg (8.22) This equation has the following interpretation.18) can be rewritten vB (t) m t− l vg (ω0 ) e−α(ω0 )l cos ω0 t − ω0 l vph (ω0 ) = m (t − τg (ω0 )) e−α(ω0 )l cos (ω0 (t − τph (ω0 ))) = (8. the envelope in B is an attenuated and delayed replica of the envelope in A.21) and of the corresponding phase delay τph (ω0 ) = β (ω0 ) l/ω0 = l/vph (ω0 ).7).15) hold. Obviously. 5. This assumption implies that the group velocity (and the group delay) are linear functions of frequency.26) We see clearly that the quadratic phase term causes a distortion. 2012) ω ϕg ω0 ϕf β Figure 8.27) m (t) = exp − 2 2T0 The standard deviation of the gaussian T0 can be used as a conventional measure of the duration of the pulse. Making use of the integral √ +∞ δ2 2 2 π − (2α)2 e (8.29) .25) d2 β dω 2 ω0 and we will assume α (ω) = 0 for simplicity. In this section we will discuss the distortions caused by a transfer function with constant magnitude and a phase curve that is non linear but can be approximated by a parabola.16) we find vB (t) Re ej(ω0 t−β0 l) 1 1 2 2π +∞ 1 2 [M (ω + ω0 ) + M (ω − ω0 )] e−j 2 β0 l(ω−ω0 ) · 0 ·ej (t−β0 l)(ω−ω0 ) dω (8. we will assume that the propagation constant can be expressed in the form β (ω) = β0 + β0 (ω − ω0 ) + where β0 = β (ω0 ) β0 = dβ dω 1 β0 (ω − ω0 )2 2 β0 = ω0 (8.Renato Orta . but the computation can no longer be carried out for a generic envelope m (t). in which t2 (8. 8. 8.3 Geometrical interpretation of phase and group velocity Distortions In the previous section we have seen that a quasi monochromatic signal is not distorted (more precisely. The simplest case for which an analytic expression can be obtained is that of a gaussian pulse. its envelope is not distorted) when the group velocity is constant on the bandwidth of the signal itself. Hence.6. By repeating with minor modifications the computations that lead to (8. see Fig.Transmission Line Theory (Nov.28) e−α x ejδx dx = α −∞ we obtain the spectrum M (ω) M (ω) = √ 2πT0 exp − 127 2 T0 ω 2 2 (8. Transmission Line Theory (Nov. as for the second. Note that the algebraic term can be rewritten as  − 1 1 −1 2 2   2 2 β0 l T0 β0 l β0 l exp j arctan = 1+j = 1+ = 2 2 2 2   T0 T0 T0 T0 + jβ0 l = 1+ 1 2 −4 β0 l 2 T0 exp −j 1 arctan 2 β0 l 2 T0 (8.26) is the sum of two terms. For the same reason. a short pulse has a large bandwidth and viceversa. Envelope of a gaussian pulse and definition of its conventional duration T0 −1 Note that the spectrum of the envelope is still gaussian with standard deviation T0 : in accordance with the uncertainty principle for the Fourier transform.31) 2 − (t − β0 l) 2 2 (T0 + jβ0 l) (8.28) with α= 1 2 2 T0 + jβ0 l δ = t − β0 l we find vB (t) Re T0 exp {j (ω0 t − β0 l)} exp 2 T0 + jβ0 l (8.26) in the form: vB (t) Re ej(ω0 t−β0 l) = Re ej(ω0 t−β0 l) 1 1 2 2π 1 1 2 2π +∞ M (Ω) e−j 2 β0 lΩ ej (t−β0 l)Ω dΩ 2 1 = −∞ +∞ e− 2 (T0 +jβ0 l)Ω ej (t−β0 l)Ω dΩ 1 2 2 (8.6.33) If we separate magnitude and phase. the lower limit of the integral can be shifted from 0 to −∞ without changing its value. 2012) m(t) 1 e-1/2 t T0 Figure 8. Rewrite (8.32) Transform now the expression in the curly brackets in such a way that the real part is obtained simply.Renato Orta .34) . so that ω0 T0 >> 1 and the first term gives a completely negligible contribution. In normal applications the pulse duration is much larger of the carrier period. we arrive at the following final expression vB (t) = T0 T (l) 1 2 exp − (t − l/vg (ω0 ))2 2T (l)2 128 cos (ϕ (t)) (8. The integrand in (8.30) −∞ By applying (8. Note that ld increases with the initial pulse duration T0 . Indeed. moreover.7. if |β0 | decreases. it gets distorted because its standard deviation increases. it is possible to compress a pulse by exploiting the dispersivity of the line on which it propagates. Even if the pulse retains the gaussian shape.36) Note that the signal at the far end of the line is still gaussian: this is related to the invariance of gaussians with respect to Fourier transforms and to the fact that by expanding the phase constant k(ω) to the second order also the line transfer function turns out to be a gaussian.34) is made to propagate from z = l to z = 0. Obviously it is necessary that the signal to be compressed has a frequency modulation (chirp) and that the sign of β0 is appropriate. depicted in Fig. Moreover it increases if the line dispersivity becomes smaller. hence if the signal (8. Note. Distorted gaussian pulse at the line end. 2012) Figure 8. Observe the spurious frequency modulation where the phase ϕ (t) is l ϕ (t) = ω t − vph (ω0 ) (t − l/vg (ω0 ))2 + 2 2T0 √ 3(l/ld )sign (β0 ) 1 − arctan 2 2 1 + 3 (l/ld ) √ 3l ld (8.Renato Orta . The quantity ld is defined doubling distance because T (ld ) = 2T0 . Note that the transmission line is a symmetric device. Consider now the phase term and note the quadratic dependance on t − l/vg (ω0 ). 129 . Compute the instantaneous frequency √ t − l/vg (ω0 ) 3(l/ld )sign(β0 ) dϕ ω (t) = = ω0 + (8.e. called chirp. the output signal will be vA (t) = exp − (t − l/vg (ω0 ))2 2 2T0 cos(ω0 t) In other words.35) and T (l) = T0 1+3 l ld 2 ld = 2 √ T0 3 |β0 | (8. The output signal is affected by a spurious frequency modulation. i. that the maximum value of the envelope becomes smaller and smaller during the propagation: it is simple to verify that the energy in the pulse does not depend on the length of the line l. in accordance with the fact that the line has been assumed to be lossless.7. the compression is obtained by removal of the frequency modulation. albeit with imaginary variance. 8. because its bandwidth becomes narrower. 2 The envelope moves at the group velocity but its variance increases from T0 to T 2 (l).Transmission Line Theory (Nov.37) 2 dt T0 1 + 3 (l/ld )2 It changes linearly and increases or decreases depending on the sign of β0 . 8. Basically.Transmission Line Theory (Nov.Renato Orta . At the other extreme. the pulse duration increases during the propagation. Intersymbol interference in a digital link on a dispersive optical fiber. consider an optical fiber link using a digital modulation such that the transmission of a pulse is associated to the logical value 1 and the absence of a pulse to the value 0. Received pulses are so distorted that the transmitted word is no longer recognizable The considerations made for the gaussian pulse hold qualitatively for any pulse waveform.36) 1 1 BT ≤ BT max = = (8.4 Digital communication As an example of the results presented above. since the pulse form changes. The general characteristics of the plot are easily explained. recalling (8. hence the distortion is small. 8. i. suppose that the receiver is able to recognize two pulses provided their time separation is greater than the pulse duration multiplied by a factor r typical of the receiver. 2012) z=0    t z=l t Figure 8. The condition of good operation can be written 1 ≥ rT (l) BT that is. however. 8. If T0 is small. there is an optimum pulse duration that can be found by setting to zero the derivative of (8.38) T0opt = |β0 | l BT max (T0opt ) = √ 2r 130 1 |β0 | l .8. Fig. as shown in Fig. we can try to select the pulse duration in such a way as to maximize BT max . hence the bit rate must be kept very low if the intersymbol interference is to be avoided.e.9 shows a plot of BT max (T0 ). in order to avoid the interference already at the transmitter side.38) 2 rT (l) β0 l rT0 1 + 2 T0 If the characteristics of the line and of the receiver are specified. Let BT be the bit rate. Clearly. if T0 is large the doubling length is large. however the bit rate is trivially low. an intersymbol interference can take place such that pulses are no longer recognized by the receiver. To quantify the phenomenon. the doubling length ld is also small. the number of bits transmitted per second. it is difficult to define precisely the pulse duration.8. If in the course of the propagation pulses suffer a distortion and increase their duration. 25). i. but that the analysis carried out above is no longer applicable.ω) with (ω) exp(−jkz) − √ ω k = ω LC = vph Y∞ V0− (ω) exp(jkz) Y∞ = 131 C L (8.5.5 Mismatched ideal transmission lines In the previous sections we have discussed the behavior of transmission lines alone.Transmission Line Theory (Nov. We know that the general solution in the ω-domain is V (z. A remarkable property of silica optical fibers is that the parameter β0 is zero at the wavelength λ = 1. Indeed. so that the doubling length becomes infinite. In this section we examine the effects of load and generator mismatch but. 8.ω) = V0+ (ω) exp(−jkz) + V0− (ω) exp(jkz) = Y∞ V0+ (8. here we obtain the desired result by the Fourier transform technique.39) I (z.55µm. non dispersive. based on a suitable change of variable. but only on those of the fiber. 8. and for this reason the band centered around λ = 1. in order to proceed by small steps. the pulse duration increases by the factor 2 in the propagation between transmitter and receiver. it is always convenient to operate the link at this wavelength. ordinary silica fibers show their minimum attenuation (about 0.9.1 General solution of transmission line equations As an example of the technique that we are going to use. Various technological solutions have been devised to unify at the same wavelength the properties of low losses and low dispersion. but if β0 = 0 it is necessary to take into account at least the cubic term.40) . this does not mean that the pulse are not distorted.3µm is called second window for its importance in the optical communications. centered at λ = 1. we assume first that the line is ideal.3µm.Renato Orta . In section 1.e. which gives rise to a different type of distortion. Unfortunately.4 we presented the classical d’Alembert solution. let us obtain first of all the general solution of the time domain transmission line equations. In any case. in the assumption that the terminations were matched. 2012) BTmax T0opt Figure 8.2 dB) in the third window. Obviously. In these √ conditions. T0 Plot of the maximum bit rate BT max on a digital link versus the pulse duration Note that T0opt does not depend on the receiver characteristics. if β0 = 0 it is possible to neglect the higher order terms in (8. B Basic circuit.Renato Orta . Moreover. 8. the same expressions of section 1.10. we obtained vB (t) in the form of an inverse Fourier transform: vB (t) = 1 2π with TV (ω) = VB (ω) ejωt dω = 1 2π +∞ TV (ω) E (ω) ejωt dω ZA (ω) 1 e−jωτ (1 + ΓB (ω)) ZA (ω) + Zg (ω) 1 + ΓA (ω) 132 (8. In turn. Note that V0+ (ω) and V0− (ω) are two arbitrary constants with respect to z ma can certainly depend on the parameter ω.t) = Y∞ v0 t − z vph − − Y∞ v0 t + z vph (8.1.41) ± where v0 are the inverse Fourier transforms of V0± (ω). Note that the general solution is clearly constituted by two waves propagating in opposite directions.2 Mismatched ideal lines Consider now the main problem of this section. as for the current.ω) exp(jωt)dω = −∞ z vph + = v0 t − + V0− (ω) exp j t + ω z vph − + v0 t + z vph ω dω = z vph (8. In the ω-domain this behavior is indicated by the fact that the two wave components have a phase proportional to ω. we have obtained again. shown in Fig. by the Fourier transform technique.Transmission Line Theory (Nov. this is a consequence of the fact that the phase velocity is frequency independent. In section 8. 8.t) = = 1 = 2π +∞ 1 2π V0+ (ω) exp(−j −∞ +∞ +∞ 1 2π z z ω) + V0− (ω) exp(+j ω) exp(jωt)dω = vph vph V0+ (ω) exp j t − −∞ V (z.10. 2012) l Rg e(t) + Z∞ . hence frequency independent. since Y∞ does not depend on ω. Note that for simplicity we have assumed that both the generator internal impedance and the load impedance are pure resistances. it is straightforward to write + i (z.42) Hence. vf RL A Figure 8. comprising a mismatched ideal line.43) −∞ (8. Suppose we want to compute the load voltage vB (t).5. Compute then the time domain voltage as v (z. Their explicit expression can be defined when the load and generator are specified.4.44) . 49) and taking the inverse Fourier transform termwise. the factor (1 + ΓB ) originates from the fact that the total voltage in B is the sum of the forward and backward components. because in this case |Γg ΓB exp(−j2ωτ )| < 1. it is zero for negative argument. Substituting the expansion into (8. Obviously. The first displays the dynamic evolution of the phenomenon.47) and note that it can be expanded by the binomial theorem: (1 − Γg ΓB exp(−j2ωτ ))−1 = 1 + Γg ΓB exp(−j2ωτ ) + Γ2 Γ2 exp(−j4ωτ ) + . This is easily explained. i. g B (8. Hence. we are interested to compute vB (t) for 0 ≤ t ≤ tmax . tmax − (2n + 1)τ is negative and this and the subsequent terms do not give any contribution.47) Note that because of the assumptions we made on load and generator.e.46) ZA + Zg 2 1 − ΓA Γg and. within a certain observation window.e. g B 2 (8. Hence.Renato Orta . we obtain VB (ω) = E(ω) 1 − Γg 1 (1 + ΓB ) exp(−jωτ ) 2 1 − Γg ΓB exp(−j2ωτ ) (8. i. In reality. by noting that each term of the sum 1 − Γg n n Γg ΓB e(t − (2n + 1)τ ) (1 + ΓB ) 2 (8.50) Apparently the solution is given in the form of an infinite series.Transmission Line Theory (Nov. 2012) where E(ω) is the Fourier transform of the open circuit generator voltage e(t) and τ = l/vph is the transit time on the line. Define Rg = Z ∞ 1 + Γg 1 − Γg ZA = Z∞ 1 + ΓA 1 − ΓA (8...e. g B (8.. of stationary states. Moreover. It is convenient to eliminate the latter to obtain a homogeneous expression. if n is sufficiently large.52) ..47) we find VB (ω) = E(ω) 1 − Γg (1 + ΓB ) exp(−jωτ )· 2 · 1 + Γg ΓB exp(−j2ωτ ) + Γ2 Γ2 exp(−j4ωτ ) + . The previous equations contain both reflection coefficients and impedances. Γg and ΓB are frequency independent. the function e(t) is causal. i.51) represents a wave that has travelled 2n + 1 times the length AB in the forward and backward direction.. We find Z∞ 1 − Γg = 2 Rg + Z ∞ 133 (8. with n + 1 reflections at the far end B (load) and n at the near end A (generator). To compute the inverse Fourier transform. in general.48) The expansion is certainly convergent if the load is passive. two alternative routes can be followed. recalling that ΓA = ΓB exp(−j2ωτ ). The lattice diagram Consider the last fraction in (8. for a fixed time t ≤ tmax .45) where Γg is the voltage reflection coefficient of the generator internal impedance. It can be verified that the following relation holds ZA 1 − Γg 1 + ΓA = (8.. which give rise to two radically different forms of writing the solution. the second yields a description in terms of resonances. we obtain: vB (t) = 1 − Γg (1 + ΓB ) e(t − τ ) + Γg ΓB e(t − 3τ ) + Γ2 Γ2 e(t − 5τ ) + . To interpret the first factor (1 − Γg )/2 it is convenient to rewrite it in terms of impedances. there is only a finite number of terms that contribute. 50) can be rewritten vB (t) = (1 + ΓB ) vA0 (t − τ ) + Γg ΓB vA0 (t − 3τ ) + Γ2 Γ2 vA0 (t − 5τ ) + . Lattice diagram for the circuit of Fig. which allows to write directly the expression of the transient response without computing first Zg e(t) + ZL A B z 1-Γg e(t) 2 ΓB Γg ΓB Γg t Figure 8. 8. the generator cannot ”know” if the line is infinite or not. shown in Fig. In this light. 8. we can introduce the voltage in A at the time t = 0+ Z∞ e(t) vA0 (t) = Rg + Z ∞ so that (8.11 in which the line is infinitely long.10. even if it has reached the far end B. called “lattice diagram”.10 134 . Definition of the surge impedance This function is clearly the partition factor vA (t) /e(t) for the circuit of Fig.Transmission Line Theory (Nov. We can say that Z∞ is the input impedance of the line but only for t ≤ 2τ and this justifies the name of ”surge impedance” that sometimes is used to denote Z∞ . 2012) Rg e(t) + Figure 8.12. because in this case the signal launched from A.. but only for t ≤ 2τ . This circuit is applicable also in the case of Fig.8..53) On the basis of this interpretation it is possible to draw a space-time plot. g B (8. produces an echo that reaches A no sooner than t = 2τ . Hence.12.11. 8. before this time.Renato Orta . i.3 0.7 0. two different situations can appear. 2012) Time domain response Max [ Vg(t) ] = 1 1 VL(t) (nat) 0. The voltage vA (t) is written immediately vA (t) = 1 − Γg {e(t) + ΓB (1 + Γg )e(t − 2τ )+ 2 +Γ2 Γg (1 + Γg )e(t − 4τ ) + . Suppose that the open circuit voltage of the generator is a signal of duration T0 . Having introduced (and proved) the lattice diagram method for the computation of vB (t).5 t / T0 0. Its length is deducible from the delay τ of the pulse front. 8. overlap partially: 2 this is called a reverberation condition. on the transit time τ . for successive values of n. .1 0.14 show the two conditions in the case e(t) is a rectangular pulse with value 1V. B 135 (8. 2 0 Zg = 10Ω. the frequency response and then evaluating the inverse Fourier transform of it.54) . Load voltage in the multiple echo conditions (τ > Z∞ = 150Ω. ZL = 300Ω) 1 1 T .15 shows the relevant diagram. for successive values of n.9 Figure 8.4 0. . are disjoint: this 2 is called a multiple echo conditions. Figs.Renato Orta .e.13.Transmission Line Theory (Nov.13 and 8.8 0. Note that the successive echoes have decreasing amplitude.2 0. Fig.5 -1 0 0. − If τ < 1 T0 the supports of the functions e(t − (2n + 1)τ ).6 0. we can use it for the computation of the voltage at the near end A and in a generic intermediate point C.5 0 -0. 8. − If τ > 1 T0 the supports of the functions e(t − (2n + 1)τ ). Depending on the line length. The continuous line is the plot of the load voltage vB (t) when the transmission line is present. because the common ratio of the geometrical series is smaller than one. The dashed line is the plot of the voltage on the load if this were directly connected to the generator. 2 0. Lattice diagram showing the computation of the voltages in various points of the line (A.4 0. Load voltage in the reverberation conditions (τ < 2 T0 .8 0. ZL = 300Ω) A C B z τC 1-Γg (1+ΓB)e(t-τ) 2 1-Γg ΓB(1+Γg)e(t-2τ) 2 1-Γg ΓgΓB(1+ΓB)e(t-3τ) 2 1-Γg Γ2 Γ (1+Γ )e(t-4τ) g B g 2 t Figure 8.3 0.15. C).5 t / T0 0. Z∞ = 150Ω.Renato Orta .14.5 -1 0 0.7 0. Zg = 10Ω.9 1 1 Figure 8. 2012) Time domain response Max [ Vg(t) ] = 1 1 VL(t) (nat) 0. 136 .5 0 -0.Transmission Line Theory (Nov.6 0.1 0. B. We know that the same 137 . Not so easy is the case of iC (t). Indeed.55) and noting that this expression consists of forward and backward waves. The easiest method is just to apply Ohm’s law: vB (t) (8. The impedance then can be considered as the transfer function of the LTI system. Indeed. B (8. . Recall that impedance is a frequency domain concept.54).56) iB (t) = RL with vB (t) given by (8.16. Sometimes we can be interested in computing the current in the load iB (t). the solution is readily found by reconsidering eq.59) i. .59) are substituted with their expressions in terms of impedances. .52).57) Since this is a general formula. . it should hold also if C −→ B: iB (t) = 1 − Γg {e(t − τ ) − ΓB e(t − τ )+ 2Z∞ +ΓB Γg e(t − 3τ ) − Γ2 Γg e(t − 3τ ) + .55) where τC = lAC /vph is the travel time from A to C. } (8.58) 1 − Γg (1 − ΓB ) {e(t − τ ))+ 2Z∞ +ΓB Γg e(t − 3τ ) + . 8. the identity of the two expression is readily proved. .e.Transmission Line Theory (Nov. B = (8. The relevant impedance relations are i+ (t) = C + vC (t) Z∞ i− (t) = − C − vC (t) Z∞ hence we can write the desired current as iC (t) = 1 − Γg {e(t − τc ) − ΓB e(t − (2τ − τc ))+ 2Z∞ +ΓB Γg e(t − (2τ + τc )) − Γ2 Γg e(t − (4τ − τc )) + . . we must clarify the actual meaning of an impedance in time domain. .(8. Note that the forward and backward voltages in C are not simultaneous and this explains the absence of a factor of the type 1 + ΓB o 1 + Γg . .56). However. introduced in connection with eq.50). iB (t) = This expression is apparently different from (8. where C is an intermediate point. B (8. The current at the near end iA (t) can be found similarly. introduced to characterize the linear behavior of a one port device: VL (ω) = ZL (ω)IL (ω) This equation has the circuit translation and the input-output system interpretation of Fig. 2012) whereas the voltage vC (t) is given by: vC (t) 1 − Γg {e(t − τc ) + ΓB e(t − (2τ − τc ))+ 2 +ΓB Γg e(t − (2τ + τc )) + Γ2 Γg e(t − (4τ − τc )) + . It is convenient to analyze in greater detail the definition of the characteristic impedance of the line as surge impedance. but if the reflection coefficients in (8. by applying the Kirchhoff loop law: e(t) − vA (t) iA (t) = Rg where vA (t) is computed by (8.Renato Orta . here it is not possible to apply Kirchhoff laws.(8. Transmission Line Theory (Nov. that the terminated line behaves as a simple resistor of value Z∞ for 0 < t < 2τ . also from the mathematical point of view. without delay. . However. if the one port device is characterized by its reflection coefficient ΓL (ω). Analogously. 138 . and we recover the standard Ohm’s law ∞ RL δ(t − t )iL (t )dt = RL iL (t) vL (t) = −∞ The reason is that a resistor does not store energy (as capacitors and inductors do). in general. 2012) IL VL I L (ω ) ZL Z L (ω ) VL (ω ) Figure 8. In terms of that. the model equation is ∞ vL (t) = zL (t) ∗ iL (t) = zL (t − t )iL (t )dt −∞ This is the correct definition of impedance in time domain and certainly. zL (t) = vL (t)/iL (t).Renato Orta . system is characterized in time domain by its impulse response z(t). let us compute the time domain input impedance of an ideal line with travel time τ . such as changes of characteristic impedance. the input-output equation is − + VL (ω) = ΓL (ω)VL (ω) in frequency domain and − vL (t) = ∞ −∞ + γL (t − t )vL (t )dt where γL (t). zL (t) = RL δ(t) since it is the inverse transform of a constant. L ZA = Z∞ from which we obtain easily (note that ΓL is frequency independent) zA (t) = Z∞ δ(t) + 2ΓL δ(t − 2τ ) + 2Γ2 δ(t − 4τ ) + 2Γ3 δ(t − 6τ ) + . lumped loads connected in series and parallel. the lattice diagram complexity increases exponentially and the method becomes useless. terminated by the resistive load RL . 1 + ΓA = Z∞ (1 + ΓA )(1 + ΓA + Γ2 + Γ3 + . expressed in the basis of forward and backward voltage waves instead of total voltage and current. . Definition of impedance of a one port device: circuit (left) and LTI system (right) points of view. hence the system is without memory and the response is instantaneous. etc. obtained as inverse Fourier transform of the transfer function ZL (ω). in the case of a (frequency independent) ideal resistor RL . the inverse transform of ΓL (ω). . is the impulse response of the system. L L From this expression it is evident. As an application.16. If the circuit contains more discontinuities. .) = A A 1 − ΓA 2 = Z∞ 1 + 2ΓL exp(−j2ωτ ) + 2ΓL exp(−j4ωτ ) + 2Γ3 exp(−j6ωτ ) + . . . in the case ΓB Γg < 0 (i. ±2 . Compute now the residues in the poles ωn .Transmission Line Theory (Nov. we find o Rn = lim ω→ωn ω − ωn 1 1 = lim = ω→ωn −Γg ΓB e−j2ωτ (−j2τ ) 1 − Γg ΓB e−j2ωτ j2τ (8. in order to apply the residue theorem. Z∞ belongs to the interval [Rg . coherently with the absence of gain in the system. The added half circle does not contribute to the integral because of Jordan’s lemma. The singularities of the integrand.Renato Orta . located in those ω values in which the denominator of (8. We see here clearly the two ways in which the frequency of a free oscillation can be defined: either as a pole of the transfer function or as the frequency for which the loop gain has magnitude one and phase n2π.e. without singularities in any finite region of the complex ω-plane.60) The integration path runs along the real axis as indicated in Fig. Observe that when ΓB Γg > 0 (i. 2012) ℑm ω  ℜe ω Figure 8. have all the same imaginary part.62) with n = 0. 8. We must evaluate vB (t) = 1 2π +∞ −∞ E(ω) exp(jωt) (1 − Γg ) (1 + ΓB ) exp(−jωτ ) dω 2 1 − Γg ΓB exp(−j2ωτ ) (8.17. which is non negative. . Integration path in the complex ω-plane and position of the singularities of the integrand function Solution in terms of resonances Consider again (8.47) and compute directly the inverse transform integral by the complex analysis methods. ±1. but can be transformed into a closed path by the addition in the upper halfplane of a half circle with radius tending to infinity. Remember that |ΓB Γg | ≤ 1. i.61) is ωn = = 1 [log |ΓB Γg | + j (arg (ΓB Γg ) + n · 2π)] = 2jτ 1 1 (arg (ΓB Γg ) + n · 2π) − j log |ΓB Γg | 2τ 2τ (8.63) If we assume that E(ω) is an entire function. for arbitrary integers n. are indicated by a cross.e. which happens if e(t) has finite duration. On the contrary. As well known.17. infinite in number.61) It is important to note that the quantity Γg ΓB exp(−j2ωτ ) is the system loop gain. if Rg and RL are both greater or smaller than Z∞ ). the poles identify the free oscillations. RB ]). The solution of (8.e. They are simple poles.60) vanishes: 1 − Γg ΓB exp(−j2ωτ ) = 0 (8. so that these singularities. then there are no other singularities and the response 139 . the poles are in symmetrical positions with respect to the imaginary axis and thei separation is π/τ . beyond those of E(ω). Applying the de l’Hˆspital rule. the pole for n = 0 has zero real part. . as in printed circuit boards (PCB) there are several lines on the same board. the dynamical response of the system is represented in terms of the system normal modes. different types of distortion arise. characterized by an input capacity.18.Transmission Line Theory (Nov. It is an alternative representation that is completely equivalent to that of (8. A more accurate model describes them as multiconductor transmission lines. in which cross-talk effects appear. which we have considered separately. which in general can be studied only by sophisticated numerical techniques 140 . if the loads that terminate the lines are non linear.Renato Orta . • Often. The resistance in parallel to C is Req = (RL Z∞ ) / (RL + Z∞ ). are simultaneously present. 8. 2012) Z∞ Figure 8. hence the time constant of the RC group is CReq . • Finally.50) in terms of multiple reflections. we can mention: • The load which terminates the line is not frequency independent. A typical example is that of a line connected to a logical gate.64) When the solution is expressed in this way. as shown in Fig.18. • If the line is not ideal. These lines can be considered as independent only in a first approximation. both the signal incident on the load and the successive echoes suffer distortions in the course of their propagation. 8. Limiting ourselves to a short list. In general we can say that the echoes have rising and descending fronts smoother than those of the incident pulse. This implies that the various echoes have a different shape one from the other and from the incident signal. C RL Transmission line connected to a logical gate vB (t) can be written +∞ vB (t) = 2πj Res(Integranda.5. ωn ) = n 1 (1 − Γg ) (1 + ΓB ) E(ωn )e+jωn (t−τ ) 2τ n=−∞ (8.3 Real interconnections In a real world interconnection problem the various distortions mechanisms. since the load behaves essentially as a low pass filter. C. Tripathi. 81-44. Pozar. [6] G. Collin . [7] A. CRC 1992 [4] D. Matrix computations. Needham. 1983. Ramo. Boca Raton. Baltimore: The Johns Hopkins University Press. Visual complex analysis. 141 . [5] R. E. Magnusson. 1997. Transmission lines and wave propagation. [2] T. Reading: Addison-Wesley.K. Papoulis. New York. Microwave engineering. Alexander. G. V. H Golub.R. C. van Duzer. McGraw-Hill Book Company Inc. “The Fourier integral and its applications”. Th. Oxford: Clarendon. 1990. Foundations for microwave engineering. Whinnery.. New York: McGraw-Hill 1992. F. 1994. J. pp. Van Loan.C. Wiley.Bibliography [1] S. M. [3] P. New York 1962. 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