Electromagnetic Waves and Antennas Exercise bookSophocles J. Orfanidis1 Davide Ramaccia2 Alessandro Toscano2 1 Department of Electrical & Computer Engineering Rutgers University, Piscataway, NJ 08854
[email protected] www.ece.rutgers.edu/~orfanidi/ewa 2 Department of Applied Electronics, University "Roma Tre" via della Vasca Navale, 84 00146, Rome, Italy
[email protected] [email protected] S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Table of Contents Chapter1 Maxwell's Equations ................................................................................... 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 Exercise.......................................................................................................... 1 Exercise.......................................................................................................... 6 Exercise........................................................................................................ 12 Exercise........................................................................................................ 16 Exercise........................................................................................................ 18 Exercise........................................................................................................ 20 Exercise........................................................................................................ 25 Exercise........................................................................................................ 29 Exercise........................................................................................................ 30 1.10 Exercise........................................................................................................ 32 1.11 Exercise........................................................................................................ 42 1.12 Exercise........................................................................................................ 54 1.13 Exercise........................................................................................................ 56 D. Ramaccia and A. Toscano S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Chapter1 Maxwell's Equations 1.1 Exercise Prove the vector algebra identities: a) A × ( B × C ) = B ( A ⋅ C ) - C ( A ⋅ B ) It is possible to write the vectors in the form: ⎧A = A x x + A yy + Az z ˆ ˆ ˆ ⎪ ⎪ ˆ ˆ ˆ ⎨B = B x x + B y y + Bz z ⎪ ˆ ˆ ˆ ⎪C = C x x + C y y + C z z ⎩ (1.1.1) and to use the follow relationship: ˆ x U × V = Ux Vx ˆ y Uy Vy ˆ z Uz = Vz (1.1.2) ˆ ˆ ˆ = x U y Vz − U z Vy − y ( U x Vz − U z Vx ) + z U x Vy − U y Vx Now we can prove the algebra identities with simply mathematical substitutions: ( ) ( ) )) = (1.1.3) ˆ ˆ ˆ A × ( B × C ) = A × x B y C z − Bz C y − y ( B x C z − Bz C x ) + z B x C y − B y C x = ) ( ( ) ˆ − y ( A x ( B x C y − B y C x ) − A z ( B y C z − Bz C y ) ) ˆ + z ( − A x ( B x C z − Bz C x ) − A y ( B y C z − Bz C y ) ) ˆ x A y B x C y − B y C x + A z ( B x C z − Bz C x ) (( ) ( Expanding the terms in (1.1.3), we have: ( ) ˆ + y ( A x B y C x − A x B x C y + A z B y C z − A z Bz C y ) ˆ + z ( A x Bz C x − A x B x C z − A y B y C z + A y B z C y ) ˆ + x A y B x C y − A y B y C x + A z B x C z − A z Bz C x A × ( B ×C ) = (1.1.4) Let us write eq. (1.1.4) in matrix form, separating the terms with the minus sign and the terms with the plus sign: D. Ramaccia and A. Toscano Pag. 1 1.7) ( B x C y A z − B x C z A y ) − ( B y C x A z − B y C z A x ) + ( Bz A x C y − Bz A y C x ) = ( Bx C y A z + B y C z A x + Bz C x A y ) − ( Bx C z A y + B y C x A z + Bz C y A x ) = ↑ ( A x B y C z + A y Bz C x + A z B x C y ) − ( A x Bz C y + A y B x C z + A z B y C x ) ˆ ˆ ˆ C ⋅ (A × B) = C ⋅ x A y Bz − A z B y − y ( A x Bz − A z B x ) + z A x B y − A y B x (( ) ( )) = (1.1.9) order them If we compare the last row of each expression.J.1.5) ⎥ 0 ⎥ ⎦ Note that the elements of the diagonal of each matrix are zero.1) and (1.1. Toscano Pag.6) = B ( AxCx + AyCy + AzCz ) − C( Ax Bx + AyBy + AzBz ) = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ = +AxCx Bx x + Byy + Bzz + AyCy Bx x + Byy + Bzz + AzCz Bx x + Byy + Bzz − = B ( A ⋅ C) − C( A ⋅ B) b) A ⋅ ( B × C ) = B ⋅ ( C × A ) = C ⋅ ( A × B ) Using relationships (1.8) order them ( C x A y Bz − C x A z B y ) − ( C y A x Bz − C y A z B x ) + ( C z A x B y − C z A y B x ) = ( C x A y Bz + C y A z B x + C z A x B y ) − ( C x A z B y + C y A x Bz + C z A y B x ) = ↑ ( A x B y C z + A y Bz C x + A z B x C y ) − ( A x Bz C y + A y B x C z + A z B y C x ) (( ) ( )) = (1.1.1. we note that they are identical so the algebra identity is verified. Orfanidis – Electromagnetic Waves and Antennas ⎡ 0 ⎢ A × ( B × C ) = ⎢Bx A yC y ⎢ ⎢ Bx A zCz ⎣ ByA xCx 0 ByA zCz Bz A x C x ⎤ ⎡ 0 ⎥ ⎢ Bz A yC y ⎥ − ⎢C x A y B y ⎥ ⎢ 0 ⎥ ⎢ C x A z Bz ⎦ ⎣ Exercises Chapter 1 C yA x Bx 0 C y A z Bz Cz A x Bx ⎤ ⎥ C z A y B y ⎥ (1. 2 .S.2). Ramaccia and A. we can write: ˆ ˆ ˆ A ⋅ ( B × C ) = A ⋅ x B y Cz − Bz C y − y ( B x C z − Bz C x ) + z B x C y − B y C x ( A x B y C z − A x Bz C y ) − ( A y B x C z − A y Bz C x ) + ( A z B x C y − A z B y C x ) = ( A x B y C z + A y Bz C x + A z B x C y ) − ( A x Bz C y + A y B x C z + A z B y C x ) ˆ ˆ ˆ B ⋅ (C × A) = B ⋅ x C y A z − C z A y − y ( C x A z − C z A x ) + z A x C y − A y C x (( ) ( )) = (1.1. D. Each term can be filled with the product of the three component with the same subscript ( a ii = A i B i C i ) : ⎡Bx AxCx ByAxCx BzAxCx ⎤ ⎡Cx Ax Bx CyAx Bx CzAx Bx ⎤ ⎢ ⎥ ⎢ ⎥ A× ( B×C) = ⎢Bx AyCy ByAyCy BzAyCy ⎥ − ⎢Cx AyBy CyAyBy CzAyBy ⎥ = ⎢ ⎥ ⎢ ⎥ ⎢ Bx AzCz ByAzCz BzAzCz ⎥ ⎢ Cx AzBz CyAzBz CzAzBz ⎥ ⎣ ⎦ ⎣ ⎦ ( ) ( ) ( ) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ − Ax Bx ( Cx x + Cyy + Czz ) − AyBy ( Cx x + Cyy + Czz ) − AzBz ( Cx x + Cyy + Czz ) = (1. 2). Toscano Pag. 3 .10) Let us begin considering the first case: D.1.J. we can write: 2 2 ˆ ˆ ˆ A × B + A ⋅ B = x A y Bz − A z B y − y ( A x Bz − A z B x ) + z A x B y − A y B x ( ) ( ) 2 + + A x B x + A y B y + A z Bz ⎛ ⎜ ⎝ ( ) 2 = ( A y Bz − A z B y ) 2 2 + ( A x Bz − A z B x ) + A x B y − A y B x 2 ( ) 2 ⎞ ⎟ + A x B x + A y B y + A z Bz ⎠ 2 ( ) 2 = ( A y Bz − A z B y ) + ( A x Bz − A z B x ) + A x B y − A y B x 2 ( ) + ( A x B x + A y B y + A z Bz ) 2 2 = A 2 B2 + A 2 B2 − 2A y Bz A z B y + A 2 B2 + A 2 B2 − 2A x Bz A z B x + z y x z z x y z A 2 B2 + A 2 B2 − 2A x B y A y Bx + A x Bx + A y B y + A z Bz x y y x ( ) 2 = 2 2 2 2 A 2 Bz + A z B2 − 2A y Bz A z B y + A 2 Bz + A z B2 − 2A x Bz A z B x + y y x x A 2 B2 + A 2 B2 − 2A x B y A y Bx + A 2 B2 + A 2 B2 + A 2 B2 + x y y x x x y y z z 2A x B y A y B x + 2A x Bz A z Bx + 2A x B y A y Bx ↑ cancel the opposites = 2 2 2 2 2 2 A 2 Bz + A z B2 + A 2 Bz + A z B2 + A 2 B2 + A 2 B2 + A 2 B2 + A 2 B2 + A z B z = y y x x x y y x x x y y ( A2x + A2y + A2 )( B2x + B2y + B2 ) = A 2 B 2 z z ˆ ˆ ˆ ˆ d) A = n × A × n + (n ⋅ A)n ˆ ˆ Does it make a difference whether n × A × n is taken to mean ( n × A ) × n or n × ( A × n ) ? ˆ ˆ ˆ ˆ ˆ The unit vector ncan be expressed as follow: ˆ ˆ ˆ ˆ ⎧n = n x x + n y y + n z z ⎪ ⎨ 2 2 2 ˆ ⎪ n = nx + n y + nz = 1 ⎩ (1.S.1) and (1. Ramaccia and A.1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 c) A×B + A ⋅B 2 2 = A 2 B 2 Using relationships (1.1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ˆ ˆ ˆ ˆ ˆ ˆ ( n × A ) × n = ⎡x ( n y Az − n z A y ) − y ( n x Az − n z A x ) + z ( n x A y − n y A x )⎤ × n = ⎣ ⎦ ( ) ˆ − y ⎡( n y A z − n z A y ) n z − ( n x A y − n y A x ) n x ⎤ + ⎣ ⎦ ˆ + z ⎡( n y A z − n z A y ) n y − ( n z A x − n x A z ) n x ⎤ = ⎣ ⎦ ˆ + x ⎡( n z A x − n x A z ) n z − n x A y − n y A x n y ⎤ + ⎣ ⎦ ˆ z +x ⎡n 2 Ax − n x n z Az − n x n y A y + n 2 A x ⎤ + y ⎣ ⎦ 2 ˆ − y ⎡n y n z Az − n z A y − n 2 A y + n y n x A x ⎤ + x ⎣ ⎦ ˆ y +z ⎡n 2 Az − n z n y A y − n z n x Ax + n 2 Az ⎤ x ⎣ ⎦ (1.11) And now consider the second case: ˆ ˆ ˆ ˆ ˆ ˆ n × ( A × n ) = n × ⎡x A y n z − Az n y − y ( A x n z − Az n x ) + z A x n y − A y n x ⎤ = ⎣ ⎦ ( ) ( ) ( ) ˆ −y ⎡ n x ( A x n y − A y n x ) − n z ( A y n z − A z n y ) ⎤ + ⎣ ⎦ ˆ +z ⎡n x ( Az n x − A x n z ) − n y ( A y n z − Az n y )⎤ = ⎣ ⎦ ˆ +x ⎡Ax n 2 − A y n y n x − Az n z n x + Ax n 2 ⎤ + y z⎦ ⎣ 2 ˆ −y ⎡ A x n x n y − A y n 2 − A y n z + A z n z n y ⎤ + x ⎣ ⎦ 2 2⎤ ˆ +z ⎡Az n x − Ax n x n z − A y n y n z + Az n y ⎣ ⎦ ˆ +x ⎡n y A x n y − A y n x − n z ( A z n x − A x n z )⎤ + ⎣ ⎦ (1.1. 4 .J. Toscano Pag. ˆ ˆ ˆ ˆ The second term of the identity can be written as: D.1.S.12) It is very easy to show that ( n × A ) × n = n × ( A × n ) . Ramaccia and A. Ramaccia and A.14) D.1. we obtain: ˆ ˆ ˆ ˆ n × A × n + (n ⋅ A)n = ˆ +x ⎡A x n 2 − A y n y n x − Az n z n x + A x n 2 ⎤ + y z⎦ ⎣ ˆ −y ⎡ A x n x n y − A y n 2 − A y n 2 + A z n z n y ⎤ + x z ⎣ ⎦ ˆ +z ⎡Az n 2 − A x n x n z − A y n y n z + Az n 2 ⎤ + x y⎦ ⎣ ˆ x +x ⎡n 2 A x + n x n y A y + n x n z Az ⎤ + ⎣ ⎦ ˆ +y ⎡n y n x A x + n 2 A y + n y n z Az ⎤ + y ⎣ ⎦ 2 ˆ +z ⎡n z n x A x + n z n y A y + n z Az ⎤ ⎣ ⎦ ↑ ˆ change signs in parentheses at first y and add = ˆ +x ⎡A x n 2 − A y n y n x − Az n z n x + A x n 2 + n 2 A x + n x n y A y + n x n z Az ⎤ + y z x ⎣ ⎦ 2 2 2 ˆ +y ⎡A y n x + A y n z − A x n x n y − Az n z n y + n y n x Ax + n y A y + n y n z Az ⎤ + ⎣ ⎦ ˆ +z ⎡Az n 2 − A x n x n z − A y n y n z + Az n 2 + n z n x A x + n z n y A y + n 2 Az ⎤ = x y z ⎣ ⎦ 2 ˆ ˆ ˆ + xA x ⎡ n 2 + n z + n 2 ⎤ + + yA y ⎡ n 2 + n 2 + n 2 ⎤ + + zA z ⎡ n 2 + n 2 + n 2 ⎤ = A x⎦ z y⎦ y z⎦ ⎣ y ⎣ x ⎣ x (1.13) Adding the two results.J. Orfanidis – Electromagnetic Waves and Antennas ˆ ˆ (n ⋅ A)n = n x A x + n y A y + n z A z ˆ +x ⎡n x n x A x + n y A y + n z Az ⎣ ˆ +y ⎡n y n x A x + n y A y + n z Az ⎣ Exercises Chapter 1 ˆ ˆ ˆ )( n x x + n y y + n z z ) = ( ( )⎤ + ⎦ ( )⎤ + ⎦ ˆ + z ⎡n z ( n x A x + n y A y + n z A z )⎤ = ⎣ ⎦ ˆ x +x ⎡n 2 A x + n x n y A y + n x n z A z ⎤ + ⎣ ⎦ ˆ +y ⎡n y n x A x + n 2 A y + n y n z Az ⎤ + y ⎣ ⎦ 2 ˆ +z ⎡n z n x A x + n z n y A y + n z Az ⎤ ⎣ ⎦ (1. Toscano Pag.S.1. 5 . e 3 ) : ˆ ˆ ˆ (Green's first identity) (Green's second identity) ˆ ˆ ˆ e1 ∂φ e2 ∂φ e3 ∂φ ⎧ ⎪∇φ = h ∂q + h ∂q + h ∂q 1 1 2 2 3 3 ⎪ ⎪ 1 ⎡ ∂ ⎛ h 2 h3 ∂φ ⎞ ∂ ⎛ h1h3 ∂φ ⎞ ∂ ⎛ h1h 2 ∂φ ⎞ ⎤ ⎪∇2 φ = ⎢ ⎜ ⎟⎥ ⎜ ⎟+ ⎜ ⎟+ h1h 2 h3 ⎣ ∂q1 ⎝ h1 ∂q1 ⎠ ∂q 2 ⎝ h 2 ∂q 2 ⎠ ∂q3 ⎝ h3 ∂q3 ⎠ ⎥ ⎪ ⎢ ⎦ ⎪ ⎤ 1 ⎡ ∂ ∂ ∂ ⎪ ∇⋅F = ( F1h 2 h3 ) + ( F2 h1h3 ) + ( F3h1h 2 )⎥ ⎢ ⎪ h1h 2 h3 ⎣ ∂q1 ∂q2 ∂q3 ⎦ ⎪ ⎪ ˆ ˆ ˆ ⎪ h1e1 h 2e2 h3e3 ⎨ ∂ ∂ ∂ ⎪∇ × F = 1 = ⎪ h1h 2 h3 ∂q1 ∂q 2 ∂q3 ⎪ h1F h 2 F2 h3F3 ⎪ 1 ⎪ ˆ ˆ ⎤ e ⎡ ∂ ⎤ e ⎡ ∂ ∂ ∂ ⎪ + 1 ⎢ ( h3F3 ) − ( h 2 F2 )⎥ + 2 ⎢ ( h1F1 ) − ( h3F3 )⎥ + ⎪ ∂q3 ∂q1 h 2 h3 ⎣ ∂q 2 ⎦ h1h3 ⎣ ∂q3 ⎦ ⎪ ⎪ ˆ ⎤ e ⎡ ∂ ∂ + 3 ⎢ ( h 2 F2 ) − ( h1F1 )⎥ ⎪ ∂q2 h1h 2 ⎣ ∂q1 ⎪ (1. All vector components are presented with respect to the normalized base ( e1 . ∇ × ( φ A ) = ( ∇φ ) × A + φ∇ × A 6. ∇ ⋅ ( ∇ × A ) = 0 7.2 Exercise Prove the vector analysis identities: 1. ∇ ⋅ ( φ∇ψ − ψ∇φ ) = φ∇ 2 ψ − ψ∇ 2 φ 4.J. ∇ × ( ∇ × A ) = ∇ ( ∇ ⋅ A ) − ∇ 2 A First of all we have to express the operator ∇ in general orthogonal coordinates in four common applications. 6 . ∇ ⋅ ( φ∇ψ ) = φ∇ 2 ψ + ∇φ ⋅ ∇ψ 3. ∇ ⋅ ( φ A ) = ( ∇φ ) ⋅ A + φ∇ ⋅ A 5. e 2 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. h 2 .S.1) ⎦ ⎩ where ( h1 .2. ∇ × ( ∇φ ) = 0 2. h 3 ) are the metric coefficients. ∇ ⋅ A × B = B ⋅ ( ∇ × A ) − A ⋅ ( ∇ × B ) 8. Ramaccia and A. Toscano Pag. For common geometries they are defined as follow: D. so each term in the ∂q i ∂q j ∂q j ∂q i ⎛ ∂φ ∂ψ ∂ 2 ψ ⎞ ⎛ ∂φ ∂ψ ∂ 2 ψ ⎞ ⎛ ∂φ ∂ψ ∂ 2ψ ⎞ =⎜ +φ +⎜ +φ +⎜ +φ ⎟ ⎟ ⎟= ⎜ ∂q1 ∂q1 ∂q1 ⎟ ⎜ ∂q 2 ∂q 2 ∂q 2 ⎟ ⎜ ∂q 3 ∂q 3 ∂q 3 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎛ ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ ⎞ ⎛ ∂φ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ⎞ =φ ⎜ + + + + = φ∇ 2 ψ + ∇φ ⋅ ∇ψ ⎟+ ⎜ ∂q1 ∂q 2 ∂q 3 ⎟ ⎜ ∂q1 ∂q1 ∂q 2 ∂q 2 ∂q 3 ∂q 3 ⎟ ⎠ ⎝ ⎠ ⎝ • Identity n° 3 First of all we expand the sum inside parentheses: ∂ψ ∂ψ ∂ψ ⎧ ˆ ˆ ˆ ⎪φ∇ψ = φe1 ∂q + φe 2 ∂q + φe3 ∂q ⎪ 1 2 3 ⎨ ⎪ψ∇φ = ψe ∂φ + ψe ∂φ + ψe ∂φ ˆ1 ˆ2 ˆ3 ⎪ ∂q1 ∂q 2 ∂q 3 ⎩ D. h 3 = 1 ⎪ h = 1. h 3 = 1) : • Identity n° 1 ˆ e1 ∂ ∇ × ( ∇φ ) = ∂q1 ˆ e2 ∂ ∂q 2 ˆ e3 ∂ ∂q3 = ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ⎛ ∂φ ⎞ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎝ ∂q1 ⎠ ⎝ ∂q 2 ⎠ ⎝ ∂q3 ⎠ ⎡ ⎛ ∂ ∂φ ⎛ ∂ ∂φ ∂ ∂φ ⎞ ∂ ∂φ ⎞ ⎤ ˆ ˆ − − ⎢e1 ⎜ ⎟ − e2 ⎜ ⎟ +⎥ ⎝ ∂q1 ∂q3 ∂q3 ∂q1 ⎠ ⎥ ⎢ ⎝ ∂q 2 ∂q3 ∂q3 ∂q 2 ⎠ =⎢ ⎥=0 ⎛ ∂ ∂φ ∂ ∂φ ⎞ ⎢ +e3 ⎜ ⎥ ˆ − ⎟ ⎢ ⎥ ∂q1 ∂q 2 ∂q 2 ∂q1 ⎠ ⎝ ⎣ ⎦ For the property of linearity of the derivate operator parentheses vanishes and also the result. • Identity n° 2 ⎛ ∂ψ ∂ψ ∂ψ ⎞ ˆ ˆ ˆ ∇ ⋅ ( φ∇ψ ) = ∇ ⋅ ⎜ e1 φ + e2 φ + e3 φ ⎟= ∂q1 ∂q 2 ∂q 3 ⎠ ⎝ ∂ ⎛ ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞ ∂ ⎛ ∂ψ ⎞ = ⎜φ ⎟= ⎜φ ⎟+ ⎜φ ⎟+ ∂q1 ⎝ ∂q1 ⎠ ∂q 2 ⎝ ∂q 2 ⎠ ∂q 3 ⎝ ∂q 3 ⎠ ∂ ∂ ∂ ∂ φ= φ .S. 7 .2) For simplicity. h 3 = 1 ⎪ ⎨ h1 = 1. Orfanidis – Electromagnetic Waves and Antennas ⎧ h1 = 1. the proves are done using rectangular coordinates ( h1 = 1. h 2 = 1. Ramaccia and A. h 2 = r. h = r.J.2. h = r sin ϑ 2 3 ⎩ 1 Exercises Chapter 1 (rectangular coordinates) (cylindrical coordinates) (spherical coordinates) (1. Toscano Pag. h 2 = 1. Ramaccia and A. Toscano Pag. 8 .J.S. Orfanidis – Electromagnetic Waves and Antennas so Exercises Chapter 1 ˆ ( φ∇ψ − ψ∇φ) = e1 ⎜ φ Now we can apply the dot product: ⎛ ∂ψ ⎛ ∂ψ ⎛ ∂ψ ∂φ ⎞ ∂φ ⎞ ∂φ ⎞ ˆ ˆ −ψ −ψ −ψ ⎟ ⎟ + e2 ⎜ φ ⎟ + e3 ⎜ φ ∂q1 ⎠ ∂q 2 ⎠ ∂q3 ⎠ ⎝ ∂q1 ⎝ ∂q 2 ⎝ ∂q3 ⎡ ∂ ⎛ ∂ψ ∂φ ⎞ ∂ ⎛ ∂ψ ∂φ ⎞ ∂ ⎛ ∂ψ ∂φ ⎞ ⎤ ∇ ⋅ ( φ∇ψ − ψ∇φ ) = ⎢ −ψ −ψ −ψ ⎜φ ⎟⎥ = ⎜φ ⎟+ ⎜φ ⎟+ ∂q1 ⎠ ∂q 2 ⎝ ∂q 2 ∂q 2 ⎠ ∂q 3 ⎝ ∂q 3 ∂q 3 ⎠ ⎦ ⎢ ⎥ ⎣ ∂q1 ⎝ ∂q1 ⎛ ∂φ ∂ψ ∂ 2 ψ ∂ψ ∂φ ∂ 2 φ ⎞ ⎛ ∂φ ∂ψ ∂ 2 ψ ∂ψ ∂φ ∂2φ ⎞ = +⎜ +φ − −ψ +φ − −ψ ⎟+⎜ ⎟+ ⎜ ∂q1 ∂q1 ∂q1 ∂q1 ∂q1 ∂q1 ⎟ ⎜ ∂q 2 ∂q 2 ∂ q 2 ∂q 2 ∂ q 2 ∂q 2 ⎟ ⎝ ⎠ ⎝ ⎠ ⎛ ∂φ ∂ψ ∂ 2 ψ ∂ψ ∂φ ∂2φ ⎞ +⎜ +φ − −ψ = ⎟ ⎜ ∂q 3 ∂q 3 ∂q 3 ∂q 3 ∂q 3 ∂q 3 ⎟ ↑ ⎝ ⎠ cancel opposite terms in parentheses ⎛ ∂2ψ ∂2ψ ∂2ψ ⎞ ⎛ ∂2φ ∂2φ ∂2φ ⎞ =φ ⎜ + + + + = φ∇ 2 ψ − ψ∇ 2 φ ⎟ − ψ⎜ ⎜ ∂q1 ∂q 2 ∂q 3 ⎟ ⎜ ∂q1 ∂q 2 ∂q 3 ⎟ ⎟ ⎝ ⎠ ⎝ ⎠ • Identity n°4 ⎡ ∂ ⎤ ∂ ∂ ˆ ˆ ˆ ∇ ⋅ ( φA ) = ∇ ⋅ ( φA1e1 + φA 2 e2 + φA3e3 ) = ⎢ ( φA1 ) + ( φA 2 ) + ( φA3 )⎥ = ∂q 2 ∂q3 ⎣ ∂q1 ⎦ ⎡⎛ ∂A ∂φ ⎞ ⎛ ∂A 2 ∂φ ⎞ ⎛ ∂A3 ∂φ ⎞ ⎤ = ⎢⎜ φ 1 + A1 + A2 + A3 ⎟⎥ = ⎟ + ⎜φ ⎟ + ⎜φ ∂q1 ⎠ ⎝ ∂q 2 ∂q 2 ⎠ ⎝ ∂q3 ∂q3 ⎠ ⎥ ⎢⎝ ∂q1 ⎣ ⎦ ⎛ ∂φ ∂φ ∂φ ⎞ ⎛ ∂A1 ∂A 2 ∂A3 ⎞ = ⎜ A1 + A2 + A3 + + ⎟ = ( ∇φ ) ⋅ A + φ∇ ⋅ A ⎟ +φ ⎜ ∂q 2 ∂q3 ⎠ ⎝ ∂q1 ∂q 2 ∂q3 ⎠ ⎝ ∂q1 • Identity n° 5 ˆ e1 ∇ × ( φA ) = ∂ ∂q1 φA1 ˆ e2 ∂ ∂q 2 φA 2 ˆ e3 ∂ = ∂q 3 φA 3 D. This choise is obligated by the fact that if first we calculated the divergence of the vector A . Cross product with the vector B would be impossible. the results would be a scalar.J. Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ⎡ ∂ ⎤ ⎡ ∂ ⎤ ⎡ ∂ ⎤ ∂ ∂ ∂ ˆ ˆ ˆ +e1 ⎢ ( φA3 ) − ( φA 2 )⎥ + e2 ⎢ ( φA1 ) − ( φA3 )⎥ + e3 ⎢ ( φA 2 ) − ( φA1 )⎥ = ∂q3 ∂q1 ∂q 2 ⎣ ∂q1 ⎦ ⎣ ∂q 2 ⎦ ⎣ ∂q3 ⎦ ⎡⎛ ∂φ ∂A3 ⎞ ⎛ ∂φ ∂A 2 ˆ φ⎟ − ⎜ =e1 ⎢⎜ A3 + A + ⎜ ∂q3 2 ∂q3 ∂q 2 ⎠ ⎝ ⎢⎝ ∂q 2 ⎣ ⎞⎤ ⎡⎛ ∂φ ∂A3 ⎞ ⎤ ∂A ⎞ ⎛ ∂φ ˆ φ ⎟ ⎥ + e 2 ⎢⎜ φ ⎟⎥ + A1 + 1 φ ⎟ − ⎜ A3 + ⎟ ∂q3 ⎠ ⎝ ∂q1 ∂q1 ⎠ ⎥ ⎢⎝ ∂q3 ⎣ ⎦ ⎠⎥ ⎦ ⎡⎛ ∂φ ∂A 2 ⎞ ⎛ ∂φ ∂A ⎞ ⎤ ˆ φ⎟ − ⎜ + e3 ⎢ ⎜ A2 + A1 + 1 φ ⎟ ⎥ = ∂q1 ⎠ ⎜ ∂q 2 ∂q 2 ⎟ ⎥ ⎢⎝ ∂q1 ⎝ ⎠⎦ ⎣ ⎡ ∂φ ⎤ ⎡ ∂φ ⎤ ⎡ ∂φ ⎤ ∂φ ∂φ ∂φ ˆ ˆ ˆ =e1 ⎢ A3 − A 2 ⎥ + e2 ⎢ A1 − A 3 ⎥ + e3 ⎢ A2 − A1 ⎥ + ∂q3 ∂q1 ∂q 2 ⎣ ∂q1 ⎦ ⎣ ∂q 2 ⎦ ⎣ ∂q3 ⎦ ⎡ ∂A ⎡ ∂A ⎡ ∂A ∂A ⎤ ∂A 2 ⎤ ∂A ⎤ ˆ ˆ ˆ φ ⎥ + e 2 ⎢ 1 φ − 3 φ ⎥ + e3 ⎢ 2 φ − 1 φ ⎥ = ( ∇φ ) × A + φ∇ × A +e1 ⎢ 3 φ − ∂q3 ⎦ ∂q1 ⎦ ∂q 2 ⎦ ⎣ ∂q1 ⎣ ∂q 2 ⎣ ∂q3 • Identity n° 6 ⎡ ⎛ ∂A ⎛ ∂A 3 ∂A1 ⎞ ⎛ ∂A 2 ∂A1 ⎞ ⎤ ∂A 2 ⎞ ˆ ˆ ˆ ∇ ⋅ ( ∇ × A ) = ∇ ⋅ ⎢e1 ⎜ 3 − − − ⎟ − e2 ⎜ ⎟ + e3 ⎜ ⎟⎥ = ⎢ ⎝ ∂q 2 ∂q 3 ⎠ ⎝ ∂q1 ∂q 2 ⎠ ⎥ ⎝ ∂q1 ∂q 3 ⎠ ⎣ ⎦ ⎡ ∂ ⎛ ∂A 3 ∂A 2 − =⎢ ⎜ ⎢ ∂q1 ⎝ ∂q 2 ∂q 3 ⎣ ⎞ ∂ ⎛ ∂A 3 ∂A1 ⎞ ∂ ⎛ ∂A 2 ∂A1 ⎞ ⎤ − − ⎟− ⎜ ⎟+ ⎜ ⎟⎥ = ⎠ ∂q 2 ⎝ ∂q1 ∂q 3 ⎠ ∂q 3 ⎝ ∂q1 ∂q 2 ⎠ ⎥ ⎦ ⎡ ∂ ∂A 3 ∂ ∂A 2 ∂ ∂A 3 ∂ ∂A1 ∂ ∂A 2 ∂ ∂A1 ⎤ =⎢ − − − + − ⎥=0 ⎢ ∂q1 ∂q 2 ∂q1 ∂q 3 ∂q 2 ∂q1 ∂q 2 ∂q 3 ∂q 3 ∂q1 ∂q 3 ∂q 2 ⎥ ⎣ ⎦ For the linearity of the derivate operator ∂ ∂ ∂ ∂ φ= φ .S. Toscano Pag. ∂q i ∂q j ∂q j ∂q i • Identity n°7 To evaluate the expression ∇ ⋅A×B . 9 . we have to calculate first the cross product and then the divergence of vector A × B . so the term in brackets is null. So we have: D. J.S. Ramaccia and A. Toscano Pag. 10 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ˆ ˆ ∇ ⋅ ( A × B ) = ∇ ⋅ ⎡e1 ( A 2 B3 − A3 B2 ) − e 2 ( A1B3 − A 3 B1 ) + e3 ( A1B2 − A 2 B1 ) ⎤ = ⎣ˆ ⎦ = = ∂ ∂ ∂ ( A 2 B3 − A3 B2 ) − ( A1B3 − A3 B1 ) + ( A1B2 − A 2 B1 ) = ∂q1 ∂q 2 ∂q3 ∂ ∂ ∂ ∂ ∂ ∂ ( A 2 B3 ) − ( A3 B2 ) − ( A1B3 ) + ( A3 B1 ) + ( A1B2 ) − ( A 2 B1 ) = ∂q1 ∂q1 ∂q 2 ∂q 2 ∂q 3 ∂q 3 ⎛ ∂A 2 ∂B ⎞ ⎛ ∂A3 ∂B ⎞ ∂B ⎞ ⎛ ∂A = ⎜ B3 + A 2 3 ⎟ − ⎜ B2 + A 3 2 ⎟ − ⎜ B3 1 + A1 3 ⎟ + ∂q1 ∂q1 ⎠ ⎝ ∂q1 ∂q1 ⎠ ⎝ ∂q 2 ∂q 2 ⎠ ⎝ ⎛ ∂A3 ∂B ⎞ ⎛ ∂A1 ∂B ⎞ ⎛ ∂B ∂A 2 ⎞ + ⎜ B1 + A 3 1 ⎟ + ⎜ B2 + A1 2 ⎟ − ⎜ A 2 1 + B1 ⎟= ∂q 2 ⎠ ⎝ ∂q 3 ∂q3 ⎠ ⎝ ∂q 3 ∂q3 ⎠ ⎝ ∂q 2 ⎛ ∂A ⎛ ∂A1 ∂A3 ⎞ ⎛ ∂A 2 ∂A1 ⎞ ∂A 2 ⎞ − − =B1 ⎜ 3 − ⎟ + B2 ⎜ ⎟ + B3 ⎜ ⎟+ ⎝ ∂q1 ∂q 2 ⎠ ⎝ ∂q 2 ∂q 3 ⎠ ⎝ ∂q3 ∂q1 ⎠ ⎛ ∂B ⎛ ∂B ⎛ ∂B ∂B ⎞ ∂B ⎞ ∂B ⎞ + A1 ⎜ 2 − 3 ⎟ + A 2 ⎜ 3 − 1 ⎟ + A 3 ⎜ 1 − 2 ⎟ = ⎝ ∂q 2 ∂q1 ⎠ ⎝ ∂q3 ∂q 2 ⎠ ⎝ ∂q1 ∂q3 ⎠ =B ⋅ ( ∇ × A ) − A ⋅ ( ∇ × B ) • Identity n° 8 ⎛ ⎛ ∂A ⎛ ∂A 3 ∂A1 ⎞ ⎛ ∂A 2 ∂A1 ⎞ ⎞ ∂A 2 ⎞ ˆ ˆ ˆ ∇ × ( ∇ × A ) = ∇ × ⎜ e1 ⎜ 3 − − − ⎟ − e2 ⎜ ⎟ + e3 ⎜ ⎟⎟ = ⎜ ⎟ ⎝ ∂q1 ∂q 2 ⎠ ⎠ ⎝ ∂q 2 ∂q 3 ⎠ ⎝ ∂q1 ∂q 3 ⎠ ⎝ ⎛ ∂ ⎛ ∂A 2 ∂A1 ⎞ ∂ ⎛ ∂A1 ∂A 3 ⎞ ⎞ ˆ − − =e1 ⎜ ⎜ ⎟⎟ + ⎜ ⎟− ⎜ ∂q ⎟ ∂q 2 ⎠ ∂q 3 ⎝ ∂q 3 ∂q1 ⎠ ⎠ 2 ⎝ ∂q1 ⎝ ⎛ ∂ ⎛ ∂A 2 ∂A1 ⎞ ∂ ⎛ ∂A 3 ∂A 2 ⎞ ⎞ ˆ − e2 ⎜ − − ⎜ ⎟⎟ + ⎜ ⎟− ⎜ ∂q ∂q ∂q 2 ⎠ ∂q 3 ⎝ ∂q 2 ∂q 3 ⎠ ⎟ 1⎝ 1 ⎝ ⎠ ⎛ ∂ ⎛ ∂A1 ∂A 3 ⎞ ∂ ⎛ ∂A 3 ∂A 2 ⎞ ⎞ ˆ + e3 ⎜ − − ⎜ ⎟− ⎜ ⎟⎟ = ⎜ ∂q ∂ q ∂q1 ⎠ ∂q 2 ⎝ ∂q 2 ∂q 3 ⎠ ⎟ ⎝ 1⎝ 3 ⎠ ⎛ ∂ ∂A 2 ∂ 2 A1 ∂ 2 A1 ∂ ∂A 3 ⎞ ˆ − − + =e1 ⎜ ⎟+ ⎜ ∂q 2 ∂q1 ∂q 2 ∂q 3 ∂q 3 ∂q1 ⎟ ⎝ ⎠ ⎛ ∂2A2 ∂ ∂A1 ∂ ∂A 3 ∂ 2 A 2 ⎞ ˆ − e2 ⎜ − − + ⎟+ ⎜ ∂q1 ∂q1 ∂q 2 ∂q 3 ∂q 2 ∂q 3 ⎟ ⎝ ⎠ ⎛ ∂ ∂A1 ∂ 2 A 3 ∂ 2 A 3 ∂ ∂A 2 ⎞ ˆ + e3 ⎜ − − + ⎟= ⎜ ∂q1 ∂q 3 ∂q1 ∂q 2 ∂q 2 ∂q 3 ⎟ ⎝ ⎠ D. Toscano Pag.S. 11 . Ramaccia and A.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ⎛ ∂ ∂A 2 ⎛ ∂ ∂A1 ⎛ ∂ ∂A1 ∂ ∂A 3 ⎞ ∂ ∂A 3 ⎞ ∂ ∂A 2 ⎞ ˆ ˆ ˆ + − + =e1 ⎜ ⎟ − e2 ⎜ − ⎟ + e3 ⎜ ⎟+ ∂q 2 ∂q1 ∂q 3 ∂q1 ⎠ ∂q1 ∂q 2 ∂q 3 ∂q 2 ⎠ ∂q1 ∂q 3 ∂q 2 ∂q 3 ⎠ ⎝ ⎝ ⎝ ⎛ ∂ 2 A1 ∂ 2 A1 ⎞ ⎛ ∂2 A2 ∂2 A2 ˆ1 ⎜ − ˆ2 ⎜ − + +e ⎟−e ⎜ ∂q 2 ⎜ ∂q1 ∂q 3 ⎟ ∂q 3 ⎝ ⎠ ⎝ ⎛ ∂ 2 A3 ∂ 2 A3 ⎞ ⎞ ˆ3 ⎜ − − ⎟= ⎟+e ⎟ ⎜ ∂q1 ∂q 2 ⎟ ⎠ ⎝ ⎠ ⎞⎤ ⎟⎥ − ⎠⎦ ⎡ ∂ ⎛ ∂A 2 ∂A 3 ⎞ ⎤ ⎡ ∂ ⎛ ∂A1 ∂A 3 ⎞ ⎤ ⎡ ∂ ⎛ ∂A1 ∂A 2 ˆ ˆ ˆ + + + =e1 ⎢ ⎜ ⎟⎥ + e2 ⎢ ⎜ ⎟ ⎥ + e3 ⎢ ⎜ ∂q 3 ⎝ ∂q1 ∂q 2 ⎢ ∂q1 ⎝ ∂q 2 ∂q 3 ⎠ ⎥ ⎢ ∂q 2 ⎝ ∂q1 ∂q 3 ⎠ ⎥ ⎣ ⎣ ⎦ ⎣ ⎦ ⎛ ∂ 2 A1 ∂ 2 A1 ⎞ ⎛ ∂2 A2 ∂2 A2 ˆ1 ⎜ ˆ2 ⎜ −e + + ⎟−e ⎜ ∂q 2 ⎜ ∂q 3 ⎟ ∂q 3 ⎝ ⎠ ⎝ ∂q1 ⎛ ∂ 2 A3 ∂ 2 A3 ⎞ ⎞ ˆ3 ⎜ + + ⎟ = ∇ (∇ ⋅ A ) − ∇2 A ⎟−e ⎟ ⎜ ∂q1 ⎟ ∂q 2 ⎠ ⎠ ⎝ D. Apply the integrated form of Gauss's law to the same volume element and show the boundary condition D1z − D 2 z = ρ s = lim Δ z → 0 ( ρΔ z ) . the second term in the right–hand side may be assumed to go to zero.3. and related by the Kelvin–Stokes theorem. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. The identity demonstrated by Stokes is the follow: ∫∫ ( ∇ × F ) dS = ∫ S c(S) F⋅d (1. Show that this leads to the boundary condition H 1y − H 2 y = − J sx . Applying the integrated form of Ampère's law to the infinitesimal face abcd. Toscano Pag. J sx ≡ lim Δ z → 0 ( J x Δ z ) . In the limit Δz → 0 .J.1) 2. Ampère's circuital law relates the magnetic field to its electric current source. The forms are equivalent. 12 .3. Ramaccia and A.3. and that these two boundary conditions can be combined vectorially into: ˆ n × (H1 − H 2 ) = Js (1.3) So applying (1. Fig. The axes are oriented such that ˆ ˆ n = z . Solution • Question n° 1 In its historically original form.3.2) 3.3. the integral form and the differential form. 1.3) to the second Maxwell's equation. shows that H 1x − H 2 x = J sy . such that its upper half lies in medium ε1 and its lower half in medium ε2 . The law can be written in two forms.S. we obtain the Ampere's law in integral form with few simply steps: D. 1.3 Exercise Consider the infinitesimal volume element ΔxΔyΔz shown below. show that H 2 y − H 1y = J x Δ z + ∂D x Δz ∂t (1. whereas the first term will be non–zero and may be set equal to the surface current density. Similarly. that is.1: Infinitesimal volume element between two medium. tangent to the curved line c that bounds the surface S.J. So we obtain: ˆ − ∫ H1 ⋅ dz − a O1 ∫ O1 b ˆ ˆ H 2 ⋅ d z + ∫ H 2 ⋅ dy + b c O2 ∫ c ˆ H 2 ⋅ dz + ∫ O2 d ˆ ˆ H1 ⋅ dz − ∫ H1 ⋅ dy = d a = J x Δ zΔ y + ∂D x Δ zΔ y ∂t H 1 and H 2 are constant inside each medium.3. For simplicity.S. the component Jx and D x . so the line integrals can be written as: − H1z ∂D x Δz Δz Δz Δz − H1y Δy = J x ΔzΔy + ΔzΔy − H 2z + H 2y Δy + H 2z + H1z ∂t 2 2 2 2 ∂D x Δz Δy ∂t i. Choose an counterclockwise path so that. H 2 y Δ y − H 1y Δ y = J x Δ z Δ y + H 2 y − H 1y = J x Δ z + ∂D x Δz ∂t • Question n° 2 D. On the contrary. Ramaccia and A. i. So the integral on that part of the path needs to be decomposed into two integral with different arguments. Note that the z–parallel sides have the first half in the medium 1 and the second in medium 2.4) where is the infinitesimal vector. one for each infinitesimal side of the rectangular abcd. respectively. and we have to define the sense ˆ of integration. using the right–hand rule. Toscano Pag. that has area S = ΔzΔy and perimeter p = 2Δz + 2Δy .3.e.3. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ∇× H = J + Integrate terms of the identity over an opened surface S: ∂D ∂t ∂D ˆ ⋅ n dS ∂t ˆ ˆ ∫∫ ( ∇ × H ) ⋅ ndS = ∫∫ J ⋅ ndS + ∫∫ S S S and apply the Stokes theorem: ∫ c(S) ˆ H ⋅ d = ∫∫ J ⋅ ndS + ∫∫ S S ∂D ˆ ⋅ ndS ∂t (1. 13 .4) can be decomposed into a sum of four integral expression. to solve the right–side of (1. The left–hand side of (1.4) we have to identify the correct component of J and D that flows through the face abcd.e. Now we can consider the infinitesimal face abcd. denote the points of contact between mediums along the segments ab and cd with O1 and O 2 . the normal is x. 7) where S1 and S2 are portions of S in the medium 1 and medium 2. one for each side of parallelepiped ΔxΔyΔz : D.3. The forms are equivalent. (1. respectively in the medium 1 and medium 2.S. Now consider the volume V = ΔxΔyΔz .6) is a simple volume integral.5) to the third Maxwell's equation. we obtain the Gauss's law in integral form with few simply steps: ∇⋅ D = ρ Integrate terms of the identity over a volume V: ∫∫∫ ( ∇ ⋅ D ) dV = ∫∫∫ ρ dV V V and apply the divergence theorem: ∫∫ S(V) ˆ D ⋅ ndS = ∫∫∫ ρdV = Qin V (1. it can be written in two forms. we can subtract vectorially these two boundary conditions: ˆ ˆ ˆ ˆ y ( H1x − H 2x ) − x H1y − H 2y = Jsx x + Jsy y ˆ n × ( H1 − H 2 ) = J s ( ) ˆ ˆ where n = z . (1.3.2).7) can be decomposed into several surface integrals. Ramaccia and A.5) So applying (1.3.6) can be decomposed into two integrals with arguments D1 and D2 .3.3. and related by the divergence theorem: ∫∫∫ ( ∇ ⋅ F ) dV = ∫∫ V S(V) ˆ F ⋅ ndS (1. respectively and ρ is considered constant inside the volume V. In order to obtain eq.1) is reduced in H 1y − H 2 y = − J sx H 1x − H 2 x = J sy . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 and similarly In the limit Δz → 0 .3. 14 .3. The right–hand side of (1.3. So we have: ˆ ˆ ∫∫ ( D1 ⋅ n ) dS1 + ∫∫ ( D2 ⋅ n ) dS2 = ∫∫∫ ρdV = ρΔxΔyΔz S1 S2 V (1.3.J. Like Ampère's circuital law. (1. The terms on the right–hand side of eq. Toscano Pag. The left–hand side of (1. the integral form and the differential form. the eq.6) ˆ where n is the outgoing unit vector normal to the closed surface S that bounds the volume V. • Question n° 3 Gauss's law relates the electric field to its electric charge sources. Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 D1z ΔxΔy + D1y Δz Δz Δz Δz Δx + D1x Δy − D1y Δx − D1x Δy − 2 2 2 2 Δz Δz Δz Δz Δx + D 2x Δy − D 2y Δx − D 2x Δy = ρΔxΔyΔz 2 2 2 2 − D 2z ΔxΔy + D 2y i. the amount ρΔz collapses in ρs which is the surface electric charge density. D.S.J. 15 . Toscano Pag.e. D1z − D2z = ρΔz In the limit Δz → 0 . 4.3) Solution First of all.4.5) and we obtain: D.4.4 Exercise Show that time average of the product of two harmonic quantities Α(t) = Re ⎡ Ae jω t ⎤ and ⎣ ⎦ B (t) = Re ⎡ Be jω t ⎤ with phasors A.4.S.4. Toscano Pag.2) (1.4. (1. (1.4) and substitute eq. (1.4. B is given by: ⎣ ⎦ A(t) B (t) = 1 T T 0 ∫ A(t) B (t)dt = 2 Re ⎡ AB ⎣ 1 ∗⎤ ⎦ (1. Then show that the time–averaged values of the cross and dot products of two time–harmonic vector quantities A (t) = Re ⎡ A e jω t ⎤ and B (t) = Re ⎡ Be jω t ⎤ can be ⎣ ⎦ ⎣ ⎦ expressed in terms of the corresponding phasors as follows: 1 A(t) × B(t) = Re ⎡ A × B∗ ⎤ ⎦ 2 ⎣ 1 A(t) ⋅ B(t) = Re ⎡ A ⋅ B∗ ⎤ ⎦ 2 ⎣ (1.1) where T = 2π ω is one period.4.4.4.4) into eq.6) Substitute eq.1): A(t) B (t) = 1 T T ∫ 0 A(t) B (t)dt = 1 AB cos (ω t + ϕ1 ) cos (ω t + ϕ 2 ) dt T∫ 0 T (1.6) into eq. (1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1.J.5) Now we have to use Euler's formula: ⎧ e = cos x + jsin x ⎪ ⎨ − jx = cos x − jsin x ⎪e ⎩ jx ⇔ ⎧ e jx + e − jx ⎪ cos x = 2 ⎪ ⎨ jx − jx ⎪sin x = e − e ⎪ 2j ⎩ (1. we express the harmonic quantities A(t) and B(t) in their extended form: ⎧ A(t) = A cos (ω t + ϕ1 ) ⎪ ⎨ ⎪ B (t) = B cos (ω t + ϕ 2 ) ⎩ (1. 16 . Ramaccia and A. J. Orfanidis – Electromagnetic Waves and Antennas T ⎛ e jω t eϕ1 + e− jω t e−ϕ1 1 AB ⎜ ⎜ T∫ 2 ⎝ 0 Exercises Chapter 1 ⎞ ⎟ dt = ⎟ ⎠ A(t) B(t) = − jω t −ϕ1 T jω t ϕ1 e e jω t eϕ2 + e− jω t e−ϕ2 AB e e + e = dt = 2T ∫ 2 0 T ϕ −ϕ − ϕ −ϕ − ϕ +ϕ ϕ +ϕ AB e2 jω t e( 1 2 ) + e( 1 2 ) + e ( 1 2 ) + e−2 jω t e ( 1 2 ) dt = = 2 2T ∫ ( )( ⎞⎛ e jω t eϕ2 + e− jω t e−ϕ2 ⎟⎜ ⎟⎜ 2 ⎠⎝ ) = ABcos (ϕ1 − ϕ2 ) AB 1 1 cos (ϕ1 − ϕ2 ) dt = T = ABcos (ϕ1 − ϕ2 ) = Re ⎡ AB∗ ⎤ ∫ ⎦ 2T 2 2 ⎣ 2T 0 0 T Operating in similar way. Ramaccia and A.S. we can demonstrate the time–averaged values of the cross and dot products of two time–harmonic vector quantities. Toscano Pag. so: a×b 1 1 Re[AB∗ ] = Re[aA × bB∗ ] = Re[A × B∗ ] 2 2 2 • Dot Product A(t) ⋅ B (t) = = 1 T T 0 ∫ ( A(t) ⋅ B (t) ) dt = 1 a⋅b jω t jω t jω t jω t ∫ Re[Ae ]a ⋅ Re[Be ]bdt = T ∫ Re[Ae ] Re[Be ]dt = T 0 0 T T a⋅b 1 1 Re[AB*] = Re[aA ⋅ bB∗ ] = Re[ A ⋅ B∗ ] 2 2 2 D. • Cross Product A (t) × B (t) = 1 T T 0 ∫ ( A(t) × B (t) ) dt = A(t) × B(t) = 1 a×b jω t jω t jω t jω t ∫ Re[Ae ]a × Re[Be ]b dt = T ∫ Re[Ae ] Re[Be ]dt T 0 0 T T The result of integral is note by previous exercise. 17 . Solution • Question n° 1 Using the identity ∇ ⋅ ( E × H ) = H ⋅ ( ∇ × E ) − E ⋅ ( ∇ × H ) and Maxwell's equations.2) which states that the net time–averaged power floating into a volume is dissipated into heat.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1.1) and integrating over a volume V bounded by closed surface. Toscano Pag. (1.5.5. show the time–averaged form of energy conservation: − ∫∫ S(V) 1 1 ˆ Re[E × H∗ ] ⋅ ndS = ∫∫∫ Re ⎡E ⋅ J∗ ⎤ dV tot ⎦ 2 2 ⎣ V (1. 3. we have: ∇ ⋅ E × H ∗ = H ∗ ⋅ ( ∇ × E ) − E ⋅ ∇ × H ∗ = H ∗ ⋅ ( − j ω B ) − E ⋅ J * − j ω D* = = − jωμ H ⋅ H ∗ − E ⋅ J ∗ tot ( ) ( ) ( ) • Question n° 2 Integrate over a volume V the right–hand side of eq. show that the integrals in (1.2) are zero and provide an interpretation.5 Exercise Assuming that B = μH : 1. 2.5.5. Show that Maxwell's equations ∇ × E = − jω B ∇ × H = J + jω D ∇⋅D = ρ ∇⋅B = 0 imply the following complex–value version of Poynting's theorem: ∇ ⋅ E × H ∗ = − jωμ H ⋅ H ∗ − E ⋅ J ∗ tot ( ) (1. (1.5. Ramaccia and A.1) where J tot = J + jω D .S. For a lossless dielectric. Extracting the real–parts of both sides of eq.1) and apply the divergence's theorem: ∫∫∫ ( ∇ ⋅ ( E × H V ∗ )) dV = ∫∫ S(V) ˆ ( E × H∗ ) ndS and now calculate the time–averaged value: D. 18 . we obtain: T ⎤ 1 ⎡ ⎢ ∫∫∫ − jωμ H ⋅ H∗ − E ⋅ J ∗ dV ⎥dt = tot T ∫⎢ ⎥ 0⎣ V ⎦ ⎡1 T ⎤ ⎢ ∫ − jωμ H ⋅ H∗ − E ⋅ J ∗ dt ⎥ dV = tot ∫∫∫ ⎢ T ⎥ V ⎣ 0 ⎦ 1 ⎡1 ∗ ∗ ⎤ ∫∫∫ ⎢ 2 Re ⎡ − jωμ H ⋅ H ⎤ − 2 Re ⎡E ⋅ J tot ⎤ ⎥ dV ⎣ ⎦ ⎣ ⎦⎦ ⎣ ( ( ) ) (1.5) The minus sign is been associated with the left–hand side because it identifies the quantity of energy that goes in the volume V –while the Poynting's vector is defined outgoing from V– and the right–hand side represents the energy dissipated as heat. So: ∫∫∫ 2 Re ⎡ E ⋅ ( − jωε E ⎣ V 1 ∗ ) ⎤ dV = 0 ⎦ (1.5.6) being the real part of jωε E ⋅ E∗ zero. Moreover zero for the right–hand side of (1.5. Orfanidis – Electromagnetic Waves and Antennas T ⎤ 1 ⎡ ∗ ˆ ⎢ E × H n dS ⎥dt T ∫ ⎢ ∫∫ ⎥ 0 ⎣ S(V) ⎦ Exercises Chapter 1 ( ) Invert the order of integrals: ⎡1 T ⎤ 1 ∗ ∗⎤ ˆ ⎡ ˆ ⎢ ∫∫ ⎢ T ∫ E × H dt ⎥ ndS = ∫∫ 2 Re ⎣E × H ⎦ ndS ⎥ S(V) ⎣ 0 S(V) ⎦ ( ) (1.5. • Question n° 3 Inside a lossless dielectric. Ramaccia and A.S.J. D.5. It is correct because electromagnetic wave pass through the dielectric. implies that not all the energy remains inside the volume.4) V The real part of j ωμ H ⋅ H ∗ is zero because the product H ⋅ H ∗ = H 2 is real and so the quantity j ωμ H ⋅ H ∗ is imaginary. Exactly in steady state the quantity of energy ingoing is equal to the outgoing one. the current density J is zero while the displacement current D is simply equal to ε E . that represents the quantity of energy ingoing the volume bounded by the surface S.3) In similar way on left–hand side. Toscano Pag. Only the term associated with the heat survives and we can write: − ∫∫ S(V) 1 ⎡ 1 ˆ Re E × H∗ ⎤ ndS = ∫∫∫ Re ⎡E ⋅ J∗ ⎤ dV tot ⎦ ⎦ 2 ⎣ 2 ⎣ V (1. 19 .5.5). 2.3) where f x .S.6. (1. We can write: ˆ (∇ × H ) × B = ⎢x ⎜ ⎡ ⎛ ∂H z ∂H y ⎞ ⎛ ∂H z ∂H x ˆ − − ⎟ − y⎜ ∂z ⎠ ∂z ⎝ ∂x ⎢ ⎝ ∂y ⎣ ˆ x ˆ y ⎛ ∂H y ∂H x ⎞ ⎤ ⎞ ˆ + z⎜ − ⎟⎥ × B = ⎟ ∂y ⎠ ⎥ ⎠ ⎝ ∂x ⎦ ˆ z ∂H y ⎞ ⎛ ∂H ⎛ ∂H z ∂H x ⎞ ⎛ ∂H y ∂H x ⎞ = ⎜ z − − − ⎟ −⎜ ⎜ ⎟ ∂z ⎠ ∂z ⎟ ⎝ ∂x ∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y Bx By Bz (1.2) because it is easy to note from the left–hand side that it is the cross product of the second Maxwell's equation (i.6.J.2).e.1) ( J × B )x + ⎛ ⎜ ∂D 1 ⎞ ⎛ ⎞ ˆ × B ⎟ = ∇ ⋅ ⎜ μ Hx H − x μ H2 ⎟ 2 ⎝ ∂t ⎠x ⎝ ⎠ (1.6. 1.6. Solution • Question n° 1 Let us begin with eq. and Tx ˆ is defined to be the vector(equal to Maxwell's stress tensor acting on the unit vector x): ˆ Tx = ε Ex E + μ Hx H − x 1 ε E 2 + μ H2 2 ( ) 3.4) D.2).6. ∇× H = J + ∂D ) with the vector ∂t B and then we extract the x–component.6 Exercise Assuming that D = ε E and B = μH . 20 . Toscano Pag. So we have to demonstrate the right–hand side of eq. Write similar equations of y. that is. (1.6.1) and (1. G x are the x–components of the vectors f = ρE + J × B and G = D × B . z components. (1.2) where the subscript x means the x–component.6. Show that Maxwell's equations imply the following relationships: ˆ ρ Ex + ⎜ D × ⎟ = ∇ ⋅ ⎜ ε Ex E − x ε E2 ⎟ ∂t ⎠ x 2 ⎝ ⎝ ⎠ ⎛ ∂B ⎞ ⎛ 1 ⎞ (1. the momentum density. From eq. The quantity G x is interpreted as the field momentum (in the x–direction) per unit of volume.6. derive the following relationship that represent momentum conservation: fx + ∂G x = ∇ ⋅ Tx ∂t (1. Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. Toscano Pag.5) the term H x ( ∇ ⋅ H ) and the couple of terms ± H x ∂ H x .6. (1. (1.6. because they're both zero: ∂x ( ∇ × H ) × B x −component = ∂H y ⎡ ∂H z ∂H x ∂H x ∂H x ⎤ + Hz −H y + Hy + Hx +⎥ ⎢ −H z ∂x ∂z ∂x ∂y ∂x ⎥ ⎢ =μ⎢ ⎥= ∂H y ∂H z ∂H x ∂H x ⎢+Hx ⎥ + Hx + Hx −H x ⎢ ⎥ ∂y ∂z ∂x ∂x ⎣ ⎦ (1.6) Let us consider the only emphasized terms of eq.6.5) From the forth Maxwell's equation (i.J.6): ∂H y ∂H y ⎡ ∂H z ∂H x ⎤ ∂H z ∂H x ⎤ 1 ⎡ −μ ⎢Hz + Hy + Hx + 2H y + 2H x ⎥ = − μ ⎢ 2H z ⎥= ∂x ∂x ∂x ⎦ ∂x ∂x ∂x ⎦ 2 ⎣ ⎣ 2 1 ⎡ ∂H 2 ∂H y ∂H 2 ⎤ 1 x⎥ ⎢ z + ˆ − μ + = −x μ∇H 2 ∂x ∂x ⎥ 2 ⎢ ∂x 2 ⎣ ⎦ and now consider the remaining terms: ⎡ ∂H y ∂H x ∂H x ∂H x ∂H z ∂H x ⎤ + Hy + Hx + Hx + Hx + Hx = ⎥ ∂z ∂y ∂x ∂y ∂z ∂x ⎦ ↑ order them ⎞⎤ ⎟⎥ = ⎠⎥ ⎦ μ ⎢Hz ⎣ ⎡ ∂H y ∂H x ⎛ ∂H x ⎞ ⎛ ∂H x ∂H z = μ ⎢ 2H x + ⎜ Hx + Hy + Hx ⎟ + ⎜ Hz ∂x ∂y ∂y ⎠ ⎝ ∂z ∂z ⎢ ⎝ ⎣ ⎡ ∂H 2 ∂ H y H x ∂ (Hx Hz ) ⎤ ⎥ = μ∇ ⋅ H x H x x + H y y + H z z ˆ ˆ ˆ =μ⎢ x + + ∂y ∂z ⎢ ∂x ⎥ ⎣ ⎦ ( ) ( ( ) ) = μ∇ ⋅ ( H x H ) So we have that eq. ∇ ⋅ B = 0 ) and the constitutive relation B = μH .7) that is the right–hand side of eq. Orfanidis – Electromagnetic Waves and Antennas Now we consider only the x–component.6.S. (1. we can add to eq.6.e. 21 . (1.6.6.6) can be written as: 1 ˆ ( ∇ × H ) × B x−component = μ∇ ⋅ ( Hx H ) − x μ∇H2 = 2 1 ⎛ ⎞ ˆ = ∇ ⋅ ⎜ μ H x H − x μ H2 ⎟ 2 ⎝ ⎠ (1. D. writing Bi = μ Hi : Exercises Chapter 1 ⎛ ( ∇ × H ) × B x −component = μ ⎢−H z ⎜ ⎡ ⎢ ⎣ ∂H y ⎡ ∂H z ∂H x ∂H x ⎤ = μ ⎢ −H z + Hz − Hy + Hy ⎥ ∂x ∂z ∂x ∂y ⎦ ⎣ ∂H z ∂H x − ∂z ⎝ ∂x ⎛ ∂H y ∂H x ⎞ ⎤ ⎞ ⎟ − H y ⎜ ∂x − ∂y ⎟ ⎥ = ⎠ ⎝ ⎠⎥ ⎦ (1. Ramaccia and A.2). So we have to demonstrate the right–hand side of eq. We can apply the cross product to the first Maxwell's equation: D × ( ∇× E) = −D × ∂B ∂t From the properties of the cross product. (1.9) the term E x ∇ ⋅ D and the couple of terms ±Ex ∂E x : ∂x D.1).6.6.6. (1. ∇⋅ D − ρ = 0 ). we can add to eq.e.6.8) Now we consider only the x–component. multiplying it with E x . (1. (1. it's possible to invert the order of the terms in the left– hand side and change the sign in the right–hand–side: ( ∇ × E) × D = D × Now consider the term ( ∇ × E ) × D : ˆ (∇ × E ) × D = ⎢x ⎜ ⎡ ⎛ ∂E z ∂E y ⎞ ⎛ ∂E z ∂E x ˆ − − ⎟ − y⎜ ∂z ⎠ ∂z ⎝ ∂x ⎢ ⎝ ∂y ⎣ ˆ x ˆ y ∂B ∂t ⎛ ∂E y ∂E x ⎞ ⎤ ⎞ ˆ + z⎜ − ⎟⎥ × D = ⎟ ∂y ⎠ ⎥ ⎠ ⎝ ∂x ⎦ ˆ z ∂E y ⎞ ⎛ ∂E ⎛ ∂E z ∂E x ⎞ ⎛ ∂E y ∂E x ⎞ = ⎜ z − − − ⎟ −⎜ ⎜ ⎟ ∂z ⎠ ∂z ⎟ ⎝ ∂x ∂y ⎠ ⎝ ∂x ⎠ ⎝ ∂y Dx Dy Dz (1. In fact.S.1). 22 . (1.J. we can add to eq. With these considerations.6. ∇ × E = − ∂B ) with the vector ∂t D and then we extract the x–component. the term −ρ Ex completes the left–hand side of eq. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ∂B ⎞ ⎛ Eq. changing its sign. Toscano Pag.6. (1.5). there is the term ⎜ D × ⎟ that ∂t ⎠x ⎝ suggests us the cross product of the first Maxwell's equation (i.6. writing Di = ε Ei : ⎛ ( ∇ × E ) × D x −component = ε ⎢−Ez ⎜ ⎞⎤ ⎟⎥ = ⎢ ⎠⎥ ⎣ ⎦ ∂E y ⎡ ∂E ∂E ∂E ⎤ = ε ⎢ −E z z + E z x − E y + Ey x ⎥ ∂x ∂z ∂x ∂y ⎦ ⎣ ∂E z ∂E x − ∂z ⎝ ∂x ⎡ ⎛ ∂E y ∂E x ⎞ ⎟ − E y ⎜ ∂x − ∂y ⎠ ⎝ (1.6. In the left–hand side. Ramaccia and A.9) As for eq.e. but in this case there is the term ρ and it's correct for the results that we want to obtain.9) the third Maxwell's equation (i.1) is obtained in similar way. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ( ∇ × E ) × D x −component = ∂E y ⎡ ∂E z ∂E ∂E ∂E x ⎤ + E z x −E y + Ey x + Ex +⎥ ⎢ −E z ∂x ∂z ∂x ∂y ∂x ⎥ ⎢ =ε⎢ ⎥ ∂E y ∂E z ∂E x ∂E x ⎢+E x ⎥ + Ex + Ex −E x ⎢ ⎥ ∂y ∂z ∂x ∂x ⎣ ⎦ (1.6.6.6.6. ∂t D. (1.(1.6.10): ε ⎢ −E z ∂E y ∂E y ∂E z ∂E ⎤ ∂E ∂E ⎤ 1 ⎡ − Ey − E x x ⎥ = − ε ⎢ 2E z z + 2E y + 2E x x ⎥ = ∂x ∂x ∂x ⎦ 2 ⎣ ∂x ∂x ∂x ⎦ ⎣ 2 1 ⎡ ∂E 2 ∂E y ∂E 2 ⎤ 1 z ˆ =− ε⎢ + + x ⎥ = −x ε∇E 2 ∂x ∂x ⎥ 2 ⎢ ∂x 2 ⎣ ⎦ ⎡ ⎡ ∂E y ∂E x ∂E x ∂E x ∂E z ∂E x ⎤ + Ey + Ex + Ex + Ex + Ex = ⎥ ∂z ∂y ∂x ∂y ∂z ∂x ⎦ ↑ order them and now consider the remaining terms: ε ⎢Ez ⎣ ⎡ ∂E y ⎞ ⎛ ∂E x ⎛ ∂E x ∂E x ∂E z ⎞ ⎤ = ε ⎢ 2E x + ⎜ Ey + Ex + Ex ⎥= ⎟ + ⎜ Ez ∂x ⎝ ∂y ∂y ⎠ ⎝ ∂z ∂z ⎟ ⎥ ⎠⎦ ⎢ ⎣ ⎡ ∂E 2 ∂ E x E y ∂ ( Ex Ez ) ⎤ ⎥ = ε∇ ⋅ E x E x x + E y y + E z z ˆ ˆ ˆ =ε⎢ x + + ∂y ∂z ⎢ ∂x ⎥ ⎣ ⎦ ( ) ( ( ) ) = ∇ ⋅ (ε E x E ) So we have that eq. Ramaccia and A.2) as follow: ∂B ⎞ 1 1 ⎛ ⎛ ∂D ⎞ ⎛ ⎞ ˆ ˆ × B ⎟ = ∇ ⋅ ⎜ ε E x E + μ Hx H − x ε E2 − x μ H2 ⎟ ρ E x + ( J × B )x + ⎜ D × ⎟ + ⎜ ∂t ⎠ x ⎝ ∂t 2 2 ⎝ ⎠x ⎝ ⎠ ∂ 1 1 ⎛ ⎞ ˆ ˆ ρ E x + ( J × B )x + ( D × B ) = ∇ ⋅ ⎜ ε E x E + μ H x H − x ε E2 − x μ H2 ⎟ ∂t 2 2 ⎝ ⎠ x It is easy to note the presence of f x .1).6.(1.10) can be written as: ( ∇ × E) × D x−component 1 ˆ = ∇ ⋅ ( ε E x E ) − x ε∇E2 = 2 1 ⎛ ⎞ ˆ = ∇ ⋅ ⎜ ε E x E − x ε E2 ⎟ 2 ⎝ ⎠ (1. 23 .1) and (1. • Question n° 2 The identity (1.S.3) is obtained adding eq.6.(1.6.J. Toscano Pag.11) that is the right–hand side of eq.10) Let us consider the only emphasized terms of eq. ∂G x and Tx as defined in the text of the exercise. 12) and (1.S. we can derive the relationship that represents momentum conservation for y and z–component: fy + fz + ∂G y ∂t = ∇ ⋅ Ty (1.14) (1.6.12) (1.2) can be written for the y and z–component as follow: ⎡⎛ ∂B ⎞ ⎛ 1 2 ⎞ ˆ ⎢ ⎜ D × ∂t ⎟ − ρ E y = ∇ ⋅ ⎜ y 2 ε E − ε E y E ⎟ ⎠y ⎝ ⎠ ⎢⎝ ⎢ ⎛ ∂D ⎞ ⎛ 1 ⎞ ˆ ⎢( J × B ) y + ⎜ × B ⎟ = ∇ ⋅ ⎜ y μ H2 − μ Hy H ⎟ ⎝ ∂t ⎠y ⎝ 2 ⎠ ⎢ ⎣ ⎡⎛ ∂B ⎞ 1 2⎞ ⎛ ˆ ⎢ ⎜ D × ∂t ⎟ + ρ E z = ∇ ⋅ ⎜ ε E z E − z 2 ε E ⎟ ⎠z ⎝ ⎠ ⎢⎝ ⎢ 1 ⎛ ∂D ⎞ ⎛ ⎞ ˆ × B ⎟ = ∇ ⋅ ⎜ μ Hz H − z μ H2 ⎟ ⎢( J × B )z + ⎜ 2 ⎝ ∂t ⎠z ⎝ ⎠ ⎣ (1.6. it is possible to demonstrate that the relationships (1.6. Orfanidis – Electromagnetic Waves and Antennas • Question n° 3 Exercises Chapter 1 Operating in the similar way to question n°1.15) ∂G z = ∇ ⋅ Tz ∂t where ⎧ ⎪f y = ( J × B ) y − ρ E y ⎪ ⎪ ⎛∂ ⎞ ⎨G y = ⎜ ( D × B ) ⎟ ⎝ ∂t ⎠y ⎪ ⎪ 1 ⎪Ty = y ε E 2 + μ H 2 − ε E y E − μ H y H ˆ ⎩ 2 ⎧ ⎪f z = ( J × B ) z + ρ E z ⎪ ⎪ ⎛∂ ⎞ ⎨G z = ⎜ ( D × B ) ⎟ ⎝ ∂t ⎠z ⎪ ⎪ 1 2 2 ˆ ⎪Tz = ε E z E + μ H z H − z ε E − μ H ⎩ 2 ( ) ( ) D. (1.6.6.6.J.1) and (1. Ramaccia and A.6. 24 .13) as in question n°2.6. Toscano Pag.13) From eq. The first term of eq. we have ℑ −1 {ε (ω )} = ℑ −1 ⎪ 2 ⎧ ε 0ω p ⎨ε 0 + 2 ω0 − ω 2 + jωγ ⎪ ⎩ ⎫ ⎪ ⎬= ⎪ ⎭ ⎧ ⎫ 1 ⎪ ⎪ 2 = ℑ−1 {ε 0 } + ε 0ωp ℑ−1 ⎨ ⎬ 2 2 ⎪ ⎪ ⎩ ω0 − ω + jωγ ⎭ (1.1) where ω p is the so–called plasma frequency of the material defined by: Ne2 ωp = ε0m and γ measures the rate of collisions per unit of time.J. Toscano Pag. (1.3) where ℑ−1 denotes the inverse Fourier transform operator. as typically the case in practice.7.7. Solution For the linearity of Fourier transform. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1.7.S.4) because the Fourier transform of Dirac delta function is: ℑ {δ (t)} = +∞ −∞ ∫ δ (t)e − jω t dt = e − jω t t =0 =1 D. (1.2) Show that the casual and stable time–domain dielectric response of eq.7. Ramaccia and A.7 Exercise Consider the permittivity of a dispersive material ε (ω ) = ε 0 + 2 ε 0ωp 2 ω0 − ω 2 + jωγ (1.7.3) is constant. Discuss the solution for the case γ 2 > ω0 .1) is given as follows: ⎧ε ( t ) = ε 0δ ( t ) + ε 0 χ ( t ) ⎪ 2 ⎨ ωp −γ t/2 e sin (ω0 t ) u ( t ) χ (t) = ⎪ ω0 ⎩ 2 where u ( t ) is the unit–step function and ω0 = ω0 − γ 2 4 . and we must assume that γ < 2ω0 . so it's easy to calculate: ℑ−1 {ε 0 } = ε 0δ (t) (1. 25 . (1.7. 26 .S. we can rewrite eq. So we 2 have to find the solutions of the equation ω 0 − ω 2 + j ω γ = 0 in ω and we obtain: ω1.3) is more complicate and it is necessary to simplify the argument.7.5) as follow. consider: ℑ je − jωi t u(t) .2 = Assuming that 2 − jγ ± −γ 2 + 4ω0 −2 (1.7) Now the problem is only to transform the trivial expression 1 (ω − ω i ) and then to apply the result to eq. Ramaccia and A. γ ω1. and we obtain: D.5) 2 −γ 2 + 4ω0 = 2ω0 and that 2ω0 > γ . (1. we can reduce the denominator in the product of two polynomials of first degree. To solve this problem.7. (1.J.7. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 The second term of eq. (1. { } where u(t) is the unit step function. First of all. 1 ⎧− 1 ⎫ ⎧ ⎫ 2ω0 2ω0 ⎪ 1 ⎪ −1 ⎪ + ⎨ 2 ⎬=ℑ ⎨ ⎬ 2 ⎪ ω0 − ω + jωγ ⎪ ⎪ (ω − ω1 ) (ω − ω2 ) ⎪ ⎩ ⎭ ⎩ ⎭ (1.2 = j ∓ ω0 2 (1.7.7.6) where it's important to note that ± has been substituted by ∓ because of the minus sign of the denominator. Now we can write: ⎧ 1 ⎪ ℑ−1 ⎨ 2 2 ⎪ ω 0 − ω + jωγ ⎩ ⎫ ⎧ ⎫ ⎧ ⎫ 1 A B ⎪ ⎪ ⎪ −1 ⎪ −1 ⎪ + ⎬=ℑ ⎨ ⎬=ℑ ⎨ ⎬ ⎪ (ω − ω1 )(ω − ω 2 ) ⎪ ⎪ (ω − ω1 ) (ω − ω 2 ) ⎪ ⎪ ⎩ ⎭ ⎩ ⎭ ⎭ where A and B are two constant that we calculate applying the method of weighted residuals: A = lim 1 1 1 1 = = =− 2ω0 ω →ω1 (ω − ω2 ) (ω1 − ω2 ) ⎛ γ ⎞ γ ⎜ j − ω0 − j − ω0 ⎟ 2 ⎝ 2 ⎠ 1 1 1 1 B = lim = = = ω →ω2 (ω − ω1 ) (ω2 − ω1 ) ⎛ γ ⎞ 2ω0 γ ⎜ j + ω0 − j + ω0 ⎟ 2 ⎝ 2 ⎠ so: ℑ −1 ⎪ .7. Toscano Pag.7). 6) in eq.7.7.7.7. Toscano .7.12) ω0 (ω0 t ) u ( t ) = ε 0δ ( t ) + ε 0 χ ( t ) Pag.4) and (1. (1.7.11). we have: j ⎡e 2ω0 ⎣ 1 jω1t −e jω2 t ⎤ ⎛ γ ⎞ ⎡ j⎛ jγ −ω ⎞t j j +ω t ⎤ ⎢e ⎜ 2 0 ⎟ − e ⎜ 2 0 ⎟ ⎥ u ( t ) = ⎠ ⎝ ⎠ u t = j ⎝ ⎦ ( ) 2ω0 ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ 1 t ⎡ −γ t −γ + jω t ⎤ − jω0 t = − e 2e 0 ⎥ u (t) = j ⎢e 2 e 2ω0 ⎢ ⎥ ⎣ ⎦ 1 = = = 1 2ω0 1 2 ω0 1 −γ je t 2 ⎡e − jω0t − e+ jω0t ⎤ u ( t ) = ⎣ ⎦ t 2 (1. we have: (1.10) Substituting the solutions (1. (1.7.S.7. Ramaccia and A. we have: ⎧ 1 ⎪ 2 ℑ−1 {ε (ω )} = ℑ−1 {ε 0 } + ε 0ωp ℑ−1 ⎨ 2 2 ⎪ ω0 − ω + jωγ ⎩ = ε 0δ ( t ) + ε 0 t 2 ω p −γ 2 Sin e (1.7.7.7.9) in eq. (1.11) −γ je ⎡ −2 jSin (ω0 t ) ⎤ u ( t ) = ⎣ ⎦ −γ ω0 e t 2 Sin (ω0 t ) u ( t ) ⎫ ⎪ ⎬= ⎪ ⎭ Using the result in (1. Orfanidis – Electromagnetic Waves and Antennas ℑ je− jωi t u ( t ) = j ∫ e− jωi t u ( t ) e jω t dt = j ∫ e ( −∞ Exercises Chapter 1 +∞ { } +∞ j ω −ωi ) t dt = = j = = 1 ⎡e j(ω −ωi )t ⎤ = ⎥0 ⎣ ⎦ j (ω − ωi ) ⎢ 0 +∞ (1.7).10).J.8) 1 ⎡ j(ω −ωi )t ⎤ − j ω −ωi ) t −e ( ⎢e ⎥= (ω − ωi ) ⎣ t →+∞ t →0 ⎦ 1 1 [ 0 − 1] = − (ω − ωi ) (ω − ωi ) So it's possible to assume that: ⎧ 1 ⎫ jω t ℑ−1 ⎨ ⎬ = − je i u ( t ) ω − ωi ⎭ ⎩ Using eq. 27 D.9) 1 ⎫ 2ω0 2ω0 ⎪ 1 1 ℑ ⎨ + je jω1t u ( t ) − je jω2t u ( t ) = ⎬= 2ω0 ⎪ (ω − ω1 ) (ω − ω2 ) ⎪ 2ω0 ⎩ ⎭ 1 ⎡ jω1t = − e jω2t ⎤ u ( t ) j e ⎣ ⎦ 2ω0 −1 ⎪ ⎧− 1 (1. Toscano Pag.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 D.S. 28 . Ramaccia and A. So: 1 ≈ 8.8.S.8 Exercise Show that the plasma frequency for electrons can be expressed in the simple numerical form: fp = 9 N (1. 602 ⋅ 10 − 19 C and its mass is about 9...1) where fp is in Hz and N is the electron density in electrons/m3.J. 602 ⋅10−19 ) 2 With N = 1012 the plasma frequency of the ionosphere is: iono fp = 9 1012 = 9 ⋅ 10 6 = 9MHz D. ε 0 is the permittivity of vacuum and m is the mass of electron.85 ⋅ 10−12 (1.14 9.. Ramaccia and A. 29 .10 ⋅ 10−31 ⋅ 8.3) The charge of an electron is 1. What is fp for the ionosphere if N = 1012 ? Solution Plasma frequency id defined as fp = 1 2π Ne 2 ε0m (1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. 988925.8.8. Toscano Pag. 2 ⋅ 3. 85 ⋅ 10 − 12 F m .10 ⋅ 10 − 31 Kg .2) where e is the electron charge. The electric permittivity is about 8. So we have to demonstrate the follow identity: 1 2π e2 =9 ε0m (1. 9. Solution Eq.9. t) = ω p ε 0 E (r . t) + γρ (r . (1.1) by γ : 1 2 ωp γ ρ (r.2) is obtained dividing eq. (1.9 Exercise Show that the relaxation equation 2 ρ (r . t) + ρ (r . t) = 0 (1.9. t) + ω p ρ (r . (1.J. t) = 0 2 where ω p γ = σ ε 0 . Ramaccia and A. The Ohm's law (1. t) = 0 ε0 (1.3) −∞ ∫ t e −γ (t − t ') ε 0 E (r .2) is reduced to eq.9. γ is the measurement of collisions per unit of time and ω p is the plasma frequency. t) γ → 0 and the eq.9. 30 .3).4) takes the instantaneous form J = σ E . t) −∞ ∫ t e − γ (t − t ') dt ' So D. t) + σ ρ (r . t) = 2 ωp (1.9. Show also that in this limit. highlighting we have to solve only the integral of an exponential: 2 J (r . Ohm's law J (r . (1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. t) + γ ρ (r. from which the naive relaxation constant τ relax = ε 0 σ was derived.9.2) Then show that it reduces to the naive relaxation equation ∂ρ σ + ρ =0 ∂t ε in the limit τ = 1 γ → 0 .4) can be written. t)dt ' (1.S.9.9. t) + ρ (r. Toscano Pag. can be written in the following form in term of dc–conductivity 2 σ = ε 0ω p γ = Ne 2 m γ : 1 γ ρ (r . It's easy to note that if τ = 1 γ → 0 .1) where ρ is the charge density in the conductor. then the term ρ (r. Toscano Pag.S. t) (1.9.J. Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas t − γ (t − t ') Exercises Chapter 1 1 ⎡ − γ (t − t ') ⎤ t 1 1 = 1 − 0] = e ⎣ ⎦ −∞ γ [ γ γ −∞ ∫ e dt ' = 1 γ −∞ ∫ t e − γ (t − t ') d[ −γ (t − t ')] = and we can write: J (r. t) = 2 ωp ε 0 γ E(r.5) D. 31 . t) = σ E(r. 10. and (1. the above equations of motion read: e E x + ωB v y − γ v x m e v y = E y − ωB v x − γ v y m vx = (1. B = z B . To solve this system.2) where ωB = eB m is the cyclotron frequency.1) ˆ with the collisional term included. Show that in component form. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. Show that: J x ± jJ y = σ ± (ω ) E x ± jE y ( ) (1.3) 3. ˆ ˆ ˆ ˆ 1.3) is: J x (t) ± jJ y (t) = ∫ σ ± (t − t ') E x (t ') ± jE y (t ') dt ' 0 t ( ) (1.10.10.5) where σ ± ( t ) = γσ 0 e − γ t e ∓ j ω B t u ( t ) is the inverse Fourier transform of σ ± (ω ) and u(t) is the unit–step function.10. Ramaccia and A.10.4 gauss = 0. Rewrite eq. Define the induced currents as J = Nev . Show that the time–domain version of eq.4 × 10 − 4 T esla ? 2. that is the dc value of the 4.4) where σ ± (ω ) = conductivity. What is the cyclotron frequency in Hz for electrons in the Earth' magnetic field B = 0. show that the solution is: e E x ± jE y v x ± jv y = m γ + j (ω ± ω B ) ( ) (1. 5. γσ 0 γ + j ( ω ± ωB ) with σ 0 = N e 2 m γ . 32 .10.5) in component form: D. The equation of motion of conduction electrons in a constant external magnetic field is mv = e(E + v × B) − mγ v that E = x E x + y E y and v = x v x + y v y . (1. Assume the magnetic field is in the z–direction. Assuming harmonic time– dependence. work with the combinations v x ± jv y .10 Exercise Conductors and plasmas exhibit anisotropic and birefringent behavior when they are in the presence of an external magnetic field.J.S. Toscano Pag. (1.10. Ramaccia and A. The conduction charges will tend to accumulate either on the right or the left side of the conductor. then no steady current can flow in this direction.6) in the special case E x (t) = E x u(t) and E y (t) = E y u (t) . E y = bE x . Toscano Pag.10.J.10. where E x and E y are constants. (1. which depends on the sign of the electric charge e.S. 1. 1. depending on the sign of b.7) Fig. This implies that if an electric field is applied in the x–direction. 33 . 6. Orfanidis – Electromagnetic Waves and Antennas t Exercises Chapter 1 J x (t) = ∫ ⎡σ xx (t − t ')E x (t ') + σ xy (t − t ')E y (t ') ⎤ dt ' ⎣ ⎦ J y (t) = ∫ ⎡σ yx (t − t ')E x (t ') + σ yy (t − t ')E y (t ') ⎤ dt ' ⎣ ⎦ 0 0 t (1.1. If the conductor has finite extent in the y–direction. an electric field will develop across the y–ends of the conductor. σ yy (t) . as show in Fig. What is the numerical value of b for electrons in copper if B is 1 gauss? 7. or negative electrons for n–type.10. For a collisionless plasma ( γ = 0 ). positive holes for p–type.10. where b = ωB γ . e. show that its dielectric behavior is determined from D x ± jD y = ε ± (ω ) E x ± jE y .8) D. This is the Hall effect and is used to determinate the sign of the conduction charges in semiconductors.1: Conductor with finite extent in y–direction. where 2 ⎛ ⎞ ωp ⎜1 − ⎟ ε ± (ω ) = ε 0 ⎜ ω ( ω ± ωB ) ⎟ ⎝ ⎠ ( ) (1. Evaluate eq. σ xy (t) .10.g. (1.6) will be: Jx = σ 0 Jy = σ0 E x + bE y 1 + b2 E y − bE x 1 + b2 (1. σ yx (t) .10. J y = 0 . and show that after a long time the steady–state version of eq.6) and identify the quantities σ xx (t) .10. 4 × 108 = 1.10 × 10 − 31 Kg ) in the Earth's magnetic field is: fB = 1. 602 × 0. 602 × 10 − 19 C . divide eq. 602 × 10−19 × 0.J. (1.10 • Question n°2 Assuming harmonic time dependence means that v i ( t ) = v i e jω t ⇒ v i ( t ) = jω v i e jω t so we have: e E x + ωB v y − γ v x m e jω v y = E y − ω B v x − γ v y m jω v x = (1. m = 9.10. the plasma exhibits birefringence. 4 × 10−4 2 × 3.9) Now combine the equations: ⎛ e ⎞ jω v x ± jv y = ⎜ E x + ωB v y − γ v x ⎟ ± ⎝m ⎠ ( ) ⎛ e ⎞ j ⎜ E y − ωB v x − γ v y ⎟ ⎝m ⎠ that is jω vx ± jvy = D. Thus.10 × 10 −31 = 1.10) Pag.10.12MHz 2 × 3.10.S. The cyclotron frequency for electrons ( e = 1. Ramaccia and A. Toscano ( ) e Ex ± jE y + ωB vy ∓ jvx − γ vx ± jvy m ( ) ( ) ( ) (1.1) by m: v= and expand the terms e (E + v × B) − γ v m ˆ ˆ vx x + v y y = e ˆ ˆ ˆ ˆ ˆ ˆ Ex x + E y y + v y Bx − vx By − γ vx x + vy y m ( ) ( ) Now it is possible to separate x and y–component as follow: e ⎧ vx = E x + v y B − γ vx ⎪ ⎪ m ⎨ ⎪v = e E − v B − γ v x y ⎪ y m y ⎩ ( ) ( ) ⇒ e eB ⎧ vx = Ex + v y − γ vx ⎪ ⎪ m m ⎨ ⎪v = e E − eB v − γ v y ⎪ y m y m x ⎩ where is easy to note the cyclotron frequency ωB . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 where ω p is the plasma frequency.14 × 9. 34 . Solution • Question n°1 First of all.14 × 9. we obtain ∓ jωB v x ± jv y and the constant C = ∓ jωB .15) ⎪ jω + ( γ ± jω B ) ⎪ ⎪ ω + j ( γ ± jω B ) ⎪ ⎩ ω + ω± ⎭ ⎩ ⎭ ⎩ ⎭ D.10.11) Substituting eq.10) becomes: ( ) ( ) ( ) jω vx ± jvy = and we obtain: ( ( ) ) e Ex ± jE y ∓ jωB vx ± jv y − γ vx ± jvy m ( ) ( ) ( ( ) ) jω v x ± jv y ± jωB v x ± jv y + γ v x ± jv y = e ( v x ± jv y ) (γ + j (ω ± ωB ) ) = m ( E x ± jE y ) e E x ± jE y = m γ + j (ω ± ω B ) ( ) ( ) e E x ± jE y m ( • Question n°3 v x ± jv y ) ( ) (1.10. we have: J x ± jJ y = Ne v x ± jv y where we can identify σ ± (ω ) as ( ) Ne2 m E ± jE y = γ + j (ω ± ωB ) x ( ) (1.14) Eq. (1.10. Toscano Pag.10.14) can be written as: γσ 0 ℑ−1 ⎨ ⎧ ⎪ ⎫ ⎫ 1 1 1 ⎫ ⎪ ⎪ ⎪ −1 ⎧ −1 ⎧ ⎬ = jγσ 0 ℑ ⎨ ⎬ = jγσ 0 ℑ ⎨ ⎬ (1.13) • Question n°4 We have to calculate the Fourier transform: ⎧ ⎫ γσ 0 ⎪ ⎪ ℑ−1 {σ ± (ω )} = ℑ−1 ⎨ ⎬ ⎪ ⎪ ⎩ γ + j (ω ± ωB ) ⎭ (1.S.11) in the expression for the induced currents ( J = Nev ). (1. γm (1. 35 . (1. Ramaccia and A.12) Ne2 γσ 0 m σ ± (ω ) = = γ + j (ω ± ωB ) γ + j (ω ± ωB ) where σ 0 = Ne 2 is the dc value of the conductivity. (1. So eq.10.10) there is the term ωB v y ∓ jv x that we have to express in the form C vx ± jvy .10. If we take out of the parentheses ±j .J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 In Eq.10.10.10. where the constant C is to be found. (1.7 (eq.16) = γσ 0e − jω± t u ( t ) = γσ 0e −γ t e ∓ jωB t u ( t ) where u ( t ) is the unit step function.15) is already calculated in exercise 1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 where ω ± = j ( γ ± j ω B ) .10.10. So we have: σ ± ( t ) = jγσ 0 ℑ−1 ⎨ ⎧ 1 ⎫ − jω t ⎬ = jγσ 0 − je ± u ( t ) = ω + ω± ⎭ ⎩ ( ) (1. Now we can write eq.10.10.7.19) ⎣ ⎦ 0 t ( ) ( ) 2 jJ y (t) = ∫ ⎡σ + (t − t ') E x (t ') + jE y (t ') − σ − (t − t ') E x (t ') − jE y (t ') ⎤ dt ' (1.21) 2 ⎝ ⎠ ⎦ ⎤ ⎛ σ (t − t ') + σ − (t − t ') ⎞ j⎜ + ⎟ E y (t ') ⎥ dt ' (1. (1.10.S.J. we obtain: 2J x (t) = ∫ ⎡σ + (t − t ') E x (t ') + jE y (t ') + σ − (t − t ') E x (t ') − jE y (t ') ⎤ dt ' (1.22) 2j ⎝ ⎠ ⎦ and it's easy to identify the follow quantities: D.10.9)).10.20) ⎣ ⎦ 0 t ( ) ( ) Manipulating the expression in the brackets. It's easy to note that the inverse Fourier transform in (1. (1.10. Ramaccia and A.12) in the time–domain version: J x (t) ± jJ y (t) = ∫ σ ± (t − t ') E x (t ') ± jE y (t ') dt ' 0 t ( ) (1.10. 36 .18) ( ) Combining them.10. we have: ⎡ ⎛ σ (t − t ') + σ − (t − t ') ⎞ J x (t) = ∫ ⎢ ⎜ + ⎟ E x (t ') + 2 ⎠ ⎣⎝ 0 t ⎡ ⎛ σ (t − t ') − σ − (t − t ') ⎞ J y (t) = ∫ ⎢ ⎜ + ⎟ E x (t ') + 2j ⎠ ⎣⎝ 0 t ⎤ ⎛ σ (t − t ') − σ − (t − t ') ⎞ j⎜ + ⎟ E y (t ') ⎥ dt ' (1. Toscano Pag.17) in its two component as follow: J x (t) + jJ y (t) = ∫ σ + (t − t ') E x (t ') + jE y (t ') dt ' J x (t) − jJ y (t) = ∫ σ − (t − t ') E x (t ') − jE y (t ') dt ' 0 0 t t ( ) (1.17) • Question n°5 It's possible to decompose eq. 10. the integral I11 results: ⎛ 1 ⎞ γσ 0 I11 = −γσ 0Ex ⎜ Ex ⎟= ⎝ − jω+ ⎠ (γ + jωB ) In the same way we can solve I12 and obtain: (1. i.21) and divide it in two integrals: (1. Let's start with I1 substituting E x (t) = E x u(t) and the definition of σ xx (t) . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 σ + (t) + σ − (t) ⎧ ⎪σ xx ( t ) = 2 ⎪ ⎛ σ + (t) − σ − (t) ⎞ ⎪σ xy ( t ) = j ⎜ ⎟ ⎪ 2 ⎝ ⎠ ⎪ ⎨ ⎪σ yx ( t ) = ⎛ σ + (t) − σ − (t) ⎞ ⎜ ⎟ ⎪ 2j ⎝ ⎠ ⎪ ⎪σ ( t ) = ⎛ σ + (t) + σ − (t) ⎞ ⎜ ⎟ ⎪ yy 2 ⎝ ⎠ ⎩ • Question n°6 Consider the expression of Jx (t) in eq. (1. Toscano (1.24) ⎛ 1 ⎞ γσ 0 I12 = −γσ 0Ex ⎜ Ex ⎟= − jω− ⎠ (γ − jωB ) ⎝ D.S.10.23) ⎛ σ (t − t ') + σ − (t − t ') ⎞ ⎛ σ + (t − t ') − σ − (t − t ') ⎞ J x (t) = ∫ ⎜ + ⎟ E x (t ')dt ' + j∫ ⎜ ⎟ E y (t ')dt ' = 2 2 ⎝ ⎠ ⎝ ⎠ 0 0 = I1 + I 2 t t I1 and I2 can be solved separately.J.e.10. Ramaccia and A. t → ∞ . So we obtain: ⎛ σ (t − t ') + σ − (t − t ') ⎞ I1 = ∫ ⎜ + ⎟ E x u(t ')dt ' = 2 ⎝ ⎠ 0 t t ⎤ 1 1⎡ = ⎢ ∫ σ + (t − t ')E x u(t ')dt ' + ∫ σ − (t − t ')E x (t ')dt '⎥ = ( I11 + I12 ) 2⎢ ⎥ 2 0 ⎣0 ⎦ t Now we solve separately I11 and I12. 37 . substituting the definitions of σ + (t) and σ − (t) respectively: I11 = ∫ σ + (t − t ')E x u(t ')dt ' = ∫ γσ 0e − jω+ (t − t ') u ( t − t ') E x u(t ')dt ' = 0 0 t − jω+ (t − t ') t t = γσ 0 E x ∫ e 0 dt ' = −γσ 0 E x ∫ e − jω+ (t − t ')d(t − t ') = 0 t ⎡e = −γσ 0 E x ⎢ ⎢ − jω + ⎣ − jω+ (t − t ') ⎤ t ⎛ 1 − e − jω+ t = −γσ 0 E x ⎜ ⎥ ⎜ ⎥0 ⎦ ⎝ − jω+ ⎞ ⎟ ⎟ ⎠ After long time.25) Pag.10. J.24) and (1.28) ⎜ ⎟= 2 ⎟ 2 ⎝ (γ + jωB ) (γ − jωB ) ⎠ 2 ⎜ γ 2 + ωB ⎝ ⎠ jγσ 0 E y ⎛ − 2 jωB ⎞ ω σ ⎜ 2 ⎟ = 2 B 02 E y 2 2 ⎜ γ + ωB ⎟ γ + ωB ⎝ ⎠ Now it is possible to write Jx (t) in steady state when a constant electric field is applied: J x (t) = γ 2 E x + ωB E y γ 2σ 0 ω σ E x + 2 B 02 E y = σ 0 = 2 2 γ 2 + ωB γ + ωB γ 2 + ωB γ 2 E x + ωB E y γ2 2 γ 2 + ωB γ2 = σ0 E x + bE y 1 + b2 = σ0 where b = ωB γ .10. we have the solution of integral I1: γσ 0E x ⎞ 1 1 ⎛ γσ 0E x + I1 = (I11 + I12 ) = ⎜ ⎟= 2 2 ⎝ (γ + jωB ) (γ − jωB ) ⎠ = = γσ 0E x ⎛ 2 ⎞ γσ 0E x ⎛ γ − jωB + γ + jωB 1 1 + ⎜ ⎜ ⎟= 2 (γ + jωB ) (γ − jωB ) ⎠ 2 ⎜ γ 2 + ωB ⎝ ⎝ ⎞ ⎟ = (1.10.10. So the results are known: I 21 = I 22 γσ 0 Ey (γ + jωB ) γσ 0 = Ey (γ − jωB ) (1.27) It is easy to calculate I2 as: I2 = = = j ( I21 − I22 ) = 2 ⎞ γσ 0 j ⎛ γσ 0 Ey − Ey ⎟ = ⎜ 2 ⎝ (γ + jωB ) (γ − jωB ) ⎠ jγσ 0 E y ⎛ ⎞ jγσ 0 E y ⎛ γ − jωB − γ − jωB ⎞ 1 1 − ⎜ ⎟ = (1. we can solve integral I2: ⎛ σ (t − t ') − σ − (t − t ') ⎞ I2 = j∫ ⎜ + ⎟ E y u(t ')dt ' = 2 ⎝ ⎠ 0 t t ⎤ j j⎡ = ⎢ ∫ σ + (t − t ')E y u(t ')dt ' − ∫ σ − (t − t ')E yu(t ')dt '⎥ = ( I21 − I22 ) 2⎢ ⎥ 2 0 ⎣0 ⎦ t The integrals I21 and I22 have the same structure of integrals I11 and I12.10. (1.S. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Now combining eq. Ramaccia and A.26) ⎟ ⎠ γσ 0E x ⎛ 2 γ 2σ 2γ ⎞ ⎜ 2 ⎟ = 2 0 2 Ex ⎜ γ + ω2 ⎟ γ + ω ⎝ B⎠ B As for integral I1.25). Toscano Pag. D. 38 .10. Ramaccia and A.10. (1. 39 . Toscano Pag.31) ⎟= 2 ⎟ 2 j ⎝ (γ + jωB ) (γ − jωB ) ⎠ 2j ⎜ γ 2 + ωB ⎝ ⎠ ω σ γσ 0 E x ⎛ − 2 jωB ⎞ ⎜ 2 ⎟ = − 2 B 02 E x 2 2 j ⎜ γ + ωB ⎟ γ + ωB ⎝ ⎠ = As for integral I3.10.e.S.22) and divide it in two integrals: ⎛ σ (t − t ') − σ − (t − t ') ⎞ ⎛ σ + (t − t ') + σ − (t − t ') ⎞ J y (t) = ∫ ⎜ + ⎟ E x (t ')dt ' + ∫ ⎜ ⎟ E y (t ')dt ' = 2j 2 ⎝ ⎠ ⎝ ⎠ 0 0 = I3 + I 4 t t I3 and I4 can be solved separately. i.29)and (1. (1.10.29) ⎛ 1 ⎞ γσ 0 I32 = −γσ 0E x ⎜ Ex ⎟= ⎝ − jω− ⎠ (γ − jωB ) Now combining eq. Let's start with I3 substituting E x (t) = E x u(t) and the definition of σ yx (t) .10.10. substituting the definitions of σ + (t) and σ − (t) respectively: I31 = ∫ σ + (t − t ')E x u(t ')dt ' = ∫ γσ 0e − jω+ (t − t ') u ( t − t ' ) E x u(t ')dt ' = 0 0 t − jω+ (t − t ') t t = γσ 0E x ∫ e 0 dt ' = −γσ 0E x ∫ e− jω+ (t − t ')d(t − t ') = 0 t ⎡e = −γσ 0E x ⎢ ⎢ − j ω+ ⎣ − jω+ (t − t ') ⎤ t ⎛ 1 − e− jω+ t = −γσ 0E x ⎜ ⎥ ⎜ ⎥0 ⎦ ⎝ − jω+ ⎞ ⎟ ⎟ ⎠ After long time. we have the solution of integral I3: I3 = (1. So we obtain: ⎛ σ (t − t ') − σ − (t − t ') ⎞ I3 = ∫ ⎜ + ⎟ E x u(t ')dt ' = 2j ⎝ ⎠ 0 = t t ⎤ 1 1⎡ ⎢ ∫ σ + (t − t ')E x u(t ')dt ' − ∫ σ − (t − t ')E x (t ')dt '⎥ = ( I31 − I32 ) 2j ⎢ ⎥ 2j 0 ⎣0 ⎦ t Now we solve separately I31 and I32. t → ∞ . the integral I31 results: ⎛ 1 ⎞ γσ 0 I31 = −γσ 0E x ⎜ Ex ⎟= ⎝ − jω+ ⎠ (γ + jωB ) In the same way we can solve I32 and obtain: (1.30).30) γσ 0 E x ⎞ 1 1 ⎛ γσ 0E x − (I31 − I32 ) = ⎜ ⎟= 2j 2 j ⎝ (γ + jωB ) (γ − jωB ) ⎠ ⎞ γσ 0E x ⎛ γ − jωB − γ − jωB ⎞ γσ E ⎛ 1 1 = 0 x⎜ − ⎜ ⎟ = (1.J. we can solve integral I4: D.10. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Consider now the expression of J y (t) in eq. S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ⎛ σ (t − t ') + σ − (t − t ') ⎞ I4 = ∫ ⎜ + ⎟ E y (t ')dt ' = 2 ⎝ ⎠ 0 t t ⎤ 1 1⎡ = ⎢ ∫ σ + (t − t ')E y u(t ')dt ' + ∫ σ − (t − t ')E y u(t ')dt '⎥ = ( I41 + I42 ) 2⎢ ⎥ 2 0 ⎣0 ⎦ t The integrals I41 and I42 have the same structure of integrals I31 and I32. So the results are known: I 41 = I 42 γσ 0 Ey (γ + jωB ) γσ 0 = Ey (γ − jωB ) (1.10.32) It is easy to calculate I4 as: I4 = = 1 ( I41 − I42 ) = 2 ⎞ γσ 0 1 ⎛ γσ 0 Ey + Ey ⎟ = ⎜ 2 ⎝ (γ + jωB ) (γ − jωB ) ⎠ ⎞ ⎟ = (1.10.33) ⎟ ⎠ γσ 0E y ⎛ 2 ⎞ γσ 0 E y ⎛ γ − jωB + γ + jωB 1 1 + ⎜ ⎜ ⎟= 2 2 ⎜ γ 2 + ωB ⎝ (γ + jωB ) (γ − jωB ) ⎠ ⎝ γσ 0E y ⎛ − 2 γ ⎞ γ 2σ 0 E = ⎜ ⎟= 2 2 y 2 ⎜ γ 2 + ωB ⎟ γ 2 + ω B ⎝ ⎠ Now it is possible to write J y (t) in steady state when a constant electric field is applied: γ 2 E y − ωB E x ωBσ 0 γ 2σ 0 J y (t) = − 2 E + 2 E = σ0 = 2 x 2 y 2 γ + ωB γ + ωB γ 2 + ωB γ 2 E y − ωB E x = σ0 γ2 2 γ 2 + ωB γ2 = σ0 E y + bE x 1 + b2 where b = ωB γ . • Question n°7 To solve this point it is necessary to obtain the expression of σ ± (ω ) for a collisionless plasma because it is note the relationship: ε ± (ω ) = ε 0 + σ ± (ω ) jω (1.10.34) The definition of the conductibility σ ± (ω ) has just obtained in this exercise, i.e. eq. (1.10.13). We have only to set γ = 0 for indicating that there is no collision between the electrons and the medium structure: D. Ramaccia and A. Toscano Pag. 40 S.J. Orfanidis – Electromagnetic Waves and Antennas Ne2 m σ ± (ω ) = j ( ω ± ωB ) Exercises Chapter 1 (1.10.35) Now we substitute eq. (1.10.35) in (1.10.34) and we have: Ne2 2 ⎛ ⎞ ωp m ⎜1 − ⎟ =ε ε ± (ω ) = ε 0 − ω ( ω ± ωB ) 0 ⎜ ω ( ω ± ωB ) ⎟ ⎝ ⎠ 2 where ε 0ωp = (1.10.36) Ne2 . m The numerical value of b for electrons in copper can be find out using: b = ω Bγ −1 = eB mγ where e = 1, 6 × 10 − 19 C , m = 9,1 × 10 − 31 Kg , γ = 4,1 × 1013 s − 1 . If B = 1 gauss = 10 − 4 Tesla , then b= 1,6 × 10−19 × 10−4 9,1 × 10−31 × 4,1 × 1013 = 1,6 × 105 = 4288 9,1 × 4,1 The result is different from the one in the text which is 43. D. Ramaccia and A. Toscano Pag. 41 S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1.11 Exercise This problem deals with various properties of the Kramers–Kronig dispersion relations for the electric susceptibility, given by: χ (ω ' ) dω ' χ r (ω ) = ℘ ∫ i π −∞ ω '− ω 1 +∞ χ (ω ' ) 1 dω ' χ i (ω ) = − ℘ ∫ r π −∞ ω '− ω +∞ (1.11.1) where ℘ denotes the "principal value" and χ (ω ) = χ r (ω ) + j χ i (ω ) is the Fourier transform of χ ( t ) . Because the time–response χ ( t ) is real–valued, its Fourier transform χ (ω ) will satisfy the Hermitian symmetry property χ ( −ω ) = χ ∗ (ω ) , which is equivalent to the even symmetry of its real part, χ r ( −ω ) = χ r (ω ) , and the odd symmetry of its imaginary part, χ i ( − ω ) = − χ i (ω ) . 1. Using the symmetry properties, show that eq. (1.11.1) can be written in the folded form: ∞ ω ' χ (ω ' ) 2 χ r ( ω ) = ℘∫ 2 i 2 dω ' π 0 ω ' −ω χ i ( ω ) = − ℘∫ 2 dω ' π 0 ω ' −ω2 2. Using the definition of the principal–value integrals, show the following integral: ℘∫ ∞ 2 ∞ ωχ r (ω ' ) (1.11.2) dω ' 2 2 0 ω ' −ω =0 (1.11.3) dx 1 a+x Hint: You may use the following indefinite integral: ∫ a 2 − x 2 = 2a ln a − x . 3. Using eq. (1.11.3), show that the relations (1.11.2) may be rewritten as ordinary integrals (without the ℘ instruction) as follows: χ r (ω ) = 2 ∞ π ∫ 0 ω ' χ i (ω ' ) − ωχ i (ω ) ω '2 − ω 2 ∞ dω ' χ i (ω ) = − 2 π ∫ 0 ωχ r (ω ' ) − ωχ r (ω ) ω '2 − ω 2 (1.11.4) dω ' . Hint: You will need to argue that the integrands have no singularity at ω ' = ω . 4. For a simple oscillator model of dielectric polarization, the susceptibility is given by: D. Ramaccia and A. Toscano Pag. 42 J.11.S.11. ⎦ 2ω0 ⎣ Solution The Cauchy's principal value is defined for the integration of a function f ( x ) with limited value in the interval [ a. Toscano Pag.5) as γ →0. χi are given as follows. 5. which can be found in standard tables of integrals: 2 ∞ 0 ∞ 0 π ∫ 1 − 2 y 2 C os θ + y 4 dy = 2 π ∫ 1 − 2 y 2 C os θ + y 4 y 2 dy = 1 2 (1 − C os θ ) where S in (θ 2 ) = γ ( 2ω 0 ) .This is a special case of Reimann's principal value: ℘∫ f ( x ) dx = lim a b x 0 −ς1 ς1 →0+ ∫ a f ( x ) dx + lim ς 2 → 0 + x +ς 0 2 ∫ b f ( x ) dx where ς1 and ς 2 are independent of each other. Consider the limit of Eq. • Question n°1 D. and that they still satisfy the Kramers–Kronig relations: χ r (ω ) =℘ 2 ωp ω0 − ω +℘ 2 ωp ω0 + ω . for which you may assume that 0 < γ < 2ω0 : 2 ∞ π ∫ 0 dx (ω02 − x 2 ) 2 = + γ 2x 2 1 2 γω 0 . so: ℘∫ f ( x ) dx = lim a b x 0 −ς a ς →0+ ∫ f ( x ) dx + ∫ f ( x ) dx x 0 +ς b In the case of the exercise the singularity is at ω ' = ω . Show that in this case the functions χr . (1. χ i (ω ) = 2 πωp ⎡δ (ω − ω0 ) − δ (ω + ω0 ) ⎤ .5) + γ 2ω 2 Show that for this model the quantities χ r (ω ) and χ i (ω ) satisfy the modified Kramers– Kronig relationships (1. 2 ∞ π ∫ 0 x 2 dx (ω02 − x 2 ) 2 = 1 + γ 2x 2 γ Indeed. 43 . show that these integrals may be reduced to the following ones. whereas in the Cauchy's principal value ς1 = ς 2 . Orfanidis – Electromagnetic Waves and Antennas 2 ωp 2 ω0 − ω 2 + jγω Exercises Chapter 1 χ ( ω ) = χ r ( ω ) − jχ i (ω ) = = 2 2 ω p ω0 − ω 2 = ( ( 2 ω0 −ω 2 ) ) 2 −j +γ ω 2 2 (ω02 − ω 2 ) 2 γωω p 2 (1. Hint: You may use the following definite integrals. b ] except for the x = x 0 .11.4). Ramaccia and A. Toscano Pag. 44 . Orfanidis – Electromagnetic Waves and Antennas Let us decompose χ r (ω ) as follow χ r (ω ) = χ r + (ω ) + χ r − (ω ) = 1 Exercises Chapter 1 π ℘∫ 0 +∞ 0 χ i (ω ' ) χ (ω ' ) 1 dω ' + ℘ ∫ i dω ' ω '− ω π −∞ ω '− ω and we change the variable in the second term ( ω ' → −ω ' ): +∞ 0 χ (ω ') χ ( −ω ' ) 1 1 χ r (ω ) = ℘ ∫ i dω ' + ℘ ∫ i d ( −ω ' ) π 0 ω '− ω π +∞ −ω '− ω ↑ Simmetry property of χi = +∞ 0 χ (ω ') − χi (ω ' ) 1 1 = ℘∫ i dω ' + ℘ ∫ d ( −ω ' ) = π ω '− ω π −ω '− ω χ (ω ') χ (ω ' ) 1 1 = ℘∫ i dω ' + ℘ ∫ i dω ' π ω '− ω π −ω '− ω 0 +∞ +∞ ⎛ χ (ω ') χi (ω ') ⎞ 1 = ℘∫ ⎜ i + ⎟ dω ' = π 0 ⎝ ω '− ω ω '+ ω ⎠ 0 +∞ +∞ 0 ↑ Invert integral's limits = +∞ ⎛ (ω '+ ω ) χi (ω ') + (ω '− ω ) χi (ω ') ⎞ 1 = ℘∫ ⎜ ⎟ dω ' = π 0⎝ ω '2 − ω 2 ⎠ +∞ ⎛ ω ' χ (ω ' ) + ωχ (ω ' ) + ω ' χ (ω ' ) − ωχ (ω ' ) 1 i i i i = ℘∫ ⎜ 2 2 π 0⎜ ω' −ω ⎝ ⎞ ⎟ dω ' = ⎟ ⎠ = ℘∫ 2 +∞ 0 ω ' χi (ω ') ω '2 − ω 2 π dω ' In the same way. Ramaccia and A. it's possible to decompose χ i (ω ) : χ i (ω ) = χ i + (ω ) + χ i − (ω ) = − ℘ ∫ π 0 1 +∞ 0 χ r (ω ' ) χ (ω ' ) 1 dω ' − ℘ ∫ r dω ' ω '− ω π −∞ ω '− ω and we change the variable in the second term ( ω ' → −ω ' ): D.J.S. we write: ℘∫ ∞ ∞ ⎡ω −ς dω ' dω ' ⎤ ⎥= = lim ⎢ ∫ + ∫ ω '2 − ω 2 ς →0 ⎢ 0 ω '2 − ω 2 ω +ς ω '2 − ω 2 ⎥ 0 ⎣ ⎦ dω ' ∞ ⎤ ⎡ ω + ω ' ω −ς 1 ω +ω' lim ⎢ ln = + ln ⎥= 2ω ς →∞ ⎢ ω − ω ' 0 ω − ω ' ω +ς ⎥ ⎣ ⎦ (1. Toscano Pag. 45 .11.2) have a singularity in ω ' = ω and we have to use the principal–value for solving the integral.S. that is zero.2).J. Orfanidis – Electromagnetic Waves and Antennas +∞ 0 χ (ω ') χ ( −ω ' ) 1 1 χ i (ω ) = − ℘ ∫ r dω ' − ℘ ∫ r d ( −ω ' ) π 0 ω '− ω π +∞ −ω '− ω Exercises Chapter 1 ↑ Simmetry property of χ r = 0 +∞ χ (ω ') χ (ω ' ) 1 1 =− ℘∫ r dω ' − ℘ ∫ r d ( −ω ' ) = −ω '− ω π ω '− ω π χ (ω ') χ (ω ' ) 1 1 =− ℘∫ r dω ' + ℘ ∫ r dω ' π ω '− ω π −ω '− ω 0 +∞ +∞ ⎛ χ (ω ' ) χ r (ω ' ) ⎞ 1 =− ℘∫ ⎜ r − ⎟ dω ' = π 0 ⎝ ω '− ω ω '+ ω ⎠ 0 +∞ +∞ 0 ↑ Invert integral's limits = ⎛ (ω '+ ω ) χ i (ω ' ) − (ω '− ω ) χ i (ω ' ) ⎞ = ℘∫ ⎜ ⎟ dω ' = π 0⎝ ω '2 − ω 2 ⎠ 1 = ℘∫ ⎜ π 0⎜ ⎝ 1 +∞ ⎛ +∞ ω ' χi (ω ') + ωχi (ω ' ) − ω ' χ i (ω ' ) + ωχ i (ω ') ⎞ ω '2 − ω 2 ⎟ dω ' = ⎟ ⎠ +∞ ωχ (ω ') 2 = ℘ ∫ 2i 2 dω ' π 0 ω' −ω • Question n°2 Using the definition of the principal–value integrals and the hint. we will have the integrand that will not diverge.6) = = ⎡ 2ω − ς 1 2ω + ς ⎤ lim ⎢ ln − ln 1 + ln 1 − ln − ⎥= 2ω ς →∞ ⎣ ς ς ⎦ ⎡ 2ω − ς 1 2ω + ς ⎤ lim ⎢ ln − ln − ⎥=0 ς ς ⎦ 2ω ς →∞ ⎣ • Question n°3 The integrands of equations (1. But if we also introduce a singularity at the numerator in ω ' = ω . we obtain: D. (1.11.3).11. (1. Ramaccia and A. to eq. Besides subtracting eq.11. So we have: χ r (ω ) = ∞ ⎤ 2 ⎡ ⎛ ω ' χi (ω ') ωχ i (ω ) ⎞ ⎢℘∫ ⎜ 2 − 2 dω ' ⎥ ⎟ π ⎢ 0 ⎝ ω ' − ω2 ω ' − ω2 ⎠ ⎥ ⎣ ⎦ ∞ ⎤ 2 ⎡ ⎛ ωχ (ω ') ωχ (ω ) ⎞ χi (ω ) = − ⎢℘∫ ⎜ 2r 2 − 2r 2 ⎟ dω '⎥ π ⎢ 0⎝ω' −ω ω' −ω ⎠ ⎥ ⎣ ⎦ (1. Ramaccia and A. Let us denote the denominator of χ i (ω ) as Den[ω] . where χi (ω ) = ( 2 γωωp 2 ω0 − ω 2 ) 2 + γ 2ω 2 where we have substituted ω ' → x to distinguish better the arguments of the integral.5) inside the modified Kramers–Kronig relationships (1.8) Substitute the quantity χ i (ω ) expressed in eq. 46 .S. So we have that: χ r (ω ) = = = 2 2 ⎡ γ xωp γωωp ⎤ ⎢x ⎥ dx = −ω π ∫ x 2 − ω 2 ⎢ Den[x] Den[ω ] ⎥ 0 ⎣ ⎦ 2 ∞ 1 2 2γωp ∞ π π ⎡ x2 ω2 ⎤ ∫ x 2 − ω 2 ⎢ Den[x] − Den[ω ] ⎥ dx = ⎢ ⎥ ⎣ ⎦ 0 1 ⎡ x 2Den[ω ] − ω 2 Den[x] ⎤ ∫ x 2 − ω 2 ⎢ Den[x]Den[ω ] ⎥ dx ⎢ ⎥ ⎣ ⎦ 0 1 2 2γωp ∞ D.11. the second term can be multiplied by a constant and we choose 2ωχ (ω ) π . Toscano Pag.11. Orfanidis – Electromagnetic Waves and Antennas ∞ ∞ ω ' χ (ω ' ) 2 dω ' χ r (ω ) = ℘∫ 2 i 2 dω ' −℘∫ 2 2 π 0 ω' −ω 0 ω' −ω ∞ ∞ ωχ (ω ' ) 2 dω ' χ i (ω ) = − ℘∫ 2r 2 dω ' −℘∫ 2 2 π 0 ω' −ω 0 ω' −ω Exercises Chapter 1 Being equal to zero.11.7) Because now the integrands don't present a singularity in ω ' = ω .J. we can cancel ℘ instruction and obtain: ∞ 2 ⎛ ω ' χ i (ω ' ) − ωχ i (ω ) ⎞ ⎜ ⎟ dω ' π ∫⎝ ω '2 − ω 2 ⎠ 0 χ r (ω ) = χ i (ω ) = − • Question n°4 2 ⎛ ωχ r (ω ' ) − ωχ r (ω ) ⎞ ⎜ ⎟ dω ' π ∫⎝ ω '2 − ω 2 ⎠ 0 ∞ (1.11. (1.4) for χ r (ω ) : χ r (ω ) = 2 ∞ π ∫ 0 xχ i ( x ) − ωχ i (ω ) x2 − ω2 dx. 47 . So we have: D.5). Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Now we can expand the numerator x 2 D en[ω ] − ω 2 D en[x ] inside the integral using the extended form for Den[x] and Den[ω] : 2 2 ⎡ 2 ⎤ ⎡ 2 ⎤ x 2Den[ω ] − ω 2Den[x] = x 2 ⎢ ω0 − ω 2 + γ 2ω 2 ⎥ − ω 2 ⎢ ω0 − x 2 + γ 2 x 2 ⎥ = ⎣ ⎦ ⎣ ⎦ 4 2 4 2 = x 2 ⎡ω0 + ω 4 − 2ω0 ω 2 + γ 2ω 2 ⎤ − ω 2 ⎡ω0 + x 4 − 2ω0 x 2 + γ 2 x 2 ⎤ = ⎣ ⎦ ⎣ ⎦ 4 2 4 2 = x 2ω0 + x 2ω 4 −2x 2ω0 ω 2 + x 2γ 2ω 2 − ω 2ω0 − ω 2 x 4 +2ω 2ω0 x 2 −ω 2γ 2 x 2 = 4 4 4 = x 2ω0 + x 2ω 4 − ω 2ω0 − ω 2 x 4 = ω0 x 2 − ω 2 − ω 2 x 2 x 2 − ω 2 = 4 = x 2 − ω 2 ω0 − ω 2 x 2 ( ) ( ) ( )( ) ( ) ( ) and substitute it inside the integral: ⎡ x 2 − ω 2 ω 4 − ω 2x 2 ⎤ 0 ⎢ ⎥ χ r (ω ) = ⎢ ⎥ dx = Den[x]Den[ω ] π ∫ x2 − ω2 ⎢ ⎥ 0 ⎣ ⎦ 2 2 ∞ 4 4 ⎤ γωp 2 ∞ ω0 − ω 2 x 2 γωp ⎡ 2 ∞ ω0 2 ω 2x 2 ⎢ ∫ dx = dx − ∫ dx ⎥ = = Den[ω ] π ∫ Den[x] Den[ω ] ⎢ π Den[x] π 0 Den[x] ⎥ 0 ⎣ 0 ⎦ 2 2γωp ∞ 1 ( )( ) 2 ∞ ⎤ γωp ⎡ 4 2 ∞ 1 2 x2 ⎢ω0 ∫ dx − ω 2 ∫ dx ⎥ = Den[ω ] ⎢ π Den[x] π 0 Den[x] ⎥ 0 ⎣ ⎦ Now it is possible to use the note results suggested by the text of the exercise: 2 2 2 2 ⎤ ωp ω0 − ω γ ωp ⎡ 4 1 2 1 ⎢ω0 ⎥= −ω = χ r (ω ) = 2 Den[ω ] ⎢ Den[ω ] γ ⎥ γ ω0 ⎣ ⎦ ( ) = 2 2 ωp ω0 − ω 2 (ω02 − ω 2 ) ( ) 2 + γ 2ω 2 that is exactly the expression of χ r (ω ) for a simple oscillator model of dielectric polarization in eq. Ramaccia and A. Toscano Pag.S. Now we have to demonstrate the dual expression for χ i (ω ) : χi (ω ) = − π∫ 0 ∞ 2 ωχ r ( x ) − ωχ r (ω ) x2 − ω2 dx where χ r (ω ) = 2 2 ωp ω0 − ω 2 (ω02 − ω 2 ) ( ) 2 + γ 2ω 2 Let us use the same notation as above.11.J. (1. 2 2 Now we can expand the numerator ω 0 − x 2 Den[ω ] − ω 0 − ω 2 Den[x] inside the integral using ( ) ( ) the extended form for Den[x] and Den[ω] : (ω02 − x 2 ) Den[ω ] − (ω02 − ω 2 ) Den[x] = (ω02 − x 2 ) ⎡⎢⎣(ω02 − ω 2 ) + γ 2ω 2 ⎤⎥⎦ − (ω02 − ω 2 ) ⎡⎢⎣(ω02 − x 2 ) + γ 2x 2 ⎤⎥⎦ = (ω02 − x 2 ) ⎡⎣ω04 + ω 4 − 2ω02ω 2 + γ 2ω 2 ⎤⎦ − (ω02 − ω 2 ) ⎡⎣ω04 + x 4 − 2ω02x 2 + γ 2x 2 ⎤⎦ = 2 2 2 4 2 4 2 6 +ω0 + ω 4ω0 − 2ω0 ω 2 + γ 2ω 2ω0 − x 2ω0 − x 2ω 4 +2x 2ω0 ω 2 −γ 2 x 2ω 2 + 6 2 4 2 4 2 −ω0 − x 4ω0 + 2ω0 x 2 − γ 2 x 2ω0 + ω 2ω0 + x 4ω 2 −2ω0 x 2ω 2 +γ 2 x 2ω 2 = ( ) ( ) ( ) ( ) ( ) 2 4 2 4 −ω0 ( x 2 − ω 2 )( x 2 + ω 2 ) + 2ω0 ( x 2 − ω 2 ) − γ 2ω0 ( x 2 − ω 2 ) − ω0 ( x 2 − ω 2 ) + x 2ω 2 ( x 2 − ω 2 ) = ( x 2 − ω 2 ) ⎡⎢⎣−ω02 ( x 2 + ω 2 ) − γ 2ω02 + ω04 + x 2ω 2 ⎤⎥⎦ 2 4 2 4 −ω0 x 4 − ω 4 + 2ω0 x 2 − ω 2 − γ 2ω0 x 2 − ω 2 − ω0 x 2 − ω 2 + x 2ω 2 x 2 − ω 2 = and substitute it inside the integral: D. Toscano Pag. Ramaccia and A.J. 48 .S. Orfanidis – Electromagnetic Waves and Antennas ∞ Exercises Chapter 1 χi (ω ) = − 2 π ∫ 0 2 2 ⎡ ω2 ω2 − x2 ωp ω0 − ω 2 ⎢ω p 0 −ω Den[x] Den[ω ] x2 − ω2 ⎢ ⎢ ⎣ 1 ( ) ( ) ⎤⎥ dx = ⎥ ⎥ ⎦ =− 2 2ωωp ∞ π ⎡ ω 2 − x 2 Den[ω ] − ω 2 − ω 2 Den[x] ⎤ 0 ⎢ 0 ⎥ dx ∫ x2 − ω2 ⎢ ⎥ Den[x]Den[ω ] 0 ⎢ ⎥ ⎣ ⎦ 1 ( ) ( ) 2 where Den[x] = ω0 − x 2 ( ) 2 2 + γ 2 x 2 and Den[ω ] = ω 0 − ω 2 ( ) 2 + γ 2ω 2 . 11. Toscano .5). • Question n° 5 Let us compare the first integrals in x and y respectively: 2 ∞ ∞ 0 π ∫ 0 dx (ω02 − x 2 ) 2 ≡ + γ 2x 2 2 π ∫ 1 − 2y 2 cos θ + y 4 dy They differ only in the denominator. (1. substituting y → x : (ω02 − x 2 ) that is: 2 + γ 2 x 2 = 1 − 2x 2 C os θ + x 4 4 2 ω0 + x 4 − 2ω0 x 2 + γ 2 x 2 = 1 − 2x 2C os θ + x 4 4 2 ω 0 − 2x 2 ⎜ ω 0 − ⎛ ⎜ ⎝ γ2⎞ 4 2 4 ⎟ + x = 1 − 2x C os θ + x 2 ⎟ ⎠ (1.11) Pag.11.10) (1. (1. 49 D. so we can compare them to find the condition of equality. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 χi (ω ) = − 2 2ωωp ∞ π ∫ 0 ⎡ x 2 − ω 2 ⎡ −ω 2 x 2 + ω 2 − γ 2ω 2 + ω 4 + x 2ω 2 ⎤ ⎤ 0 0 ⎢ 0 ⎥⎥ ⎢ ⎣ ⎦ ⎢ ⎥ dx = Den[x]Den[ω ] x2 − ω2 ⎢ ⎥ ⎣ ⎦ 1 ( ) ( ) 2 2 2 2 2 4 2 2 2 ωωp 2 ∞ −ω0 x + ω − γ ω0 + ω0 + x ω dx = =− Den[ω ] π ∫ Den[x] 0 2 2 ⎞ ⎤ ωωp ⎡ 2 ∞ ⎛ 2 x 2 1 1 1 2 2 2 2 4 2 x ⎢ =− − ω0 ω − γ ω0 + ω0 +ω ⎜ −ω0 ⎟ dx ⎥ = Den[ω ] ⎢ π ∫ ⎜ Den[x] Den[x] Den[x] Den[x] Den[x] ⎟ ⎥ ⎠ ⎦ ⎣ 0⎝ ( ) =− 2 2 2 ⎤ 2 ωωp ⎡ ω0 ω0 ω 2 γ 2 ω0 ω4 ω2 ⎥ ⎢− − − + 02 + = 2 2 Den[ω ] ⎢ γ γω0 γ ⎥ γ ω0 γ ω0 ⎣ ⎦ 2 2 ⎤ γωωp γωωp ⎥= = ⎥ Den[ω ] ω 2 − ω 2 2 + γ 2ω 2 ⎦ 0 2 2 2 ωωp ⎡ ω0 ω 2 γ 2 ω0 ω 2 ⎢ + + − − = γ γ γ γ Den[ω ] ⎢ γ ⎣ ( ) that is exactly the expression of χ i (ω ) for a simple oscillator model of dielectric polarization in eq.11.9): 4 2 ω 0 − 2 ω 0 x 2 C os θ + x 4 = 1 − 2 x 2 C o s θ + x 4 (1.11.11. (1.J. Ramaccia and A.10) in eq.S. we have: γ2 γ2 ⎛θ ⎞ C os θ = 1 − 2Sin 2 ⎜ ⎟ = 1 − 2 2 = 1 − 2 ⎝2⎠ 4ω0 2ω0 Now it is possible substitute eq.9) Using the suggested relation Sin (θ 2 ) = γ ( 2ω 0 ) and the relation C os θ = 1 − 2Sin (θ 2 ) .11. 11). Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 In order to match the right– and left–side of eq.11. we have: D. (1.11. (1.12) χr0 (ω ) and χi0 (ω ) represent the real and imaginary part of χ 0 (ω ) for γ It is easy to note that the real part converges as: 2 ωp (ω02 − ω 2 ) 0 χ r (ω ) = lim = 2 γ →0 2 2 2 2 2 ω0 − ω 2 (ω0 − ω ) + γ ω 2 → 0. • Question n° 6 Starting from the expression of χ (ω ) as in eq.5) and applying the limit as γ → 0 we get: χ 0 (ω ) = lim χ (ω ) = lim γ →0 2 ωp 2 γ →0 ω0 − ω 2 + jγω 2 = 2 ωp (ω02 − ω 2 ) γω 2 = lim − j lim ωp 2 2 2 2 γ →0 (ω02 − ω 2 ) + γ 2ω 2 ω0 − ω γ →0 (ω02 − ω 2 ) + γ 2ω 2 (1. Toscano Pag. respectively.11.14) So. 50 . it is necessary that ω0 → 1 and it is correct because the integrals in y are a reduced form of the integrals in x. Ramaccia and A. Dividing and multiplying numerator and denominator by ω 2 and π respectively.11.13) 2 2 ωp ωp ⎤ 1 ⎡ ⎢℘ ⎥ =℘ 2 = −℘ ω0 + ω ⎥ ω0 − ω 2 2ω0 ⎢ ω0 − ω ⎣ ⎦ 2 ωp 2 For what concerns χ i0 (ω ) = ω p lim γ →0 ( γω 2 ω0 −ω 2 2 ) we have to note that it is very similar to +γ ω 2 2 the following definition of the Dirac delta function: δ ( x ) = lim 1 2 ε ε →0+ π x +ε2 (1. (1.J.11.11. Indeed the result of integrals in y can be written as follow: 1 2 (1 − C os θ ) = 1 ⎛ ⎛ γ 2 ⎞⎞ ⎟ 2 ⎜1 − ⎜1 − 2⎟ ⎜ ⎜ 4ω0 ⎟ ⎟ ⎝ ⎠⎠ ⎝ = 1 γ2 2 2ω0 = 1 γ ω0 →1 that is exactly the result of integrals in x when ω0 → 1 . we first manipulate χi0 (ω ) in order to apply eq.S.14). (1. assuming ε → γ and x→ 2 ω0 − ω 2 .11..11.15) has the same form of eq.11. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 γω 2 χ i0 (ω ) = ωp lim γ →0 ( ω2 2 ω0 − ω 2 ) 2 = + γ ω 2 2 ω = 2 πωp 2 ω2 (1.11.18) In order to apply (1. .16) In order to write (1. .J.11.11. x 2 . 51 D.17) to (1.19) Pag.14). Ramaccia and A. consider a function f ( x ) with n zeros: ⎧f ( x ) = 0 ⎪ ⎨ ⎪f ' ( x ) ≠ 0 ⎩ in x i = x1 . we apply the following two properties of the Dirac delta function: 1. (1.16) in a simpler form. we have: δ x2 − a2 = ( ) d 2 x − a2 dx ( δ (x − a) ) x =a + d 2 x − a2 dx ( δ (x + a) ) x=−a = = 1 1 1 δ (x − a) + δ (x + a) = ⎡δ ( x − a ) + δ ( x + a ) ⎤ ⎦ 2a 2 −a 2a ⎣ 2. if we consider f ( x ) = x 2 − a 2 .11.S.17) i =1 As an example. So it is possible to write: ω χ i0 (ω ) = 2 πω p ω δ⎜ 2 ⎛ ω0 − ω 2 ⎜ ω ⎝ ⎞ ⎟ ⎟ ⎠ (1. Toscano . Consider a function g ( x ) : g ( x )δ ( x − x 0 ) = g ( x 0 )δ ( x − x 0 ) (1. − ω 0 } (1. x n then δ (f ( x )) = ∑ n δ ( x − xi ) f ' ( xi ) (1..11.11.15) ω 2 γ →0 π ⎛ 2 ω0 − ω 2 ⎞ lim 1 ⎜ ⎜ ⎝ γ ω 2 ⎟ +γ ⎟ ⎠ The limit γ → 0 of eq.16). first of all we have to evaluate the zeros of the function f (ω ) = 2 ω0 − ω 2 : ω f (ω ) = 0 ⇔ ω = {+ ω 0 . (1.11. 22) inside the modified Kramers–Kronig relationship (1.J. Substitute the quantity χi0 (ω ) expressed in eq. then.20) 2 2⎞ 2 2 ⎛ ω −ω ⎛ 1 ⎞ ω −ω = − ⎜ 0 2 ⎟ − ⎜ ⎟ 2 ω = −2 − 0 2 ⎜ ω ⎟ ⎝ω ⎠ ω ⎝ ⎠ ( ) ( ) Now it is possible to write: χi0 2 2 πωp ⎛ ω0 − ω 2 ⎞ δ⎜ ⎟= (ω ) = ω ⎜ ω ⎟ ⎝ ⎠ = 2 πωp δ (ω − ωi ) = ω i=1 f ' (ωi ) ∑ 2 ⎡ ⎢ 2 πωp ⎢ δ (ω − ω0 ) δ (ω + ω0 ) ⎢ + = ω ⎢ ω2 −ω2 ω2 −ω2 −2 − 0 2 0 ⎢ −2 − 0 2 0 ω0 ω0 ⎢ ⎣ = 2 πωp ⎤ ⎥ ⎥ ⎥= ⎥ ⎥ ⎥ ⎦ (1. (1.11. 52 .4) for χ r (ω ) : χ r (ω ) = 2 ∞ π ∫ 0 xχ 0 ( x ) − ωχ 0 (ω ) i i x2 − ω2 dx where we have substituted ω ' → x to distinguish better the arguments of the integral.11. Orfanidis – Electromagnetic Waves and Antennas and its first derivate respect to ω Exercises Chapter 1 : 2 2 d ω0 − ω 2 d d ⎛ ω0 − ω 2 ⎞ 2 2 d (1 ω ) ⎛ 1 ⎞ +⎜ ⎟ = f ' (ω ) = f (ω ) = ⎜ ⎟ = ω0 − ω ⎟ dω dω ⎜ ω dω dω ⎝ω ⎠ ⎝ ⎠ (1. Ramaccia and A.11. apply the second property of the Dirac delta function (1.11. So: D.21) ⎡δ (ω − ω0 ) + δ (ω + ω0 ) ⎤ ⎦ 2ω ⎣ 2 πωp ⎡ 1 1 ⎤ ⎢ ω δ ( ω − ω 0 ) + ω δ ( ω + ω0 ) ⎥ = 2 ⎣ ⎦ 2 πωp ⎡ 1 ⎤ 1 δ ( ω + ω0 ) ⎥ = ⎢ δ ( ω − ω0 ) − ω0 2 ⎣ ω0 ⎦ 2 πωp Let us.11.18): χ i0 (ω ) = = = (1.11.S.22) ⎡δ (ω − ω0 ) − δ (ω + ω0 ) ⎤ ⎦ 2ω0 ⎣ This expression still satisfies the Kramers–Kronig relations. Toscano Pag. 14). (1. the limit γ → 0 to χ i (ω ) should be applied.11.11.12): χ r0 So: (ω ) = lim 2 2 ωp ω0 − ω 2 γ →0 (ω02 − ω 2 ) ( ( ) 2 + γ 2ω 2 χ i0 (ω ) = − ∫ π 0 2 ∞ ⎡ ⎤ 2 2 2 2 2 ω p ω0 − ω 2 ⎢ ω p ω0 − x ⎥ − lim ⎢ ⎥ dx 2 2 2 2 2⎥ x 2 − ω 2 γ →0 ⎢ ω 2 − x 2 2 + γ 2 x 2 ω0 − ω +γ ω ⎣ 0 ⎦ ω ( ) ) ( ( ) ) and.11. exchanging the limit with the integral. which is given by eq. Toscano Pag. which is real negative).11. Ramaccia and A. D. Then.23) 0 The Kramers–Kronig relationship (1. we can write: ⎧ ⎪ 2∞ ω χ i0 (ω ) = lim ⎨ − ∫ 2 γ →0 ⎪ π x − ω 2 0 ⎩ ⎡ ⎤ ⎫ 2 2 2 2 2 ω p ω0 − ω 2 ⎢ ω p ω0 − x ⎥ ⎪ − ⎢ ⎥ dx ⎬ 2 2 2 2 2 2 2 2 2 2⎥ ⎪ ⎢ ω0 − x +γ x +γ ω ω0 − ω ⎣ ⎦ ⎭ ( ( ) ) ( ( ) ) Now it is easy to note that the argument of the limit has already been solved in the solution of question n°4 of this exercise resulting in χ i (ω ) . So we can write: χ r (ω ) = ℘ 2 ωp ω0 ω0 ω02 − ω 2 =℘ 2 ωp ω02 − ω 2 = χ 0 (ω ) r (1.11.6). (1. Also the integral I2 vanishes because it is over the real–positive values of x while the Dirac delta function is always zero on this interval (it is not zero only for x = −ω0 . 53 .J.4) for χ i (ω ) with χ r (ω ) = χ r (ω ) can be written using 0 the definition of χr (ω ) as in eq.S. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 χ r (ω ) = 2∞ 2 π ωp 1 ∫ x 2 − ω 2 ⎡ x (δ ( x − ω0 ) − δ ( x + ω0 ) ) − ω (δ (ω − ω0 ) − δ (ω + ω0 ) )⎤ dx = ⎣ ⎦ π 2 ω0 0 ∞ ⎡∞ ⎤ x x ⎢∫ 2 δ ( x − ω0 ) dx − ∫ 2 δ ( x + ω0 ) dx − ⎥ 2 2 2 ⎥ ωp ωp ⎢ 0 x − ω 2 0 x −ω = [ I1 + I2 + I3 + I4 ] ⎢ ⎥= ∞ ω0 ⎢ ∞ ω ⎥ ω0 ω δ (ω − ω0 ) dx + ∫ 2 δ (ω + ω0 ) dx ⎥ ⎢+ ∫ 2 x −ω2 x − ω2 ⎢ 0 ⎥ 0 ⎣ ⎦ The integrals I3 and I4 vanishe as demonstrated in (1. So the convolution in time domain is also the corresponding operation of the product in frequency domain and vice versa. respectively.S. we have that χ (ω ) = ℑ{ χ ( t )} = ℑ{χ ( t ) u ( t )} = ℑ{χ ( t )} ∗ ℑ{u ( t )} = 1 χ (ω ) ∗ U (ω ) 2π where χ (ω ) and U (ω ) are the Fourier transforms of χ ( t ) and u ( t ) .12 Exercise Derive the Kramers–Kronig relationship: χ (ω ) = +∞ χ (ω ' ) ℘∫ dω ' πj ω −ω' 1 (1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. producing a third function that is typically viewed as a modified version of one of the original functions. it represents the transformation obtained when a signal passes through a black–box system with a known impulse response. In similar way.12. In signal theory. that is. Solution The convolution is a mathematical operation on two functions f and g. Toscano Pag. The Fourier transformation of Heaviside step function is: U (ω ) =℘ 1 + πδ (ω ) jω So we have: 1 χ ( ω ) = ℘ ∫ χ ( ω ' ) U ( ω − ω ' ) dω ' 2π = = 1 1 1 ℘ ∫ χ (ω ' ) π dω ' + 2π j (ω − ω ') 2π −∞ +∞ −∞ −∞ +∞ +∞ −∞ +∞ ∫ χ ( ω ' ) δ ( ω − ω ' ) dω ' = 1 1 1 ℘ ∫ χ (ω ') dω ' + χ ( ω ) 2π j 2 (ω − ω ' ) D.J. 54 . It is defined as: +∞ −∞ +∞ −∞ ( f ∗ g )( t ) ∫ f (τ ) g ( t − τ ) d τ = ∫ g (τ ) f ( t − τ ) dτ Using the causality condition χ ( t ) = χ ( t ) u ( t ) . in frequency domain the output of the system is the product of the Fourier transformations of the input signal and the impulse response. Ramaccia and A. expressing it as the convolution of the Fourier transforms of χ ( t ) and u ( − t ) .1) −∞ by starting with the causality condition χ ( t ) u ( − t ) = 0 and translating it to the frequency domain. πj (ω − ω ' ) 1 −∞ D. 55 . Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 Rearranging terms and canceling a factor of 1 2 .J. Toscano Pag. we obtain the Kramers–Kronig relation in its complex–value form: χ (ω ) = +∞ χ (ω ' ) ℘∫ dω ' .S. Using the Lorentz transformations: E'⊥ = γ ( E ⊥ + cβ × B ⊥ ) H '⊥ = γ ( H ⊥ − cβ × D⊥ ) 1 ⎛ ⎞ B '⊥ = γ ⎜ B ⊥ − β × E ⊥ ⎟ c ⎝ ⎠ E'/ / B '/ / 1 ⎛ ⎞ D'⊥ = γ ⎜ D⊥ + β × H ⊥ ⎟ c ⎝ ⎠ H '/ / D'/ / (1. the constitutive relations are assumed to be the usual ones.13 Exercise An isotropic homogeneous lossless dielectric medium is moving with uniform velocity v with respect to a fixed coordinate frame S.13.2) (1.J. β c = v c 2 and γ = 1 following form in the fixed frame S: .4). that is.13.S. Considering the first equation of set (1. Ramaccia and A. we can substitute to all of component with the superscript with the correspondent definition given by the Lorentz transformation: D.1) = E/ / = B/ / 1− β 2 = H// = D/ / where cβ = v . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 1. show that the constitutive relations take the D = ε E + av × ( H − ε v × E ) B = μ H − av × ( E + μ v × H ) (1.13. D ' = ε E ' and B' = μH' .3) where a = εμ − ε 0 μ0 . 56 . Toscano Pag.13. In the frame S' moving with dielectric.4) where the subscripts and // indicate the component perpendicular and parallel at the velocity vector v . 1 − εμ v 2 Solution It is possible to express the constitutive relations D ' = ε E ' and B' = μH' as follow: ( ) B ' = B'⊥ + B'/ / = μ ( H '⊥ + H '/ / ) D ' = D'⊥ + D'/ / = ε E'⊥ + E'/ / (1.13. 57 . Orfanidis – Electromagnetic Waves and Antennas D'⊥ + D'/ / = ε E'⊥ + E'/ / ⎛ ⎝ 1 ⎞ ⎠ Exercises Chapter 1 ( ) (1.6) The relation between the parallel components of vector D and vector E is the same in the two frame S and S'.13. so we have to evaluate it using the set (1. we have: B '⊥ = μ H '⊥ ⎛ ⎝ ⎞ ⎠ γ ⎜ B ⊥ − β × E ⊥ ⎟ = μ γ ( H ⊥ − cβ × D ⊥ ) 1 B ⊥ − β × E ⊥ = μ H ⊥ − μ cβ × D ⊥ c 1 B ⊥ = μ H ⊥ + β × E ⊥ − μ cβ × D ⊥ c 1 c and express it as function of velocity v: B⊥ = μH⊥ + 1 c2 v × E⊥ − μ v × D⊥ (1. (1.7) in eq.13. Toscano Pag.13.13. Considering only the perpendicular component.7) Now we can substitute eq.6) and obtain: 1 1 ⎛ ⎞ D⊥ + 2 v × H ⊥ = ε E⊥ + ε v × ⎜ μ H ⊥ + 2 v × E⊥ − μ v × D⊥ ⎟ c c ⎝ ⎠ 1 ε D⊥ = ε E⊥ − 2 v × H ⊥ + εμ v × H ⊥ + 2 v × v × E⊥ − εμ v × v × D⊥ c c 1⎞ ε ⎛ D⊥ + εμ v × v × D⊥ = ε E⊥ + ⎜ εμ − 2 ⎟ v × H ⊥ + 2 v × v × E⊥ c ⎠ c ⎝ D. Ramaccia and A. In the fixed frame S the constitutive relation B = μH is not valid.S.J. On the contrary.1) and the constitutive relation B' = μH' in the frame S' where it is valid. (1.13.5) γ ⎜ D⊥ + β × H ⊥ ⎟ + D / / = ε ⎡γ ( E ⊥ + cβ × B ⊥ ) + E / / ⎤ ⎣ ⎦ c γ D ⊥ + β × H ⊥ + D / / = εγ ( E ⊥ + cβ × B ⊥ ) + ε E / / c c γ γ γ D ⊥ + β × H ⊥ + D / / = εγ ( E ⊥ + cβ × B ⊥ ) + ε E / / Now it is possible to substitute cβ = v and β c = v c 2 and separate the component parallel and perpendicular at the velocity vector v : 1 ⎧ ⎪ D⊥ + 2 v × H ⊥ = ε E⊥ + ε v × B ⊥ c ⎨ ⎪D = ε E // ⎩ // ( ⊥ −component) ( // − component) (1. the perpendicular component depends on both electric and magnetic field.13. (1.13.13. Toscano Pag. Now we can identify with X the unknown quantity and we can impose that: ε ⎧ E ⊥ + XE ⊥ = ε E ⊥ ⎪ 2 ⎪ 1 − εμ v ⎪ ⎨ εμ − ε 0 μ 0 ε ⎪ v × v × E ⊥ + X E ⊥ = −ε v × v × E⊥ 2 2 ⎪ c 1 − εμ v 1 − εμ v 2 ⎪ ⎩ ( ) ( ) ( ) which can be simplified as: D. Ramaccia and A. so: D⊥ = (1 − εμ v ) 2 ε E ⊥ + av × H ⊥ + c 1 − εμ v 2 ( ε 2 ) v × v × E⊥ (1.2).S.13. (1.13. that is: ε ε ⎧ E⊥ + v × v × E⊥ ⎪ 2 2 1 − εμ v c 1 − εμ v 2 ⎪ ⎪ ⎨ ⎪ε E − ε εμ − ε 0 μ 0 v × v × E ⊥ ⎪ ⊥ 1 − εμ v 2 ⎪ ⎩ ( ) ( ) ( ) where we already substituted (1 − εμ v ) 2 εμ − ε 0 μ 0 =a.13. (1. To find it. we will compare the our expression and the expression suggested in eq. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 The double cross product v × v × D ⊥ can be evaluated using the BAC-CAB rule: v × v × D⊥ = v ( v ⋅ D⊥ ) − D⊥ ( v ⋅ v ) = 0 − v 2 D⊥ = − v 2 D⊥ so 1 ⎞ ε ⎛ D⊥ − εμ v2D⊥ = ε E⊥ + ⎜ εμ − 2 ⎟ v × H ⊥ + 2 v × v × E⊥ c ⎠ c ⎝ (1 − εμ v2 ) D⊥ = ε E⊥ + (εμ − ε 0μ0 ) v × H⊥ + cε2 v × v × E⊥ D⊥ = (1. It is possible to think that probably we have to add and subtract a unknown quantity.9) and eq.9) Comparing eq.8) (1 − εμ v ) 2 ε E⊥ + ( εμ − ε 0μ0 ) v × H (1 − εμ v ) 2 ⊥ + c 1 − εμ v2 2 ( ε ) v × v × E⊥ It is easy to note that the term v × H ⊥ is multiplied by the coefficient a.2). we can note that the coefficient of E⊥ and v × v × E⊥ are different.J. 58 . 13. 59 .13. (1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 1 ε ⎧ +X =ε ⎪ 1 − εμ v 2 ⎪ ⎪ ⎨ εμ − ε 0 μ0 ε v2 ⎪− + X = +ε v 2 ⎪ c 2 1 − εμ v 2 1 − εμ v 2 ⎪ ⎩ ( ) (1. Ramaccia and A.13.10).S.13.10) ( ) ( ) where in the second equation is present the term − v 2 because of v × v × E ⊥ = − v 2 E ⊥ . So finally we can write: D⊥ = = ( ε 1 − εμ v2 ) E⊥ + av × H ⊥ + c2 1 − εμ v2 ( ε ) v × v × E⊥ + XE⊥ − XE⊥ = (1 − εμ v ) 2 ε E⊥ + av × H ⊥ + c 1 − εμ v 2 ( ε 2 ) v × v × E⊥ + ε (1 − εμ v ) 2 εμ v2 E⊥ − ε (1 − εμ v ) 2 εμ v2 (1.J.12) It is easy to note that X has the same module. Toscano Pag.14) c2 1 − εμ v2 c 1 − εμ v 2 ( ( ε ε 2 ) ) v × v × E⊥ + ε v × v × E⊥ − ε ( εμ v 2 εμ 1 − εμ v 2 2 ) (1 − εμ v ) (1 − εμ v2 ) =ε ε ⎛ 1 ⎞ ⎜ 2 − εμ ⎟ v × v × E⊥ ⎝c ⎠ (1.13.13.13.15) ( ε 0 μ0 − εμ 1 − εμ v 2 ) v × v × E⊥ = −ε av × v × E⊥ D.13.13.9). From the first of (1.11) and from the second: X= c 2 1 − εμ v 2 ( ε v2 2 2 ) (1 − εμ v2 ) (1 − εμ v2 ) ⎡ ε 0 μ0 + εμ −ε 0 μ0 ⎤ = ε ⎣ ⎦ ε v2 εμ − ε 0 μ0 = ε v2 ⎡1 ⎤ ⎢ 2 + εμ − ε 0 μ0 ⎥ = ⎣c ⎦ 2 =ε (1 − εμ v ) v (1 − εμ v2 ) εμ v (1. This confirms our argument of finding a quantity to add and subtract at eq. but opposite sign. we obtain that: X =ε − ( ε 1 − εμ v 2 ) =ε +1 − εμ v 2 −1 ( 1 − εμ v 2 ) = −ε ( εμ v 2 1 − εμ v 2 ) (1.13) E⊥ Now we can simplify eq. (1.13) as follow: ( ε 2 1 − εμ v ) E⊥ − ε ( εμ v 2 2 1 − εμ v ) E⊥ = (1 − εμ v ) 2 ε (1 − εμ v 2 ) E ⊥ = ε E ⊥ E⊥ = v × v × E⊥ = (1. S. Ramaccia and A.13. In the same way. 60 .13. Toscano Pag.18) D.13.13.13.13).14) and (1.13.13.13.13. Orfanidis – Electromagnetic Waves and Antennas Substituting (1.15) in eq.16).16) Now using the second equation of (1. it is possible write the expression of vector D in the fixed frame S: D = D⊥ + D/ / = ε E ⊥ + a v × H ⊥ − ε av × v × E ⊥ + ε E / / = = ε ( E ⊥ + E / / ) + av × H ⊥ − ε av × v × E ⊥ + av × H / / − ε a v × v × E / / = (1. but we kwon that the relations are dual and we can obtain eq.13.J. (1.3).13.17) and eq. (1.3) just operating a changing of variables as follow: ⎧D → B ⎪E → H ⎪ ⎨ ⎪v × H → −v × E ⎪ε → μ ⎩ So: D = ε E + av × ( H − ε v × E ) → B = μ H − av × ( E + μ v × H ) (1. (1. (1.13. we can demonstrate eq. (1.6) and eq.17) ↑ null ↑ null = ε E + av × H − ε av × v × E = ε E + a v × ( H − ε v × E ) It is easy to note that eq.2) are identical. (1. we obtain: D ⊥ = ε E ⊥ + av × H ⊥ − ε av × v × E ⊥ Exercises Chapter 1 (1. edu www. University "Roma Tre" via della Vasca Navale.edu/~orfanidi/ewa 2 Department of Applied
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[email protected] Waves and Antennas Exercise book Chapter 2: Uniform Plane Waves Sophocles J.com alessandro. 84 00146.com .ece. Piscataway.toscano@gmail. Italy davide. Orfanidis1 Davide Ramaccia2 Alessandro Toscano2 1 Department of Electrical & Computer Engineering Rutgers University.rutgers. ........................... 24 2...16 ExerciseEquation Section (Next) .......................................................13 ExerciseEquation Section (Next) .... 9 ExerciseEquation Section (Next) .............. 18 2..... 20 2......................................................................................7 2..........8 2................................................... 1 2................... 4 ExerciseEquation Section (Next) .... 26 2..........1 2............. 23 2.........15 ExerciseEquation Section (Next) ........................................... 14 ExerciseEquation Section (Next) ................................5 2..............................4 2................... 40 2..........................................................................................14 ExerciseEquation Section (Next) ..............................12 ExerciseEquation Section (Next) ................ 11 ExerciseEquation Section (Next) ................................................................................................................................... 39 2............ 12 ExerciseEquation Section (Next) ............................................................. 21 2... 42 .......... Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Table of Contents Chapter 2 Uniform Plane WavesEquation Chapter 2 Section 1 ................................................... 11 ExerciseEquation Section (Next) ........... 1 ExerciseEquation Section (Next) .................................................19 ExerciseEquation Section (Next) ................................................................... 28 2.......3 2........S...............11 ExerciseEquation Section (Next) ...2 2.............. 37 2....18 ExerciseEquation Section (Next) ........................... 3 ExerciseEquation Section (Next) .........................10 ExerciseEquation Section (Next) ...............................................17 ExerciseEquation Section (Next) ............6 2..........................................9 Exercise ........................J........................................................................................ ............................20 ExerciseEquation Section (Next) ........................ 48 2...... 44 2........................29 ExerciseEquation Section (Next) .......................S.... 58 2................... 69 ......26 ExerciseEquation Section (Next) .....................................25 ExerciseEquation Section (Next) ..................... Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2........28 ExerciseEquation Section (Next) . 49 2..........27 ExerciseEquation Section (Next) ...J.............................................. 66 2.......... 46 2........................................... 53 2...................................................21 ExerciseEquation Section (Next) .......................................................................................24 ExerciseEquation Section (Next) ........................23 ExerciseEquation Section (Next) ................................................................................................. 55 2.......................................................22 ExerciseEquation Section (Next) . 62 2............... 1.S.1. t ) = 0 ⎜ ∂z c ∂t ⎟ ⎝ ⎠ and the forward-backward equations: ∂E + 1 ∂E + =− c ∂t ∂z 1 ∂E − ∂E − =+ c ∂t ∂z (2.J. ∂ζ∂ξ ∂E+ = 0. Toscano Pag.1) (2.2) become in these variables: ∂ 2E = 0. Show that the wave equation: ⎛ ∂2 1 ∂2 ⎞ ⎜ 2 − 2 2 ⎟ E ( z. we have to evaluate the derivates: D. ξ ) of the independent variables ζ = z − ct and ξ = z + ct .1 Exercise A function E ( z. Ramaccia and A. 1 .1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Chapter 2 Uniform Plane WavesEquation Chapter 2 Section 1 2. ζ ξ Solution First of all. ∂ξ ∂E− =0 ∂ζ (2.3) Thus. t ) may be thought of as a function E (ζ . E+ may depend only on and E− only on . 1.1. (b) and (d) of (2. consequently ∂E − =0 ∂ζ (2.1.1.4).2) as follow: ∂E + ∂E =− + ∂ξ ∂ξ that is verified only when ∂E + =0 ∂ξ (2. (2. Orfanidis – Electromagnetic Waves and Antennas ⎧ ∂ζ ⎪ ∂z ⎪ ⎪ ∂ζ ⎪ ∂t ⎨ ⎪ ∂ξ ⎪ ∂z ⎪ ∂ξ ⎪ ⎩ ∂t ∂ (z − ct) = 1 ⇒ ∂ζ = ∂z ∂z ∂ = (z − ct) = −c ⇒ ∂ζ = −c∂t ∂t ∂ = (z + ct) = 1 ⇒ ∂ξ = ∂z ∂z ∂ = (z + ct) = + c ⇒ ∂ξ = + c∂t ∂t = Exercises Chapter 2 (a) (b) (2. ξ ) = 0 ∂ζ∂ξ ⎞ ⎟ E (ζ .5) In the same way.1. using the relationships (c) and (d) of (2.1) to obtain: ⎛ ∂2 ∂2 + ⎜ ⎜ ∂ζ∂ξ ∂ζ∂ξ ⎝ that is: 2 ∂2 E (ζ .4) (c) (d) so.1.S. the backward equation in (2. we have −c2∂t = ∂ζ∂ξ . we can rewrite the forward equation in (2.2) becomes: ∂E − ∂E =− − ∂ζ ∂ζ and.1. Ramaccia and A. Toscano Pag.1.4).1.J. (a) and (c) of (2. multiplying eq. 2 .4). we have ∂ 2 z = ∂ζ∂ξ and. ξ ) = 0 ⎟ ⎠ Using the relationships (a) and (b) of (2. multiplying eq.6) D.4). Now we can substitute them inside eq.1. The ⎦ pulse is launched towards the positive z–direction. t ) and H ( z. D. that is. t ) = xE 0 ⎡ u ( t ) − u ( t − T ) ⎤ . 2.1 shows the expanding wave–fronts at time t and t + Δt . which implies that E ( z. by E ( 0.1) Fig. or. t ) = F ( z − ct ) = F ( 0 − c ( t − z c ) ) . t ) = E ( z − ct. where u ( t ) is the unit step function and E0 is a constant. t − z c ) = xE 0 ⎡ u ( t − z c ) − u ( t − z c − T ) ⎤ ⎦ H ( z.2. t ) is completely determined by E ( z. z ≤ ct . at the time t the wave–front has propagated only up to the position z = ct .2 ExerciseEquation Section (Next) A source located at z = 0 generates an electromagnetic pulse of duration of T seconds. t − z c ) Using this property . Toscano Pag. t ) : E ( z. alternatively. Ramaccia and A.S. we have E ( z.2.2. 3 . we find for the electric and magnetic fields: ˆ ⎣ E ( z. Fig. Determine expressions for E ( z. given by ˆ ⎣ E ( 0. t ) = E 1 ˆ ˆ z × E ( z. Solution For a forward–moving wave. the non–zero values of the fields are restricted to t − z c ≥ 0 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.1: Expanding wave–font at time t and t + Δt . 2. t ) and sketch them versus z at any given t. Because of the unit step.0 ) = E ( 0.J. t ) = E ( 0. t ) = y 0 ⎡ u ( t − z c ) − u ( t − z c − T ) ⎤ ⎦ Z0 Z0 ⎣ (2.0 ) or. H ( x. y. y.e. z ) e jω t .3 ExerciseEquation Section (Next) Show that for a single–frequency wave propagating along z–direction the corresponding transverse fields E ( z ) . But we are in presence of an uniform plane waves propagating along z–direction. y. H ( z ) satisfy the system of equations: ∂ ⎡ E ⎤ ⎡ 0 = ˆ⎦ ⎣ ∂z ⎢ H × z ⎥ ⎢− jωε ⎣ − jωμ ⎤ ⎡ E ⎤ ˆ⎦ 0 ⎥ ⎢H × z ⎥ ⎦⎣ (2. The Maxwell's equations can be written in the form: ⎧ ∂ H( x. the term e jωt can be simplified: ⎧∇ × E ( x. c = 1 με . So the electric and magnetic field can be expressed as E ( x. z) E y (x. respectively. y. y. i. z are the partial derivates. z ) e jω t = − jωμ H ( x. z ) e jω t = jωε E ( x.z) e =−μ ⎪ ∂t ⎨ ⎪ ∂ E( x. y.y. y. y. Solution For a single–frequency wave.J. z) E z (x. y. Ramaccia and A.z) e =−μ ∂t ⎩ ( ) ) (2. z ) e jω t ⎪ ⎨ ⎪∇ × H ( x. y. the z– D. 4 .z) ejωt jωt ⎪∇×E( x.y. z.3) ( Evaluating the derivate in the right–hand side of both equations. y. z ) e jω t . and Z0 = μ ε .3. Show that the following similarity transformation diagonalizes the transition matrix.S.3. y.z) ejωt jωt ⎪∇×H( x. and discuss its role in decoupling and solving the above system in terms of forward and backward waves: ⎡1 Z0 ⎤ ⎡ 0 ⎢ ⎥⎢ ⎣1 − Z0 ⎦ ⎣ − jωε − jωμ ⎤ ⎡1 Z0 ⎤ ⎥ ⎥⎢ 0 ⎦ ⎣1 − Z0 ⎦ −1 ⎡ − jk =⎢ ⎣ 0 0⎤ ⎥ jk ⎦ (2. we can assume a time–dependence as e jω t . Toscano Pag. z) where ∂ i with i = x. t ) = E ( x. so the electric filed vector lies on the x–y plane. y components of the vector entries. z ) e jω t ⎩ The curl of the electric (or magnetic) field can be written as determinant of the following matrix: ˆ x ∇×E = ∂x ˆ y ∂y ˆ z ∂z E x (x. t ) = H ( x.1) where the matrix is meant to apply individually to the x.y.3.2) where k = ω c . z. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.y. 4) (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 component of the electric field is null( Ez = 0 ).(2. Toscano Pag.9) and (2. so (2.S. (2.3.3. y.J. eq.3. y.3.e. y.8) can be written as follow: ∂E ˆ = − jωμ H × z ∂z (2.3. (2.4) is similar to: ˆ x ∇×E = 0 ∂E x (x. (2.7) needs only to invert the cross product at the left–hand side and to change the sign at the right–hand side: ∂H ˆ × z = − jωε E ∂z (2.10) to obtain (2. D.3. 5 . y.8) and. y. y. z) ∂z ˆ z 1 0 (2. the left–hand side simplifies into: ∂E ⎞ ∂E ∂E ∂E ∂E ∂E ⎛ ˆ ˆ ˆ ˆ ˆ⎜ˆ ˆ (z ⋅ z) − z ⎛ z ⋅ ⎞ = − z z = ⎜z× ⎟× z = ⎟ ∂z ⎠ ∂z ∂z ∂z ⎝ ⎝ ∂z ⎠ ∂z where we used the condition ∂zEz = 0 for a plane wave.3.5) ˆ that is simply the cross–product of z and ˆ z× ˆ z× ∂E ( x. The curl becomes: ˆ ˆ ˆ x y z ∇×E = 0 0 ∂z E x (x. y.3.3. y.9) On the contrary.z ) (2. z) 0 The only applicable derivate is ∂ z .3. using BAC-CAB rule.z ) = + jωε E ( x. y. the derivates along x and y are null. z) ∂z ˆ y 0 ∂E y (x.7) ˆ Consider eq.3. So eq.6) (2.6) and apply the cross–product with z to both of side: ∂E ⎞ ⎛ ˆ ˆ ˆ ⎜z× ⎟ × z = − jωμ H × z ∂z ⎠ ⎝ (2. z ) : ∂z ∂E ( x.3. Ramaccia and A.10) Now it is easy to write in matrix form eq.z ) ∂z ∂H ( x. i. z) E y (x.z ) ∂z = − jωμ H ( x.3.1).3. and also in each plane the vector has constant amplitude. 3. ⎧ − jωμν12 = α1ν 11 ⎨ ⎩ − jωεν11 = α1ν 12 ⎧− jωμν 22 = α 2ν 21 ⎨ ⎩− jωεν 21 = α 2ν 22 (2. 2 and νi = ⎜ i1 ⎟ ⎝ν i2 ⎠ Expanding it two different systems of equations. The eigenvectors are found from the eigenvalues that are the roots of the characteristic polynomial: Det ( A − α I ) = 0 ⎡ 0 where A = ⎢ ⎣ − jωε So we have: − jωμ ⎤ .2).3.13). The diagonal matrix is simply: ⎡α1 0 ⎤ ⎡ − jk 0 ⎤ ⎢ 0 α ⎥ = ⎢ 0 + jk ⎥ ⎦ 2⎦ ⎣ ⎣ (2. The left–hand side is composed by the product of the matrix A and two matrixes for the base change. so it is diagonalizable.3.3.J. I is the identity matrix and α are the eigenvalues. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 The transition matrix has to be diagonalized and we need of eigenvectors to create the matrix for the base change.3. 0 ⎥ ⎦ ⎛⎡ 0 Det ⎜ ⎢ ⎝ ⎣ − jωε − jωμ ⎤ ⎡1 0⎤ ⎞ ⎛ −α ⎥ − α ⎢ 0 1 ⎥ ⎟ = Det ⎜ − jωε 0 ⎦ ⎣ ⎦⎠ ⎝ − jωμ ⎞ 2 2 ⎟ = α + ω με = 0 −α ⎠ which gives: α1 = − jω με = − jk α 2 = + jω με = + jk It has two separate eigenvalues. Toscano Pag.3. These matrixes are made putting in row the eigenvectors ν1 and ν 2 calculated as follow: A ⋅ ν i = αi ν i ⎛ν ⎞ with i = 1. Ramaccia and A.12) and (2.13) Solving (2. one for each eigenvalue. 6 .12) (2.S. we have: ⎡ 0 ⎢ − jωε ⎣ ⎡ 0 ⎢ − jωε ⎣ − jωμ ⎤ ⎛ν11 ⎞ ⎛ν11 ⎞ ⎥ ⎜ν ⎟ = α1 ⎜ν ⎟ 0 ⎦ ⎝ 12 ⎠ ⎝ 12 ⎠ − jωμ ⎤ ⎛ν 21 ⎞ ⎛ν 21 ⎞ ⎥ ⎜ν ⎟ = α 2 ⎜ν ⎟ 0 ⎦ ⎝ 22 ⎠ ⎝ 22 ⎠ or equivalently.11) that is the right–hand side of (2. we find that the eigenvectors are given by: D. It can be simplified to obtain: ⎡1 Z0 ⎤ ⎡ 0 ⎢1 − Z ⎥ ⎢ − jωε 0⎦⎣ ⎣ 1 2 ⎤ ⎡ − jk 0 ⎤ − jωμ ⎤ ⎡ 1 2 ⎥ ⎢1 ( 2Z ) −1 ( 2Z ) ⎥ = ⎢ 0 + jk ⎥ 0 ⎦⎣ 0 0 ⎦ ⎣ ⎦ (2.S. 7 .i. In order to verify that eq. (2. Orfanidis – Electromagnetic Waves and Antennas ⎛ 1 ⎞ ⎛ 1 ⎞ ν1 = ⎜ ⎟ . j) ) where Minor ( P.17) D.i.15): P −1 −1 ⎡1 Z0 ⎤ =⎢ ⎥ ⎣1 − Z0 ⎦ = 1 Cij (P) Det [ P ] ( ) T where Cij (P) is the matrix of cofactors of P.J.3.3.3. Ramaccia and A.16) and now we can write: ⎡1 Z0 ⎤ P −1 = ⎢ ⎥ ⎣1 − Z0 ⎦ −1 = 1 ⎡ − Z0 −2Z0 ⎢ −1 ⎣ 12 ⎤ − Z0 ⎤ ⎡ 1 2 ⎥ = ⎢1 ( 2Z ) −1 ( 2Z ) ⎥ 1 ⎦ ⎣ 0 0 ⎦ So: ⎡1 Z0 ⎤ ⎡ 0 PAP −1 = ⎢ ⎥⎢ ⎣1 − Z0 ⎦ ⎣ − jωε ⎡ − jωε Z0 =⎢ ⎣ jωε Z0 12 ⎤ − jωμ ⎤ ⎡ 1 2 ⎥ ⎢1 ( 2Z ) −1 ( 2Z ) ⎥ = 0 ⎦⎣ 0 0 ⎦ 12 ⎤ − jωμ ⎤ ⎡ 1 2 ⎢ ⎥= − jωμ ⎥ ⎣1 ( 2Z0 ) −1 ( 2Z0 ) ⎦ ⎦ ⎡ ⎛ ωε Z0 ωμ ⎞ ⎛ ωε Z0 ωμ ⎞ ⎤ + − ⎢ − j⎜ ⎟ − j⎜ ⎟⎥ 2Z0 ⎠ 2Z0 ⎠ ⎥ ⎝ 2 ⎢ ⎝ 2 =⎢ ⎥ ⎢ + j⎛ ωε Z0 − ωμ ⎞ + j⎛ ωε Z0 + ωμ ⎞ ⎥ ⎜ ⎟ ⎜ ⎟ ⎢ ⎝ 2 2Z0 ⎠ 2Z0 ⎠ ⎥ ⎝ 2 ⎣ ⎦ where Z0 = μ ε . Toscano Pag.3.2) is correct. j) is defined as follow: Cij (P) = ( −1) i+ j Det ( Minor( P .14) Now we can write the matrix for the base change as: T ⎡ ν1 ⎤ ⎡ν 11 ν 12 ⎤ ⎡1 Z0 ⎤ ⎢ ⎥=⎢ = ν ⎥ ⎢1 − Z0 ⎥ ⎢ T⎥ ν ⎦ ⎣ ν 2 ⎦ ⎣ 21 22 ⎦ ⎣ (2.3.3. j) represents the matrix obtained by P cancelling the i–th row and j–th column. The cofactor in position ( i. we have to calculate the inverse of the matrix (2. It is easy to evaluate it: ⎡−Z Cij (P ) = ⎢ 0 ⎣ − Z0 −1⎤ 1⎥ ⎦ (2.15) where the superscript T indicate the vector transpose. ν 2 = ⎜ ⎟ ⎝ Z0 ⎠ ⎝ − Z0 ⎠ Exercises Chapter 2 (2. we are able to express the electric field in term of forward and backward wave. Orfanidis – Electromagnetic Waves and Antennas where k = ω με and (2.2) allow us to decouple the electric and the magnetic field as in (2.20). in this ˆ ˆ E = Z0 H × k = Z0 H × z ˆ ˆ where k = z . as follow: Exercises Chapter 2 (2.2).18) (2. Toscano Pag.20) So using (2.3. (2.3.3.3.19) and.19).17) can be substituted in (2. being the electromagnetic wave propagating along z–direction. The diagonal matrix given in (2.3. 8 .3. Ramaccia and A. D.3.1) as follow: ∂ ⎡ E ⎤ ⎡ − jk 0 ⎤ ⎡ E ⎤ = ˆ⎦ ⎣ ˆ⎦ ∂z ⎢ H × z ⎥ ⎢ 0 + jk ⎥ ⎢ H × z ⎥ ⎣ ⎦⎣ that is ⎧∂ ⎪ ∂z E = − jkE ⎪ ⎨ ⎪ ∂ H × z = + jkH × z ˆ ˆ ⎪ ∂z ⎩ case vacuum.19) The electric and magnetic field are related by the characteristic impedance of the medium.20) in the set (2.3.J.17) is equivalent to (2.3.3.21) It is possible to note that the diagonalization (2.S.3.3. using the relationship (2. we obtain: ⎧∂ ⎪ ∂z E = − jkE ⎪ ⎨ 1 ∂ 1 ⎪ E=+ jkE Z0 ⎪ Z0 ∂z ⎩ (2.3. 2) 789. The wavelength λ can be expressed via the frequency of the wave in Hertz. and violet having the nominal wavelengths of 700.J. Since the propagation factor e − jkz accumulates a phase of k radians per meter. we can calculate the frequency range for the electromagnetic visible spectrum: 380 × 10−9 < λ < 780 × 10−9 that is: ⇒ c 380 × 10−9 >f > c 780 × 10−9 (2. orange. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. Ramaccia and A. Toscano Pag.1). Solution The wavelength λ is the distance by which the phase of the sinusoidal wave changes by 2π radians. we have by definition that kλ = 2π . 530.6 THz The frequencies of the colours are: Colour Wavelength (nm) Frequency (THz) D. What is this in THz? In particular. blue. as follows: λ= 2π 2π c 2π c c = = = k ω 2π f f (2. 470.4.5 THz > f > 384.4. yellow.4. 610.4 ExerciseEquation Section (Next) The visible spectrum has the wavelength range 380–780 nm.S. determine the frequencies of red.1) Using the relation (2. f = ω 2π . 9 . 590. and 420 nm. green. 4. 10 .3 Exercises Chapter 2 Table 2. Ramaccia and A. Orfanidis – Electromagnetic Waves and Antennas Red Orange Yellow Green Blue Violet 700 610 590 530 470 420 428.1: Wavelength and frequency of colours in the visibility region.5 491.3 714.J. Toscano Pag.8 508. D.5 566.0 638.S. it is easy to obtain the result: f= c 3 × 108 20 × 10 −6 λ = = 15 × 1012 = 15 THz 2. Toscano Pag.2 cm f 2.1). (2.4.45 GHz is: λ= c 3 × 108 = = 0.6 ExerciseEquation Section (Next) What is the wavelength in meters or cm of a wave with the frequencies of 10 KHz.5 ExerciseEquation Section (Next) What is the frequency in THz of a typical CO2 laser (used in laser surgery) having the far infrared wavelength of 20 µm? Solution Using eq. 10 MHz. (2.S. Ramaccia and A.45 GHz? Solution Using eq.43 GHz The wavelength in cm of the typical microwave oven frequency of 2.45 × 109 D.1).030 m = 30 mm f 10 × 1012 3 × 108 21 × 10−2 λ(10KHz) = λ(10MHz) = λ(10GHz) = The frequency in GHz of the 21–cm hydrogen line observed in the cosmos is: f= c λ = = 1. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.122 m = 12.4. and 10 GHz? What is the frequency in GHz of the 21–cm hydrogen line observed in the cosmos? What is the wavelength in cm of the typical microwave oven frequency of 2. we have: c 3 × 108 = = 30000 m = 30 Km f 10 × 103 c 3 × 108 = = 30 m f 10 × 106 c 3 × 108 = = 0. 11 .J. 12 . So H(z.1) The cross–product in the left–hand side of eq.7.) By inserting E ( z. Toscano Pag. Determine also the magnetic field H ( z. (2. determine the k–ω relationship as a consequence of these equations. Assuming valid the constitutive relation B = μ H : ˆ ∇ × xE 0e jω t − jkz = − μ ∂H ∂t (2.7. but you don't yet know the relationship between k and ω (you may assume they are both positive. t) has to depend by z and t in the same way as E ( z. t ) into Maxwell's equations. t ) in the first Maxwell's equation.7 ExerciseEquation Section (Next) ˆ Suppose you start with E ( z. Repeat the problem if ˆ E ( z. t ) = yE 0e jω t − jkz . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. t ) = xE 0e jω t + jkz and ˆ E ( z. Ramaccia and A.1) can be expanded as follow: ˆ yjkE 0e jω t − jkz = − μ ∂H ∂t It is easy to note that H (z. t) = yH0e jω t − jkz : D. t ) because the ˆ variables t and z are presented only in the exponential. Solution Consider the source–free Maxwell's equations: ∂B ⎧ ⎪ ∇ × E = − ∂t ⎪ ⎨ ⎪ ∇ × H = ∂D ⎪ ∂t ⎩ and now substitute the expression of E ( z.J. t ) = xE 0e jω t − jkz . t ) and verify that all Maxwell's equations are satisfied.S. Ramaccia and A.2) Assuming valid the constitutive relation D = ε E . t) in the second Maxwell's equation: ˆ ˆ ∇ × yH0e jω t − jkz = ε xE 0 ∂e jω t − jkz ∂t (2.S.2) and (2.J. Toscano Pag.3) and the derivate in the right–hand side can be expanded as follow: ˆ x ∂ ∂ ˆ ˆ H 0e jω t − jkz + z H 0e jω t − jkz = jωε xE 0e jω t − jkz ∂z ∂x ( ) ( ) which gives: kH0 = −ωε E0 (2. Orfanidis – Electromagnetic Waves and Antennas ˆ ˆ yjkE 0 e jω t − jkz = − μ yH 0 ∂ jω t − jkz e ∂t Exercises Chapter 2 ˆ ˆ yj kE 0 e jω t − jkz = − μ y H 0 jω e jω t − jkz which gives: kE0 = −ωμ H0 (2. substitute the expression of E ( z.7.3) The cross–product in the left–hand side of eq.4) we can find the k–ω relation: ⎧ kE 0 = −ωμ H 0 ⎨ ⎩ kH 0 = −ωε E 0 k 2 E 0 H 0 = ω 2 με E 0 H 0 and.7.4) Thanks to the relationships (2.7.7.7. 13 . consequently: k = ±ω με (2. t ) and H (z. (2.7.7.5) D. S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.8 ExerciseEquation Section (Next) Determine the polarization types of the following waves, and indicate the direction, if linear, and sense of rotation, if circular or elliptic: ⎡a. ⎢ ⎢ b. ⎢ ⎢c. ⎢ ⎢d. ⎣ ˆ ˆ E = E 0 ( x + y ) e− jkz ˆ ˆ E = E 0 x − 3y e − jkz ˆ ˆ E = E 0 ( jx + y ) e− jkz ˆ ˆ E = E 0 ( x − 2 jy ) e − jkz ⎡e. ⎢ ⎢f . ⎢ ⎢g. ⎢ ⎢ h. ⎣ ˆ ˆ E = E 0 ( x − y ) e− jkz E = E0 ( ) ( ˆ ˆ 3x − y e − jkz ) ˆ ˆ E = E 0 ( jx − y ) e jkz ˆ ˆ E = E 0 ( x + 2 jy ) e jkz (2.8.1) Solution The polarization of a plane wave is defined to be the direction of the time–varying real–valued field E ( z, t ) = ℜe ⎡E ( z ) e jω t ⎤ where E ( z ) = E0e ± jkz . At any fixed point z, the vector E ( z, t ) may ⎣ ⎦ be along a fixed linear direction or it may be rotating as a function of t, tracing a circle or an ellipse. Consider the following expression for the electric field: ˆ ˆ E ( z, t ) = xA x e jφx + yA ye ( jφy )e jω t ± jkz ˆ = xA x e ( j ω t ± jkz+φx ) ˆ + yA y e j ω t ± jkz+φy ( ) (2.8.2) where Ax and Ay are real–positive quantities. Extracting the real part for each component, we find the corresponding real–valued x,y components: E x ( z, t ) = A x cos (ω t ± kz + φx ) E y ( z, t ) = A y cos ω t ± kz + φ y ( ) (2.8.3) The sign of kz is defined by the direction of propagation of the wave: forward–moving fields have the minus sign, e.g. −kz , backward–moving fields the plus sign, e.g. +kz . In order to determine the polarization type of the waves, we consider the time–dependence of these fields at some fixed point along z–axis. For convenience we choose z = 0 : E x ( z, t ) = A x cos (ω t + φx ) E y ( z, t ) = A y cos ω t + φ y ( ) (2.8.4) The parameters Ax, Ay, φx , φy allow us to determine the type of polarization: Linear polarization ( φa = φb = 0 or φa = 0 , φb = −π ): the two components Ex , Ey are in phase and the electric field vector oscillates along a straight line. It is of interest the direction, D. Ramaccia and A. Toscano Pag. 14 S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 respect the x–axis, along which the electric field oscillates with angular frequency ω . It is directly related to the amplitudes of the components Ex , Ey : ϑ = arctan ⎜ Ey slope Ay/Ax ⎛ Ay ⎞ ⎟ ⎝ Ax ⎠ (2.8.5) Ex Ex Ey slope -Ay/Ax Fig. 2.8.1:Directions along the electric field oscillates in linear polarization. Elliptical polarization ( φx − φy = ± π 2 ): Ex and Ey have different amplitudes and are in quadrature phase because one is always 90° out of phase respect to other. Circular polarization ( A x = A y and φx − φy = ± π 2 ): this is a particular case of elliptical polarization when the amplitudes of the components are equal. The sign of the relative phase φ = φx − φy suggests the sense of rotation: counter–clockwise ( φ = − π 2 ) and clockwise ( φ = + π 2 ) and consequently, according to the direction of propagation, left or right elliptical polarization (or circular in particular cases). E x ( t ) = A cos ω t E y ( t ) = A cos (ω t + π 2 ) = A sin ω t Fig. 2.8.2: Counter–clockwise rotation of the electric field vector. E x ( t ) = A cos ω t E y ( t ) = A cos (ω t − π 2 ) = −A sin ω t D. Ramaccia and A. Toscano Pag. 15 S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Ex(t) ωt Ey(t) Fig. 2.8.3: Clockwise rotation of the electric field vector. To decide whether this represents a right or left polarization, we use the IEEE convention. Curl the fingers of your left and right hands into a fist and point both thumbs towards the direction of propagation. If the fingers of your right (left) hand are curling in the direction of rotation of the electric field, then the polarization is right (left) polarized. y E right–polarized x z E y left–polarized x z y E -z left–polarized E -z y right–polarized x x Fig. 2.8.4: Left and right circular polarizations. ˆ ˆ Let us solve the exercise for the case (a): E ( z ) = E 0 ( x + y ) e − jkz . First of all we have to express the field in its real–valued form in z = 0 , in order to obtain an expression similar to eq. (2.8.4): ˆ ˆ E ( t, z ) = E 0ℜe ⎡( x + y ) e ( ⎢ ⎣ so j ω t − kz ) ⎤ ⎥ ⎦ ˆ ˆ = E 0 x cos (ω t − kz ) + E 0 y cos (ω t − kz ) E x ( t, z ) = E0 cos (ω t − kz ) E y ( t, z ) = E0 cos (ω t − kz ) D. Ramaccia and A. Toscano Pag. 16 S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 It is easy to note that it is a forward–moving wave, because of the term −kz , and linear polarized, being φx = φy = 0 . Using eq. (2.8.5), the direction ϑ of electric field vector is 45°. ˆ ˆ On the contrary, let us solve the exercise for the case (c): E = E 0 ( jx + y ) e − jkz . Its real–valued form in z = 0 is: ˆ ˆ j ω t −kz ) ⎤ = E 0ℜe ⎡ xe jπ 2 + y e j(ω t −kz ) ⎤ = ˆ ˆ E ( t, z ) = E 0ℜe ⎡( jx + y ) e ( ⎢ ⎥ ⎢ ⎥ ⎣ ⎦ ⎣ ⎦ ˆ ˆ = E 0 x cos (ω t − kz + π 2 ) + E 0 y cos (ω t − kz ) ( ) so E x ( t, z ) = E0 cos (ω t − kz + π 2 ) E y ( t, z ) = E0 cos (ω t − kz ) The wave is still forward–moving, but the relative phase φ = φx − φy = π 2 , so it is in general elliptical polarized. In this case E x = E y , i.e. A x = A y , then it is circular polarized. According to Fig. 2.8.2 and Fig. 2.8.3, the sense of rotation is counterclockwise. Now we apply the IEEE convention and the find that the field is right–circular polarized. Table 2.8.1 contains the results of the exercise for each given electric field: # a b c d e f g h Expression Polarization Type Linear Linear Circular Elliptical Linear Linear Circular Elliptical Direction/ Sense of Rotation 45° -60° Counter–clockwise Counter–clockwise -45° -30° Clockwise Clockwise ˆ ˆ E = E 0 ( x + y ) e − jkz ˆ ˆ E = E 0 x − 3y e − jkz ( ) ˆ ˆ E = E 0 ( jx + y ) e − jkz ˆ ˆ E = E 0 ( x − 2 jy ) e − jkz ˆ ˆ E = E 0 ( x − y ) e − jkz E = E0 ( ˆ ˆ 3x − y e − jkz ) ˆ ˆ E = E 0 ( jx − y ) e jkz ˆ ˆ E = E 0 ( x + 2 jy ) e jkz Table 2.8.1: Results of exercise n° 2.8. D. Ramaccia and A. Toscano Pag. 17 S.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.9 ExerciseEquation Section (Next) A uniform plane wave, propagating in the z–direction in vacuum, has the following electric field: ˆ ˆ E ( z, t ) = 2xCos (ω t − kz ) + 4ySin (ω t − kz ) (2.9.1) 1. Determine the vector phasor representing E ( z, t ) in the complex form E = E0e jω t − jkz . 2. Determine the polarization of this electric field (linear, circular, elliptic, left–handed, right–handed). 3. Determine the magnetic field H ( z, t ) in its real–valued form. Solution • Question n°1 First of all, we need to manipulate (2.9.1) in order to obtain an expression with components similar to: E x ( z, t ) = A x cos (ω t ± kz + φx ) E y ( z, t ) = A y cos ω t ± kz + φy ( ) (2.9.2) So we can write: E x ( z, t ) = 2cos (ω t − kz ) E y ( z, t ) = 4cos (ω t − kz − π 2 ) from which, we can obtain the complex–valued electric field: ˆ E ( z, t ) = x 2e ( j ω t − jkz ) (2.9.3) ˆ + y 4e ( j ω t − jkz −π 2 ) ˆ ˆ = 2x + 4ye − jπ ( 2 ) e jωt− jkz = ˆ ˆ = ( 2x − j4y ) e ( j ω t − kz ) • Question n°2 In (2.9.3) it is easy to note that the relative phase φ = φx − φy = π 2 , so according to Fig. 2.8.2 and Fig. 2.8.3, the sense of rotation is counter–clockwise. The field is forward–moving and, using the IEEE convention, the field is right–elliptical polarized, being E x ≠ E y . • Question n°3 Using the relation: ˆ E = Z0 ( H × z ) where Z0 is the characteristic impedance of vacuum, we find: D. Ramaccia and A. Toscano Pag. 18 S. t ) x + E y ( z. Ramaccia and A.J. 19 . t ) = z × E x ( z. t ) x = Z0 1 ˆ ˆ ( 2 cos (ω t − kz ) y − 4 cos (ω t − kz − π 2 ) x ) Z0 ( ) ( ) D. t ) y − E y ( z. Toscano Pag. Orfanidis – Electromagnetic Waves and Antennas H ( z. t ) = = = Exercises Chapter 2 1 1 ˆ ˆ ˆ ˆ z × E ( z. t ) y = Z0 Z0 1 ˆ ˆ E x ( z. right.10. we need to manipulate (2.10 ExerciseEquation Section (Next) A uniform plane wave propagating in vacuum along the z–direction has real–valued electric field components: E x = cos (ω t − kz ) . Ramaccia and A. E y = 2sin (ω t − kz ) (2. t ) = xe ( j ω t − jkz ) ˆ + y 2e ( j ω t − kz ) j ω t − jkz −π 2 ) ˆ ˆ = x + 2ye − jπ ( 2 ) e jωt− jkz = ˆ ˆ = ( x − j2y ) e ( • Question n°2 The polarization of the wave is elliptical because the module of x.). linear. t ) = 2cos (ω t − kz − π 2 ) from which.1) in order to obtain an expression with components similar to (2. It is also right polarized because the relative phase φ = φx − φy = π 2 . y components are different. Determine the numerical values of the complex–valued coefficients A.2): E x ( z.S. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. Solution • Question n°1 First of all.J. 2.10.1) ˆ ˆ 1. Toscano Pag. 20 . Determine the polarization of this wave (left. etc. we can obtain the complex–valued electric field: ˆ E ( z. Explain your reasoning. D. t ) = cos (ω t − kz ) E y ( z. B and the correct sign of the exponent. Its phasor form has the form E = ( Ax + By ) e ± jkz .9. etc. Ramaccia and A.3) D. the sense of rotation is clockwise. z.3. E ( z. (2. For the case (a). we have to write E ( z ) in its real–valued form as a function of t.11.8.S. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.2 and Fig.) and the Solution • Case (a) First of all. 21 .J. Toscano Pag. one given in its real–valued form. • Case (b) In this case. using the IEEE convention. circular. E ( z ) = ⎡(1 + j) x − (1 − j) y ⎤ e − jkz ⎣ direction of propagation. 2. 2. t ) = 2cos (ω t + kz ) from which. determine the field in its real–valued form as a function of t.11 ExerciseEquation Section (Next) Consider the two electric fields.11. we rewrite the first field of (2. so according to Fig.2) ˆ + y 2e ( j ω t + jkz ) ˆ = xe − jπ ( 2 ˆ + 2y e jω t + jkz = ) ˆ ˆ = ( 2 y − jx ) e ( j ω t + kz ) In (2.1) as follow: E x ( z. left. t ) = xe ( j ω t + jkz −π 2 ) (2. determine the field in its phasors form. and the other.8. For the case (b). z. we can obtain the complex–valued electric field: ˆ E ( z.11.11. The field is backward–moving and. t ) = ℜ e ⎡( (1 + j) x − (1 − j) y ) e ( ⎢ ⎥ ⎣ ⎦ ˆ ˆ j ω t −kz ) ⎤ = = ℜ e ⎡( (1 + j) x − (1 − j) y ) e ( ⎢ ⎥ ⎣ ⎦ ˆ ˆ = ℜ e ⎡( (1 + j) x − (1 − j) y ) ( cos (ω t − kz ) + jsin (ω t − kz ) ) ⎤ = ⎣ ⎦ ˆ ˆ ˆ ˆ = ( x − y ) cos (ω t − kz ) − ( x + y ) sin (ω t − kz ) = ˆ ˆ ˆ ˆ = ( x − y ) cos (ω t − kz ) − ( x + y ) cos (ω t − kz − π 2 ) = ˆ ˆ = x ( cos (ω t − kz ) − cos (ω t − kz − π 2 ) ) − y ( cos (ω t − kz ) + cos (ω t − kz − π 2 ) ) (2.2) it is easy to note that the relative phase φ = φx − φy = −π 2 . t ) = cos (ω t + kz − π 2 ) E y ( z. t ) = x sin (ω t + kz ) + 2y cos (ω t + kz ) ˆ ˆ⎦ b. right. the field is right–elliptical polarized ( E x ≠ E y ). in its phasor form: ˆ ˆ a.11. So: ˆ ˆ j ω t −kz ) ⎤ = E ( z. determine the polarization of the wave (linear.1) For both cases. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 It is necessary to apply to (2. we obtain: ˆ ˆ E ( z. The field is forward–moving and. the sense of rotation is clockwise.8. the field is left–circular polarized ( E x = E y ).2 and Fig. t ) = x ( cos (ω t − kz ) − cos (ω t − kz − π 2 ) ) − y ( cos (ω t − kz ) + cos (ω t − kz − π 2 ) ) = ˆ ˆ = x ( cos (ω t − kz ) + cos (ω t − kz + π 2 ) ) − y ( cos (ω t − kz ) + cos (ω t − kz − π 2 ) ) ˆ ˆ = 2x ( cos (ω t − kz + π 4 ) cos ( − π 4 ) ) − 2y ( cos (ω t − kz − π 4 ) cos (π 4 ) ) = ˆ ˆ = 2x cos (ω t − kz + π 4 ) − 2y cos (ω t − kz − π 4 ) (2.5) The electric field components are: E x ( z.3) the sum–to–product identity or Prosthaphaeresis formula: cos α + cos β = 2 cos α +β 2 cos α −β 2 (2. so according to Fig.6).3.4) and. 2.6) Using (2. because cos ( x ) = − cos ( x ± π ) . we obtain: 1 + j = 2e jπ 4 − (1 − j) = 2e j3π 4 and.11. consequently E ( z. t ) = ℜ e ⎡ ⎢ ⎣ ( ˆ ˆ j ω t − kz ) ⎤ = 2e jπ 4 x + 2e j3π 4 y e ( ⎥ ⎦ ) ˆ ˆ j ω t − kz ) ⎤ = = 2ℜ e ⎡ e jπ 4 x + e j3π 4 y e ( ⎢ ⎥ ⎣ ⎦ ˆ ˆ = 2xcos (ω t − kz + π 4 ) + 2y cos (ω t − kz + 3π 4 ) ( ) D. using the IEEE convention.8.J. the exercise can be solved writing the complex amplitude of each component in the form: 2 2 ⎧ ⎪ a + jb = Me jϕ where ⎨M = a + b ⎪ϕ = arctan b a ⎩ (2.S. t ) = 2 cos (ω t − kz + π 4 ) E y ( z.11. t ) = − 2 cos (ω t − kz − π 4 ) = 2 cos (ω t − kz + 3π 4 ) and it is easy to note that the relative phase φ = φx − φy = − π 2 .11. In similar way.11. 22 . Toscano Pag.11. 2. Ramaccia and A. 2) ˆ j ω t −kz −π 4 ) + ye j(ω t −kz −π 4 ) = xe− jπ ˆ ˆ E ( z. t ) = cos (ω t − kz − π 4) from which.12. Determine the corresponding magnetic field H ( t. D.J. etc. right. 23 . z ) = z × ⎡ x cos (ω t − kz − π 4 ) + y sin (ω t − kz + π 4 ) ⎤ = ⎦ Z0 Z0 1 ˆ ˆ ⎡ y cos (ω t − kz − π 4 ) − x sin (ω t − kz + π 4 ) ⎤ ⎦ Z0 ⎣ In (2.) of this wave. z ) given in its real–valued form. z ) = x cos (ω t − kz − π 4 ) + y sin (ω t − kz + π 4 ) 1. we rewrite (2. (2.8. Determine the polarization type (left.S. (2.12. t ) = cos (ω t − kz − π 4) E y ( z. Ramaccia and A. we can obtain the complex form as eq. 2.2) ( 4 ˆ + ye− jπ 4 ) e jωt− jkz = Using the relation ˆ E = Z0 ( H × z ) where Z0 is the characteristic impedance of vacuum.12.12 ExerciseEquation Section (Next) A uniform plane wave propagating in the z–direction ha the following real–valued electric field: ˆ ˆ E ( t.2) it is easy to note that the relative phase φ = φx − φy = 0 . linear. Determine the complex–phasor form of this electric field. z ) = = • Question n°3 1 1 ˆ ˆ ⎣ˆ ˆ z × E ( t.12. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. tilted by 45° with respect to the x–axis.1) as follow: E x ( z. Toscano Pag. so the wave is linear polarized. 3. t ) = xe ( = • Question n°2 1 ˆ ˆ j ω t −kz ) (1 − j) ( x + y ) e ( 2 (2. we find: H ( t.1) Solution • Question n°1 First of all. so it is inside the interval [ 0. the sense of rotation is counter–clockwise. using the IEEE convention.13 ExerciseEquation Section (Next) Determine the polarization type (left. In this case the electric field doesn't propagate along the z direction.8. π 2] and. Toscano ( ˆ 2 − 2y sin ω t − k ( x + z ) ) ( 2 ) 2 ⎤= ⎥ ⎦ )) Pag. 2. the field is left–circular polarized.J.) and the direction of propagation of the following electric fields given in their phasor form: a) b) c) ˆ ˆ E ( z ) = ⎡ 1 + j 3 x + 2y ⎤ e jkz ⎣ ⎦ ( ) ˆ ˆ⎦ E ( z ) = ⎡(1 + j) x − (1 − j) y ⎤ e− jkz ⎣ ˆ ˆ ˆ − jk x + z E ( z ) = ⎡ x − z + j 2y ⎤ e ( ) ⎣ ⎦ 2 Solution • Case (a) We have to writing the complex amplitude of each component Ex . The field is backward–moving and. Ramaccia and A. etc.3. Ey in the form Ae jϕ . So we have to identify a new coordinate system in order to apply the well–known steps to solve the problem. Express the electric field in the real–valued form: − jk x + z 2⎤ ˆ ˆ ˆ E ( t. linear.S.6): (1 + j 3 ) = so: (1)2 + ( 3 e ) 2 jarctan ( 31 ) = 2e jπ 3 ˆ E ( z ) = 2 ⎡ xe jπ ⎣ 3 ˆ + y ⎤ e jkz ⎦ The relative phase φ = φx − φy = π 3 . z ) = ℜe ⎡ x − z + j 2y e jω t e ( ) ⎢ ⎥= ⎣ ⎦ ˆ ˆ ˆ = ℜe ⎡ x − z + j 2y cos ω t − k ( x + z ) 2 + jsin ω t − k ( x + z ) ⎢ ⎣ ( ( ) )( ( ) ( ˆ ˆ = ( x − z ) cos ω t − k ( x + z ) D.11. right. 24 .11.8. but it is tilted with respect to the z axis and lies on the z-x plane. according to Fig. • • Case (b) Case (c) See case (b) of exercise n° 2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. being E x = E y .2 and Fig. using (2. 2. Ramaccia and A. y.2 and Fig. 2.1: Rotation of the coordinate system.3. 2.8. z z 45° x x Fig. so according to Fig. D. the sense of rotation is clockwise.S. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 This wave does not propagate along the z–direction. Toscano Pag. z ') where the y axis is the same because E ( t. y. The field is forward–moving and .13. being E x = E y . the field is left–circular polarized .8. 25 . z) → (x '. z ) = 2x 'cos (ω t − kz ') − 2y sin (ω t − kz ' ) = ˆ ˆ = 2x 'cos (ω t − kz ' ) − 2y cos (ω t − kz '− π 2 ) = ˆ ˆ = 2x 'cos (ω t − kz ' ) + 2y cos (ω t − kz '+ π 2 ) = The relative phase φ = φx − φy = −π 2 . z ) has constant phase for any y. using the IEEE convention. The expression of x' and z' are given by: ⎧x + z ⎪ 2 = z' ⎪ ⎨ ⎪x − z = x' ⎪ 2 ⎩ z '− x ' ⎧ ⎪z = ⎪ 2 ⎨ ⎪ x = x '+ z ' ⎪ 2 ⎩ ⇒ and we can rewrite the electric field in the new coordinate system: ˆ ˆ E ( t. and consequently the plane with constant phase are not identified for any constant value of z. So we have to apply a change of coordinate system (x. 2.J. A z determinant of the following matrix: ˆ x A × B = Ax Bx ˆ y Ay By ˆ z Az Bz ( ) and B = B x . b .14. B y . z ) = Ax cos (ω t − kz + φa ) + By cos (ω t − kz + φb ) Show that: (2.14 ExerciseEquation Section (Next) Consider a forward–moving wave in its real–valued form: ˆ ˆ E ( t. Ramaccia and A.14. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.14.5) D. and we have: ˆ Det ( M ) = z ⎡ABcos ϑ Δb cos ϑ a − ABcos ϑ Δa cos ϑ b ⎤ = ⎣ ⎦ ˆ = zAB ⎡cos ϑ Δb cos ϑ a − cosϑ Δa cos ϑ b ⎤ ⎣ ⎦ cos (θ − ϕ ) + cos (θ + ϕ ) 2 (2.3).4) can be written as: (2.14. z ) = A cos ϑ Δa A cos ϑ a ˆ y Bcos ϑ Δb Bcos ϑ b ˆ z 0 = Det ( M ) 0 where ϑ Δi = ω ( t + Δt ) − k ( z + Δz ) + φi and ϑ i = ω t − kz + φi with i = a. z ) = ABz sin (φa − φb ) sin (ωΔt − kΔz ) (2. consequently. Toscano Pag. z + Δz ) × E ( t. A y . we can write: ˆ x E ( t + Δt.14.3) Using (2. (2.4) The expression inside the brackets can be simplified using the product–to–sum identity for cosine: cos θ cos ϕ = and.2) Solution The cross product of two vectors A = A x .14.J.1) ˆ E ( t + Δt. z + Δz ) × E ( t.14. 26 .S. B z ( ) is defined as the (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 ˆ Det ( M ) = zAB ⎡ cos ϑ Δb cos ϑ a − cos ϑ Δa cos ϑ b ⎤ = ⎣ ⎦ ⎡ 1 ⎤ ⎢ + 2 cos (ωΔt − kΔz + φb − φa ) + cos ( 2ω t − 2kz + ωΔt − kΔz + φb + φa ) + ⎥ ˆ = zAB ⎢ ⎥= ⎢ − 1 cos (ωΔt − kΔz + φ − φ ) + cos ( 2ω t − 2kz + ωΔt − kΔz + φ + φ ) ⎥ a b b a ⎢ 2 ⎥ ⎣ ⎦ 1 ˆ ⎡ = z AB ⎣ cos (ωΔt − kΔz + φb − φa ) − cos (ωΔt − kΔz + φa − φb ) ⎤ ⎦ 2 ( ( ) ) The expression inside the brackets can be still simplified using now the product–to– sum identity for sine: sin θ sin ϕ = and. consequently.6) 1 ˆ Det ( M ) = z AB ⎡cos (ωΔt − kΔz + φb − φa ) − cos (ωΔt − kΔz + φa − φb ) ⎤ = ⎣ ⎦ 2 1 ⎡ ⎤ ˆ = z AB ⎣cos (ωΔt − kΔz − (φa − φb ) ) − cos (ωΔt − kΔz + (φa − φb ) ) ⎦ = (2.J. we have: cos (θ − ϕ ) − cos (θ + ϕ ) 2 (2. Toscano Pag. Ramaccia and A.14.14.S.7) 2 ˆ = zABsin (ωΔt − kΔz ) sin (φa − φb ) D. 27 . show that the semi–axes A'. B' are given by the equations: A = ′2 A 2 − B2τ 2 1−τ 2 . 28 .6) Then.15.4) Then show that the following condition is satisfied. where τ = tan θ : ( A2 − B2τ 2 )( B2 − A2τ 2 ) = A2B2 sin2 φ 2 (1 −τ 2 ) Using this property.S.3) 1 ⎥ B′2 ⎥ ⎦ Show that the required angle θ is given by tan 2θ = 2AB A 2 − B2 cos φ (2. one must determine the tilt angle θ such that the following matrix condition is satisfied: ⎡ cos θ ⎢ − sin θ ⎣ ⎡ 1 sin θ ⎤ ⎢ A 2 ⋅⎢ cos θ ⎥ ⎢ cos φ ⎦ − ⎢ AB ⎣ − cos φ ⎤ AB ⎥ ⎡ cos θ ⎥⋅ 1 ⎥ ⎢ sin θ ⎣ 2 ⎥ B ⎦ ⎡ 1 ⎢ ′2 − sin θ ⎤ A = sin 2 φ ⎢ cos θ ⎥ ⎢ ⎦ ⎢ 0 ⎣ ⎤ 0 ⎥ ⎥ (2. 2.15.1) to be equivalent to the rotated one with components E′ = E x cos θ + E y sin θ x E′y = E y cos θ − E x sin θ (2.15.15.J.15. B = ′2 B2 − A 2τ 2 1 −τ 2 (2.15. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.15 ExerciseEquation Section (Next) Show that in order for the polarization ellipse E2 x A 2 + E2 y B 2 −2 ExEy AB cos φ = sin 2 φ (2. Toscano Pag.15.1: General polarization ellipse. Ramaccia and A.5) (2.2) Fig. transform these equations into the form: D. Finally. A ′B′ = AB sin φ (2. As suggested from Fig. show that A'.11) Eq.15. (2.7) ( ) 2 + 4A 2 B2 cos 2 φ where s = sign(A − B) . 2. it is possible to obtain its inverse: ⎡ E′ ⎤ ⎡ cos θ x ⎢ E′ ⎥ = ⎢ ⎣ y ⎦ ⎣ − sin θ (2.15.1 the minor axis is B′ and the major axis is A′ and the equation of the tilted ellipse can be rewritten as: ⎡ 1 ⎢ 2 ⎡ E′x E′y ⎤ ⋅ ⎢ A′ ⎣ ⎦ ⎢ ⎢ 0 ⎣ ⎤ 0 ⎥ ⎡ E′ ⎤ x ⎥⋅⎢ ⎥ =1 1 ⎥ ⎣ E′y ⎦ ⎦ B′2 ⎥ (2. The ellipse is not rotated with respect to the axes E′ .1.9) and.11) represents the tilted ellipse shown in Fig. E′y and it is possible to define new values of the minor and major axis in x this rotated coordinate system. 29 .15. we can substitute ⎡ 1 sin θ ⎤ ⎢ A 2 ⋅⎢ cos θ ⎥ ⎢ cos φ ⎦ − ⎢ AB ⎣ cos φ ⎤ AB ⎥ ⎡cos θ ⎥⋅ 1 ⎥ ⎢ sin θ ⎣ 2 ⎥ B ⎦ − − sin θ ⎤ ⎡ E′x ⎤ 2 ⎥ ⋅ ⎢ E′ ⎥ = sin φ cos θ ⎦ ⎣ y ⎦ (2.S.15.15.15. B' satisfy the relationships: A′2 + B′2 = A 2 + B2 .9): ⎡ cos θ ⎡ E′x E′y ⎤ ⋅ ⎢ ⎣ ⎦ − sin θ ⎣ sin θ ⎤ ⎡ E x ⎤ ⎥⋅⎢ ⎥ cos θ ⎦ ⎣ E y ⎦ ⇒ ⎡ E x ⎤ ⎡ cos θ ⎢E ⎥ = ⎢ ⎣ y ⎦ ⎣ sin θ − sin θ ⎤ ⎡ E′ ⎤ x ⎥ ⋅ ⎢ E′ ⎥ (2.15. 2. from the matrix form of (2.15.J.1) can be written in matrix form as follow: ⎡ 1 ⎢ 2 ⎡Ex E y ⎤ ⋅ ⎢ A ⎣ ⎦ ⎢ cos φ ⎢ − AB ⎣ − cos φ ⎤ AB ⎥ ⎡ E x ⎤ ⎥ ⋅ ⎢ ⎥ = sin 2 φ 1 ⎥ ⎣E y ⎦ ⎦ B2 ⎥ (2.15.10) into (2.10) cos θ ⎦ ⎣ y ⎦ Noting that the first vector in (2.2).12) Multiplying left and right side of (2.8) Solution The polarization ellipse in eq.15.15.9) is the transposed of the one in (2.10). Ramaccia and A. (2. Toscano Pag.15. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2 A′ = B′ = 1 2 s A + B2 + 2 2 1 2 s A + B2 − 2 2 ( ) ( A 2 + B2 ) ( A 2 + B2 ) + 4A 2 B2 cos 2 φ (2.15.15.15.15.12) by sin 2 φ D. 15.15.15.17) Eq.15.15.14) 1 ⎥ ⎦ B′2 ⎥ The relationship (2. it can be manipulated in order to demonstrate eq.16) Let us now divide both side of (2.13) and comparing eq.15) can be written as: ⎡ 1 ⎢ 2 ⎢ A ⎢ cos φ ⎢ − AB ⎣ − cos φ ⎤ AB ⎥ ⎡ cos θ ⎥⋅ 1 ⎥ ⎢ sin θ ⎣ 2 ⎥ B ⎦ ⎡ 1 ⎢ 2 ⎢ A ⎢ cos φ ⎢ − AB ⎣ − sin θ ⎤ 2 ⎡ cos θ ⎥ = sin φ ⎢ sin θ cos θ ⎦ ⎣ (2.S. (2. Ramaccia and A. (2.11) and eq.15.4).15. defining τ = tan θ .15.17) represents the following linear system: D. (2.J. (2. Firstly. 30 . we note they are equal if and only if: ⎡ cos θ ⎢ − sin θ ⎣ − cos φ ⎤ AB ⎥ ⎡ cos θ ⎥⋅ 1 ⎥ ⎢ sin θ ⎣ 2 ⎥ B ⎦ ⎤ 0 ⎥ ⎥ (2. we get: − cos φ ⎤ ⎡ 1 ⎥ ⎡1 −τ ⎤ ⎡1 −τ ⎤ ⎢ A′2 AB ⎥⋅⎢ = sin 2 φ ⎢ ⎥⋅⎢ 1 ⎥ ⎣τ 1 ⎥ ⎦ ⎣τ 1 ⎦ ⎢ 0 ⎢ ⎦ ⎣ B2 ⎥ ⎤ 0 ⎥ ⎥ 1 ⎥ ⎦ B′2 ⎥ (2.16) by cos θ and. Toscano Pag.15. (2.15.15) ⎡ cos θ = sin 2 φ ⎢ ⎣ − sin θ Since: ⎡ cos θ ⎢ − sin θ ⎣ sin θ ⎤ cos θ ⎥ ⎦ −1 ⎡ 1 −1 ⎢ 2 sin θ ⎤ A′ ⎥ ⋅⎢ cos θ ⎦ ⎢ ⎢ 0 ⎣ ⎡ cos θ =⎢ ⎣ sin θ − sin θ ⎤ cos θ ⎥ ⎦ ⎡ 1 − sin θ ⎤ ⎢ A′2 ⋅⎢ cos θ ⎥ ⎢ ⎦ ⎢ 0 ⎣ ⎤ 0 ⎥ ⎥ 1 ⎥ ⎦ B′2 ⎥ eq.14) is fundamental to solve the whole exercise. Left multiplying both side by: −1 ⎡ cos θ ⎢ − sin θ ⎣ we obtain: ⎡ cos θ ⎢ − sin θ ⎣ sin θ ⎤ cos θ ⎥ ⎦ −1 sin θ ⎤ cos θ ⎥ ⎦ ⎡ cos θ ⋅⎢ ⎣ − sin θ ⎡ 1 sin θ ⎤ ⎢ A 2 ⋅⎢ cos θ ⎥ ⎢ cos φ ⎦ − ⎢ AB ⎣ − cos φ ⎤ AB ⎥ ⎡cos θ ⎥⋅ 1 ⎥ ⎢ sin θ ⎣ 2 ⎥ B ⎦ − sin θ ⎤ = cos θ ⎥ ⎦ ⎤ 0 ⎥ ⎥ 1 ⎥ B′2 ⎥ ⎦ (2.13).15.15. Orfanidis – Electromagnetic Waves and Antennas ⎡ 1 ⎢ ′2 A sin 2 φ ⎡ E′x E′y ⎤ ⋅ ⎢ ⎣ ⎦ ⎢ ⎢ 0 ⎣ ⎡ 1 sin θ ⎤ ⎢ A 2 ⋅⎢ cos θ ⎥ ⎢ cos φ ⎦ − ⎢ AB ⎣ Exercises Chapter 2 ⎤ 0 ⎥ ⎡ E′ ⎤ ⎥ ⋅ ⎢ x ⎥ = sin 2 φ 1 ⎥ ⎣ E′y ⎦ B′2 ⎥ ⎦ ⎡ 1 ⎢ ′2 − sin θ ⎤ = sin 2 φ ⎢ A cos θ ⎥ ⎢ ⎦ ⎢ 0 ⎣ (2. Toscano .15. Ramaccia and A.15.19) (c) (d) It is easy to show that (2.15.18)(c) and (2.15.S.19) (d) are equivalent.18) (c) (d) Substituting (2.15. Orfanidis – Electromagnetic Waves and Antennas ⎧ 1 cos φ sin 2 φ = ⎪ 2 −τ AB A ′2 ⎪A ⎪ 1 cos φ sin 2 φ ⎪ +τ = AB ⎪ B2 B′2 ⎨ sin 2 φ ⎪ cos φ τ − + 2 =τ ⎪ AB B A ′2 ⎪ ⎪ cos φ τ sin 2 φ − − 2 =τ ⎪ B′2 ⎩ AB A ⎧ 1 cos φ sin 2 φ −τ = ⎪ 2 AB A A ′2 ⎪ ⎪ 1 cos φ sin 2 φ ⎪ +τ = ⎪ B2 AB B′2 ⎨ ⎪ − cos φ + τ = τ ⎛ 1 − τ cos φ ⎞ ⎜ 2 ⎪ AB B2 AB ⎟ ⎝A ⎠ ⎪ cos φ ⎞ ⎪ cos φ τ ⎛ 1 ⎪ − AB − 2 = τ ⎜ 2 + τ AB ⎟ A ⎝B ⎠ ⎩ cos φ τ cos φ ⎞ ⎛ 1 + 2 = τ ⎜ 2 −τ AB B AB ⎟ ⎝A ⎠ cos φ τ τ cos φ − + 2 = 2 −τ 2 AB B AB A τ cos φ τ cos φ − − 2 = − 2 −τ 2 AB A AB B − cos φ τ cos φ ⎞ ⎛ 1 − 2 = −τ ⎜ 2 + τ AB A AB ⎟ ⎝B ⎠ Exercises Chapter 2 (a) (b) (2.15.18)(d) leads to: (a) (b) (2. In fact: (c) − ⇒ ⇒ ⇒ (d) Starting from (2. or (d): cos φ τ cos φ ⎞ ⎛ 1 + 2 = τ ⎜ 2 −τ AB B AB ⎟ ⎝A ⎠ cos φ τ τ cos φ − + 2 = 2 −τ 2 AB B AB A τ τ cos φ cos φ − 2 = −τ 2 + AB AB B2 A − ⎛ A 2 − B2 ⎞ cos φ τ⎜ 2 2 ⎟= 1 −τ 2 ⎜ A B ⎟ AB ⎝ ⎠ ⇒ ⇒ ⇒ ( ) Pag.15.18)(b) in (2.J.15.18)(a) and (2.19) (c) and (2. 31 D.15.19) (c). D= ( A 2 − B2 ) 2 + 4A 2 B2 cos 2 φ = ( A 2 + B2 ) 2 − 4A 2 B2 sin 2 φ . The choice between s = 1 and −s = 1 does not matter because both are solution of eq. Orfanidis – Electromagnetic Waves and Antennas ⎛ A 2 B 2 ⎞ cos φ τ AB ⎟ cos φ =⎜ 2 = 2 1 − τ 2 ⎜ A − B2 ⎟ AB A − B2 ⎝ ⎠ Exercises Chapter 2 (2.15. (2.15.15.22) τ1. This choice is according to the fact that in general one defines the major axis with A and the minor axis with B and we have that s = 1 . Toscano Pag.15.15. τ s identifies the two solution for s = ±1 .J.22).2 = ⎛ A 2 − B2 ⎞ −⎜ 2 2 ⎟ ± ⎜ A B ⎟ ⎝ ⎠ ⎛ A 2 − B2 ⎞ cos 2 φ +4 2 2 ⎜ 2 2 ⎟ ⎜ A B ⎟ A B ⎝ ⎠ = cos φ 2 AB 2 ⎛ A 2 − B2 ⎞ 1 −⎜ 2 2 ⎟ ± 2 2 ⎜ A B ⎟ A B ⎠ = ⎝ ( A 2 − B2 ) cos φ 2 AB 2 2 + 4A 2 B2 cos 2 φ = (2.23) = where: • • • B2 − A 2 ± ( A 2 − B2 ) + 4A 2 B2 cos 2 φ 2ABcos φ = B2 − A 2 + sD = τs 2ABcos φ s = ±1 .15.15.19)(c). can be viewed as a quadratic equation in τ : ⎛ A 2 − B2 ⎞ cos φ cos φ +τ ⎜ 2 2 ⎟ − =0 ⎜ A B ⎟ AB AB ⎝ ⎠ 2τ 1−τ 2 tan θ 1 − tan θ 2 =2 τ 1−τ 2 = 2AB A 2 − B2 cos φ (2. we can set the value of s as: s = sign ( A − B) (2. So.20) can be written as tan 2θ = which is the same of eq.15.4). for simplicity. (2. Ramaccia and A. or (d).20) It is known that: tan 2θ = 2 so (2. Eq. 32 .15. (2. D.21) τ2 with its solution given by: (2.S.24) which is the same value set in the text of the exercise. In fact: D.15. useful to simplify the expressions when substituting τ s in eq. τ −s satisfy the identities (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 The values τ s . B′ and A.18) (a) and (b). B. if we substitute them inside (2.15. We want to remark that the direct substitution is not a good choice because it leads to an expression difficult to be managed.19) (a) or (b).15. but easier to be understood.19) (c) and (2. (2.15.19) (d) and so. Ramaccia and A.19) and let us diagonalize the matrix: ⎡ 1 ⎢ 2 ⎢ A ⎢ cos φ ⎢ − AB ⎣ − cos φ ⎤ AB ⎥ ⎥ 1 ⎥ B2 ⎥ ⎦ ⎡ 1 ⎢ ′2 ⎢A ⎢ ⎢ 0 ⎣ ⎤ 0 ⎥ ⎥ 1 ⎥ B′2 ⎥ ⎦ ⇒ diagonalize ⇒ The eigenvalues are the roots of the characteristic polynomial: ⎛⎡ 1 ⎜⎢ 2 Det ⎜ ⎢ A ⎜ ⎢ cos φ ⎜ ⎢ − AB ⎝⎣ that is ⎛ 1 ⎞⎛ 1 ⎞ cos 2 φ − λ ⎟⎜ 2 − λ ⎟ − 2 2 = 0 ⎜ 2 ⎝A ⎠⎝ B ⎠ A B =0 B2 A 2 B2 ⎛ A 2 + B2 ⎞ 1 λ 2 − λ ⎜ 2 2 ⎟ + 2 2 1 − cos 2 φ = 0 ⎜ A B ⎟ A B ⎝ ⎠ A 2 B2 A2 1 − ⇒ ⇒ ⇒ ⇒ − cos φ ⎤ ⎞ ⎥ AB ⎡1 0 ⎤ ⎟ ⎥−λ ⎢ ⎥⎟ = 0 1 ⎥ ⎣0 1 ⎦ ⎟ ⎟ ⎦ B2 ⎥ ⎠ λ − λ + λ2 − cos 2 φ ( ) λ 2 ( A 2 B2 ) − λ ( A 2 + B2 ) + (1 − cos 2 φ ) = 0 λ 2 ( A 2 B2 ) − λ ( A 2 + B2 ) + sin 2 φ = 0 Finally we get: (2. with s = ±1 (2.25) λs ( = A 2 + B2 + s ) ( A 2 + B2 ) 2 − 4A 2 B2 sin 2 φ 2A 2 B2 = A 2 + B2 + sD 2A 2 B2 .26) λ± s are the eigenvalues.S. Let us now stop to consider the linear system (2.15.15. Toscano Pag.15. So it is better to follow a longer way to solve the problem.J. we are able to eliminate φ in the definition of A′ . 33 . Orfanidis – Electromagnetic Waves and Antennas 1 A2 1 Exercises Chapter 2 −τs cos φ 1 B2 − A 2 + sD cos φ 1 B2 − A 2 + sD A 2 + B2 − sD = 2− = 2− = = λ−s AB A AB 2AB cos φ A 2A 2 B2 2A 2 B2 cos φ 1 B2 − A 2 + sD cos φ 1 B2 − A 2 + sD A 2 + B2 + sD = 2− = = λs −τs = 2− AB B AB 2AB cos φ B2 B 2A 2 B2 2A 2 B2 (2.15.15.15.27) or in a more compact form: cos φ ⎧ 1 ⎪ 2 − τ s AB = λ−s ⎪A ⎨ ⎪ 1 − τ cos φ = λ s s ⎪ AB ⎩ B2 (2.30) became: sin 2 φ λs λ−s sin 2 φ λs = sin 2 φ sin φ A 2 B2 2 1 λs = A 2 B2λs = ⎡ A 2 + B2 + sD ⎤ 2⎣ ⎦ B′2 = λs λ−s λ−s = sin 2 φ sin φ A 2 B2 2 λ−s = A 2 B2λ−s 1 = ⎡ A 2 + B2 − sD ⎤ ⎦ 2⎣ (2.15.30) λs λ−s = ( A 2 + B2 ) = ( = A ′2 = A 2 + B2 A 2 + B2 + sD A 2 + B2 − sD ⋅ = 2A 2 B2 2A 2 B2 2 − D2 4A 4 B4 = (2.28) Comparing (2.19) (a) and (2.15. Ramaccia and A.15.29) and consequently ⎧ 2 sin 2 φ sin 2 φ λ = ⎪ A′ = λ−s λs λ−s s ⎪ ⎨ 2 2 ⎪ 2 sin φ sin φ λ B′ = = ⎪ λs λs λ−s −s ⎩ The product λs λ−s is given by: (2.15. 34 .28) with (2.32) D.19) (b). we can write: ⎧ sin 2 φ = λ−s ⎪ ⎪ A ′2 ⎨ 2 ⎪ sin φ ⎪ ′2 = λs ⎩ B (2.15.31) ) ( 2 − A 2 + B2 ) 2 + 4 A 2 B2 sin 2 φ 4 A 4 B4 = sin 2 φ A 2 B2 and consequently the equations (2. Toscano Pag.15.S.15.J. J.(2.15.34) which can be substituted in (2.15. Ramaccia and A.15.15.33) 2 1 −τ s ( A 2 − B2 )τ s ABcos φ = 2 1−τs (2.(2.37) D. Toscano Pag.15. multiplying A′ and B′ .15. (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 giving the expressions for A′ and B′ as required by the exercise [see eq. we obtain: A′2 + B′2 = 2 A 2 − B2τ s 2 1 −τ s 2 B2 − A 2τ s 2 1 −τ s + = = A 2 2 − B2τ s +B 2 2 1−τs 2 − A 2τ s ( A2 + B2 ) (1 − τ s2 ) = A2 + B2 = 2 1 −τ s (2.32): A′2 = A 2 B2λs B′2 = A 2 B2λ−s we can substitute the other value of λ± s as in eq. that is: 1 cos φ ⎧ ⎪λ−s = 2 − τ s AB ⎪ A ⎨ ⎪λ = 1 + τ cos φ ⎪ s B2 s AB ⎩ and we have: cos φ ⎞ ⎛ 1 = A 2 + τ s ABcos φ A ′2 = A 2 B2 ⎜ 2 + τ s AB ⎟ ⎝B ⎠ cos φ ⎞ ⎛ 1 2 B′2 = A 2 B2 ⎜ 2 − τ s ⎟ = B − τ s ABcos φ AB ⎠ ⎝A From eq.35) 2 1 −τs = B 2 2 − A 2τ s 2 1 −τs Eq. Adding A′ and B′ .7)].15. 35 .15. we obtain: A′2 B′2 = A 2 B2 ( ) 2 λs λ−s = A 2 B2 ( ) 2 sin 2 φ A B 2 2 = A 2 B2 sin 2 φ ⇒ A′B′ = AB sin φ (2.21) we have also: 2τ s = 2AB A 2 − B2 cos φ ⇒ (2.15. In the same way.15.33): A = A + τ s ABcos φ = A + τ s ′2 2 2 ( A2 − B2 )τ s = A2 − B2τ s2 (A 2 1−τs 2 B′2 = B2 − τ s ABcos φ = B2 − τ s AB − B τs 2 ) 2 1−τs (2. (2.S.35) gives the expression of A′ and B′ as required by the exercise [see eq.36) and. starting from (2. (2.28).15.15.6)]. 36) and (2.S.37) are equal to eq. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 It is easy to note that eq. 36 .8) in the text of the exercise.15. Toscano Pag.15. (2.15.J. Ramaccia and A. (2. D. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.16. Then prove the more general relationship: (2.S. Ramaccia and A.3) and we have: ˆ⎣ E ( 0 ) × E ( t ) = z ⎡ ABcos φa cos (ω t + φb ) − ABcos φb cos (ω t + φa ) ⎤ = ⎦ ˆ = zAB ⎡cos φa cos (ω t + φb ) − cos φb cos (ω t + φa ) ⎤ ⎣ ⎦ (2. so: ˆ E (0) × E ( t ) = z AB ⎡cos (ω t − φ ) − cos (ω t + φ ) ⎤ ⎦ 2 ⎣ D. show the cross–product equation: ˆ E ( 0 ) × E ( t ) = zABsin φ sin ω t where φ = φa − φb .16.16 ExerciseEquation Section (Next) ˆ ˆ Considering the electric field E ( t ) = xA cos (ω t + φa ) + yBcos (ω t + φb ) . 37 . we can write: ˆ x E ( 0) × E ( t ) = ˆ y ˆ z A cos φa B cos φb 0 A cos (ω t + φa ) B cos (ω t + φb ) 0 (2. i. Toscano Pag.2) Solution Using (2. (2.16.16.4) can be written as: cos (θ − ϕ ) + cos (θ + ϕ ) 2 (2.e.J. cos (α ) = cos ( −α ) .4) The expression inside the brackets can be simplified using the product–to–sum identity for cosine: cos θ cos ϕ = and. (2.16. consequently.16.5) ˆ E ( 0 ) × E ( t ) = zAB ⎡cos φa cos (ω t + φb ) − cos φb cos (ω t + φa ) ⎤ ⎣ ⎦ ⎡ 1 ⎤ ⎢ + 2 cos (φa − ω t − φb ) + cos (φa + ω t + φb ) + ⎥ ˆ = zAB ⎢ ⎥= ⎢ − 1 cos (φ − ω t − φ ) + cos (φ + ω t + φ ) ⎥ b a b a ⎢ 2 ⎥ ⎣ ⎦ AB ˆ ⎡cos (φa − ω t − φb ) − cos (φb − ω t − φa ) ⎤ = =z ⎦ 2 ⎣ AB ˆ =z ⎡cos ( −ω t + φ ) − cos ( −ω t − φ ) ⎤ ⎦ 2 ⎣ ( ( ) ) The cosine is an even function.1) ˆ E ( t1 ) × E ( t 2 ) = zABsin φ sin (ω ( t 2 − t1 ) ) Discuss how linear polarization can be explained with the help of this result.3).14. 38 .16. (2.J.7) The more general relationship (2. we have: cos (θ − ϕ ) − cos (θ + ϕ ) 2 (2.1): ˆ x E ( t1 ) × E ( t 2 ) = Det A cos (ω t1 + φa ) A cos (ω t 2 + φa ) ˆ ˆ y z B cos (ω t1 + φb ) 0 = B cos (ω t 2 + φb ) 0 ˆ = zAB ⎡ cos (ω t1 + φa ) cos (ω t 2 + φb ) − cos (ω t1 + φb ) cos (ω t 2 + φa ) ⎤ = ⎣ ⎦ ⎡ ⎤ AB ⎢ cos (ω t1 + φa − ω t 2 − φb ) + cos (ω t1 + φa + ω t 2 + φb ) + ⎥ ˆ =z = 2 ⎢ − cos (ω t + φ − ω t − φ ) − cos (ω t + φ + ω t + φ ) ⎥ 1 b 2 a 1 b 2 a ⎦ ⎣ AB ⎡ cos (ω ( t1 − t 2 ) + φ ) − cos (ω ( t1 − t 2 ) − φ ) ⎤ = ˆ =z ⎦ 2 ⎣ AB ⎡ cos (ω ( t 2 − t1 ) − φ ) − cos (ω ( t 2 − t1 ) + φ ) ⎤ = ˆ =z ⎦ 2 ⎣ ˆ = zABsin φ sin (ω ( t 2 − t1 ) ) (2. represented by the relative phase φ = φa − φb . consequently. D.16.8) When the electric field is linear polarized.16. the electric field vector. sampled in any t . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 The expression inside the brackets can be still simplified using now the product–to– sum identity for sine: sin θ sin ϕ = and.2) can be proven in the same way of eq.16. so the angle between the two vectors E ( t1 ) and E ( t 2 ) .16. Toscano Pag. Ramaccia and A.S. is always zero and the cross–product is null at any Δt = t 2 − t1 .6) ˆ E ( 0) × E ( t ) = z AB ⎡cos (ω t − φ ) − cos (ω t + φ ) ⎤ = ⎦ 2 ⎣ ˆ = zABsin φ sin ω t (2. is always along a fixed direction. 2 k cηc = ωμ and k c = ω 2 με c for the complex–valued quantities k c .1) where ε c = ε ′ − jε ′′ is the complex value of permittivity of a lossy media and k c = β − jα .4) and extract the real part: ⎡ β − jα ⎤ β − ℜe ⎡ηc 1 ⎤ = ℜe ⎢ ⎥= ⎣ ⎦ ⎣ ωμ ⎦ ωμ D.17.17.S.2) Solution From the first property.3) 1 ηc = ωμ (2. Ramaccia and A.17. ηc of ηc = μ εc (2. let us express the characteristic impedance as: ηc = and invert it: − ηc 1 = ωμ kc kc (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.4) Now it is possible to substitute k c → β − jα in eq.J.17 ExerciseEquation Section (Next) Using the properties equation: k c = ω με c . Toscano Pag.17. show the following relationships: ωε ′′ β − ℜe ⎡ηc 1 ⎤ = = ⎣ ⎦ 2α ωμ (2.17. 39 . (2. 18.18.1): D. starting from its definition in (2.18.18.17. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.4) ′ ′′ where ε d = ε d − jε d is the permittivity of dielectric and σ its conductivity.1) θ θ⎞ −1 2 ′⎛ k c = β − jα = ω με d ⎜ cos − jsin ⎟ ( cos θ ) 2 2⎠ ⎝ μ ηc = η ′ − jη ′′ = d ε′ θ θ⎞ ⎛ 12 ⎜ cos + jsin ⎟ ( cos θ ) 2 2⎠ ⎝ (2.1) and the relationship (2. we can write: k c = ω με c = ω μ ( ε ′ − jε ′′ ) = = ω με ′ (1 − jtan θ ) = ω με ′ (1 − jtan θ ) The complex–valued permittivity ε c is also defined as 12 (2.5) ( ) 12 ( cos θ )−1 2 = ⎛ − jθ ⎞ θ θ ′ ⎜ e 2 ⎟ ( cos θ )−1 2 = ω με d ⎛ cos − jsin ⎞ ( cos θ )−1 2 ′ ⎜ = ω με d ⎟ ⎜ ⎟ 2 2⎠ ⎝ ⎝ ⎠ In the same way it is possible to express ηc as in (2.3) ′ ′′ ε c = ε ′ − jε ′′ = ε d − j ⎜ ε d + ⎛ ⎝ σ⎞ ω⎟ ⎠ (2. Ramaccia and A.17.1).18.2).18.S. So in (2.18.18. Toscano Pag.2) Solution Using the definition of k c in (2. 40 .J.18 ExerciseEquation Section (Next) Show that for a lossy medium the complex–valued quantities k c and ηc may be expressed as follows. in terms of the loss angle θ defined in: τ = tan θ = ′′ ε ′′ σ + ωε d = ′ ε′ ωε d (2.3) we can ′ substitute ε ′ → ε d and tan θ → sin θ cos θ to obtain: sin θ ⎞ ′ ⎛ k c = ω με d ⎜ 1 − j ⎟ cos θ ⎠ ⎝ 12 = 12 cos θ − jsin θ ⎞ ′ ⎛ = ω με d ⎜ ⎟ cos θ ⎝ ⎠ ′ = ω με d ( cos θ − jsin θ ) ′ = ω με d e − jθ = 12 12⎛ 1 ⎞ ⎜ ⎟ ⎝ cos θ ⎠ = (2. 6) μ − jθ e ′ εd ( ) −1 2 ( cos θ )1 2 = μ ′ εd θ θ⎞ ⎛ 12 ⎜ cos + jsin ⎟ ( cos θ ) 2 2⎠ ⎝ D. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 ηc = μ μ μ = = = εc ε ′ − jε ′′ ε ′ − jε ′ tan θ 12 ⎞ μ ⎛ 1 = ⎜ ′ ε d ⎝ 1 − jtan θ ⎟ ⎠ = ⎞ μ ⎛ cos θ = ⎜ ′ ε d ⎝ cos θ − jsin θ ⎟ ⎠ 12 (2. Toscano Pag.J.18.S. Ramaccia and A. 41 . 19.367 2 = −1 = 428. α 2 = 73.19.13 nepers m . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.14 m (2.19. we have: 1 ε c = ( 65 − j25 ) ε 0 = 652 + 252 e jArc tan( −25 65) 2 ε c = ( 75 − j25 ) ε 0 = 752 + 252 e jArc tan( −25 75) 69. respectively. we calculate the wavenumbers: k1 c = β − jα = 51.18 ( cos ( 0. The complex dielectric constants of the mashed potatoes and carrots are ε c = ( 65 − j25) ε 0 and ε c = ( 75 − j25) ε 0 .3) The corresponding attenuation constants and penetration depths are: α 1 = 78.161) − jsin ( 0.1835 ) ) = 421 − j78.322 2 = −1 = 456.31 65 − j25 428.45 GHz is: k 0 = ω μ0ε 0 = 2π f 2π × 2.1835 ) − jsin ( 0.64e −j 0.19 ExerciseEquation Section (Next) It is desired to reheat frozen mesh potatoes and frozen cooked carrots in a microwave oven operating at 2.322 The free–space wave number of a microwave at 2.45 × 109 rad = = 51. δ 1 =1 α 1 = 12.161) ) = 450 − j73.31 79. respectively.14 nepers m .18e 2 kc − j0.8 cm δ 2 =1 α 2 = 13.06e − j0.1) so. Toscano Pag.31 8 c0 m 3 × 10 Using k c = ω μ0ε c = ω μ0ε 0 ( ε c ε 0 ) = ω μ0ε 0 ε c ε 0 = k 0 ε c ε 0 . Determine the penetration depth and assess effectiveness of this heating method. Ramaccia and A. 42 .13 m 51. Solution First of all we have to express the complex–values of the permittivity as follow: 2 2 ⎧ ⎪ a + jb = Me jϕ where ⎨M = a + b ⎪ϕ = arctan b a ⎩ (2. determine the attenuation of the electric field (in dB and absolute units) at a depth of 1 cm from the surface of the food. using the superscripts 1 and 2 to indicate the permittivity of potatoes and carrots.23e − j0.1835 51. Moreover.31 75 − j25 456.J.64e − j0.23 ( cos ( 0.161 0.06e −j (2.2) = β − jα = 51.31 69.7 cm D.S.45 GHz.367 79. J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 This heating method is effective because the penetration depths are bigger than the dimension of mesh potatoes and carrots.68 dB dB 2 AdB ( z = 1 cm ) = 8.4) = 0.686 z δ 1 = 8.8 = 0. D. the fields at a depth of 1 cm are 92. The attenuation of the electric field (in dB and absolute units) at a depth of 1 cm from the surface of the food is: A1 ( z = 1 cm ) = 8.19.686 12.19.686 13.925 = 0.5) 20 Thus.686 z δ 2 = 8. Ramaccia and A. 43 .S.5% and 93% of their values at the surface.930 A1 = 10 − 2 AdB (2. The energy in the electromagnetic waves reheat successfully the foods.63 dB A1 − dB A1 = 10 20 (2. Toscano Pag.7 = 0. 2) α ωμσ The conductibility of the copper is 5. The RF interference inside the enclosure is required to be at least 50 dB down compared to its value outside.686 z δ (2.20.20.5) δ= 8.20. α and δ in a good conductor are: β =α = δ= 1 = ωμσ 2 2 = π fμσ = 1 π fμσ (2. What is the minimum thickness of the copper shield in mm? Solution The parameters β . Toscano Pag. 44 .S.3) where the frequency f is expressed in Hertz. 50 50 0.20.20. Ramaccia and A.20.686 z δ ≥ 50dB ⇒ z≥ (2.1) (2.0661× f −1 2 7 −7 π fμσ π × 4π × 10 × 5.8 × 10 (2.J.0661× f −1 2 = 0.4) and its minimum value over the frequency range of interest is at least 50 dB. The attenuation in dB is: A dB ( z ) = 8.686 8. we have: A dB ( z ) = 8.3805 × f −1 2 (2.20. So inverting the eq.20 ExerciseEquation Section (Next) We wish to shield a piece of equipment from RF interference over the frequency range from 10 kHz to 1 GHz by enclosing it in a copper enclosure. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. so the skin depth at frequency f is: δ= 1 1 = f −1 2 = 0.8 ×107 Siemens/m.5) can be plotted as function of frequency in the range 10 kHz–1 GHz: D.4) with the assumption that AdB ( z ) ≥ 50dB .686 The inequality (2. z =3. D. 2. On the contrary at low frequencies the thickness is more.20.S.1: Thickness of copper shield in mm for 50dB of attenuation. that represents the minimum thickness of the shield in order to satisfy the attenuation limit. exactly. 0. at 10 kHz. exactly.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Fig.8 mm.012 mm of copper are sufficient. The high frequency interference is attenuated of 50dB using a copper shield with thickness very low. Ramaccia and A. at 1 GHz. Toscano Pag. 45 . 3805 × f −1 2 (2.686 The inequality (2.21 ExerciseEquation Section (Next) In order to protect a piece of equipment from RF interference.686 z δ (2.5 ×107 S m .686 z δ ≥ 50dB ⇒ z≥ (2. 46 .S. So inverting the eq.5 × 10 0.21.0661× f −1 2 = 0.21. 50 50 0. Over what frequency range can this shield protect our equipment assuming the same 50dB attenuation requirement of the previous problem? Solution First of all we have to evaluate the skin depth as function of the frequency f: δ= The attenuation in dB is: 1 1 = f −1 2 7 −7 π fμσ π × 4π × 10 × 3. The conductivity of aluminium is 3.1) A dB ( z ) = 8. Toscano Pag.2) and its minimum value over the frequency range of interest is at least 50 dB.21. we have: A dB ( z ) = 8. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.686 8.21.21. we construct an enclosure made of aluminium foil (you may assume a reasonable value for its thickness).1: Thickness of aluminium shield in mm for 50dB of attenuation. 2. Ramaccia and A.3) can be plotted as function of frequency in the range 10 kHz–1 GHz: Fig.J.0851× f −1 2 (2.2) with the assumption that AdB ( z ) ≥ 50dB . D.21.3) δ= 8. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 A typical thickness of aluminium is about 1 mm. D. Ramaccia and A. and this shield reduces by 50 dB only electric fields with frequency greater than 150 kHz.S.J. 47 . Toscano Pag. t ) = E ( 0. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. by E ( 0. we have E ( z. t ) is completely determined by E ( z. t ) = where Hi = Ei Z0 with i = 1. t ) and H ( z. t ) = x E1e jω1t + E 2e jω2t Determine the fields E ( z.22.0 ) or.22.22 ExerciseEquation Section (Next) A uniform plane wave propagating towards the positive z–direction in empty space has an electric field at z = 0 that is a linear superposition of two components of frequencies ω1 and ω2 : ˆ E ( 0. t ) : E ( z. 2 .2) where ki = ωi c with i = 1.0 ) = E ( 0. t − z c ) = x E1e ˆ = x E1e jω1t e jk1z + E 2e jω1t e jk 2z ( ( jω1( t + z c ) + E 2e jω2 ( t + z c ) ) )= ) (2.22. t ) = F ( z − ct ) = F ( 0 − c ( t − z c ) ) . t ) at any point z. t ) = E ( z − ct. and the magnetic field is: H ( z. t − z c ) Using this property . ( ) (2. 2 .3) D. Ramaccia and A. 1 ˆ ˆ z × E ( z. t ) = y H1e jω1t e jk1z + H 2e jω1t e jk 2z Z0 ( (2. alternatively. Toscano Pag. 48 . which implies that E ( z.J.S. we find for the electric and magnetic fields: ˆ E ( z.1) Solution For a forward–moving wave. say at z = z1 and z = z2 are related by the matrix equation: ⎡ E ( z1 ) ⎤ ⎡ cos k ⎢ ⎥=⎢ ⎣η0 H ( z1 ) ⎦ ⎢ jη sin k ⎣ where jη −1 sin k ⎤ ⎡ E ( z 2 ) ⎤ ⎥⎢ ⎥ cos k ⎥ ⎣η0 H ( z 2 ) ⎦ ⎦ (2. 49 . 4) Let Z ( z ) = location z.J. and we multiplied the magnetic field by η0 = μ0 ε 0 in order to give it E (z) 1 and Y ( z ) = be the normalized wave impedance and admittance at Z(z) η0 H ( z ) the same dimensions as the electric field.2) 3) Assuming μ = μ0 and ε = n 2ε 0 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. show that the fields at two different z–locations.23. E ( z ) = xE ( z ) and H ( z ) = yH ( z ) . 1 + jη Z ( z 2 ) tan k Y ( z1 ) = 1 + jη −1Y ( z 2 ) tan k Y ( z 2 ) + jη tan k (2. Z ( z) = E ( z) ηH ( z) ? Solution • Question n° 1 Assuming an harmonic time dependence e jωt .S.23.23.3) = z2 − z1 . consisting of the forward and backward components: E ( z ) = E + e − jkz + E − e jkz H (z) = ( E+e− jkz − E−e jkz ) η 1 (2. Toscano Pag. Ramaccia and A. that is. Show the relationships at the location z1 and z2 : Z ( z1 ) = Z ( z 2 ) + jη −1 tan k . 2) Show that the time–averaged energy flux in the z–direction is independent of z and is given by: 1 1 2 2 Pz = ℜe E ( z ) H∗ ( z ) = E+ + E− 2 2η { } ( ) (2.1) 1) Verify that these expressions satisfy all of Maxwell's equations. the Maxwell's equations can be written as follow: D.23 ExerciseEquation Section (Next) An electromagnetic wave propagating in a lossless dielectric is described by the electric and ˆ ˆ magnetic fields.4) What would be these relationships if had we normalized to the medium impedance.23. so that n is the refractive index of the dielectric. we obtain: ˆ ˆ ˆ ∇ × xE ( z ) = −yjkE ( z ) = − jyω με E ( z ) = = − jω μ 2ε ε ˆ ˆ y E ( z ) = − jωμ y E (z) = μ μ E (z) (2. isotropic and uniform medium. Orfanidis – Electromagnetic Waves and Antennas ⎧∇ × E = − jωB ⎪ ⎪∇ × H = jω D + J ⎨ ⎪∇ ⋅ D = ρ ⎪∇ ⋅ B = 0 ⎩ Exercises Chapter 2 (2.11) D. So inserting (2.9) ˆ = − jωμ y η = − jωμ H ( z ) It is very simple to verify the second Maxwell's equation in the same way.7 we know the k − ω relationship and as consequence also the expression of the η= μ ε (2.23.5) becomes: ⎧∇ × E = − jωμ H ⎪∇ × H = jωε E ⎪ ⎨ ⎪∇ ⋅ E = 0 ⎪∇ ⋅ H = 0 ⎩ Now it is possible to verify the first Maxwell's equation in set (2.23. The third and fourth equation are the divergence of the electric and magnetic field respectively: ˆ ∇ ⋅ E = ∇ ⋅ xE ( z ) = ∂ x E ( z ) = 0 ˆ ∇ ⋅ H = ∇ ⋅ yH ( z ) = ∂ y H ( z ) = 0 • Question n° 2 (2.23.6): ˆ x ˆ ∇ × xE ( z ) = ∂ x E (z) ˆ y ∂y 0 ˆ z ˆ ˆ ∂ z = y ∂ z E ( z ) − z∂ y E ( z ) = 0 (2.5) In a source less.8) where μ = μ0μr .S.23.23.10) (2.7) From exercise 2. Ramaccia and A.6) ) ( ˆ = −yjk ( E + e− jkz − E − e jkz ) characteristic impedance η : k = ω με .7). Toscano Pag.23. ε = ε 0ε r . ˆ ˆ = y∂ z E + e− jkz + E −e jkz = y − jkE + e− jkz + jkE − e jkz = ( ) (2. linear. So the set (2.23.23.23.J. 50 .23.8) in (2.23. the quantities ρ and J are zero and the constitutive relations B = μ H and D = ε E are valid. 15) becomes: E ( z1 ) = E ( z 2 ) cos k + j η0H ( z 2 ) n sin k (2. Using the Euler's formula e jx = cos x + jsin x .S. (2.12) Substituting (2.23.23.17) can be written in the matrix form as (2.16) and eq.1) in eq.23.23. (2. • Question n° 4 D.13) ( ) ( ) • Question n° 3 Consider the expression for E ( z1 ) and multiply it by the neutral term e jkz2 e− jkz2 : E ( z1 ) = E + e − jkz1 e jkz 2 e − jkz 2 + E − e jkz1 e jkz 2 e − jkz 2 consequently.15) The term that multiplies cos k is simply E ( z 2 ) and the term that multiplies sin k is simply η H ( z 2 ) . (2.J.23.14) where = z2 − z1 .23.16) In the same way it is possible to write H ( z1 ) as a function of E ( z 2 ) and H ( z 2 ) : η0 H ( z1 ) = jnE ( z 2 ) sin k + η0 H ( z 2 ) cos k Now eq.23. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 The energy flux can be evaluated as dot product of the Poynting vector and the unit vector along the z–direction: 1 ˆ ˆ Pz = P ⋅ z = ℜe ⎡E × H∗ ⎤ ⋅ z ⎣ ⎦ 2 (2. Since the characteristic impedance of the medium η can be written as: η= μ0 η η = 0 = 0 n ε 0ε r εr where n is the refractive index. Ramaccia and A. we have: ⎡ E∗ e+ jkz − E∗ e− jkz ⎤ − 1 ⎢ E e− jkz + E e jkz x × + ˆ ˆ ˆ Pz = ℜe y⎥ ⋅ z = + − ⎢ ⎥ 2 η ⎢ ⎥ ⎣ ⎦ 1 ˆ ˆ = ℜe ⎡ E + e − jkz + E − e jkz E∗ e+ jkz − E∗ e− jkz z ⎤ ⋅ z = + − ⎢ ⎥ ⎣ ⎦ 2η 1 1 2 2 2 2 E+ − E− = ℜe ⎡ E + − E − ⎤ = ⎢ ⎥ 2η ⎣ ⎦ 2η ( ) ( ) ( )( ) (2.23. 51 .23.12).17) Pag. the eq. Toscano (2.14) can be written as: E ( z1 ) = E + e− jkz2 ( cos k + jsin k ) + E − e jkz2 ( cos k − jsin k = E + e− jkz 2 + E − e jkz2 cos k + j E + e− jkz2 − E − e jkz 2 sin k ( ) ( ) )= (2.23.23. (2. (2.23. E ( z1 ) = E + e − jkz 2 e jk + E − e jkz 2 e − jk (2.23. eq.3). 20): Z ( z 2 ) + jtan k ⎧ ⎪ Z ( z1 ) = 1 + jZ ( z 2 ) tan k ⎪ ⎨ ⎪Y ( z ) = Y ( z 2 ) + jtan k 1 ⎪ 1 + jY ( z 2 ) tan k ⎩ D.19) and (2.18) (2. Toscano Pag.23. Orfanidis – Electromagnetic Waves and Antennas The normalized wave impedance at location z1 is: Z ( z1 ) = Exercises Chapter 2 η0 H ( z1 ) E ( z1 ) and we can substitute the electric and magnetic field with their respective relationships (2.S. So: sin k n Z ( z1 ) = jnE ( z 2 ) sin k + η0 H ( z 2 ) cos k Dividing the numerator and denominator by η0 H ( z 2 ) cos k .16) and (2.20) = n −1 1+ j tan k Z ( z2 ) 1 + jn −1Y ( z 2 ) tan k If we had normalized Z ( z ) and Y ( z ) to the medium impedance.23.23. 52 . (2.23.17).23.19) Y ( z1 ) = 1 + jnZ ( z 2 ) tan k 1 = = Z ( z1 ) Z ( z 2 ) + jn −1 tan k 1 + jn tan k Z ( z2 ) = Y ( z 2 ) + jn tan k (2.23.23. we have: E ( z 2 ) cos k sin k + jn −1 η0 H ( z 2 ) cos k cos k Z ( z 2 ) + jn −1 tan k = Z ( z1 ) = E ( z 2 ) sin k 1 + jnZ ( z 2 ) tan k 1 + jn η0 H ( z 2 ) cos k The admittance Y ( z1 ) is the inverse of Z ( z1 ) : E ( z 2 ) cos k + j η0 H ( z 2 ) (2.J. simply we have to cancel the term n of refractive index inside eq. Ramaccia and A. 24. y.24. z′ ) is a rotated coordinate system with respect to a fix coordinate system ( x.J.24 ExerciseEquation Section (Next) Show that the time–averaged energy density and Poynting vector of the obliquely moving wave: ˆ ˆ E ( r. 2. t ) = [ x′A + y′B] ⋅ ⎡ x′A∗ + y′B∗ ⎤ = AA∗ + BB∗ = A + B ⎣ ⎦ 1 1 1 2 2 ∗ ˆ ˆ ˆ ˆ H ( r. t ) = 2 [ y′A − x′B] ⋅ ⎡ y′A∗ − x′B∗ ⎤ = 2 AA∗ + BB∗ = 2 A + B ⎣ ⎦ η η η ( ) ( ) and now we can substitute them inside the definition of the time–averaged energy density: D.S. y′.2) 1 1 2 2 2 2 ∗ ⎤ ˆ′ 1 ˆ ˆ A + B = ( z cos θ + x sin θ ) A +B P = ℜe ⎡ E × H = z ⎣ ⎦ 2 2η 2η w= ( ) ( ) ( ) ˆ ˆ ˆ where z′ = ( z cos θ + x sin θ ) is the unit vector in the direction of propagation. Show that the energy transport velocity is v = P ˆ = cz′ . Toscano Pag. t ) = [ x′A + y′B] e jω t − jkz′ H ( r. Ramaccia and A. 53 . are given by: 1 1 ⎡1 ⎤ 1 2 2 ℜe ⎢ ε E ⋅ E∗ + μ H ⋅ H∗ ⎥ = ε A + B 2 2 ⎣2 ⎦ 2 (2. t ) = 1 η ˆ ˆ [ y′A − x′B] e jω t − jkz′ (2. t ) ⋅ H ( r. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.24. t ) ⋅ E ( r. z ) as shown in Fig. w Solution The dot products E ⋅ E∗ and H ⋅ H ∗ can be evaluated as follow: 2 2 ∗ ˆ ˆ ˆ ˆ E ( r.1) where ( x ′.1: Rotation of coordinate system. 4) in the definition of th energy transport velocity. D. we obtain: P = w ˆ z′ 1 1 2 2 A +B 1 1 1 2η ˆ ˆ ˆ ˆ = z′ = z′ = z′ = cz′ ηε 1 μ με 2 2 ε A +B ε 2 ε ( ) ( ) v= ( ) ( ) where c = 1 με is the speed of light in the free space.24. (2. Ramaccia and A. Toscano Pag.24.4) P = ℜe ⎡E × H∗ ⎤ = z′ ⎣ ⎦ 2 2η 2η Substituting eq.24. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 ⎡1 1 1 1 ⎡1 ⎤ 1 2 2 2 2 ⎤ w = ℜe ⎢ ε E ⋅ E∗ + μ H ⋅ H∗ ⎥ = ℜe ⎢ ε A + B + 2 μ A + B ⎥ = 2 2 ⎣2 ⎦ 2 2η ⎢2 ⎥ ⎣ ⎦ 1 1 ⎡1 2 2 2 2 ⎤ 1 2 2 = ℜe ⎢ ε A + B + ε A + B ⎥ = ε A + B 2 2 ⎣2 ⎦ 2 On the contrary the cross product E × H ∗ can be written as: ˆ x′ E× H = ∗ ( ) ( ) ( ) ( ) ( ) (2.24. 54 .3) and (2.J. 1 1 1 2 2 2 2 ˆ ˆ ˆ A + B = ( z cos θ + x sin θ ) A + B (2.3) ˆ y′ B A∗ ˆ z′ A −B ∗ η η ⎛ A2 B2 ⎞ 1 ⎟ = z′ A 2 + B 2 ˆ ˆ + 0 = z′ ⎜ ⎜ η η ⎟ η ⎝ ⎠ 0 ( ) and. consequently.S. we can find the magnetic field as follow: H ( x.25. t ) = ∂ x 0 − jk ( x + z ) ˆ y ∂y E ( x. z. Toscano Pag. y. y. z. t ) − jωμ (2. working with Maxwell's equations.4) where E ( x. ˆ 2.25. z. z ) into Maxwell's equations.2) Solution • Question n° 1 From the first Maxwell's equation. if we started with a magnetic field given by: ˆ H ( x.S. 55 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.25.J.1) 1. z.25. y. t ) and the ˆ propagation direction k . k= ω0 c0 (2. t ) = E 0 e jω t e 2 .25 ExerciseEquation Section (Next) A uniform plane wave propagating in empty space has electric field: ˆ E ( x. determine the electric field E ( x. z. What is the direction of propagation and its unit vector k ? 3. z. t ) ˆ z ∂z 0 (2. t ) = The cross product has to be evaluated: 1 ∇× E ( x.25.5) and consequently D. work out an expression for the corresponding magnetic field H ( x. t ) = yH 0e jω t e − jk ( 3z − x 2 ) (2. Inserting E ( x. t ) = − x ∂ ⎛ jω t − jk ( x + z ) ⎜ E 0e e ∂z ⎝ ⎞ + z ∂ ⎛ E e jω t e − jk( x + z ) ⎟ ˆ ⎜ 0 ∂x ⎝ ⎠ 2 2 ⎞= ⎟ ⎠ 2 − jk ⎛ jω t − jk ( x + z ) ˆ = −x ⎜ E 0e e 2⎝ ⎞ + z − jk ⎛ E e jω t e − jk( x + z ) ⎟ ˆ ⎜ 0 ⎠ 2⎝ ⎞ ⎟ ⎠ (2.25. z. z ) . z ) = yE0e jω t e − jk( x + z ) 2 . The determinant of matrix (2. y. z. Ramaccia and A.4) is: 2 ˆ ∇ × E ( x.3) ˆ x ∇ × E ( x. t ) ⎥ = − jωμ ⎣ ⎦ Exercises Chapter 2 ⎤ jω με 1 ⎡ jω με ˆ ˆ = E ( x.6) η 2 ˆ −x E ( x. • Question n° 2 The direction of propagation can be found as cross product between the direction of oscillation of the electric and magnetic field. η = μ ε and z′ = ( z − x ) 2 . Ramaccia and A. 2. Orfanidis – Electromagnetic Waves and Antennas H ( x. z. z. So: ˆ x ˆ ˆ ˆ k = y × z′ = 0 1 − 2 ˆ y 1 0 ˆ z 0 = 1 2 ˆ ˆ ( x + z ) = x′ ˆ 2 (2. t ) (2. z. t ) = H 0e jω t e − jk ( 3z − x 2 ) .8) where H ( x. z.25.J. t ) ˆ z ∂z 0 (2. z. 56 .25. z. z. z. t ) η 2 = E ( x.25. t ) = 1 jωε ˆ x ∂x 0 ˆ y ∂y H ( x. We can assume a new coordinate system aligned with the components of the electromagnetic wave as depicted in Fig.25.7) • Question n° 3 Using the inverse form of the second Maxwell's equation. t ) = 1 jωε ∇ × H ( x. we have: E ( x. z. t ) − z 2 E ( x. Fig. z.1: Rotation of the coordinate system. z.25. 2. z. t ) η ˆ z′ ˆ ˆ ˆ where k = ω με . t ) − z E ( x. Toscano Pag.1. t ) = − jωμ 1 ⎡ jk jk ⎤ ˆ ˆ ⎢ x 2 E ( x. t ) ⎥ = ⎢x − jω μ ⎢ 2 2 ⎥ ⎣ ⎦ ˆ =z E ( x.S. So: D. t ) = = 1 ∇ × E ( x. z. t ) ⎥ = ⎢−x jωε ⎣ 2 2 ⎦ ⎤ jω με 1 ⎡ − jω 3 με ˆ ˆ H ( x. Toscano Pag. t ) x′ 2 ˆ ˆ 3x + z .25. z. t ) + z H ( x. z. Orfanidis – Electromagnetic Waves and Antennas E ( x. η = μ ε and x′ = (2. We can assume a new coordinate system aligned ) ) with the components of the electromagnetic wave as depicted in Fig. z.2: Rotation of the coordinate system. z. t ) 2 ˆ where k = ω με .25.J.S. 2. z. So: ˆ x ˆ ˆ ˆ k = y × z′ = 0 3 ˆ y ˆ z ˆ ˆ ˆ 1 0 = x − 3z = − z ′ 0 1 D. Ramaccia and A. t ) ⎥ = H ( x. z. The direction of propagation can be found as cross product between the direction of oscillation of the electric and magnetic field. t ) = = 1 ⎡ ∂ ⎛ jω t − jk ( ˆ ⎢ −x ⎜ H 0e e jωε ⎢ ∂ z ⎝ ⎣ 3z − x 2 ⎞ Exercises Chapter 2 ) ∂ ⎛ jω t − jk ( ˆ ⎟ + z ⎜ H 0e e ∂x ⎝ ⎠ 3z − x 2 ⎞ ⎤ ) ⎟⎥ = ⎠⎥ ⎦ ⎤ 1 ⎡ − j 3k jk ˆ ˆ H ( x. 57 . z.9) ( ( 1 ˆ ˆ ˆ 3x + z = η H ( x. t ) + z = ⎢ −x 2 2 jω ε ⎢ ⎥ ⎣ ⎦ 1 = η H ( x. show that the polarization vectors at the input.26.S. and after the first and second polarizer. a2 with respect to the x– and y–directions.3) and ⎡ cos ϕ ⎢ sin ϕ ⎣ − sin ϕ ⎤ ⎡ a1 0 ⎤ ⎡ cos ϕ cos ϕ ⎥ ⎢ 0 a2 ⎥ ⎢ − sin ϕ ⎦⎣ ⎦⎣ sin ϕ ⎤ ⎡ a1 0 ⎤ ⎡ cos θ ⎤ cos ϕ ⎥ ⎢ 0 a2 ⎥ ⎢ sin θ ⎥ ⎦⎣ ⎦ ⎦⎣ (2. 2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2. The analyzer is rotated by an angle ϕ so that the same gains a1 . so that a2 a1 .1) ˆ 2 ˆ 2 E2 = x′ a1 cos ϕ cos θ + a1a2 sin ϕ sin θ + y′ a2 cos ϕ sin θ − a1a2 sin ϕ cos θ ˆ ˆ ˆ ˆ where ( x′. Ramaccia and A.26. y ′) are related to ( x.1: Polarizer–analyzer filter combination. 1. 58 .26.26 ExerciseEquation Section (Next) A linearly polarized light wave with electric field E0 at angle θ with respect to the x–axis is incident on a polarizing filter. Show that the input light intensity is proportional to the quantity: D.26. Toscano Pag. Ignoring the phase retardance introduced by each polarizer.J. a2 now refer to the x'– and y'– directions. are: ˆ ˆ E0 = x cos θ + y sin θ ˆ ˆ E1 = xa1 cos θ + ya2 sin θ (2.26.4) 3. follow by an identical polarizer (the analyzer) whose primary axes are rotated by an angle ϕ relative to the axes of the first polarizer. as shown in Fig.2) (2. y ) as follow: ( ) ( ) ˆ ˆ ⎡ x′ ⎤ ⎡ cos ϕ sin ϕ ⎤ ⎡ x ⎤ =⎢ ⎢ y′⎥ − sin ϕ cos ϕ ⎥ ⎢ y ⎥ ⎣ˆ ⎦ ⎣ ⎦⎣ˆ⎦ 2. Explain the meaning an usefulness of the matrix operations: ⎡ a1 ⎢0 ⎣ 0 ⎤ ⎡ cos ϕ a2 ⎥ ⎢ − sin ϕ ⎦⎣ sin ϕ ⎤ ⎡ a1 cos ϕ ⎥ ⎢ 0 ⎦⎣ 0 ⎤ ⎡ cos θ ⎤ a2 ⎥ ⎢ sin θ ⎥ ⎦ ⎦⎣ (2. Assume that the amplitude attenuation through the first polarizer are a1 . The polarizer transmits primarily the x–polarization. 59 . The rotation can be expressed by the matrix (2. 0 ⎤ ⎡ cos θ ⎤ ⎡ a1 cos θ ⎤ ⎡a ˆ ˆ E1 = A ⋅ E0 = ⎢ 1 ⎥⎢ ⎥ = xa1 cos θ + ya2 sin θ ⎥=⎢ ⎣ 0 a2 ⎦ ⎣ sin θ ⎦ ⎣ a2 sin θ ⎦ (2. consequently.26. that is each component is attenuated by a factor a1 or a2 . Toscano Pag.sin ϕ ˆ = x′ sin ϕ ⎤ ⎡ a1 cos θ ⎤ = cos ϕ ⎥ ⎢ a2 sin θ ⎥ ⎦⎣ ⎦ ( 2 a1 cos ϕ cos θ ˆ + a1a2 sin ϕ sin θ + y ′ ) ( 2 a2 cos ϕ sin θ − a1a2 sin ϕ cos θ ) (2. will be given by the generalized Malus's law: I= 1 4 4 2 2 a1 + a2 cos 2 ϕ + a1 a2 sin 2 ϕ 2 ( ) (2. show that the average of the intensity of part (3) over all angles 0 ≤ θ ≤ 2π . Orfanidis – Electromagnetic Waves and Antennas 4 4 2 2 I = a1 cos 2 θ + a2 sin 2 θ cos 2 ϕ + a1 a2 sin 2 ϕ + 2 2 +2a1a2 a1 .26. Solution • Question n° 1 The electric field E1 after the polarizer is an attenuated form of the field E0 . that is incoherent. Ramaccia and A.sin ϕ sin ϕ ⎤ E1 cos ϕ ⎥ ⎦ (2.8) from which: 0 ⎤ ⎡ cos ϕ ⎡a E2 = ⎢ 1 ⎥⎢ ⎣ 0 a2 ⎦ ⎣ .6) The case a2 = 0 represents the usual Malus's law. according to x– or y–directions respectively.26.S.5) 4. a2 now refer to the x'– and y'–directions. So we can characterize the polarizer with a own attenuation matrix: ⎡a A=⎢ 1 ⎣0 0⎤ a2 ⎥ ⎦ and.2) and then the x'– and y'–components of the field E2 have to be attenuated by a factor a1 or a2 . according to x– or y–directions respectively.26. So: 0 ⎤ ⎡ cos ϕ ⎡a E2 = ⎢ 1 ⎥⎢ ⎣ 0 a2 ⎦ ⎣ .J.a2 cos ϕ sin ϕ cos θ sin θ Exercises Chapter 2 ( ( ) ) (2.7) The electric field E1 passes thought the second polarizer that is rotated of an angle by an angle ϕ so that the same gains a1 . If the input light were unpolarized.26.26.9) • Question n° 2 D. 11) • Question n° 4 The average of the intensity over all angles 0 ≤ θ ≤ 2π can be evaluated integrating the eq. 2 I = E x′ + E y′ = a1 cos ϕ cos θ + a1a2 sin ϕ sin θ ) + ( a22 cosϕ sin θ − a1a2 sin ϕ cosθ ) 4 2 2 3 = ( a1 cos 2 ϕ cos 2 θ + a1 a2 sin 2 ϕ sin 2 θ + 2a1 a2 sin ϕ sin θ cos ϕ cos θ ) + 4 2 2 3 + ( a2 cos 2 ϕ sin 2 θ + a1 a2 sin 2 ϕ cos 2 θ − 2a1a2 sin ϕ sin θ cos ϕ cos θ ) 2 2 2 2 4 4 = ( a1 cos 2 θ + a2 sin 2 θ ) cos 2 ϕ + a1 a2 sin 2 ϕ + 2a1a2 ( a1 . the third matrix represents the attenuation through the polarizer and the fourth represents the tilt of the electric field vector with respect to the x– and y–directions.4).26. In the matrix operation (2. Using these matrixes operation. −ϕ .10) that is the light intensity is proportional to the sum of the square module of the components of electric field. it is a very simple model system. • Question n° 3 The light intensity is the time–averaged energy density multiplied by the speed of light in the host medium. 60 . Toscano Pag. but the angle is opposite. that is vacuum in this case.26. shown here for simplicity.S.e.26. From exercise 2.26. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 The matrix operation (2.a2 ) sin ϕ sin θ cos ϕ cos θ 2 2 2 ( 2 = (2.24. Ramaccia and A. but without any attenuations. i.J.26. we have already demonstrated the expression of the time–averaged energy density w and so: I= 2 1 2 1 2 cε E = cε ⎛ E x′ + E y′ ⎞ ⎜ ⎟ 2 2 ⎝ ⎠ (2.26. (2.3) is simply the cascade of the matrix that we used to solve the point (1). This suggest that (2.11) and dividing it by 2π as follow: D. In fact the first two matrixes are necessary to rotate and attenuate the field E1 that passes through the analyzer.4) represents a system with another analyzer at the end tilted of an angle −ϕ . it is possible to note that there is a new matrix M at the beginning of the expression: ⎡ cos ϕ ⎢ sin ϕ ⎣ − sin ϕ ⎤ ⎡ a1 0 ⎤ ⎡ cos ϕ cos ϕ ⎥ ⎢ 0 a2 ⎥ ⎢ − sin ϕ ⎦⎣ ⎦⎣ ↑ M sin ϕ ⎤ ⎡ a1 0 ⎤ ⎡cos θ ⎤ cos ϕ ⎥ ⎢ 0 a2 ⎥ ⎢ sin θ ⎥ ⎦⎣ ⎦ ⎦⎣ This has the same structure of a rotation matrix. 12).26. we obtain: I= 1 2π 2π 0 ∫ I dθ = 2 ⎡a1 + a2 ⎤ cos ⎣ ⎦ 4 4 1 2 2 2 ϕ + a1 a2 sin 2 ϕ D. Toscano Pag.a2 sin ϕ cos ϕ + 2π ( ) ∫ cosθ sin θ dθ We can solve each integral separately as follow: 2π 2π 2π ⎧ 2π 1 + cos 2θ 1 1 2 ⎪ ∫ cos θ dθ = ∫ dθ = ∫ dθ + ∫ cos 2θ dθ = π 2 2 2 ⎪0 0 0 0 ⎪ 2π 2π 2π 2π ⎪ 1 − cos 2θ 1 1 2 dθ = ∫ dθ − ∫ cos 2θ dθ = π ⎨ ∫ sin θ dθ = ∫ 2 2 2 ⎪0 0 0 0 ⎪ 2π 2π 1 ⎪ ⎪ ∫ cos θ sin θ dθ = 2 ∫ sin 2θ dθ = 0 0 ⎩0 (2. Orfanidis – Electromagnetic Waves and Antennas 1 2π 2π 4 a1 cos2 ϕ 2π 2π 4 a2 cos2 ϕ 2π 2π 0 2π Exercises Chapter 2 2 2 a1 a2 sin 2 ϕ 2π 2π 0 I= ∫ 0 I dθ = ∫ 0 cos 2 θ dθ + ∫ 0 sin 2 θ dθ + ∫ dθ + (2. Ramaccia and A.13) and.13) in (2.26. substituting (2.26. 61 .26.J.S.12) 1 2 2 2a1a2 a1 . 1. 62 . prove the alternative Doppler formulas: f ′ = fγ (1 − β cos θ ) = f 1 − β cos θ 1 + β cos θ ′ (2.27. prove the following identity between the angle θ .27.1: Plane wave viewed from stationary and moving frames.S. Toscano Pag. We assume that the wave–vector k in S lies in the xz–plane and forma an angle θ with the z–axis as shown in Fig. k ) : ω ′ = γ (ω − β ck z ) β ⎞ ⎛ k′ = γ ⎜ k z − ω ⎟ z c ⎠ ⎝ k′ = k x x (2. prove the equivalence of the three relationships given by: cos θ ′ = cos θ − β 1 − β cos θ 1 − β 2 . They are a consequence of the Lorentz transformation of the frequency–wavenumber four–vector (ω c. Ramaccia and A.4) D. θ ′ : (1 − β cos θ )(1 + β cos θ ′ ) = (1 + β cos θ )(1 − β cos θ ′ ) = 1 − β 2 Using this identity.1) 1− β where β = v c and γ = 1 Then. First.27.27.27.J.\ ⇔ sin θ ′ = sin θ γ (1 − β cos θ ) ⇔ tan (θ ′ 2 ) tan (θ 2 ) = 1+ β (2.1) relate the apparent propagation angles θ . 2. θ ′ in the two frames that are different because of the aberration of light due to the motion.27 ExerciseEquation Section (Next) Consider an uniform plane wave propagating in vacuum as viewed from the vintage point of two coordinate frames: a fixed frame S and a frame S' moving towards the z–direction with velocity v.3) Solution The three relationships in (2. 2. Fig.27.27. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.2) γ (1 + β cos θ ′ ) =f (2. 9) So: 1− β 2 1− β 1+ β tan θ ′ 2 1 1+ β = = = = tan θ 2 γ (1 − β ) 1− β 1− β 1− β (2. with k ′ = ω ′ c .1): tan θ ′ 2 = = sin θ ′ sin θ = = 1 + cos θ ′ γ (1 − β cos θ )(1 + cos θ ′ ) sin θ = ⎛ ⎛ cos θ − β ⎞ ⎞ γ (1 − β cos θ ) ⎜ 1 + ⎜ ⎟⎟ ⎝ ⎝ 1 − β cos θ ⎠ ⎠ sin θ = = ⎛ 1 − β cos θ + cos θ − β ⎞ γ (1 − β cos θ ) ⎜ ⎟ ⎜ ⎟ 1 − β cos θ ⎝ ⎠ sin θ 1 tan θ 2 = = γ (1 − β )(1 + cos θ ) γ (1 − β ) (2.6) ω′ ⎧ ⎪ω = γ 1 − β cos θ ( ) ⎨ ⎪ω ′ sin θ ′ = ω sin θ ⎩ ⇒ ω ′ sin θ ′ = ω ′ sin θ γ (1 − β cos θ ) (2. From the first and second equation we can obtain the expression for cos θ ′ : (2.27. Orfanidis – Electromagnetic Waves and Antennas where β = v c and γ = 1 Exercises Chapter 2 1 − β 2 . we have: ⇒ ω ′ cos θ ′ = ω ′ ( cos θ − β ) (1 − β cos θ ) (2. Ramaccia and A.27.7) The half–angle formula for the tangent is in general: tan x sin x 1 − cos x = = 2 1 + cos x sin x (2. k′ = k′ sin θ ′ .4) may be rewritten in z x the form: ω ′ = ωγ (1 − β cos θ ) ω ′ cos θ ′ = ωγ ( cos θ − β ) ω ′ sin θ ′ = ω sin θ The three equations are equivalent to evaluate the angular frequency ω ′ in the moving frame S' and they relate the different angular θ . (2. θ ′ regardless of the frequency. and similarly in the frame S'. Toscano Pag. Setting k z = k cosθ . with k = ω c .27.27.27.27. k x = k sin θ .10) D.27.5) ω′ ⎧ ⎪ωγ = (1 − β cos θ ) ⎨ ⎪ω ′ cos θ ′ = ωγ ( cos θ − β ) ⎩ From the first and third equation.27. the Eqs. 63 . k′ = k′ cos θ ′ .S. we can obtain the third relationship of (2.J.8) and using it. 12) So: sin 2 θ ′ sin θ 2 (1 − β cos θ ) = = = 1 γ 2 (1 − β cos θ ) 2 (1 − β cos θ ) = (2. we can write two identity: D.J. This is reasonable if the velocity vector v is directed in the negative z–direction. (2.13) we can write that: 1− β 2 = sin 2 θ ′ sin θ 2 (1 − β cos θ )2 = (1 + β cos θ ′)(1 − β cos θ ) (2.27.27. is formally identical to the first identity.27. (2.27.11) Consequently since sin 2 θ ′ sin θ 2 (1 − β cos θ )2 = 1 − β 2 ↑ (2. relating the frequency of the wave as measured by an observer in the moving frame S' to the frequency of a source in the fixed frame S: f ′ = fγ (1 − β cos θ ) = f 1 − β cos θ 1− β 2 (2.27. (2.27.14) The second identity in eq.11) and (2.27.27. Toscano Pag.2) between the angle θ .27. i. Ramaccia and A.1): sin θ ′ = sin θ γ (1 − β cos θ ) ⇒ 1 γ 2 = 1− β 2 = sin 2 θ ′ sin θ 2 (1 − β cos θ )2 (2.27. but the angles θ . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 In the identity of eq.27.2). (2.27. that is v → − v .13) 1− β 2 1 − β 2 + β cos θ − β cos θ = = (1 − β cos θ ) (1 − β cos θ ) (1 − β cos θ ) + β ( cosθ − β ) = 1 + β ( cos θ − β ) = (1 − β cos θ ) (1 − β cos θ ) = 1 + β cos θ ′ Using (2.14). that can be expressed using the second identity in eq.e.14) to the relativistic Doppler formula. Since the term β is defined as v c and it is the square. the sign of v is negligible and the identity is valid.27.S.15) From eq. (1 + β cos θ )(1 − β cos θ ′ ) = 1 − β 2 . θ ′ are inverted. θ ′ we note that the last term is equal to 1 γ 2 .27. 64 . The alternative Doppler formulas in eqs.2) = (1 − β cos θ )(1 + β cosθ ′) we have to prove that: sin 2 θ ′ sin 2 θ (1 − β cos θ ) = 1 + β cos θ ′ (2. (2.3) are obtained applying (2. 27.16) D. 65 . Orfanidis – Electromagnetic Waves and Antennas ⎧ 1− β 2 1 = 2 ⎪(1 − β cos θ ) = (1 + β cos θ ′ ) γ (1 + β cos θ ′ ) ⎪ ⎨ −1 1 ⎪⎛ 1− β 2 ⎞ = γ = ⎜ ⎟ ⎪⎝ ⎠ (1 + β cos θ ′ )(1 − β cos θ ) ⎩ and then substitute them inside (2. Toscano Pag.15): f ′ = fγ (1 − β cos θ ) = f′ = f 1 − β cos θ 1− β 2 fγ Exercises Chapter 2 γ 2 (1 + β cos θ ′) γ (1 + β cos θ ′) 1 − β cos θ = f 1 − β cos θ =f =f 1 + β cos θ ′ (1 + β cos θ ′)(1 − β cos θ ) (2.27.J. Ramaccia and A.S. Sb .28.2) = f a γ (1 − β cos θ a ) = = fa γ a (1 − β a cos θ ) γ (1 − β cos θ b ) 1 − β cos θ b va v v 1 1 1 . show that the Doppler formula may be written in the following equivalent forms: f b = fa γ b (1 − β b cos θ ) fa 1 − β cos θ a (2. Fig. β = . Moreover. Ramaccia and A. γ a = .28. Sb.J. γb = . Sb.28. 66 .1: Propagating plane wave along the θ –direction. γ= (2. 2.28 ExerciseEquation Section (Next) We consider two reference frame Sa . vb with respect to our fixed frame S. 1 − β b cos θ cos θ b = cos θa − β 1 − β cos θa (2. Sa.28.28.5) D. If in the frame S the wave is propagating along the θ –direction shown in Fig. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.1.S. 2.4) and θa . Let fa and fb be the frequencies of the wave as measured in the frames Sa . In proving the relativistic Doppler formula: fa = f 1 − βa 1 − βb . Toscano Pag. show the following relations among these angles: cos θa = cos θ − βa .28. βb = b .1) it was assumed that the plane wave was propagating in the z–direction in all three reference frames S.3) 2 2 2 c c c 1 − βa 1 − βb 1− β v b − va where βa = and v is the relative velocity of the observer and source given by v= 1 − v b va c2 (2. Sb moving along the z–direction with velocities va . 1 − βa cos θ cos θ b = cos θ − β b .28. θb are the propagation directions in the frame Sa. fb = f 1 + βa 1 + βb ⇒ fb = fa 1 − β b 1 + βa 1 + βb 1 − βa (2. and we assume that θ = 0° in the frame S. 6) If we consider separately the frame Sa and the frame Sb.28. θ is the angle along which the wave propagates in the stationary frame and β is the ratio between the velocity v of the moving frame with respect to the fixed frame and the speed of light in vacuum. f ′ → f b ⎨ ⎩θ → θ a .28.7). 67 .28. θb expressed in (2. So D. θ ′ → θ b and γ .27.16): f ′ = fγ (1 − β cos θ ) = 1 − β cos θ 1 + β cos θ ′ γ (1 + β cos θ ′ ) f =f (2. and substitute it inside the second equation.28. in order to obtain: f b = fa γ b (1 − β b cos θ ) γ a (1 − β a cos θ ) (2.5) can be obtained from the analogue expression for the angles θ .3).7) From the first equation in (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Solution The relativistic Doppler formula relates the frequency of the wave as measured by an observer in the moving frame S' to the frequency of a source in the fixed frame S: f ′ = fγ (1 − β cos θ ) = f 1 − β cos θ 1− β 2 (2.9) where ⎧f → f a .J. So: f b = f a γ (1 − β cos θ a ) = fa 1 − β cos θ a = fa γ (1 + β cos θ b ) 1 + β cos θ b (2. he will have the sensation that the frame Sa is fixed and the frame Sb is moving.28. we can write for each frame a relativistic Doppler formula: f a = fγ a (1 − β a cos θ ) = f f b = fγ b (1 − β b cos θ ) = f 1 − β a cos θ 2 1 − βa 1 − β b cos θ 2 1 − βb (2.11) where θ ′ is the apparent angle along which the wave propagates in the moving frame. θ ′ : cos θ ′ = cos θ − β 1 − β cos θ (2.S. So it is possible to use the relationships (2.28.8) If the observer moves with the same velocity of the frame Sa.28. Toscano Pag.28. Ramaccia and A.28. we can write f as a function of fa. β are expressed in (2.10) The relationships between the angles θa . 12) we can write: cos θ = and we can substitute it in the second identity: βa + cos θa 1 + βa cos θa β a + cos θ a − βb cos θ − β b 1 + β a cos θ a cos θ b = = = 1 − β b cos θ 1 − β β a + cos θ a b 1 + β a cos θ a β a + cos θ a − β b (1 + β a cos θ a ) = 1 + β a cos θ a − β b β a + β b cos θ a β + cos θ a − β b − β b β a cos θ a = a = 1 + β a cos θ a − β b β a − β b cos θ a = = (1 − β b βa ) cos θa − ( β b − β a ) = (1 − β b βa ) − ( β b − β a ) cos θa ⎛ β − βa ⎞ cos θ a − ⎜ b ⎟ ⎝ 1 − β b β a ⎠ = cos θ a − β = 1 − β cos θ a ⎛ β − βa ⎞ 1− ⎜ b ⎟ cos θ a ⎝ 1 − βb βa ⎠ D.12) from which we can obtain the expression of cosθb as a function of cosθa . we can write the two following relationships: cos θa = cos θ − βa .S. Ramaccia and A. 1 − βa cos θ cos θ b = cos θ − β b 1 − β b cos θ (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 if we consider first the frame Sa and then the frame Sb with respect to the fixed frame S.28. 68 .28. From the first identity of (2. Toscano Pag.J. Assume that the earth has conductivity σ = 10−3 S m . and the penetration depth δ = 1 α in meters. The depth d may be determined by measuring the roundtrip time t ( h ) of the transmitted signal at successive horizontal distances h. as shown in Fig. With reference to the above figure. 4.29.2) where n r is the real part of the complex refractive index n c .1 for a buried–pipe.2 μs . Ramaccia and A.1) where E0 is the transmitted signal. You may use the "weakly lossy dielectric" approximation. permittivity ε = 9ε 0 . What is the depth d in meters? D. Toscano Pag. 3. Determine the numerical value of the wavenumber k = β − jα in meters-1.1: Section of the ground with an underground object. and h is the horizontal displacement of the antenna from the pipe. 2. Fig. You may ignore the angular response of the radar antenna and assume it emits isotropically in all directions into the ground. Determine the value of the complex refractive index nc = n r − jni of the ground at 900 MHz. 1.29. and permeability μ = μ0 . Suppose t ( h ) is measured over the range −2 ≤ h ≤ 2 meters over the pipe and its minimum recorded value is t min = 0. Show that t ( h ) is given by: t (h) = 2n r d2 + h2 c0 (2. 5.29 ExerciseEquation Section (Next) Ground–penetrating radar operating at 900 MHz is used to detect underground objects. 2. d is the depth of the pipe.S. explain why the electric field returning back to the radar antenna after getting reflected by the buried–pipe is given by: E ret E0 2 ⎡ 4 h2 + d2 = exp ⎢ − δ ⎢ ⎣ ⎤ ⎥ ⎥ ⎦ (2.29. 2. 69 . Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 2.J.29. 29.57 rad / m 0 0 ⎪ ⎪ ⎨ 4π ×10−7 σ μ0 10−3 ⎪ α= = = 0. i.11 79. The complex value of the permittivity of the ground at 900 MHz is: ε = 9ε 0 − j σ 10−3 = 9 × 8.4).0045π = 3e j0.5) = 3 ( cos 0.29. the permittivity is complex.65 − j1.3): ⎛ σ ⎞ σ k = β − jα = ω 9μ0ε 0 ⎜1 − j ⎟ = ω 9μ0ε 0 − j 2ω 9ε 0 ⎠ 2 ⎝ ⇓ μ0 9ε 0 ⎧ β = 3ω μ ε = 6π × 900 ×106 × 4π ×10−7 × 8.65 = and consequently.J.063 rad / m 2 9ε 0 2 ⎪ 9 × 8.02 • Question n° 3 Pag.87 meters.66 × 10−12 e j0.652 + 1.0022π = = ε0 8.0022π + jsin 0. So we can evaluate (2.11 79. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Solution • Question n° 1 ⎛ ′′ σ + ωε d ⎞ τ⎞ ′ ⎛ ′ k = β − jα = ω με d ⎜ 1 − j ⎟ = ω με d ⎜ 1 − j ⎟ ′ 2⎠ 2ωε d ⎠ ⎝ ⎝ In the weakly lossy case. In this case ′ ′′ ε d = 9ε 0 and ε d = σ ω .4) and in this case the relative permittivity is complex because the material is lossy. ε = ε d + jε d . i.0022π ) = 2.3) ′ ′′ ′ ′′ where ε d and ε d are the real and imaginary part of the dielectric constant ε .112 e jtan1.29.85 × 10−12 (2. and ′′ ′ τ = tan θ = ε d ε d is the loss tangent that is a convenient way to quantify the losses.e.112 e = 79. 70 D. the complex refractive index of the ground is: n= ε 79.e. Toscano .85 ×10−12 − j = ( 79. the propagation parameter k becomes: (2. • Question n° 2 The definition of the refractive index is: n= ε ε0 (2.0045π = 10−12 79.S.99 + j0.29. Ramaccia and A.11) ×10−12 = 6 ω 900 ×10 jtan1.85 ×10−12 ⎩ The corresponding penetration depth δ = 1 α = 15.85 ×10−12 = 56.66 ×10−12 e j0.29.652 + 1.65 = 10−12 79. using (2. 29. using (2. and the round trip of the wave from the antenna to the buried–pipe is 2r long.10) and.29.8) is: E ret = E 0 e −2α h 2 +d2 h 2 + d 2 −2 jβ h 2 +d 2 e (2.8) from which E ret E0 2 =e −4α h 2 +d 2 ↑ α =1 δ = e −4 h 2 +d 2 δ (2. we obtain: 2r = • Question n° 5 c0 t (h) nr ⇒ t (h) = 2n r 2n r = r d2 + h2 c0 c0 (2. Ramaccia and A. where n r is the real part of the refraction index of the ground. This values can be substituted inside (2. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 Ignoring the angular response of the radar antenna and assuming it emits isotropically in all directions into the ground.7). So the round trip is two times long and the backward signal. using (2.29. v is the constant speed in meter per second and t is the time in second.S.7) The transmitted signal reaches the object and returns back to the radar antenna after getting reflected by the buried–pipe. Toscano Pag.29. can be expressed as: E ret = E 0e −2 jkr = E 0e −2α The module of eq. 71 .29.6) and (2.11) D.29.9) • Question n° 4 The time is the ratio of distance divided by speed that explains the amount of distance covered in a given time: s = v⋅t (2.10) where s is the distance in meter.29.29. The wave propagates into the ground with velocity cg = c0 n r . the electric field into the ground is: E ground = E 0 e − jkr ↑ k = β − jα = E 0 e −α r e − jβ r (2.J.29.29.7).29. (2.6) The distance r between the antenna and the buried–pipe can be evaluated using the Pythagoras' theorem: r = h2 + d2 (2. that is the value of h is zero.29.S.2 × 10−6 = 10 m 2n r 2×3 (2. we can evaluate the depth d in meters: d= c0 3 × 108 t (h) = 0.29. Using (2.J. Orfanidis – Electromagnetic Waves and Antennas Exercises Chapter 2 It is possible to note from (2. Ramaccia and A.11).11) that the minimum roundtrip time t ( h ) is when the antenna is aligned with the pipe. Toscano Pag.12) D.29. 72 .