EE 422G Notes: Chapter 6 Instructor: CheungPage 6-1 Applications of the Laplace Transform Application in Circuit Analysis 1. Review of Resistive Network 1) Elements 2) Superposition 1 PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-2 3) KVL and KCL – Select a node for ground. Watch out for signs! 4) Equivalent Circuits Open Circuit Voltage s OC V V = = R s = Equivalent Resistance Short Circuit Current s SC I I = = Thevenin Equivalent Circuit Norton Equivalent Circuit R s = Same as before PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-3 5) Nodal Analysis and Mesh Analysis Mesh analysis (use KVL) ¹ ´ ¦ + + = − + − + = 2 2 4 1 1 1 2 1 3 2 1 2 1 1 1 ) ( ) ( S S S V I R I R V I I R I I R I R V Solve for I 1 and I 2 . 2. Characteristics of Dynamic Networks 1) Inductor 2) Capacitor ∫ ∞ − = = t C C C C d i C t v t v dt d C t i τ τ ) ( 1 ) ( or ) ( ) ( ∫ ∞ − = = t L L L L d v L t i t i dt d L t v τ τ ) ( 1 ) ( or ) ( ) ( (Use KCL) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-4 3) Operation Amplifier A general op-amp model is described above. In practice, the input resistance, R in , is very large (> 10 12 Ω) and the gain, A, is very large (>10 5 ). Thus, we will use the ideal model in the analysis: 1. Input current I i (t) = 0 (due to the large input impedence) 2. Input voltage difference v i (t) = 0 and output voltage v o (t) is dictated by external circuit (due to the large gain) Based on the ideal op-amp model, v 2 (t) = v 1 (t) (1) Also, as the op-amp does not have any input current, applying KCL at the inverting port, we have v 2 (t)/R a = (v o (t)-v 2 (t))/R b v o (t)/v 2 (t) = 1+R b /R a Plug in (1), we have v o (t)/v 1 (t) = 1+R b /R a This circuit is called Non-Inverting Amplifier. - + ≡ + - + - v o (t) v i (t) + - v i (t) - + - Av i (t) R in I i (t) Inverting input Non-Inverting input - + + - v 1 (t) R a R b v o (t) i o (t) v 2 (t) + - + - v o (t) Example: PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-5 4) Mutual Inductor – used in transformer. Two separate circuits with coupling currents. To link the two circuits together, introduce a combined current term (i 1 +i 2 ): dt di M L i i dt d M dt di M dt di L dt di M dt di M t v i i dt d M dt di M L dt di M dt di M dt di M dt di L t v 2 2 2 1 2 2 2 2 1 2 2 1 1 1 2 1 1 1 1 1 ) ( ) ( ) ( ) ( ) ( ) ( − + + = − + + = + + − = + + − = Equivalent circuit: Make sure both i 1 and i 2 point either away or toward the polarity marks to make the mutual inductance M positive. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-6 Example : Apply mesh analysis to the following circuit Using Laplace Transform ) ( ) ( 1 ) ( ) ( ) ( ) 0 ( ) ( 1 )) 0 ( ) ( ( ) ( ) ( ) ( ) ( s RI s I Cs s I Ls s RI s v s s I C i s sI L s V s V s V s V C R C L + + = + ( ¸ ( ¸ + + − = + + = − Define ‘Generalized Resistors’ (Impedances) ) ( ) ( ) ( ) ( s Z s I s V Ls s Z L L L = ⇒ = ) ( ) ( ) ( 1 ) ( s Z s I s V Cs s Z C C C = ⇒ = Both capacitor and inductor behave exactly like a resistor! R s Z s Z s V s I s RI s I s Z s I s Z s V C L s C L s + + = ⇒ + + = ⇒ ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( Everything we know about resistive network can be applied to dynamic network in Laplace domain: ¦ ¦ ¦ ) ¦ ¦ ¦ ` ¹ analysis mesh and analysis Nodal circuit Equivalent KCL and KVL ion superposit Law Ohms d Generalize PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-7 3. Laplace transform models of circuit elements. What if the initial conditions are not zero? 1) Capacitor Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE!! Z C Cv(0 - ) I(s) + - V(s) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-8 2) Inductor Alternatively, you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) 3) Resistor V(s) = RI(s) 4) Voltage and Current Sources (Don’t forget to apply Laplace Transform on them) 5) Op-Amp : same ideal model assumption Z L i(0 - )/s I(s) + - V(s) PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-9 6) Mutual Inductance (Transformers) ⇓ Laplace transform model: Obtain it by using inductance model PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-10 Example: Find Complex Norton Equivalent circuit given 0 ) 0 ( = − c v Solution 1) Compute the Short-Circuit Current sc s I s I = ) ( Straightforward to see: ) ( 2 ) ( s I s I S − = To compute I(s), apply mesh analysis on the left loop: 3 2 ) ( 2 ) ( 3 1 ) ( ) ( 3 ) ( ) 3 1 ( 1 ) ( 3 1 ) ( 1 + − = − = ⇒ + = ⇒ + = + = + Ω × = s s I s I s s I s I s s s I s s s I s I s s No need to do inverse Laplace transform as the equivalent circuit is in the s- domain. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-11 2) Find the equivalent impedance s Z Normally, we can just kill all the independent sources and combine the impedances (using resistive combination rules). However, as there is a dependent source, we need to drive it with a test voltage: ) ( 2 ) ( ) ( ) ( s I s V s I s V Z test test test s = = Mesh analysis on the left loop: 0 ) ( ) ( 3 ) ( ) 1 ( 0 ) ( 3 1 ) ( = ⇒ | ¹ | \ | − = Ω ⇒ = + Ω × s I s I s s I s s I s I So we got an interesting result: CIRCUIT OPEN 0 ) ( ⇒ ∞ = = s V Z test s Vtest(s) a Z L V test (s) + - I test (s) Z s PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-12 Example: Find the transfer function H(s) = V o (s)/V i (s) of the following circuit. Assume all initial conditions are zero. This is called the Sallen-Key circuit, which we will see again in filter design. Rewrite everything in Laplace domain, we have We recognize the op-amp configuration as a non-inverting amplifier, so we have a b R R K + =1 PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung Page 6-13 To find V o , we need V b which depends on V a . All other nodal voltages are known. Thus, we need two nodal equations: Applying KCL at node a, we have: 0 1 1 2 1 = − + − + − s C KV V R V V R V V b a b a i a 1 1 1 2 1 2 1 1 1 1 1 V R V s KC R V s C R R b a = | | ¹ | \ | + − | | ¹ | \ | + + ⇒ (1) Applying KCL at node b, we have: 0 2 2 = + − b a b sV C R V V 0 1 1 1 2 2 = | | ¹ | \ | + + − ⇒ b a V s KC R V R (2) Combining equations (1) and (2) by eliminating V a , we get: [ ] 1 ) 1 ( 1 1 1 2 1 2 2 2 2 1 2 1 + − + + + = s K C R C R C R s C C R R V V i b Since b o KV V = , we have [ ] 1 ) 1 ( 1 1 2 1 2 2 2 2 1 2 1 + − + + + = s K C R C R C R s C C R R K V V i o where a b R R K + =1 PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com EE 422G Notes: Chapter 6 Instructor: Cheung 3) KVL and KCL – Select a node for ground. Watch out for signs! 4) Equivalent Circuits Thevenin Equivalent Circuit Vs = VOC = Open Circuit Voltage Rs = Equivalent Resistance Norton Equivalent Circuit I s = I SC = Short Circuit Current Rs = Same as before Page 6-2 PDF Created with deskPDF PDF Writer .com .docudesk.Trial :: http://www. com . Characteristics of Dynamic Networks 1) Inductor v L (t ) = L d iL ( t ) dt 1 t or iL (t ) = ∫ v L (τ )dτ L −∞ 2) Capacitor iC (t ) = C d vC ( t ) dt 1 t or vC (t ) = ∫ iC (τ )dτ C −∞ Page 6-3 PDF Created with deskPDF PDF Writer .docudesk. 2.EE 422G Notes: Chapter 6 Instructor: Cheung 5) Nodal Analysis and Mesh Analysis (Use KCL) Mesh analysis (use KVL) VS 1 = R1 I1 + R2 ( I1 − I 2 ) + R3 ( I1 − I 2 ) VS 1 = R1 I1 + R4 I 2 + VS 2 Solve for I1 and I2.Trial :: http://www. Rin. the input resistance.Trial :: http://www. - In practice. Thus.EE 422G Notes: Chapter 6 Instructor: Cheung 3) Operation Amplifier Non-Inverting input + vi(t) - + Inverting input + + ≡ - Ii(t) Rin Avi(t) vi(t) + + vo(t) vo(t) A general op-amp model is described above.docudesk. is very large (>105). Page 6-4 PDF Created with deskPDF PDF Writer . applying KCL at the inverting port. Input current Ii(t) = 0 (due to the large input impedence) 2. v2(t) = v1(t) (1) Also. is very large (> 1012 ) and the gain. as the op-amp does not have any input current. Input voltage difference vi(t) = 0 and output voltage vo(t) is dictated by external circuit (due to the large gain) Example: v2(t) v1(t) + Ra + Rb io(t) + vo(t) Based on the ideal op-amp model. A. we have vo(t)/v1(t) = 1+Rb/Ra This circuit is called Non-Inverting Amplifier. we have v2(t)/Ra= (vo(t)-v2(t))/Rb vo(t)/v2(t) = 1+Rb/Ra Plug in (1). we will use the ideal model in the analysis: 1.com . introduce a combined current term (i1+i2): di di di di v1 (t ) = L1 1 − M 1 + M 1 + M 2 dt dt dt dt di d = ( L1 − M ) 1 + M (i1 + i2 ) dt dt di di di di v 2 (t ) = M 1 + M 2 + L2 2 − M 2 dt dt dt dt d di = M (i1 + i2 ) + ( L2 − M ) 2 dt dt Equivalent circuit: Page 6-5 PDF Created with deskPDF PDF Writer . To link the two circuits together.com .docudesk. Make sure both i1 and i2 point either away or toward the polarity marks to make the mutual inductance M positive.EE 422G Notes: Chapter 6 Instructor: Cheung 4) Mutual Inductor – used in transformer. Two separate circuits with coupling currents.Trial :: http://www. EE 422G Notes: Chapter 6 Instructor: Cheung Example : Apply mesh analysis to the following circuit Using Laplace Transform V ( s ) = VL ( s ) + VC ( s ) + VR ( s ) = L( sI ( s ) − i (0− )) + = ( Ls ) I ( s ) + 1 I ( s ) vC (0) + + RI ( s ) C s s 1 I ( s ) + RI ( s ) Cs Define ‘Generalized Resistors’ (Impedances) Z L ( s ) = Ls ⇒ VL ( s ) = I ( s ) Z L ( s ) 1 ⇒ VC ( s ) = I ( s ) Z C ( s ) Cs Both capacitor and inductor behave exactly like a resistor! ZC ( s) = ⇒ Vs ( s ) = Z L ( s ) I ( s ) + Z C ( s ) I ( s ) + RI ( s ) ⇒ I ( s) = Vs ( s ) Z L ( s) + Z C ( s) + R Everything we know about resistive network can be applied to dynamic network in Laplace domain: superposition KVL and KCL Equivalent circuit Nodal analysis and mesh analysis Generalized Ohms Law Page 6-6 PDF Created with deskPDF PDF Writer .com .Trial :: http://www.docudesk. com . you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) I(s) ZC Cv(0-) V(s) - + BE VERY CAREFUL ABOUT THE POLARITY OF VOLTAGE SOURCE AND THE DIRECTION OF CURRENT SOURCE!! Page 6-7 PDF Created with deskPDF PDF Writer .EE 422G Notes: Chapter 6 Instructor: Cheung 3. Laplace transform models of circuit elements.docudesk.Trial :: http://www. What if the initial conditions are not zero? 1) Capacitor Alternatively. Trial :: http://www. you can also represent it as an impedance and a parallel current source (Norton equivalent circuit) I(s) ZL i(0-)/s V(s) 3) Resistor V(s) = RI(s) - + 4) Voltage and Current Sources (Don’t forget to apply Laplace Transform on them) 5) Op-Amp : same ideal model assumption Page 6-8 PDF Created with deskPDF PDF Writer .com .docudesk.EE 422G Notes: Chapter 6 Instructor: Cheung 2) Inductor Alternatively. docudesk.EE 422G Notes: Chapter 6 Instructor: Cheung 6) Mutual Inductance (Transformers) ⇓ Laplace transform model: Obtain it by using inductance model Page 6-9 PDF Created with deskPDF PDF Writer .com .Trial :: http://www. EE 422G Notes: Chapter 6 Instructor: Cheung Example: Find Complex Norton Equivalent circuit given vc (0 − ) = 0 Solution 1) Compute the Short-Circuit Current I s ( s) = I sc Straightforward to see: I S ( s ) = −2 I ( s ) To compute I(s). apply mesh analysis on the left loop: 1 1 3 s+3 = I ( s ) × 1 Ω + 3I ( s ) = (1 + ) I ( s ) = I ( s) s s s s 1 ⇒ I (s) = s+3 2 ⇒ I s ( s ) = −2 I ( s ) = − s+3 No need to do inverse Laplace transform as the equivalent circuit is in the sdomain.docudesk.com . Page 6-10 PDF Created with deskPDF PDF Writer .Trial :: http://www. as there is a dependent source.EE 422G Notes: Chapter 6 Instructor: Cheung 2) Find the equivalent impedance Z s Normally. we need to drive it with a test voltage: Itest(s) Vtest(s) Zs + - Vtest(s) Zs = Mesh analysis on the left loop: Vtest ( s) Vtest ( s) = I test ( s ) 2 I ( s ) 3I ( s ) =0 s 3 ⇒ (1 Ω) I ( s ) = − I ( s ) s ⇒ I ( s) = 0 I ( s) × 1 Ω + So we got an interesting result: Z s = Vtest ( s ) = ∞ ⇒ OPEN CIRCUIT 0 a ZL Page 6-11 PDF Created with deskPDF PDF Writer .Trial :: http://www.com .docudesk. we can just kill all the independent sources and combine the impedances (using resistive combination rules). However. docudesk. Rewrite everything in Laplace domain.EE 422G Notes: Chapter 6 Instructor: Cheung Example: Find the transfer function H(s) = Vo(s)/Vi(s) of the following circuit.com . we have We recognize the op-amp configuration as a non-inverting amplifier. This is called the Sallen-Key circuit. which we will see again in filter design. Assume all initial conditions are zero. so we have K = 1+ Rb Ra Page 6-12 PDF Created with deskPDF PDF Writer .Trial :: http://www. we have Vo K = 2 Vi R1 R2 C1C 2 s + [R2 C 2 + R1C 2 + R1C1 (1 − K )]s + 1 where K = 1 + Rb Ra Page 6-13 PDF Created with deskPDF PDF Writer .Trial :: http://www. we need Vb which depends on Va. we need two nodal equations: Applying KCL at node a. we get: Vb 1 = 2 Vi R1 R2 C1C 2 s + [R2 C 2 + R1C 2 + R1C1 (1 − K )]s + 1 Since Vo = KVb .docudesk. we have: Vb − Va + C 2 sVb = 0 R2 ⇒− 1 1 Va + R + KC1 s Vb = 0 R2 2 (2) Combining equations (1) and (2) by eliminating Va.EE 422G Notes: Chapter 6 Instructor: Cheung To find Vo. we have: Va − Vi Va − Vb Va − KVb + + =0 1 R1 R2 C1 s 1 1 1 1 ⇒ + R R + C1 s Va − R + KC1 s Vb = R V1 1 2 2 1 (1) Applying KCL at node b. Thus.com . All other nodal voltages are known.