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S-Domain Circuit Analysis s-Domain Circuit Analysis Operate directly in the s-domain with capacitors, inductors and resistors Key feature – linearity is preserved Ccts described by ODEs and their ICs Order equals number of C plus number of L Element-by-element and source transformation Nodal or mesh analysis for s-domain cct variables Solution via Inverse Laplace Transform Why? Easier than ODEs Easier to perform engineering design Frequency response ideas - filtering 165 MAE140 Linear Circuits Element Transformations Voltage source i(t) Time domain + vS _ v(t) =vS(t) i(t) = depends on cct Transform domain V(s) =VS(s)=L(vS(t)) I(s) = L(i(t)) depends on cct Current source iS I(s) =L(iS(t)) V(s) = L(v(t)) depends on cct v(t) 166 MAE140 Linear Circuits Element Transformations contd v (t) v (t) V (s) V (s) Controlled sources 1 = µ 2 ! 1 = µ 2 i1(t) = "i2(t) ! I1(s) = "I2(s) v1(t) = ri2(t) ! V1(s) = rI2(s) i1(t) = gv2(t) ! I1(s) = gV2(s) Short cct, open cct, OpAmp relations vSC (t) = 0 ! VSC (s) = 0 iOC (t) = 0 ! IOC (s) = 0 vN (t) = vP (t) ! VN (s) = VP (s) Sources and active devices behave identically Constraints expressed between transformed variables This all hinges on uniqueness of Laplace Transforms and linearity 167 MAE140 Linear Circuits Element Transformations contd i iR iC L + + + Resistors v R vC vL vR (t) = RiR (t) iR (t) = GvR (t) V (s) = RI (s) I (s) = GV (s) R R R R 1 Z (s) = R Y (s) = Capacitors R R R dv (t) 1 t i (t) = C C v (t) = i (# )d# + v (0) C dt C C " C C 0 1 v (0) I (s) = sCV (s) ! Cv (0!) V (s) = I (s) + C C C C C sC C s 1 Inductors ZC (s) = YC (s) = sC di (t) 1 t sC v (t) = L L i (t) = v (# )d# + i (0) L dt L L " L L 0 1 i (0) V (s) = sLI (s) ! Li (0) I (s) = V (s) + L L L L L sL L s 1 ZL(s) = sL YL (s) = sL 168 MAE140 Linear Circuits Element Transformations contd i (t) IR(s) Resistor R + + VR (s) = RIR (s) R vR(t) R VR(s) - - Capacitor IC(s) i (t) IC(s) C + + + 1 1 vC(t) C VC(s) sC sC CvC (0) - - VC(s) v (0) +_ C s 1 v (0) - I (s) = sCV (s) − Cv (0−) V (s) = I (s) + C C C C C sC C s Note the source transformation rules apply! 169 MAE140 Linear Circuits Element Transformations contd 1 i (0) Inductors V (s) = sLI (s) − Li (0) I (s) = V (s) + L L L L L sL L s I (s) L I (s) + L + i (0) iL(t) sL V (s) L L sL + s V (s) L - vL(t) _ - + LiL(0) - 170 MAE140 Linear Circuits Example 10-1, T&R, 5th ed, p 456 RC cct behavior Switch in place since t=-∞, closed at t=0. Solve for vC(t). t=0 R R I1(s) I (s) + 2 v C 1 VA C Cv (0) - sC C 171 MAE140 Linear Circuits Example 10-1 T&R, 5th ed, p 456 RC cct behavior Switch in place since t=-∞, closed at t=0. Solve for vC(t). t=0 R R I1(s) I (s) + 2 v C 1 VA C Cv (0) - sC C Initial conditions vC (0) = VA s-domain solution using nodal analysis V (s) V (s) I (s) = C I (s) = C = sCV (s) 1 R 2 1 C sC t-domain solution via inverse Laplace transform V ! t V (s) = A v (t) = V e RCu(t) C 1 c A s + RC 172 MAE140 Linear Circuits Example 10-2 T&R, 5th ed, p 457 R Solve for i(t) + R V A +_ I(s) sL s VL (s) V u(t) + A _ L _ i(t) + LiL(0) - 173 MAE140 Linear Circuits Example 10-2 T&R, 5th ed, p 457 R Solve for i(t) + R V A +_ I(s) sL s VL (s) V u(t) + A _ L _ i(t) + LiL(0) - V KVL around loop A ! (R + sL)I(s) + Li (0) = 0 s L # V & VA VA % i (0) " A ( Solve i (0) $ L R' I(s) = L + L = R + R s + R s s + R s(s + L) L L # Rt Rt & VA VA " L " L Invert i(t) = % " e + iL (0)e (u (t) Amps ! $ R R ' 174 MAE140 Linear Circuits ! Impedance and Admittance Impedance (Z) is the s-domain proportionality factor relating the transform of the voltage across a two-terminal element to the transform of the current through the element with all initial conditions zero V(s) = Z(s)I(s) Admittance (Y) is the s-domain proportionality factor relating the transform of the current through a two-terminal element to the transform of the ! voltage across the element with initial conditions zero I(s) = Y(s)V(s) Impedance is like resistance Admittance is like conductance ! 175 MAE140 Linear Circuits Circuit Analysis in s-Domain Basic rules The equivalent impedance Zeq(s) of two impedances Z1(s) and Z2(s) in series is Zeq (s) = Z1(s) + Z2(s) I(s) Same current flows + Z1 V (s) Z (s)I s Z (s)I(s) Z (s)I s V(s) = 1 ( )+ 2 = eq ( ) Z2 - The equivalent admittance Yeq(s) of two admittances Y1(s) and Y2(s) in parallel is Yeq (s) = Y1(s) +Y2(s) I(s) Same voltage + V(s) I(s) = Y1(s)V (s) +Y2(s)V (s) = Yeq (s)V (s) Y1 Y2 - 176 MAE140 Linear Circuits Example 10-3 T&R, 5th ed, p 461 Find ZAB(s) and then find V2(s) by voltage division L sL A A + + v (t) +_ C R v (t) + 1 1 2 V1(s) _ R V (s) sC 2 - - B B 1 1 RLCs2 + Ls + R Z (s) = sL + R = sL + = Ω eq 1 + sC + sC RCs 1 R & Z1(s) # & R # V (s) = $ !V (s) = V (s) 2 Z (s) 1 $ 2 ! 1 %$ eq "! %RLCs + sL + R" 177 MAE140 Linear Circuits Superposition in s-domain ccts The s-domain response of a cct can be found as the sum of two responses 1. The zero-input response caused by initial condition sources, with all external inputs turned off 2. The zero-state response caused by the external sources, with initial condition sources set to zero Linearity and superposition Another subdivision of responses 1. Natural response – the general solution Response representing the natural modes (poles) of cct 2. Forced response – the particular solution Response containing modes due to the input 178 MAE140 Linear Circuits Example 10-6, T&R, 5th ed, p 466 The switch has been open for a long time and is closed at t=0. Find the zero-state and zero-input components of V(s) Find v(t) for IA=1mA, L=2H, R=1.5KΩ, C=1/6 µF t=0 + + V(s) v(t) I 1 A sL R L R C RCIA IA s sC - - 179 MAE140 Linear Circuits Example 10-6, T&R, 5th ed, p 466 The switch has been open for a long time and is closed at t=0. Find the zero-state and zero-input components of V2(s) Find v(t) for IA=1mA, L=2H, R=1.5KΩ, C=1/6 µF t=0 + + V(s) v(t) I 1 A sL R L R C RCIA IA s sC - - I A I A C 1 RLs Vzs (s) = Zeq (s) = Z (s) = = s 2 1 1 eq 1 1 2 s + s + + + sC RLCs + Ls + R RC LC sL R RI As Vzi (s) = Zeq (s)RCI A = 2 1 1 s + s + 180 MAE140 Linear Circuits RC LC Example 10-6 contd I A + I A C Vzs (s) = Zeq (s) = s 2 1 1 V(s) s + s + I A 1 RC LC sL R RCI s sC A RI As Vzi (s) = Zeq (s)RCI A = - 2 1 1 s + s + RC LC Substitute values 6000 3 "3 Vzs(s) = = + (s +1000)(s + 3000) s +1000 s + 3000 "1000t "3000t vzs(t) = [3e " 3e ]u(t) 1.5s "0.75 2.25 Vzi (s) = = + ! (s +1000)(s + 3000) s +1000 s + 3000 "1000t "3000t vzi (t) = ["0.75e + 2.25e ]u(t) What are the natural and forced responses? 181 MAE140 Linear Circuits ! Example Formulate node voltage equations in the s-domain C R2 R1 2 + + + v (t) + C R v (t) µvx(t) v (t) 1 _ 1 3 x - 2 - - 182 MAE140 Linear Circuits Example Formulate node voltage equations in s-domain C R2 R1 2 + + + v (t) + C R v (t) µvx(t) v (t) 1 _ 1 3 x - 2 - - R R2 A 1 B C D + + + 1 + V (s) V1(s) _ R V (s) µ x V (s) sC 3 x - 2 1 - - 1 C v (0) C1vC1(0) 2 C2 sC2 183 MAE140 Linear Circuits Example contd R2 R1 A B C D + + + 1 + V (s) V1(s) _ R V (s) µ x V (s) sC 3 x - 2 1 - - 1 C v (0) C1vC1(0) 2 C2 sC2 V (s) V (s) V (s) Node A: V A ( s ) = V 1 ( s ) Node D: D = µ x = µ C Node B: V (s) "V (s) V (s) "V (s) V (s) B A + B D + B R1 R2 1 sC1 V (s) "V (s) + B C " C v (0) " C v (0) = 0 1 1 C1 2 C2 sC2 Node C: "sC2VB (s) + [sC2 + G3]VC (s) = "C2vC2(0) ! 184 MAE140 Linear Circuits ! Example Find vO(t) when vS(t) is a unit step u(t) and vC(0)=0 A B C D + R R1 C 2 v (t) + - O vS(t) _ + Convert to s-domain 185 MAE140 Linear Circuits Example Find vO(t) when vS(t) is a unit step u(t) and vC(0)=0 A B C D + R R1 C 2 v (t) + - O vS(t) _ + Convert to s-domain 1 R 1 V (s) sC V (s) R2 V (s) VA(s) B C D + V (s) + - O VS(s) _ + CvC(0) 186 MAE140 Linear Circuits 1 Example R 1 V (s) sC V (s) R2 V (s) VA(s) B C D + V (s) Nodal Analysis + - O V (s) _ + S Cv (0) Node A:VA(s) = VS (s) C Node D:VD (s) = VO (s) Node C: VC (s) = 0 Node B:(G1 + sC)VB (s) ! G1VS (s) = CvC (0) Node C KCL: ! sCVB (s) ! G2VO (s) = !CvC (0) Solve for VO(s) # sG1C & # & % G ( R s V (s) = " 2 V (s) = "% 2 ) (V (s) O % ( S 1 S G1 + sC % R1 s + ( $% '( $ R1C' # & R s 1 "R 1 = "% 2 ) ( = 2 ) 1 1 % R1 s + ( s R1 s + $ R1C' R1C t − − Invert LT R2 R1C vO (t) = e u(t) R 187 MAE140 Linear Circuits 1 ! Features of s-domain cct analysis The response transform of a finite-dimensional, lumped-parameter linear cct with input being a sum of exponentials is a rational function and its inverse Laplace Transform is a sum of exponentials The exponential modes are given by the poles of the response transform Because the response is real, the poles are either real or occur in complex conjugate pairs The natural modes are the zeros of the cct determinant and lead to the natural response The forced poles are the poles of the input transform and lead to the forced response 188 MAE140 Linear Circuits Features of s-domain cct analysis A cct is stable if all of its poles are located in the open left half of the complex s-plane A key property of a system Stability: the natural response dies away as t→∞ Bounded inputs yield bounded outputs A cct composed of Rs, Cs and Ls will be at worst marginally stable With Rs in the right place it will be stable Z(s) and Y(s) both have no poles in Re(s)>0 Impedances/admittances of RLC ccts are “Positive Real” or energy dissipating 189 MAE140 Linear Circuits.
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