Chapter 13 The Laplace Transform in Circuit Analysis 13.1 13.2-3 13.4-5 13.6 13.7 13.8 Circuit Elements in the s Domain Circuit Analysis in the s Domain The Transfer Function and Natural Response The Transfer Function and the Convolution Integral The Transfer Function and the SteadyState Sinusoidal Response The Impulse Function in Circuit Analysis 1 Key points How to represent the initial energy of L, C in the s-domain? Why the functional forms of natural and steadystate responses are determined by the poles of transfer function H(s) and excitation source X(s), respectively? Why the output of an LTI circuit is the convolution of the input and impulse response? How to interpret the memory of a circuit by convolution? 2 Section 13.1 Circuit Elements in the s Domain 1. Equivalent elements of R, L, C 3 A resistor in the s domain iv-relation in the time domain: v (t ) R i (t ). By operational Laplace transform: Lv(t ) LR i (t ) R Li (t ) , V ( s ) R I ( s ). Physical units: V(s) in volt-seconds, I(s) in ampere-seconds. 4 An inductor in the s domain iv-relation in the time domain: d v (t ) L i (t ). dt By operational Laplace transform: Lv (t ) LL i(t ) L Li(t ) , V ( s ) L sI ( s ) I 0 sL I ( s ) LI 0 . initial current 5 Equivalent circuit of an inductor Series equivalent: Parallel equivalent: Thévenin Norton 6 A capacitor in the s domain iv-relation in the time domain: d i (t ) C v (t ). dt By operational Laplace transform: Li (t ) LC v(t ) C Lv(t ) , I ( s ) C sV ( s ) V0 sC V ( s ) CV0 . initial voltage 7 Equivalent circuit of a capacitor Parallel equivalent: Series equivalent: Norton Thévenin 8 Section 13.2, 13.3 Circuit Analysis in the s Domain 1. 2. 3. 4. 5. 6. Procedures Nature response of RC circuit Step response of RLC circuit Sinusoidal source MCM Superposition 9 How to analyze a circuit in the s-domain? 1. Replacing each circuit element with its s-domain equivalent. The initial energy in L or C is taken into account by adding independent source in series or parallel with the element impedance. 2. Writing & solving algebraic equations by the same circuit analysis techniques developed for resistive networks. 3. Obtaining the t-domain solutions by inverse Laplace transform. 10 Why to operate in the s-domain? It is convenient in solving transient responses of linear, lumped parameter circuits, for the initial conditions have been incorporated into the equivalent circuit. It is also useful for circuits with multiple essential nodes and meshes, for the simultaneous ODEs have been reduced to simultaneous algebraic equations. It can correctly predict the impulsive response, which is more difficult in the t-domain (Sec. 13.8). 11 Nature response of an RC circuit (1) Q: i(t), v(t) = ? Replacing the charged capacitor by a Thévenin equivalent circuit in the s-domain. KVL, algebraic equation & solution of I(s): V0 I CV0 V0 R IR, I ( s ) . 1 s sC 1 RCs s ( RC ) 12 Nature response of an RC circuit (2) The t-domain solution is obtained by inverse Laplace transform: V0 R V0 t ( RC ) 1 1 i (t ) L L e 1 s s ( RC ) R V0 t ( RC ) e u(t ). R 1 i(0+) = V0/R, which is true for vC(0+) = vC(0-) = V0. i() = 0, which is true for capacitor becomes open (no loop current) in steady state. 13 Nature response of an RC circuit (3) To directly solve v(t), replacing the charged capacitor by a Norton equivalent in the s-domain. Solve V(s), perform inverse Laplace transform: V0 V . CV0 sCV , V ( s ) 1 s ( RC ) R v (t ) L1V0 s ( RC ) 1 V0e t ( RC )u(t ) Ri (t ). 14 Step response of a parallel RLC (1) Q: iL(t) = ? iL(0-) = 0 vC (0-) = 0 15 Step response of a parallel RLC (2) KCL, algebraic equation & solution of V(s): I dc V V I dc C sCV , V ( s ) 2 . 1 1 s R sL s ( RC ) s ( LC ) Solve IL(s): V ( s) I dc ( LC ) 1 2 I L ( s) sL s s ( RC ) 1 s ( LC ) 1 3.84 107 s s (6.4 10 ) s (1.6 10 ) 2 4 9 . 16 Step response of a parallel RLC (3) Perform partial fraction expansion and inverse Laplace transform: 24 20127 20 127 I L ( s) (mA s). s s ( 32k j 24k ) s ( 32k j 24k ) iL (t ) 24u(t ) 20e j127 ( 32 k) t e e j ( 24 k ) t u(t ) c.c. 24 40e ( 32 k)t cos ( 24k)t 127 u(t ) (mA) 24 e ( 32 k)t 24cos(24k )t 32sin(24k )t u(t ) (mA). 17 Transient response due to a sinusoidal source (1) For a parallel RLC circuit, replace the current source by a sinusoidal one: ig (t ) I mcost u(t ). The algebraic equation changes: V V sI m sCV Ig 2 , 2 R sL s I m C s 2 V ( s) 2 , 2 2 1 1 s s ( RC ) s ( LC ) V I m ( LC ) 1 s I L ( s) 2 . 2 2 1 1 sL s s ( RC ) s ( LC ) 18 Transient response due to a sinusoidal source (2) Perform partial fraction expansion and inverse Laplace transform: K1 K1* K2 K 2* . I L ( s) s j s j s ( j ) s ( j ) Driving frequency Neper frequency Damped frequency iL (t ) 2 K1 cost K1 2 K 2 e t cost K 2 u(t ). Steady-state response (source) Natural response (RLC parameters) 19 Step response of a 2-mesh circuit (1) Q: i1(t), i2(t) = ? i1 (0-) = 0 i2(0-) =0 20 Step response of a 2-mesh circuit (2) MCM, 2 algebraic equations & solutions: 336 (1) 8.4 sI1 42( I1 I 2 ) s 42( I 2 I1 ) (10s 48) I 2 0 ( 2) 42 I1 336 42 8.4 s 90 10s I 2 0 42 s . 1 15 14 42 I1 42 8.4 s 336 s s s 2 s 12 . I 42 90 10s 0 7 8.4 1.4 2 s s 2 s 12 1 21 Step response of a 2-mesh circuit (3) Perform inverse Laplace transform: i1 (t ) 15 14e i2 (t ) 7 8.4e 2 t 2 t e 1.4e 12 t 12 t 336 u(t ) (42 // 48) 15 A. 42 u(t ) 15 42 48 7 A. 22 Use of superposition (1) Given 2 independent sources vg, ig and initially charged C, L, v2(t) = ? 23 Use of superposition: Vg acts alone (2) 1 2 1 Vg 1 V1 Vg V1 V1 V2 R sL ( sC ) 1 0, R sL sC V1 sCV2 R , 1 1 1 sCV 1 sC V 0. V2 V1 V2 0. 1 2 R ( sC ) 1 R2 2 24 Use of superposition (3) For convenience, define admittance matrix: 1 1 sC sC R sL V1 1 1 V 2 sC sC R2 Y11 Y12 V1 Vg R1 . Y12 Y22 V2 0 Y12 R1 V2 V. 2 g Y11Y22 Y12 25 Use of superposition: Ig acts alone (4) 1 2 Y11 Y12 V1 0 Y11 Y Y V I , V2 Y Y Y 2 I g . 12 22 2 g 11 22 12 Same matrix Same denominator 26 Use of superposition: Energized L acts alone (5) 1 Y11 Y12 V1 Y Y V 0 12 22 2 Same matrix 2 s Y12 s , V2 Y Y Y 2 . 11 22 12 Same denominator 27 Use of superposition: Energized C acts alone (6) 1 2 "" Y Y 11 12 V1 C Y Y "" C , 12 22 V2 (Y11 Y12 )C "" . V2 2 Y11Y22 Y12 The total voltage is: V2 V2 V2 V2 V2"" . 28 Section 13.4, 13.5 The Transfer Function and Natural Response 29 What is the transfer function of a circuit? The ratio of a circuit’s output to its input in the s-domain: Y ( s) H ( s) X (s) A single circuit may have many transfer functions, each corresponds to some specific choices of input and output. 30 Poles and zeros of transfer function For linear and lumped-parameter circuits, H(s) is always a rational function of s. Poles and zeros always appear in complex conjugate pairs. The poles must lie in the left half of the s-plane if bounded input leads to bounded output. Im Re 31 Example: Series RLC circuit input If the output is the loop current I: I sC 1 2 H ( s) . 1 Vg R sL ( sC ) s LC sRC 1 If the output is the capacitor voltage V: ( sC ) 1 1 V . H ( s) 2 1 Vg R sL ( sC ) s LC sRC 1 32 How do poles, zeros influence the solution? Since Y(s) = H(s) X(s), the partial fraction expansion of the output Y(s) yields a term K/(s-a) for each pole s = a of H(s) or X(s). The functional forms of the transient (natural) and steady-state responses ytr(t) and yss(t) are determined by the poles of H(s) and X(s), respectively. The partial fraction coefficients of Ytr(s) and Yss(s) are determined by both H(s) and X(s). 33 Example 13.2: Linear ramp excitation (1) Q: vo(t) = ? 50tu(t) 50/s2 34 Example 13.2 (2) Only one essential node, use NVM: Vo Vg Vo Vo 6 0, 1000 250 0.05s 10 s Vo 1000( s 5000) H ( s) 2 . 7 Vg s 6000s 2.5 10 H(s) has 2 complex conjugate poles: s 3000 j 4000. Vg(s) = 50/s2 has 1 repeated real pole: s = 0(2). 35 Example 13.2 (3) The total response in the s-domain is: Vo ( s ) H ( s )Vg ( s ) 5 104 ( s 5000) s 2 s 2 6000s 2.5 107 Ytr Yss expansion coefficients depend on H(s) & Vg(s) 5 5 104 80 5 5 104 80 10 4 104 . 2 s 3000 j 4000 s 3000 j 4000 s s poles of H(s): -3k j4k pole of Vg(s): 0(2) The total response in the t-domain: vo (t ) ytr yss 5 10 3 3, 000 t e cos(4,000t 80 ) u (t ) 10t 4 10 4 u (t ). 36 Example 13.2 (4) Steady state component yss(t) Total response = 0.33 ms, impact of ytr(t) 37 Section 13.6 The Transfer Function and the Convolution Integral 1. 2. 3. 4. Impulse response Time invariant Convolution integral Memory of circuit 38 Impulse response If the input to a linear, lumped-parameter circuit is an impulse (t), the output function h(t) is called impulse response, which happens to be the natural response of the circuit: X ( s ) L (t ) 1, Y ( s ) H ( s ) 1 H ( s ), y (t ) L1Y ( s ) L1H ( s ) h(t ). The application of an impulse source is equivalent to suddenly storing energy in the circuit. The subsequent release of this energy gives rise to the natural response. 39 Time invariant For a linear, lumped-parameter circuit, delaying the input x(t) by simply delays the response y(t) by as well (time invariant): X ( s, ) Lx (t )u(t ) e s X ( s ), Y ( s, ) H ( s ) X ( s, ) e s H ( s ) X ( s ) e sY ( s ), y (t , ) L1Y ( s, ) L1Y ( s ) t t y (t )u(t ). 40 Motivation of working in the time domain The properties of impulse response and timeinvariance allow one to calculate the output function y(t) of a “linear and time invariant (LTI)” circuit in the t-domain only. This is beneficial when x(t), h(t) are known only through experimental data. 41 Decompose the input source x(t) We can approximate x(t) by a series of rectangular pulses rec(t-i) of uniform width : sq (t ) sq (t i ) By having , rec(t-i)/ (t-i), x(t) converges to a train of impulses: x(t ) x(i ) lim rec (t i ) i 0 0 x(i ) (t i ). i 0 42 Synthesize the output y(t) (1) Since the circuit is LTI: (t ) h(t ), (t i ) h(t i ), a x (t ) a y (t ); i i i i x (i ) (t i ) i 0 x (i ) h(t i ). i 0 43 Synthesize the output y(t) (2) As , summation integration: if x(t) extends (-, ) y (t ) x ( )h(t )d x ( )h(t )d . 0 By change of variable u = t-, y (t ) x (t u )h(u )du. The output of an LTI circuit is the convolution of input and the impulse response of the circuit: y (t ) x (t ) h (t ) x ( )h(t )d x (t )h( )d . 44 Convolution of a causal circuit For physically realizable circuit, no response can occur prior to the input excitation (causal), {h(t) = 0 for t < 0}. Excitation is turned on at t = 0, {x(t) = 0 for t < 0}. y (t ) x (t ) h (t ) t x (t )h( )d . 0 45 Effect of x(t) is weighted by h(t) The convolution integral t y (t ) x (t )h( )d 0 shows that the value of y(t) is the weighted average of x(t) from t = 0 to t = t [from = t to = 0 for x(t-)]. 0 t If h(t) is monotonically decreasing, the highest weight is given to the present x(t). 46 Memory of the circuit If h(t) only lasts from t = 0 to t = T, the convolution integral T t y (t ) x (t )h( )d . 0 implies that the circuit has a memory over a finite interval t = [t-T,t]. If h(t) = (t), no memory, output at t only depends on x(t), y(t) = x(t)*(t) = x(t), no distortion. 47 Example 13.3: RL driven by a trapezoidal source (1) Q: vo(t) = ? 1 Vo 1 Vo Vi , H ( s ) . s 1 Vi s 1 1 t h (t ) L e u(t ). s 1 1 48 Example 13.3 (2) t vo (t ) vi (t )h( )d . Separate into 3 intervals: 0 49 Example 13.3 (3) Since the circuit has certain memory, vo(t) has some distortion with respect to vi(t). 50 Section 13.7 The Transfer Function and the Steady-State Sinusoidal Response 51 How to get sinusoidal steady-state response by H(s)? In Chapters 9-11, we used phasor analysis to get steady-state response yss(t) due to a sinusoidal input x (t ) A cost . If we know H(s), yss(t) must be: yss (t ) H ( j ) A cost ( ), where H ( j ) H ( s ) s j H ( j ) e j ( ) . The changes of amplitude and phase depend on the sampling of H(s) along the imaginary axis. 52 Proof x (t ) A cost A cos cos t A sin sin t. X ( s ) A cos s cos sin s A sin A . 2 2 2 2 2 2 s s s s cos sin Y ( s) H ( s) X ( s) H ( s) A Ytr ( s ) Yss ( s ), 2 2 s K1 K1* where Yss ( s) , K1 Y ( s )( s j ) s j s j s j s cos sin H ( s) A s j s j j cos sin H ( j ) Ae j H ( j ) A . 2 j 2 j ( ) j H ( j ) e Ae 1 yss (t ) L c.c. A H ( j ) cost ( ). 2( s j ) 53 Obtain H(s) from H( j) We can reverse the process: determine H( j) experimentally, then construct H(s) from the data (not always possible). Once we know H(s), we can find the response to other excitation sources. 54 Section 13.8 The Impulse Function in Circuit Analysis 55 E.g. Impulsive inductor voltage (1) Q: vo(t) = ? i1(0-)=10 A i2(0-)=0 The opening of the switch forces the two inductor currents i1, i2 change immediately by inducing an impulsive inductor voltage [v = Li'(t)]. 56 E.g. Equivalent circuit & solution in the s-domain (2) initial current V0 (100 s 30) V0 0, 10 3s 15 2s improper rational 2(6s 2 65s 150) 60 10 V0 ( s ) . 12 s ( s 5) s s5 57 E.g. Solutions in the t-domain (3) 60 10 5 t v0 (t ) L 12 12 ( t ) ( 60 10 e )u(t ). s s 5 1 To verify whether this solution vo(t) is correct, we need to solve i(t) as well. 100 s 30 4 2 I (s) , i (t ) (4 2e 5t )u (t ). 10 3s 15 2s s s 5 jump jump 58 Impulsive inductor voltage (4) The jump of i2(t) from 0 to 6 A causes i2 (t ) 6 (t ) , contributing to a voltage impulse L2i2 (t ) 12 (t ). After t > 0+, vo (t ) (15)i2 (t ) ( 2 H )i2 (t ) 15( 4 2e 5t ) 2( 10e 5t ) 60 10e 5t , consistent with that solved by Laplace transform. 59 Key points How to represent the initial energy of L, C in the s-domain? Why the functional forms of natural and steadystate responses are determined by the poles of transfer function H(s) and excitation source X(s), respectively? Why the output of an LTI circuit is the convolution of the input and impulse response? How to interpret the memory of a circuit by convolution? 60 Practical Perspective Voltage Surges 61 Why can a voltage surge occur? Q: Why a voltage surge is created when a load is switched off? Model: A sinusoidal voltage source drives three loads, where Rb is switched off at t = 0. Since i2(t) cannot change abruptly, i1(t) will jump by the amount of i3(0-), voltage surge occurs. 62 Example Let Vo = 1200 (rms), f = 60 Hz, Ra = 12 , Rb = 8 , Xa = 41.1 (i.e. La = Xa/ = 109 mH), Xl = 1 (i.e. Ll = 2.65 mH). Solve vo(t) for t > 0-. To draw the s-domain circuit, we need to calculate the initial inductor currents i2(0-), i0(0-). 63 Steady-state before the switching The three branch currents (rms phasors) are: I1 = Vo/Ra = (1200)/(12 ) = 100 A, I2 = Vo/(jXa) = (1200)/(j41.1 ) = 2.92-90 A, I3 = Vo/Rb = (1200)/(8 ) = 150 A, The line current is: I0 = I1 + I2 + I3 = 25.2-6.65 A. Source voltage: Vg = Vo + I0(jXl) = 125-11.5 V. The two initial inductor currents at t = 0- are: i2(t) = 2.92(2)cos(120t-90), i2(0-) = 0; i0(t) = 25.2(2)cos(120t-6.65), i0(0-) = 35.4 A. 64 S-domain analysis The s-domain circuit is: (I0 = 35.4 A) Vg = 12511.5 V (rms) By NVM: (Ll = 2.65 mH) (12 ) Vo Ll I 0 Vg sLl (La = 109 mH) Vo Vo 0, Ra sLa Ra Ll Vg ( s) I 0 Ra 253 Vo s Ra ( La Ll ) La Ll s 1475 866.85 86 6.85 s j120 s j120 65 Inverse Laplace transform 253 866.85 86 6.85 , Given Vo ( s ) s 1475 s j120 s j120 vo (t ) 253e 1475t 173 cos(120t 6.85 ) u (t ). 66