Unit 6 Introduction to Laplace Tranform Technique in Circuit Analyi In thi unit we conider the application of Laplace Tranform to circuit analyi. A relevant dicuion of the one-ided Laplace tranform i found in Chapter 12 of the text. The Laplace Tranform of any time function, ay f(t), which you have encountered in a co-requiite math coure will be ymbolized in thee note a L{f(t)} = 0 f(t)e t dt = F (). Thi mathematical contruct will allow the converion of many of the differential equation encountered in circuit analyi into algebraic equation. The latter may generally olved with relative eae and an invere Laplace Tranform, L 1 {F ()} = f(t), may be ued to determine the time-domain olution of the original equation. 6.1 Circuit Element To model a circuit element in the -domain we imply Laplace tranform the voltage current equation for the element terminal in the time domain. Thi give the - domain relationhip between the voltage and the current which may be modelled by an appropriate circuit. The tranformation of a voltage and current from the time domain reult in dimenion of volt-econd and ampere-econd in the -domain. Impedance i till meaured in ohm in the -domain. We will ue the paive ign convention in our -domain model. Alo, we ll ue V and I to mean V () and I(), 1
repectively. Reitor in the -Domain In the time domain Time Domain Model v = ir. Since R i a contant, in the -domain, -Domain Model V = RI (6.1) where V = L{v} and I = L{i}. Inductor in the -Domain In the time domain Time Domain Model v = L di dt (In thi model I 0 i the initial current in the inductor.) (6.2) Uing equation for differentiation when Laplace tranforming (ee Table), equation (6.2) become V = L [I I 0 ] = LI LI 0 (6.3) where I 0 = i(0 ). Of coure, if there i no initial current, I 0 = 0. We note that equation (6.3) may alo be written a I = V L + I 0 (6.4) While the inductor may be modelled in variou equivalent way in the -domain, the lat two equation immediately ugget two of thee: Serie Model Equation (6.3): 2
(L) i the -domain impedance. LI 0 i like a contant voltage whoe value depend on the initial condition. Parallel Model Equation (6.4): Thi time, I 0 / i like an independent (contant)current ource depending on initial condition. Of coure, if i(0 ) = 0, both of the above reduce to: i.e. the inductor tranform to an impedance (L). Capacitor in the -Domain In the time domain Time Domain Model i = C dv dt (6.5) (In thi model V 0 allow for the poibility of an initial voltage acro the capacitor.) Converting (6.5) via Laplace tranformation give I = C [V V 0 ] = CV CV 0 (6.6) where V 0 = v(0 ). Of coure, if there i no initial voltage, V 0 = 0. We note that equation (6.6) may alo be written a V = ( ) 1 C I + V 0 (6.7) While the capacitor may be modelled in variou equivalent way in the -domain, the lat two equation immediatly ugget two of thee: Serie Model Equation (6.7): 1/(C) i the -domain impedance. V 0 / i like a contant voltage whoe value depend on the initial condition. Parallel Model Equation (6.6): 3
Thi time, CV 0 i like an independent (contant)current ource depending on initial condition. Of coure, if v(0 ) = 0, both of the above reduce to: i.e. the capacitor tranform to an impedance (1/(C)). 6.2 -Domain Circuit Analyi 1. General In the -domain, if no energy i tored in the inductor or capacitor, the relationhip between V and I for each paive element of impedance Z i till V = IZ (6.8) In thi domain, Z R = R, Z L = L and Z C = 1/(C). Technique involving Kirchhoff Law (KVL and KCL), Node-Voltage, Meh-Current, Delta-Wye Tranformation, Thévenin, etc., etc. till hold! If there i initially tored energy, equation (6.8) may be modified by adding the appropriate independent ource in erie or parallel with the element impedance a depicted in the previou ection. 2. Application 1 Natural Repone of an RC Circuit In the circuit hown below on the left, the capacitor ha an initial voltage of V 0, and we wih to find the time-domain expreion for i and v. Method 1: Time Domain -Domain Uing KVL on the -domain circuit, we get V 0 + IZ c + IZ R = 0 = V 0 = V 0 4 = IZ c + IZ R = I C + IR.
From thi I = CV 0 1 + RC. Dividing by RC in the numerator and denominator on the right put I into a recognizable form for invere Laplace tranformation: I = V 0/R + 1 RC. The form i obviouly K + a and L 1 {I} = i = V 0 R e t/rc u(t) (6.9) Then, v = ir = V 0 e t/rc u(t) (6.10) [Remembering that u(t) = 1 for t 0 +, thi i the ame a we had before.] Method 2: We may alo find v before finding i by employing the parallel model for the capacitor a follow: Redraw the original time-domain circuit a Uing node-voltage at A: CV 0 + V 1/C + V R = 0 = V = CV 0 C + 1/R = V 0 + 1/RC. Then, v = L 1 {V } = V 0 e t/rc u(t) a in equation (6.10). 5
Application 2 Step Repone of a Parallel RLC Circuit Aumed that there i no energy tored in the circuit hown below when the witch i opened at time t = 0. We wih to find i L (t). Note that Iource = i dc u(t) and L{i dc u(t)} = I dc /. Becaue there i no initial tored energy (i.e. i L (0 ) = 0 and v C (0 ) = 0, the general form of the -domain circuit i Uing KCL at the top node, Thi implie from which V = I dc + I C + I R + I L = 0. I dc = CV + V R + V L I dc [C + 1 + 1 R L ] /C /C I dc C V = [ 2 + + ] (A). 1 RC LC However, I L = V/L o that from (A) we have I dc LC I L = [ 2 + ( ) ( 1 RC + 1 )] (B). LC Subtituting the value of R, L and C into (B) reult in I L = 384 10 5 ( 2 + 64, 000 + 16 10 8 ). Factoring the denominator allow u to expand I L uing partial fraction: I L = 384 10 5 ( + 32, 000 j24000)( + 32, 000 + j24000) = K 1 + K 2 ( + 32, 000 j24000) + K 2 ( + 32, 000 + j24000). 6
Now, Alo, K 1 = I L =0 = 384 10 5 (32, 000 2 + 24000 2 ) = 0.024. K 2 = I L ( + 32000 j24000) = 32,000+j24000 = 384 10 5 ( 32, 000 + j24000)(j48, 000) = 0.020 126.87 which immediately give K 2 = 0.020 126.87. Now, L 1 {K 1 /} = L 1 {0.024/} = 0.024u(t). On from the tranform Table, it i een that the (ditinct) complex conjugate pair tranform to 2 K e αt co(βt + θ)u(t) where here K = K 2 = 0.020 ; α = 32, 000 ; β = 24, 000 and θ = 126.87. Therefore, i L = [ 0.024 + 0.040e 32,000t co(24, 000t + 126.87 ) ] u(t) A. Again, note that the multiplier u(t) account for t 0. Note that i L (0) = 0 and i L ( ) = 0.024 A, a hould be the cae. 7
Application 3 Multiple Mehe (Tranient) Step Repone Example While multiple node-voltage or meh-current analyi lead to imultaneou differential equation in the time domain, Laplace tranform allow u to replace thee equation with imultaneou algebraic ytem in the -domain. Thi i illutrated with an example below: In the following circuit, the dc current and voltage ource are applied at the ame time. There i no initially tored energy in any of the circuit component. (a) Derive the -domain expreion for V 1 and V 2. (b) For t > 0, derive the time domain expreion for v 1 and v 2. (c) Determine v 1 (0 + ) and v 2 (0 + ). (d) Find the teady-tate value of v 1 and v 2. (a) Firt, repreent the circuit in the -domain. Next, apply the node-voltage technique to node 1 and 2: Node 1: V 1 V 2 + V 1 5 = 0 ; Node 2: V 2 V 1 + V 2 3 + V 2 (15/) 15 = 0 Solving for V 1 and V 2, we get V 1 = 5( + 3) ( 2 + 2.5 + 1) and V 2 = 2.5(2 + 6) ( 2 + 2.5 + 1). (b) Partial fraction expanion give V 1 = 15 50/3 + 0.5 + 5/3 + 2 and V 2 = 15 125/6 + 0.5 + 25/3 + 2 Now, the 1/ and 1/( + a) form are readily recognizable from the tranform table 8
o that [ v 1 (t) = L 1 {V 1 } = 15 50 3 e 0.5t + 5 ] 3 e 2t u(t) V [ v 2 (t) = L 1 {V 2 } = 15 125 6 e 0.5t + 25 ] 3 e 2t u(t) V (c) From part (b), v 1 (0 + ) = 15 50 3 + 5 3 = 0; v 2(0 + ) = 15 125 6 + 25 3 = 2.5. [Note, from the initial value theorem, v 1 (0 + ) = lim V 1 () = 15 50 3 + 5 3 = 0 and imilarly for v 2 (0 + ).] (d) Here, again, we may find v 1 ( ) and v 2 ( ) from part (b) or we may ue the final value theorem: v 1 ( ) = lim V 1 () = 15 0 + 0 = 15 V 0 v 2 ( ) = lim V 2 () = 15 0 + 0 = 15 V 0 which i what we would get uing the reult in part (b) alo. 9
Application 4 Thévenin Equivalent in the -domain (a) Given the following circuit, find the -domain Thévenin equivalent with repect to the terminal a and b. There i no initial charge on the capacitor. Firt, ketch the -domain equivalent to the left of the a-b terminal. With no load acro the a-b terminal, there i no current in the 5 Ω reitor (thi i the tricky obervation here) o that V x = 20... (1) Now, determine V Th by applying the node-voltage rule at node 1: Uing equation (1) and implifying give V Th = 20( + 2.4) ( + 2)... (2) 10
Next, we eek Z Th : The Thévenin equivalent impedance (in the -domain) may be found by applying a tet ource acro the a-b terminal while horting the independent power upply.: Apply node-voltage at node 2 while noting V x = 5I T... (3) Thu, the node-voltage technique, incorporating tet voltage V T, give: I T + V T V x 2/ + V T 0.2V x V x 1 = 0... (4) Simplifying (4) give Z Th = V T I T = 5( + 2.8) ( + 2)... (5) The Thévenin equivalent circuit i hown below (to the left of terminal a-b). The -domain load i alo hown. (b) Find I ab in the -domain for the given load. Clearly, I ab = Uing equation (2) and (5), V Th Z Th + Z L where Z L = 2 +. I ab = 20( + 2.4) [( + 6)( + 3)]. For practice uing the Laplace tranform table, you hould find the time-domain current correponding to the lat expreion. 11