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Chapter 4 Circuit Analysis with Laplace Transforms 4 10 Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition Copyright © Orchard Publications The current is now found as and thus, (4.18) b. The impedance can also be found by successive combinations of series and parallel impedances, as it is done with series and parallel resistances. For convenience, we denote the network devices as and shown in Figure 4.14. Figure 4.14. Computation of the impedance of Example 4.4 by series parallel combinations To find the equivalent impedance , looking to the right of terminals and , we start on the right side of the network and we proceed to the left combining impedances as we would combine resistances where the symbol denotes parallel combination. Then, (4.19) We observe that (4.19) is the same as (4.18). V A s () V S s 1 ----------------------------------- V A s s -------------- V A s s1 s + ----------------- ++ 0 = 1 1 s -- 1 s +   V A s V S s = V A s s 3 1 + s 3 2s 2 + ------------------------------------ V S s = Is V S s V A s 1 1 s 3 1 + s 3 2 +  V S s 2 1 + s 3 2 + V S s == = Zs V S s s 3 2 + 2 1 + Z 1 Z 2 Z 3 ,, Z 4 1 1s s s Z s Z 1 Z 2 Z 3 Z 4 a b ab || Z 3 Z 4 + || Z 2 [] Z 1 + = ss 1 s + ss1 s -------------------------- 1 + s 2 1 + 2 1 + s ---------------------------- 1 + s 3 s + 2 1 + ---------------- 1 + s 3 2 + 2 1 + = =

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Signals and Systems with MATLAB Computing and Simulink Modeling, Fourth Edition 4 11 Copyright © Orchard Publications Complex Admittance Y(s) 4.3 Complex Admittance Y(s) Consider the parallel circuit of Figure 4.15 where the initial conditions are zero. Figure 4.15. Parallel GLC circuit in s domain For the circuit of Figure 4.15, Defining the ratio as , we obtain (4.20) and thus the voltage can be found from (4.21) where (4.22) We recall that . Therefore, is a complex quantity, and it is referred to as the complex input admittance of an parallel circuit. In other words, is the ratio of the current excitation to the voltage response under zero state (zero initial condi- tions). Example 4.5 Compute and for the circuit of Figure 4.16. All values are in (ohms). Verify your answers with MATLAB. s domain GLC G + I S s () V s 1 sL ----- sC GV s 1 -----Vs sCV s ++ Is = G 1 s C   Vs = I S s Ys ----------- G 1 s C 1 Zs ---------- == I S s = G 1 s C = s σ j ω + = s domain Y s I S s
Chapter 4 Circuit Analysis with Laplace Transforms 4 12

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