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2. 2 Ohm’s law specifies that v(t) = Ri(t) Taking the Laplace transform of both sides V(s) = RI(s) The impedance Z(s) is defined as and the admittance Y(s)=I(s)/V(s) is 1/R

3. Since v(t) = L di/dt taking the Laplace transform of both sides of this equation yields V(s) = L[sI(s) − i (0−)] The impedance is Z(s)=sL and the initial condition is modeled as a voltage source in series. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.3

4. 4 Find the voltage v(t) given an initial current i(0−) =1 A. Answer: v(t) = [3.2e −8t − 1.2e −0.5t ]u(t) volts

5. Using i=C dv/dt, we can find Z(s)=1/sC and we model the initial condition as in (b) and (c) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.5

6. Determine the mesh currents i 1 (t) and i 2 (t) (assume zero initial energies) Answer: (using a computer for calculations) i 1 (t) =−96.39e −2t − 344.8e −t + 841.2e −0.15t cos 0.8529t + 197.7e −0.15t sin 0.8529t mA i 2 (t) =−481.9e −2t − 241.4e −t + 723.3e −0.15t cos 0.8529t + 472.8e −0.15t sin 0.8529t mA Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.6

7. Find v x Answer: v x = [4 + 6.864 e−1.707t − 5.864 e−0.2929t ]u(t) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.7

8. Find v(t) using source transformation (and a computer). Answer: v(t) =[5.590 ×10 −5 e −0.1023t + 2.098 cos(3t + 3.912 ◦ ) + 0.1017e −0.04885t cos(0.6573t + 157.9 ◦ )]u(t) Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.8

9. Find the frequency-domain Thévenin equivalent of the highlighted network: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.9 Method: apply a test current:

10. H(s) is the transfer function of the circuit, defined as the ratio of the output to the input. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.10

11. If we define the impulse response as h(t), then the output y(t) is related to the input x(t) via the convolution integral: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.11

12. If x(t)=v i (t)=u(t)-u(t-1) and h(t)=2e -t u(t), then by flip/slide/integrate, we find v o (t): Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.12

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14. Convolution in the time domain is multiplication in the frequency domain: Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.14

15. We can explore the properties of F(s) by graphing |F(s)| on the s plane. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.15

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19. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.19 When V s =0, the output is the natural response, and it can only be non-zero at the pole s=−R/L. Hence i n (t)=Ae -Rt/L.

20. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.20

21. Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.21

22. Synthesize a circuit that will yield the transfer function H(s) =V out /V in =10(s+2)/(s+5). Copyright © 2013 The McGraw-Hill Companies, Inc. Permission required for reproduction or display.22