DeCarlo Ch10a Solutions

# Linear Circuit Analysis: The Time Domain and Phasor Approach

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1 PROBLEM SOLUTIONS CHAPTER 10 S OLUTION 10.1 Using KCL, we can write C dv C dt + v C R = i in t ( 29 Dividing by C: dv C dt + 1 RC v C = i in t ( 29 C We know that i in t ( 29 = 20sin 400 t ( 29 mA, which can be represented by a complex exponential, i in t ( 29 = Re 20 e j 400 t e - j π /2 [ ] mA. For convenience we will simply let i in t ( 29 = 20 e j 400 t e - j π /2 mA, knowing that we must take the real part to complete our solution. The output voltage will also be reparesented as a complex exponential: v C t ( 29 = V m e j 400 t + ( 29 = V m e j 400 t e j Substituting this expression into the differential equation and canceling e j 400 t : j 400 V m e j + V m RC e j = 20 × 10 - 3 e - j π /2 C Thus V m e j 1 RC + j 400 = 20 × 10 - 3 C V m e j = - j 4000 1000 + j 400 = 3.714 ∠- 111.8 o where the values for R = 100 and C = 5 mF were substituted in. Thus, V m = 3.714 = 0 - tan - 1 400 1000 = - 111.8 ° Taking into account a 90 o phase shift we obtain v C t ( 29 = 3.714cos 400 t - 111.8 ° ( 29 = 3.714sin 400 t - 21.8 ° ( 29 V and i out t ( 29 = 18.57sin 400 t - 21.8 ° ( 29 mA S OLUTION 10.2 From KCL and component definitions:

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2 i in t ( 29 - v L 25 - i L = 0 0.1 25 di L dt + i L = i in t ( 29 di L dt + 250 i L = 250 i in t ( 29 We represent the input signal by the complex exponential: i in t ( 29 = 0.2 e j 250 t A and the unknown current can be represented as i L ( t ) = I L e j 250 t + ( 29 . Substituting this into the differential equation and canceling e j 250 t : j 250 I L e j + 250 I L e j = 50 Thus I L e j j 250 + 250 ( 29 = 50 I L e j = 50 250 + j 250 = 0.14142 ∠ - 45 o and I L = 0.141, = - 45 ° i L t ( 29 = 0.141cos 250 t - 45 ° ( 29 A S OLUTION 10.3. Construct differential equation by KVL and device definitions: v in t ( 29 - 0.5 di L dt - 200 i L = 0 di L dt + 400 i L = 2 v in t ( 29 We represent v in t ( 29 as the complex exponential function, v in t ( 29 = 20 e j 400 t V. The current in the inductor has the form: i L = I L e j 400 t + ( 29 . Substituting into the differential equation and canceling e j 400 t : j 400 I L e j + 400 I L e j = 40 Thus I L e j j + 400 ( 29 = 40 I L e j = 40 400 + j 400 = 0.070711 ∠- 45 o and I L = 0.0707, = - 45 ° , i L t ( 29 = 70.7cos 400 t - 45 ° ( 29 mA Hence, v out t ( 29 = 14.14cos 400 t - 45 ° ( 29 V S OLUTION 10.4. Construct differential equation using KVL and device definitions:
3 v in t ( 29 - v C - C dv C dt R = 0 RC dv C dt + v C = v in t ( 29 The output voltage is defined as: v out t ( 29 = v in t ( 29 - v C t ( 29 This means that finding v C is enough to be able to obtain the output voltage. The input voltage is represented by the complex exponential: v in t ( 29 = 20 e j 250 t e - j π /2 V and v C t ( 29 = V m e j 250 t + ( 29 . Substituting into the differential equation, dividing by e j 250 t , and rearranging: j 250 RCV C e j + V C e j = - j 20 V C e j j 250 RC + 1 ( 29 = - j 20 V C e j = - j 20 1 + j = 14.142 ∠- 135 o Now V out e j = - j 20 - V C e j V out e j = 10 - j 10 = 14.142 ∠ - 45 o Thus, in the time-domain, v out t ( 29 = 14.142cos 250 t - 45 ° ( 29 V S OLUTION 10.5. The circuit is identical to that of problem 10.1. Thus, RC dv s dt + v s = Ri s t ( 29 Ri s t ( 29 - RC dv s dt = v s Moreover, the complex exponential solution is given by v s ( t ) = V m e j = 223.6 e - j 63.43 o V Hence R - jRC 223.6 e - j 63.43 o = 223.6 e - j 63.43 o = 100 - j 200 i.e., R - jRC (100 - j 200) = R - 200 RC - j 100 RC = 100 - j 200

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4 Thus RC = 2 s, R = 100 + 200 RC = 500 , and C = 4 mF.
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