ece201 sp2010 solutions

- 1 -

ECE 201 Spring 2011

Homework Sets (All problems are from the 3rd edition of DeCarlo and Lin)

HW Set #1 Due W 1/12

1. Chapter 1 – prob. 1 (For part e, please use Figure P1.1b.) 2. Chapter 1 – prob. 7(a) 3. Chapter 1 – prob. 8

HW Set #2 Due F 1/14

1. Chapter 1 – prob. 11 2. Chapter 1 – prob. 13 3. Chapter 1 – prob. 17


1. Chapter 1 – prob. 19 (Be careful. Table 1.2 lists relative, not absolute, resistivities!)

2. Chapter 1 – prob. 22(a) 3. Chapter 1 – prob. 27 4. Chapter 1 – prob. 37


1. Chapter 2 – prob. 3 2. Chapter 2 – prob. 6 3. Chapter 2 – prob. 14 There are two typos:

1) In 14.(a), “Figure P2.13” should be “Figure P2.14” 2) In 14.(b), “I1 = 4A” should be “Iin = 4A”

4. Chapter 2 – prob. 17

HW Set #5 Due M 1/24

1. Chapter 2 – prob. 26 (It is the current out of the positive terminal of the source)

2. Chapter 2 – prob. 32 (a) 3. Chapter 2 – prob. 40




1. Chapter 3 – prob. 2 2. Chapter 3 – prob. 6(a)



HW Set #9 Due F 2/4


- 2 -


1. Chapter 5 – prob. 1 2. Chapter 5 – prob. 21(a) 3. Chapter 5 – prob. 26


1. Chapter 5 – prob. 41 2. Chapter 5 – prob. 43




1. Chapter 6 – prob. 9(a) and (b) 2. Chapter 6 – prob. 17 3. Chapter 6 – prob. 21 4. Chapter 6 – prob. 37



In Figure P6.53, change the symbol inside the circle on the left of the circuit to an upward arrow for is1.


1. Chapter 7 – prob. 2 (Change part (b) to read “Find and plot the instantaneous absorbed power.”)

2. Chapter 7 – prob. 4






1. Chapter 8 – prob. 4 2. Chapter 8 – prob. 5 3. Chapter 8 – prob. 9(b)


1. Chapter 8 – prob. 18 (a) and (b) 2. Chapter 8 – prob. 19 (a) and (b) 3. Chapter 8 – prob. 20 (a) and (b)


1. Chapter 8 – prob. 18 (c) - (e) 2. Chapter 8 – prob. 19 (c) - (e) 3. Chapter 8 – prob. 20 (c) and (d)

- 3 -



Note that −V0 and V1 are marked on the ordinate of Figure P8.33(a).




1. Chapter 9 – prob. 16(b) 2. Chapter 9 – prob. 18 (use C = 10mF instead of C = 8mF) 3. Chapter 9 – prob. 20


1. Chapter 9 – prob. 28 2. Chapter 9 – prob. 32(a) (compute the response only; no need to plot)



Please use the following figure in place of Figure P9.49 in the textbook.







Change answer to part b to “(b) 3R”


1. Chapter 8 – prob. 40(a) 2. The op amps in the following circuit are ideal

For vIN(t) as shown in the plot, find and plot vout(t). Label the axis.

(continue on next page…)

- 4 -


1. Add the complex numbers j / 41z 10e π= and j5 / 8

2z 5e π= . Express your answer (a) in real and imaginary parts, and (b) as a magnitude and phase. Draw z1, z2 and z1 + z2 on the complex plane. Label the real part, the imaginary part, and magnitude and the phase of z1 + z2.

2. Multiply the complex numbers z1 = 12-16 j and z2 = 1+0.75j. Express your answer (a) in real and imaginary parts, and (b) as a magnitude and phase. (c) Compare 1z 2z to 1 2z z . (d) What is the relationship between the phase of z1, the phase of z2, and the phase of z1z2?

3. For z1 and z2 given in problem 2, determine z1/z2. Express your answer (a) in real and imaginary parts, and (b) as a magnitude and phase. (c) Compare 1z and 2z to 1 2z / z . (d) What is the relationship between the phase of z1, the phase of z2, and the phase of z1/z2?

- 5 -














1. Chapter 11 – prob. 5 2. Chapter 11 – prob. 7 3. Chapter 11 – prob. 9 4. Chapter 11 – prob. 10





The last sentence before (a) should be: “This constitutes a pf of 0.866 lagging.”

HW Set #40 Not collected. Solutions will be posted.


Homework #3

Prob. 1-19

(a) 2.575 Ω 800 2.575 0.8 Ω 2.06Ω

(b) From table 1.2, 200 feet of 14-gauge nickel wire has resistance:

5.1 2.575 Ω1000 200 5.1 2.575 0.001 200 2.6265Ω

(c) !""#$ % &'!(#)

!""#$ % &'!(#) 2.06 % 2.6265 4.6865Ω

Prob. 1-22

(a) + ,- 400 sin20πt 1034- % 2 % 5 0.16 sin20πt- 8

1.28 5,6-20πtW

>> t=[0:0.0001:0.5];

>> p=1.28*(sin(20*pi*t)).^2;

>> plot(t,p);

>> grid;

>> xlabel('t (time)');

>> ylabel('p(t) (instantaneous power)');

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.2

0.4

0.6

0.8

1

1.2

1.4

t (time)

p(t)

(in

stan

tane

ous

pow

er)

Prob 1-27

(a) + 89:;

< -

+ 120 12060 240Ω

(b) + 89:; 2

< 2 -

+ 2 120 120150 192Ω

Prob 1-37

'& '&

> ?'&-

?'&-

+;- >- - ?-'&- -

-- - ?-'&- -

-

ECE 201 Spring 2010

Homework 5 Solutions

Problem 26

Since V2 = 60 V , the current through 60 Ω branch is 1 A. The resistors 90 Ω

and 180 Ω are in parallel. Their equivalent resistance is

Req =90 × 180

90 + 180= 60

Now 60 Ω and 60 Ω are in series. Thus the voltage drop across the parallel

combination of 40 Ω and 120 Ω is 120 V. Thus current through 40 Ω resistor

is 120/40=3 A. Hence,

Is = 3 + 1

= 4 A

Applying KVL around the loop containing the source and 40 Ω resistor,

Vs − 180 × 4 − 120 = 0

⇒ Vs = 840 V

Power delivered by the source is given by,

Ps = Vs × Is

= 840 × 4

= 3360 W

Problem 32(a)

Using voltage division, the following equations can be written,

R1 + R2 + Rs = 2400 (1)

R1 + R2

R1 + R2 + Rs

= 0.75 (2)

R2

R1 + R2 + Rs

= 0.25 (3)

1

Using (1) and (3), R2 = 600 Ω. Using (2), R1 = 1200 Ω. Using (1), Rs =

600 Ω.

Problem 40

The resistors 9 kΩ and 18 kΩ are in parallel. Thus their equivalent resistance

of 6 kΩ is in series with 6 kΩ. Now the combination 4 kΩ and 12 kΩ are in

parallel. The current Iin divides into I1 and I2 through the parallel resistors

in inverse ratio of their resistances. Similarly, the current I2 divides in inverse

ratio of resistances 9 kΩ and 18 kΩ in parallel. Thus we get,

I1 = 120 ×12

12 + 4= 90 mA

I2 = 120 − 90 = 30 mA

I3 = 30 ×9

9 + 18= 10 mA

Vin = 0.09 × 4 × 103

= 360 V

Psource = VinIin

= 43.2 W

2

ECE 201 Spring 2010


Problem 37

(a)

The following loop equation can be written,

−(I1 − 0.75)200 − 300I1 − (I1 + 0.1)500 = 0

⇒ I1 = 0.1 A

P = V I

Ps1 = 200(0.75 − 0.1)0.75

= 97.5 W

Ps2 = 500(0.1 + 0.1)0.1

= 10 W

(b)

Again, writing the loop equation and comparing the equation obtained with

that given in the problem,

−(I1 − 0.4)R1 − 600I1 − (I1 + 0.1)R2 = 0

(R1 + R2 + 600)I1 = 0.4R1 − 0.1R2

⇒ R1 + R2 = 1400

4R1 − R2 = 600

⇒ R1 = 400 Ω

R2 = 1000 Ω

1

Problem 42

(a)

The following loop equations can be written,

21 − 20I1 − 80(I1 − I2) = 0

24 + 80I2 − 80(I1 − I2) = 0

⇒ I1 = 0.15 A

I2 = −0.075 A

VR3= (I1 − I2)R3

= 0.225 × 80 = 18 V

PR3= 18(0.225) = 4.05 W

(b)

Again, writing the loop equations,

21 − (I1 + I2)20 − 80I2 − 24 = 0

21 − (I1 + I2)20 − 80I1 = 0

⇒ I1 = 0.225 A

I2 = −0.075 A

VR3= I1R3

= 0.225 × 80 = 18 V

PR3= 18(0.225) = 4.05 W

Problem 52

(a)

Assume clockwise loop currents I1 and I2 in the two rightmost loops respec-

tively. The following equations can then be written for the two loops,

−(I1 − Is1)R1 − rmIx − R2(I1 − I2) = 0

Ix = Is1 − I1

−R2(I2 − I1) − R3I2 − Vs2 = 0

2

Putting in matrix form,

rm − R1 − R2 R2

R2 −(R2 + R3)

I1

I2

=

(rm − R1)Is1

Vs2

(b)

Putting the respective literal values,

−80 40

40 −120

I1

I2

=

−40

40

⇒ I1 = 0.4 A

I2 = −0.2 A

(c)

VA = (1 − 0.4)100 = 60 V

VB = 0.6(40) = 24 V

(d)

Ps1 = 60 × 1 = 60 W

Ps2 = 40 × 0.2 = 8 W

(e)

Pdep = 60 × 0.6 ×−0.4 = −14.4 W

3

ECE 201 Spring 2010


Problem 9

(a)

2R and 6R in series gives 8R. 8R and 8R in parallel gives 4R. Thus 12R is in

series with Vs in the simplified circuit. Thus the Thevenin voltage is given

by

Voc =Vs

12R×

1

2× 6R

=Vs

4= 30 V

To find the Thevenin resistance, we short circuit Vs. Thus 8R and 8R are

in parallel, which gives 4R, which in turn is in series with 2R. This gives 6R

and 6R in parallel. Thus Rth = 3R = 900 Ω.

(b)

Using the Thevenin equivalent circuit to simplify our calculations,

PRL=

(

30

900 + RL

)2

RL

= 0.1875 W, 0.24 W, 0.244898 W (RL = 300 Ω, 600 Ω, 1200 Ω)

The Thevenin equivalent circuit analysis allows us to analyze the circuit

without the load once and then plug in the various load resistor values to

compute the relevant quantities. However, using earlier techniques would

require 3 separate analyses to compute these values. Hence the use of a

Thevenin equivalent reduces the effort needed to obtain the answers.

1

(c)

Using Maximum Power Transfer Theorem, for maximum power trans-

fer, RL = 900 Ω and resultant power delivered to the load is given by

Pmax = (30)2/(4 × 900) = 0.25 W .

(d)

If Vs is doubled, power becomes four times, as Pmax ∝ V 2

s . Thus power

delivered to the load becomes Pload = 0.25 × 4 = 1 W .

Problem 17

We proceed to find the Thevenin voltage Voc and Thevenin resistance Rth.

Norton current is then given by Isc = Voc/Rth and Norton resistance is same

as Thevenin resistance. Since this problem involves a dependent source, we

cannot apply the conventional method. We write KVL equations to find the

required values.

vAB − 300iA − vx = 0

vx − 200(iA − vx/18000) −4

3vx = 0

⇒ vx = −18000

29iA

vAB = −320.689655iA

vAB = RthiA + Voc

⇒ Voc = 0

Rth = −320.69 Ω

⇒ Isc = 0

Rnorton = −320.69 Ω

Problem 21

(a)

Again, we write KVL equations to find the Thevenin and Norton equivalent

circuits.

vAB − 40iA − 200(iA + kvx + vx/50) = 0

Vs − vx + 40iA − vAB = 0

2

⇒ vAB [1 + 200(k + 0.02)] = iA [240 + 8000(k + 0.02)] + 200Vs(k + 0.02)

vAB = 60iA + 18

vAB = iARth + Voc

⇒ Voc = 18 V

Rth = 60 Ω

Isc = 0.3 A

(b)

Clearly, using the previous set of equations, Voc is zero for k = −0.02 and for

this value of k Rth = 240 Ω.

Problem 37

(a)

vAB = iARth + Voc

54 = 0.01Rth + Voc

66 = 0.04Rth + Voc

⇒ Rth = 400 Ω

Voc = 50 V

Isc = 0.125 A

(b)

RL = 400 Ω, Pmax = V 2

oc/4RL = 2500/(4 × 400) = 1.5625 W .

3

P 4 (a)

P 4 (c)

0 0.005 0.01 0.015-10

-8

-6

-4

-2

0

2

4

6

8

10

t(s)

i out(t

) (m

A)

0 0.005 0.01 0.0150

10

20

30

40

50

60

70

80

90

100

t(s)

WL(t

) (n

J)

ECE 201 Spring 2010


Problem 27

(a)

Let V be the common voltage across C1 and C2. Thus

(C1 + C2)dV

dt= is(t)

C2

dV

dt= iC2(t)

⇒ iC2(t) =C2

C1 + C2

is(t)

(b)

Using KVL across the second loop and relation from part (a), we get

rmiC2(t) − (L1 + L2)di

dt= 0

vout(t) = L2

di

dt

= rm

L2C2

(L1 + L2)(C1 + C2)is(t)

Problem 38

(a)

Starting from the right end and combining the inductors in series and parallel

consecutively, we get Leq as

Leq = 4 + ((36)||(10 + ((1 + 5)||(3))))

= 4 + ((36)||(10 + ((6)||(3))))

1

= 4 + ((36)||(10 + (2)))

= 4 + ((36)||(12))

= 4 + 9

= 13 mH

(b)

Clearly, the given circuit has 1.2 mH and 0.6 mH in parallel, and this com-

bination is in series with 2.4 mH, the equivalent of which is in parallel with

7 mH. Thus

Leq = 7||(2.4 + (1.2||0.6))

= 7||(2.4 + 0.4)

= 7||2.8

= 2 mH

Problem 41

(a)

Remember that the equivalent capacitance expressions for series and par-

allel connections are opposite of that used for resistances. The equivalent

capacitance can be written as

Ceq = C1||(C2 + (C3||C4))

= 4 +6 × 3

9= 6 µF

vs(t) =1

Ceq

∫ t

−∞

is(τ) dτ

=1

6sin(104

t) V

(b)

Ceq = C1||(C2 + (C3||C4) + C5)

2

(C2 + (C3||C4) + C5) =(

1

18+

1

54+

1

10.8

)

−1

= 6 µF

⇒ Ceq = 6 + 60

= 66 µF

vs(t) =1

Ceq

∫ t

−∞

is(τ) dτ

=1

660(1 − cos(105

t)) V

3

8-18(a)

8-18(b)

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

14

16

18

20

t(s)

Vc(t

)(V

)

0 2 4 6 8 10 12 14 160

1

2

3

4

5

6

7

8

9

10

t(s)

Vc(t

)(V

)

8-19(a)

8-19(b)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

t(s)

i L(t)(

A)

0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-0.05

-0.045

-0.04

-0.035

-0.03

-0.025

-0.02

-0.015

-0.01

-0.005

0

t(s)

i L(t)(

A)

8-20(b)

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-10

-5

0

5

10

15

20

t(s)

Vc(t

)(V

)

0 0.1 0.2 0.3 0.4

-5

0

5

10

15

20

8-20 (d) Vc

I(t)

t (s)

VcI(t

) (V

)

0 0.1 0.2 0.3 0.4

-5

0

5

10

15

20

8-20 (d) Vc

II(t)

t (s)

VcII(t

) (V

)

0 0.1 0.2 0.3 0.4

-5

0

5

10

15

20

8-20 (d) Vc(t)

t (s)

Vc(t

) (V

)

0 0.1 0.2 0.3 0.40

0.1

0.2

0.3

0.4

0.5

0.6

8-20 (d) Ic

I(t)

t (s)

I cI(t

) (A

)

0 0.1 0.2 0.3 0.40

0.1

0.2

0.3

0.4

0.5

0.6

8-20 (d) Ic

II(t)

t (s)

I cII(t

) (A

)

0 0.1 0.2 0.3 0.40

0.1

0.2

0.3

0.4

0.5

0.6

8-20 (d) Ic(t)

t (s)

I c(t

) (A

)

8-18

8-19

0 5 10 1510

11

12

13

14

15

16

17

18

19

20

8-18 (c) VC

(t)

t (s)

VC

(t)

(V)

0 5 10 15-20

-19

-18

-17

-16

-15

-14

-13

-12

-11

-10

8-18 (d) VC

(t)

t (s)

VC

(t)

(V)

0 5 10 150

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10

-3 8-18 (e) IC

(t)

t (s)

I C(t

) (A

)

0 0.005 0.01

0

0.05

0.1

0.15

0.2

8-19 (c) IL(t)

t (s)

I L(t

) (A

)

0 0.005 0.01

-0.22

-0.2

-0.18

-0.16

-0.14

-0.12

-0.1

8-19 (d) IL(t)

t (s)

I L(t

) (A

)

0 0.005 0.010

5

10

15

20

25

8-19 (e) VL(t)

t (s)

VL(t

) (V

)

ECE 201 Spring 2010


Problem 31

(a)

Using voltage division,

vC(0−) =20R

50RV0

= 0.4V0

= vC(0+)

(b)

The Thevenin equivalent is given by

Voc = V0

R3

R3 + (R1||R2)

= 0.8V0

Rth = R1||R2||R3

= 4R

(c)

vC(∞) = 0.8V0

vC(0+) = 0.4V0

⇒ vC(t) = vC(∞) + [vC(t0+) − vC(∞)]e−

t−t0

RthC

= 0.8V0 − 0.4V0e−

t

4RC

1

(d)

vC(T−) = vC(T+) = 0.8V0 − 0.4V0e−1.5

= 0.71075V0

(e)

τ = RthC

= 4RC

(f)

Using the same relation as in part (c), with t0 = T ,

vC(t) = 0.71075V0e−

t−T

4RC

(g)

For simplicity, let all quantities be normalized to 1. Then the plot for vC(t)

would look as shown on the next page.

Problem 33

(a)

iL(0+) = iL(0−) =−10

80 + 20= −0.1 A

2

0 5 10 15 20 250

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

t

v C(t

)

Plot of vC

(t) for 0<=t<=4T

Figure 1: Plot of vC(t) for problem 31

(b)

Here we have to consider two intervals, 0 ≤ t < 80 ms and 80 ms ≤ t <

160 ms. For 0 ≤ t < 80 ms,

iL(t) = iL(∞) + [iL(0+) − iL(∞)]e−tR/L

=20

100+ [−0.1 − 0.2]e−25t

= 0.2 − 0.3e−25t

iL(80−) = iL(80+) = 0.2 − 0.3e−2

= 0.1594 A

Again, for the interval 80 ms ≤ t < 160 ms,

iL(t) = iL(∞) + [iL(80+) − iL(∞)]e−(t−80)R/L

= −0.1 + [0.1594 + 0.1]e−25(t−80)

= −0.1 + 0.2594e−25(t−80)

We can write vout(t) as the following,

vout(t) = vin(t) − R1iL(t)

3

= 4 + 24e−25t, 0 ≤ t < 80 ms

= −2 − 20.752e−25(t−80ms), 80 ms ≤ t < 160 ms

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16−30

−20

−10

0

10

20

30

t(seconds)

v out(t

)(vo

lts)

Plot of vout

(t) for 0<=t<=160ms

Figure 2: Plot of vout(t) for problem 33

4

9-16(b)

0 1 2 3 4 5 6 7 8-1

-0.5

0

0.5

1

1.5

2

t(s)

Vc(t

) (V

)

9-18(a)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

1

2

3

4

5

6

7

8

9

10

t(s)

Vc(t

)(V

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.4

-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

t(s)

i L(t)(

A)

9-18(b)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

2

4

6

8

10

12

t(s)

Vc(t

)(V

)

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-0.4

-0.35

-0.3

-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

t(s)

i L(t)(

A)

ECE 201 Spring 2010


Problem 43

(a)

To find the initial conditions at t=0-, we can write the following KVL equa-

tions,

20 + 40iL(0−) − 60(0.1 − iL(0−)) = 0

⇒ iL(0−) = iL(0+) = iC(0+) = −0.14 A

⇒ vC(0+) = vC(0−) = 20 − 0.14 × 40

= 14.4 V

After t=0, the circuit is a series RLC circuit with the following characteristic

equation,

s2 +

R

Ls +

1

LC= 0

s2 + 15s + 50 = 0

⇒ s1 = −5, s2 = −10

⇒ vC(t) = c1e−5t + c2e

−10t

⇒ c1 + c2 = 14.4

(5 × 10−3)(5c1 + 10c2) = 0.14

⇒ c1 = 23.2, c2 = −8.8

⇒ vC(t) = (23.2e−5t− 8.8e−10t) V

The plot of vC(t) for 0 < t ≤ 1 is shown on the next page.

1

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

5

10

15

t(seconds)

v C(t

)(V

olts

)

Plot of vC

(t) (part (a)) for 0<t<=1s

(b)

In this case, the situation after t=0 can be visualized as a series RLC circuit

with a constant input voltage source (after source transformation) which will

have the value 0.5× 60 = 30 V . The initial conditions and the characteristic

equation do not change. However, the expression for vC(t) will now have a

term due to the constant input also. Thus,

vC(t) = c1e−5t + c2e

−10t + 30

⇒ c1 + c2 + 30 = 14.4

5c1 + 10c2 = 28

⇒ vC(t) = (−36.8e−5t + 21.2e−10t + 30) V

The plot of vC(t) for 0 < t ≤ 1 follows.

2

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 114

16

18

20

22

24

26

28

30

t(seconds)

v C(t

)(V

olts

)

Plot of vC

(t) (part (b)) for 0<t<=1s

Problem 49

(a)

At t=0-, the capacitor is open circuit and the inductor is short circuit. At

t=0+, the current source is off and the voltage source is still operating. Thus

we can write the following equations,

vC(0−) = vC(0+) = 10 V

iL(0−) + 0.005 =10

1000⇒ iL(0+) = iL(0−) = 0.005 A

10 − vL(0+) − vC(0+) = 0

⇒ vL(0+) = 0

iC(0+) +10

1000= 0.005

⇒ iC(0+) = −0.005 A

3

(b)

For t > 0, we can write the following KVL and KCL equations,

10 − LdiL

dt− vC = 0

iL = CdvC

dt+

vC

R

⇒d2vC

dt2+

1

RC

dvC

dt+

vC

LC=

10

LC

Thus the characteristic equation is

s2 + 2000s + 1.087 × 107 = 0

⇒ s1,2 = −1000 ± j1000π

⇒ vC(t) = e−1000t[A cos(1000πt) + B sin(1000πt)] + 10

A = 0, B = −10

π(Using initial conditions)

⇒ vC(t) = 10 −10

πe−1000t sin(1000πt) V

iL(t) = CdvC

dt+

vC

R

= 0.005[

2 − e−1000t

1

πsin(1000πt) + cos(1000πt)

]

A

(c)

vC(t) is computed above in part (b).

(d)

vL(t) = 10 − vC(t)

=10

πe−1000t sin(1000πt) V

(e)

iC(t) = CdvC

dt

=0.005

πe−1000t[sin(1000πt) − π cos(1000πt)] A

The plots for the respective quantities follow.

4

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10−3

5

6

7

8

9

10

11

12x 10

−3

t(seconds)

i L(t)(

A)

Plot of iL(t) for 0<t<=5ms

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10−3

7.5

8

8.5

9

9.5

10

10.5

11

t(seconds)

v C(t

)(V

)

Plot of vC

(t) for 0<t<=5ms

5

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10−3

−1

−0.5

0

0.5

1

1.5

2

2.5

t(seconds)

v L(t)(

V)

Plot of vL(t) for 0<t<=5ms

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 10−3

−5

−4

−3

−2

−1

0

1

2

3x 10

−3

t(seconds)

i C(t

)(A

)

Plot of iC

(t) for 0<t<=5ms

6

ECE 201 Spring 2010


Problem 40

(a)

The following differential equation can be written using KCL at the inverting

terminal of the op amp and using the virtual ground concept,

vs(t)

R1

+vout(t)

R2

+ Cdvout(t)

dt= 0

vs(t) = −100u(t)

= −100, t ≥ 0

⇒ vout(t) =100R2

R1

(1 − e−t/R2C)

= 400(1 − e−2t) mV, t ≥ 0

Problem not from the book

If we denote the voltage at the non-inverting terminal of the op amp as

vL(t), then using the virtual ground concept, the output voltage is given

by Vout(t) = 10vL(t). Also, vL(t) = VIN(t) − 1000iL(t). The problem then

reduces to finding iL(t). Due to the square wave shape of VIN(t), the graph

for iL(t) will also be periodic with maximum and minimum values of x and

−x (say). The task is to find the value of x, which can be done using the

following equation,

iL(t) = iL(∞) + [iL(0) − iL(∞)]e−t/τ

τ =L

R

= 2 µs

x = 2 × 10−4 + [−x − 2 × 10−4]e−0.5

⇒ x = 0.489 × 10−4

1

The plot of Vout(t) for 0 ≤ t ≤ 2µs is drawn below. Note that Vout(t) will be

periodic with period 2µs.

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

x 10−6

−2.5

−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

2.5

t(seconds)

Vou

t(t)(

Vol

ts)

Plot of Vout

(t) for 0<=t<=2us. Note that it is periodic with period 2us

2

ECE 201 Spring 2010


Problem 30

(a)

Yin(jω) =1

R+

1

jωL

= 0.05 −j0.25

ω

Zin(jω) =1

Yin(jω)

=j20ω

5 + jω

(b)

IL =IinZin(jω)

jωL

= 10 ×10(i + j)

j20

= 5√

2e−jπ/4

⇒ iL(t) = 10 cos(5t − π/4) mA

Problem 40

Let the impedances of R1, C and R2, L combinations be Z1 and Z2 respec-

tively.

Z1 =[

1

500+ j(400 × 5 × 10−6)

]

−1

= 250(1 − j)

Z2 =

[

1

100+

1

j(400 × 0.125)

]

−1

= 20(1 + 2j)

1

Let the phasor current through the source be I.

I =V

270 − j210

⇒ VC = I × 250(1 − j)

=120√

2(1.0256 − j0.1282)

⇒ vC(t) = 124.032 cos(400t− 7.125) V

IL =

(

V

270 − j210

)(

R2

R2 + jωL

)

=V × 100

(270 − j210)(100 + j50)

=240√

2(0.00128205 + j0.0002564)

⇒ iL(t) = 0.3138 cos(400t + 11.3099) A

2

0 0.5 1 1.5 2 2.5 3-0.01

-0.008

-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.0111-2(a) Instantaneous Current i(t)

time (s)

curr

ent

(A)

Instantaneous Current

0 0.5 1 1.5 2 2.5 30

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

11-2(a) Instantaneous Power p(t) and Average Power pave

time (s)

pow

er

(W)

Instantaneous Power

Average Power

0 0.5 1 1.5 2 2.5 3-0.01

-0.008

-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.0111-2(b) Instantaneous Current i(t)

time (s)

curr

ent

(A)


0 0.5 1 1.5 2 2.5 30

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

11-2(b) Instantaneous Power p(t) and Average Power pave

time (s)

pow

er

(W)

Instantaneous Power

Average Power

0 0.5 1 1.5 2 2.5 3-0.01

-0.008

-0.006

-0.004

-0.002

0

0.002

0.004

0.006

0.008

0.0111-2(c) Instantaneous Current i(t)

time (s)

curr

ent

(A)


0 0.5 1 1.5 2 2.5 30

0.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

11-2(c) Instantaneous Power p(t) and Average Power pave

time (s)

pow

er

(W)

Instantaneous Power

Average Power

ECE 201 Spring 2010


Problem 5

(a)

For Figure (a),

I2

eff =1

T

∫ T

0

i2(t) dt

=1

9

[∫

3

0

9 dt +∫

6

3

16 dt

]

=25

3

⇒ Ieff = 5/√

3

For Figure (b),

I2

eff =1

4

[∫

1

0

9 dt +∫

3

2

16 dt

]

= 25/4

⇒ Ieff = 2.5

(b)

Current through RL is given by i(t) × 60/(30 + 60) = (2/3)i(t). Power

absorbed by RL is then given by the square of the effective current through

it times the resistance. Thus

P =4

9×

25

3× 30

=1000

9W

1

(c)

P =4

9×

25

4× 30

=250

3W

Problem 7

In all parts of the problem, the following identities will be used,

∫

2π/m

0

cos(mnx) dx = 0

∫

2π/m

0

sin(mnx) dx = 0

Here m and n are integers.

(a)

V2

1eff =1

T

∫ T

0

v2

1(t) dt

=1

T

∫ T

0

[102 + 40 cos(20t) + 2 cos(40t)] dt

= 102

⇒ V1eff = 10.0995

(b)

V2

2eff =1

T

∫ T

0

v2

2(t) dt

=1

T

∫ T

0

[501 + cos(4t) + 12.51 + cos(8t) + 50cos(6t) + cos(2t)] dt

= 62.5

⇒ V2eff = 7.9057

(c)

V2

3eff =1

T

∫ T

0

v2

3(t) dt

=1

T

∫ T

0

[501 + cos(4t) + 36.42771 + cos(8t) + 6.24991 − cos(8t) + . . .] dt

= 92.6776

⇒ V3eff = 9.6269

2

Problem 9

(a)

ZL =[

1

5+ j(30 × 5 × 10−3)

]

−1

= 0.8(4 − j3)

⇒ VL = Iin × ZL

=4√

2(4 − j3)

⇒ vL(t) = 20 cos(30t − 36.87) V

(b)

Pinst(t) = iin(t)vL(t)

= 100 cos(30t) cos(30t − 36.87) W

Paverage =VmIm

2cos(θv − θi)

= 50 cos 36.87

= 40 W

Problem 10

(a)

Is =50 6 − 90

6 + j12 − j4

= 5 6 − 143.13 (magnitude = 5)

(b)

Paverage = VeffIeff cos(θv − θi)

= 5 × 50 cos(−53.13)

= 150 W

3

(c)

Paverage = R|Is|2

= 6 × 25

= 150 W (same as part (b))

(d)

Is =50 6 − 90

30 + j50 − j10

= 1 6 − 143.13 (magnitude = 1)

Paverage = VeffIeff cos(θv − θi)

= 1 × 50 cos(−53.13)

= 30 W

4

ECE 201 Spring 2010


Problem 30

(a)

Zth = (R1||R2) − j/ωC

= (20 − j10) Ω

Voc = Vs

R2

R1 + R2

= 20 Vrms

For maximum power transfer, ZL = Z∗

th = (20 + j10) Ω.

(b)

Pavg =V 2

oc

4RL

= 5 W

Problem 37

(a)

Here we cannot apply the maximum power transfer theorem because RL is

fixed. The expression for average power across RL is given by

Pavg =V 2

s RL

(R + RL)2 + (ωL − 1/ωC)2

Clearly, the average power is maximum when the denominator is minimum,

which happens when the following two conditions hold,

R = 0, ωL =1

ωC

1

C =1

ω2L

= 2.5 mF

Pavgmax=

V 2

s

RL

=502

5= 500 W

(b)

Pavg =V 2

s RL

(R + RL)2 + (ωL − 1/ωC)2

For maximum Pavg ,

dPavg

dRL

= 0

⇒ R2

L = R2 + (ωL − 1/ωC)2

⇒ RL = 4.011 Ω

2

ece201 sp2010 solutions

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