Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo

ECE 202 LINEAR CIRCUIT ANALYSIS II (SP’10)Homework #6 Solution (21–24)

Problem 21

Rs = 2Ω, RL = 8Ω, R1 = 190Ω, L = 20Hvin(t) = 100 [V] for all time.

When the switch has been in position A for a long time (left figure), the current through the inductor isvin/(Rs +RL) = 10 [A] which is the initial condition for the inductor for t ≥ 0. So, iL(0−) = 10 [A].For t ≥ 0, the equivalent s-domain circuit accounting for the initial condition is shown on the right.

I1 =Vin(s)+LiL(0−)Rs +R1 +RL +Ls

=10s+5

s(s+10)

Vout(s) = R1 I1 = 19010s+5

s(s+10)

vout(t) = 95(1+19e−10t )u(t) [V]

pR1(t) = i1(t)vout(t) = v2out/R1 =

952

(19e−10t +1

)2[W]

Ediss =∫ t

0pR1(τ) dτ =

(47.5t −180.5e−10t −857.4e−20t +1038

)[J]

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo

Problem 22–23

(a) vC1(0) and vC2(0)

vC1(0) = 0 [V] , vC2(0) = 0 [V]

(b) vC1(t) and vC2(t) for 0 ≤ t ≤ 1

VC1(s) =20s

1/(C1s)R1 +1/(C1s)

=100

s(s+5)

vC1(t) = 20(1− e−5t)u(t) [V]

vC2(t) = 0 [V]

(c) vC1(1) and vC2(1)

vC1(1) = 20(1− e−5)u(t) = 19.865[V]

vC2(1) = 0 [V]

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo

(d) vC1(t) and vC2(t) for 1 ≤ t ≤ 2

I =

(vC1(1)

se−s

)1

R2 +1/(C1s)+1/(C2s)=

vC1(1)500

1s+2

e−s

VC1(s) =vC1(1)

se−s − 1

C1sI =

vC1(1)s

e−s − vC1(1)s(s+2)

e−s =vC1(1)

2

(1s+

1s+2

)e−s

vC1(t) =vC1(1)

2

(1+ e−2(t−1)

)u(t −1) = 9.933

(1+ e−2(t−1))u(t −1) [V]

VC2(s) =1

C2sI =

vC1(1)2

(1s− 1

s+2

)e−s

vC2(t) =vC1(1)

2

(1− e−2(t−1)

)u(t −1) = = 9.933

(1− e−2(t−1))u(t −1) [V]

(e) vC1(2) and vC2(2)

vC1(2) =vC1(1)

2

(1+ e−2

)= 11.277[V]

vC2(2) =vC1(1)

2

(1− e−2

)= 8.588[V]

(f) vC1(t) and vC2(t) for 2 ≤ t ≤ 4

I =

(20s− vC2(2)

s

)e−2s 1

R1 +1/(C1s)=

(20− vC1(2)

100

)e−2s 1

s+5

VC1(s) = −R1I +20s

e−2s =

(20s− 20− vC1(2)

s+5

)e−2s

vC1(t) =(

20−(20− vC1(2)

)e−5(t−2)

)u(t −2) =

(20−8.723e−5(t−2)

)u(t −2) [V]

vC2(t) = vC2(2) = 8.588[V]

(g) vC1(4) and vC2(4)

vC1(4) =(

20− (20− vC1(2))e−10

)≈ 20[V]

vC2(4) = vC2(2) = 8.588[V]

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo

(h) vC2(t) for t ≥ 4

I =

(vC1(4)− vC2(4)

se−4s

)1

R2 +1/(C1s)+1/(C2s)=

(vC1(4)− vC2(4))500

1s+2

e−4s

VC2(s) =1

C2sI +

vC2(4)s

e−4s

=12

(vC1(4)+ vC2(4)

s− vC1(4)− vC2(4)

s+2

)e−4s

vC2(t) =12

((vC1(4)+ vC2(4))− (vC1(4)− vC2(4))e

−2(t−4))

u(t −4)

=(14.294−5.7058e−2(t−4))u(t −4)

(i) Complete vout(t)

0 1 2 3 4 50

2

4

6

8

10

12

14

vout

(t) [V]

t [sec]

vout(t) =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

0 for t < 1

9.933(1− e−2(t−1)

)u(t −1) for 1 ≤ t < 2

8.588 for 2 ≤ t < 4(14.2942−5.7058e−2(t−4)

)u(t −4) for 4 ≤ t

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Purdue University, School of Electrical and Computer Engineering Prof. DeCarlo

Problem 24

For the detailed explanation of this problem, please see textbook (3rd Ed.) pg. 651. Example 13.21.

vs(t) = −2 [V] = E

When the switch is in position A, the voltage source charges up the capacitor and the charge stored on C is Q = CEWhen the switch moves to position B, all the charges in C1 have to be transferred to the feedback capacitor. Thecharge transferred to the feedback capacitor is Qf = CE = k fCvc f = −k fCvout . So it follows that at t=T andevery 2T seconds thereafter, vout gets incremented by −E/kf .

0 5 10 15 20 250

5

10

15

20

25

t [ms]

vout

(t) [V]

(a) kf=1

(b) kf=2

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