202ex2sum04sols

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EE-202

Exam II

July 9, 2004

Name: SOLUTIONS

Student ID: _________________

Division 1

INSTRUCTIONS

There are 12 multiple choice worth 5 points each; there is 1 workout

problem worth 40 points.

This is a closed book, closed notes exam. No scrap paper or

calculators are permitted. A transform table is attached to the back of the

exam.

Circle the correct answer for the multiple choice. Unclear circles will

be marked incorrect.

Nothing is to be on the seat beside you.

When the exam ends, all writing is to stop. This is not negotiable.

No writing while turning in the exam/scantron or risk an F in the

exam.

All students are expected to abide by the customary ethical standardsof the university, i.e., your answers must reflect only your own knowledge

and reasoning ability. As a reminder, at the very minimum, cheating will

result in a zero on the exam and possibly an F in the course.

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EE-202, Ex 2, page 2

MULTIPLE CHOICE.

1. A transfer function with pole-zero plot below has H ( s) = K sm+ a1

sm!1

+!+ am

sn+ b1 s

m!1+!+ b

n

for appropriate m and n and has a gain of –5 at s = 0. Then K = :

(1) –1 (2) 2 (3) –2 (4) 4

(5) 5 (6) –4 (7) –5 (8) None of above

2. Consider the circuit below in which vout

(0) = 2 V. The switch S has been closed for

a long time and moves to position B at t = 1! . Then

vout

(2) = (in V):

(1) 12 (2) 20 (3) 3 (4) 4

(5) 5 (6) 6 (7) 7 (8) None of above

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3. Consider the circuit below. A valid loop equation in a matrix formulation would be:

(1) V in = 1+ s !s

s +1

" # $

% & ' I 1 +

s

s +1 I 2 +V (2) V in = 1+ s +

s

s +1

! " #

$ % & I 1 +V

(3) (s + 2) I 1 ! I 2 = 0 (4)s +1

sV 1 = I 2 ! I 1

(5) 0 =s

s +1 I 1 +

s

s +1+1+

1

s

!

"#$

%&I 2 'V (6) (s + 2) I 1 + I 2 = 0

(7) Two of above (8) None of above

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4. The pole-zero plot of a transfer function is given below. An input which will cause

the response to be unbounded (unstable) is:

(1) e!2t

u(t ) (2) ! (t ) (3) ! '(t ) (4) e!t

u(t )

(5) cos(t )u(t ) (6) u(t) (7) sin(2t )u(t ) (8) none of above

5. If a circuit with H ( s) =( s +1)( s !1)

s( s + 4)is excited by the input v

in(t ) = 6cos(3t ) V, then

the magnitude of the output cosine in SSS is:

(1) 2/3 (2) 2 (3) 3 (4) 4

(5) 10 3 (6) 2 10 15 (7) 1.5 (8) None of above

6. For the circuit with transfer function H ( s) =( s +1)( s !1)

s( s +100)is excited by the input

vin(t ) = 3cos(100t ! 45

o) V, then the phase of the output cosine in SSS is

(approximately):

(1) 45o (2) !45o (3) 90o (4) !90o

(5) 135o (6) 0o

(7) 180o (8) None of above

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7. For the circuit below to be stable in the sense of BIBO, the complete range of a must

be:(1) 1 < a (2) 1.5 < a (3) –3 < a (4) 3 < a

(5) 1.5 > a (6) a > –1.5 (7) a < 3 (8) None of above

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8. The following is the magnitude frequency response of a transfer function H(s).

The best possible transfer function leading to this magnitude response is:

(1)( s2 +1)

( s +1)2(2)

( s2 !1)2

( s +1)2+1"

#$%&'

(3) ( s2 +1)2

( s ! 0.1)2+1"

#$%&'

2(4) ( s2 +1)2

( s + 0.1)2+1!

"#$%&

2

(5)( s +1)4

( s + 0.1)2 +1!"#

$%&

2(6)

( s2 +1)( s2 !1)

( s + 0.5)2+1"

#$%&'

( s + 0.4)2+1"

#$%&'

(7)

( s2 !1)

2

( s + 0.1)2 +1"#$

%&'

(8) None of above

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9. The circuit below is to be frequency and magnitude scaled so that the value of H(s) at

s = j5 becomes the value of Hnew(s) at s = j1000. If the final value of the capacitor is to

be 1 mF, then the new final value of the inductor is to be Lnew = (in H):

(1) 0.1 (2) 0.2 (3) 0.02 (4) 10

(5) 5 (6) 20 (7) 40 (8) none of above

10. A particular circuit has input admittance Y in(s) =s

(s ! p1)(s ! p2 )with poles

p1 = 10 and p2 = 5 . If the associated circuit is magnitude scaled by K m = 5 and

frequency scaled by K f = 10 , then the new location of p1 is:

(1) 1 (2) 2 (3) 100 (4) 20

(5) 5 (6) 50 (7) –100 (8) None of above

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11. The function f(t) has Laplace transform F (s) =2s +1

s(s + 4). The value of

f (!)

f (0+ )is:

(1) 1 (2) 2 (3) 0 (4) 0.25 (5) 0.5

(6) 0.125 (7) 8 (8) None of the above

12. The convolution, y(t ) , of h(t ) = u(t +1)! u(t ) with f (t ) = e!t

u(t ) for the time

interval !1" t " 0 is:

(1) 0 (2) 1! e!t (3) e

!t !1 (4) (1! e

!1)e!t

(5) (e!1!

1)e!t

(6)e!(t +1)

!

1 (7) 1! e

!(t +1)

(8) None of above

MC SOLUTIONS

Solution 1: H ( s) = K ( s + 2)( s !1)

( s +1) ( s +1)2 +1"#$

%&'

( H (0) = !5 = ! K ( K = 5 .

Solution 2: Let V1 denote the voltage across the first 1 F cap. Then

V 1 =

1

s

1

s+

2

s

!12

s=4

s. Hence v1(1

!

) = 4 V. For t > 1 we have

V out (s) = (4 + 2)e!s

"

1

2s=

3

se!s

. Thus vout

(2) = 3 V.

Solution 3: Loop equations are:

V in = 1+ s +s

s +1

! " #

$ % & I 1 '

s

s +1 I 2 +V , 0 = !

s

s +1 I 1 +

s

s +1+1+

1

s

"

#$%

&'I 2 !V , and

I 2 ! I 1 =s +1

sV 1 = (s +1) I 1!!"!!0 = (s + 2) I 1 ! I 2

Solution 4: u(t) by inspection.

Solution 5: H ( j3) = j3+1 j3!1 j3 j3+ 4

= 103" 5

= 23

. Hence output = 4.

Solution 6: H ( j4) =( j100 +1) ( j100!1)

j100 ( j100 +100). Hence

!Output = ! H ( j4) + 45o " 90o + 90o # 90o # 45o # 45o = 0o .

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Solution 7. 2 sV out

+ 0.5(1+ a)V out

+V out

=V in ! H ( s) =

1

2 s +1.5+ 0.5a. Hence

1.5+ 0.5a > 0 ! a > "3 .

Solution 9. K f =1000

5= 200 .

0.001=1

K f K m!!!!!K m =

1000

200= 5!!!!! Lnew =

5

2004 = 0.1 H

Solution 10. s !10 = 0!"! s

10!10 = 0!#! p1new = 100

Solution 11. f (0+ ) = lims!"

sF (s) = 2 and f (!) = lims"0

sF (s) = 0.25 . Hence f (!)

f (0+ )=

1

8.

Solution 12. y(t ) = e!(t !" )d "

!1

t

# = e!(t !" )$

%&'!1

t = 1! e

!(t +1)

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EE-202, Ex 2, page 10

WORKOUT PROBLEM I (40 POINTS): Consider the circuit below in which Rs = 8 100 Ω.

In solving this problem you must make intelligent approximations in your calculationsthat can by justified by physical reasoning. Your work is to be organized and clear.

Unclear or sloppy work will be penalized and may be considered incorrect.

(a) (8 pts) With the switch in position A, find the transfer function H 1( s)=

V out

( s)

V s1( s)

.

Then find the associated impulse response.

Solution (a): The impedance of the parallel combination of the 40 Ω resistor and the

capacitor is

8 ! 40

s

40 +8

s

=8

s + 0.2. Hence H 1(s) =

8

s + 0.2

10 +8

s + 0.2

=8

10s + 2 + 8=

0.8

s +1. Thus

the impulse response is vout

(t ) = 0.8e!t

u(t ) V.

(b) (10 pts) With the switch in position B, find the transfer function H 2( s) =V out (

s)

V s2( s)

.

Then find the associated step response.

Solution (b): Short vs1(t). A simple source transformation generates a current source of

value is2(t ) =vs2(t )

Rs

driving the parallel resistance of Rs//10 Ω in parallel with the

capacitor. However, Rs//10 approximates Rs. Therefore

(5 pts) V out (s) =

8 Rs

s

Rs +8

s

!V s2(s)

Rs

!"! H 2(s) =V out (s)

V s2(s)=

8

Rs

s +8

Rs

=100

s +100

(5 pts) Thus, V out (s) =100

s(s +100)=

1

s!

1

s +100and v

out (t ) = (1! e

!100t )u(t ) V.

(c) (22 pts) For this part suppose vs1(t ) = 10u(t ) V and v

s2(t ) = 2u(t ) V. The capacitor

IC is zero at t = 0. The switch is initially in position A. The switch moves from A to B at

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EE-202, Ex 2, page 11

t = 5 s and back to A at t = 6 s, then back to B at t = 11 s. Compute an expression for

vout

(t ) for 0 ≤ t < 11 s. Accurately sketch your plot on the graph below after suitable

labeling.

Solution (c):

Case 1. (5 pts) Switch in position A. From part (a), V out (s) = 8s(s +1)

= 8s!

8s +1

. Thus

for 0 ≤ t ≤ 5 s, vout

(t ) = 8(1! e!t

)u(t ) .

Case 2. (7 pts) Here vC (5) ! 8 V. Observe that in position B, the maximum steady

state contribution of the voltage due to vs1(t ) is

8

100

10 +8

100

! 0 relative to the contribution

of vs2 (t ) so we ignore v

s1(t ) . Using a current source model of the capacitor and the

result of part (b), we have by linearity/superposition,

V out (s) = 2100

s(s +100)+CvC (5)

8

s +100

! " #

$ % & e'5s

=2

s'

2

s +100+

8

s +100

! " #

$ % & e'5s

Therefore,

vout

(t ) = 2u(t ! 5) + 6e!100(t !5)

u(t ! 5) V

Case 3. (7 pts) Here, vC (6) ! 2 V. From Case 2, superposition, and case 1,

V out (s) =8

s !8

s +1+CvC (6)

8

s +100" # $ % & ' e!6s=

8

s !6

s +1" # $ % & ' e!6s

Therefore,

vout

(t ) = 8u(t ! 6) ! 6e!(t !6)

u(t ! 6) V

»t = 0:.05:11;

»f1 = 8*(1-exp(-t)).*(ustep(t)-ustep(t-5));»f2 = (2+6*exp(-100*(t-5))).*(ustep(t-5)-ustep(t-6));

»f3 = (8-6*exp(-(t-6))).*ustep(t-6);

»plot(t,f1+f2+f3)»grid

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EE-202, Ex 2, page 12

0 2 4 6 8 1 0 1 2

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202ex2sum04sols

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