202finalsolssp06

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EE-202

FINAL

May 5, 2006

Name: __________________________________(Please print clearly)

Student ID: _________________

CIRCLE YOUR DIVISION

DeCarlo 2:30 MWF Furgason 3:30 MWF

INSTRUCTIONS

There are 28 multiple choice worth 6 points each and

there is 1 workout problem worth 12 points.

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

permitted. A transform table will be handed out separately.

Carefully mark your multiple choice answers on the scantron form. Work on

multiple choice problems and marked answers in the test booklet will not be graded.

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 standards of 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.

Communicating with any of your classmates, in any language, by any means, for

any reason, at any time between the official start of the exam and the official end of the

exam is grounds for immediate ejection from the exam site and loss of all credit for this

exercise.

Do not open, begin, or peek inside this exam until you are instructed to do so



EE-202, Ex 1 Sp 06 page 2

MULTIPLE CHOICE.

1. Determine the Laplace transform, L f (t ){ } , of f (t ) =

d

dt 3! 2e!2t + e!t "

#$%u(t ){ } where u(t ) is

the unit step function.

(1) 3

s

!

2

s +1+

1

s + 2 (2) 3

s

+1

s +1!

2

s + 2 (3) 4

s +1!

1

s + 2

(4) 3!2s

s +1+

s

s + 2 (5) 3+

4s

s + 2!

s

s +1 (6)

4

s + 2!

1

s +1

(7) 3+s

s +1!

2s

s + 2 (8) None of these

ANSWER : (7)

2. If two signals x (t ) and y(t ) are related by the equations

dx(t )

dt + 2 y(t ) = 4! (t ) and 2 x(t )! y( z)dz

0!

t

" = 2u(t ) ,

where x (0!

) = 2 , u(t ) is the unit step function, and ! (t ) is the Dirac delta function, then X (s) is:

(1) 10

s

(2) 2

s

(3) 8

5 (4)

1.2

s

(5) 8

s

(6) 1.6

s

(7)6

5 (8) None of these

ANSWER 2: sX (s)! 2 + 2Y (s) = 4 !!"!!sX (s) + 2Y (s) = 6 , 2 X (s)!Y (s)

s=

2

s!"!4sX (s)! 2Y (s) = 4

Therefore, 5sX (s) = 10!!! X (s) =2

s.




3. The impedance, Zin, of the circuit shown below is, in ohms:

(1) 2 + 2s +2

s

(2) 2 + 2s +2a

s

(3) 2 + 2s +2 ! a

s

(4) 2 + 2s +2 + a

s

(5) 2 + 2s +2 + 2a

s

(6) 2 + 2s +

2 ! 2a

s (7) 2 + 2s +

2a ! 2

s

(8) None of these

ANSWER 3: (6) V in(s) = (2 + 2s) I R(s)+2

s( I R(s) ! aI R(s)) = 2 + 2s +

2 ! 2a

s

" # $

% & ' I R(s)

4. The value of C for which the transfer function for the op amp circuit below is

H (s) = 1+s

(s + 2)(s + 4)is C = (in F):

(1) 1 (2) 2 (3) 0.5 (4) 4

(5) 5 (6) 0.25 (7) 0.2 (8) None of above

ANSWER 4: (3)V in

1+1

Cs

= (s + 4) V out !V in( )!!"!!V out V in

= 1+s

s +1

C

# $ %

& ' ( (s + 4)

!"!C = 0.5

5. The impulse response of the circuit below is (in V):

(1) 0.25e!4t

u(t ) (2) 2.5e!0.25t

u(t ) (3) 0.25e!0.25t

u(t ) (4) 0.4e!4t

u(t )




(5) 0.025e!4t

u(t ) (6) 0.4e!0.25t

u(t ) (7) 0.025e!0.25t u(t ) (8) None of above

ANSWER 5: (2) Reflecting to the primary side, we have C eq = 2 F. By V-division on Primary side,

V prim (s) =

1

2s

2 +1

2s

V in =1

4s +1V in =

0.25

s + 0.25V in . Hence V out (s) = 10V prim(s) =

2.5

s + 0.25V in .

vout

(t ) = 2.5e!0.25t

u(t ) .

6. Consider the circuit below in which vout

(0!

) = 1 V. Then vout

(t ),!t ! 0 is (in V):

(1) e!t u(t ) (2) e!0.25t

u(t ) (3) e!10t u(t ) (4) e!4t u(t )

(5) e!2t u(t ) (6) e!

0.5t u(t ) (7) e!0.1t u(t ) (8) None of above

ANSWER 6: (1) The equivalent resistance seen by the capacitor is50

100= 0.5 Ω. Using the equivalent

current source model of an initialized cap in the s-domain, we have V out (s) =1

2s +1

0.5

! 2 =1

s +1.

Hence, vout

(t ) = e!t u(t ) V.

7. For the circuit shown below v1(0) = 0 as are the initial voltages on all other capacitors, at time t = 1!

both switches are flipped to positions A. Then v1(2) = (in V):




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

(5) 5 (6) 6 (7) 20 (8) 50

ANSWER 7: (3) Initially, from left, V 5Fl =20

10s=

2

s. From right, V 5Fr =

10

s. With switches in

position A, there are two current source excitations: 5 *2 = 10 A on left, and 5*10 = 50 A on right.

C eq = 20 F. Therefore V 1 =60

20s=

3

s.

8. A circuit with transfer function H (s) =(100 + s)(100 ! s)

s2+ s + (100)

2is excited by the input

vin(t ) = 10cos(100t ! 45o) V, then the magnitude of the output in SSS is:

(1) 100 (2) 200 (3) 300 (4) 400(5) 50 (6) 1000 (7) 2000 (8) none of above

ANSWER 8: (7) 10 ! H ( j 100) = 10 !(100 + j 100)(100 " j 100)

"1002+ j 100 + (100)

2 = 10 !

100 2( )2

100= 2000




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

0 1 2 3 4 5 6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency, rad/s

M a g n i t u d e o f H ( j w )

TextEnd

Then the best candidate for H(s) is:

(1)(s + 2)

4

(s +1)2(s + 3)

2(2)

(s2+1.8)(s

2+ 2.2)

(s + 2)4

(3)(s + j 1.8)

2(s ! j 2.2)

2

(s +1)2(s + 4)

2(4)

(s2+ 3.5)(s

2+ 4.6)

(s + j 2)2(s ! j 2)

2

(5)(s

2+1.87

2)(s

2+ 2.15

2)

(s + 0.375)2+1.5

2( ) (s + 0.62)2+ 2.5

2( )(6)

(s2+ 4)(s

2+ 9)

[(s +1)2+ 3][(s +1)

2+ 8]

(7)(s

2+ 2

2)(s

2+ 2.5

2)

(s + 0.4)2+1.5

2( ) (s + 0.6)2+ 2.5

2( )(8) none of above

ANSWER 9: (5)




10. The low pass circuit shown below has a gain of 0.707 at ! = 1 rad/s. If the circuit is to befrequency scaled so that this gain occurs ! = 100 rad/s with the restriction that the circuit is impedance

scaled so that the inductor is 40 mH, then the NEW value of C 2 is C 2,new = (in mF):

(1) 5 (2) 20 (3) 5/3 (4) 45

(5) 50 (6) 6 (7) 150 (8) none of the above

ANSWER 10: (1) K mK f

!

4

3= 0.04 !"!K m =

3

4! 4 = 3 . C 2,new =

1.5

3!100=

0.5

100= 0.005 F.

11. Given that H (s) =V out

I in

in the circuit below, the COMPLETE range gm for which the circuit is

stable is:

(1) gm > 1 (2) gm < 1 (3) gm < 0.5 (4) gm > 0.5 (5) gm < !2 (6) gm > !2

(7) gm < 2 (8) none of the above

ANSWER 11: (3) I in = 0.5V out ! gmV out + 0.5sV out = 0.5 ! gm + 0.5s( )V out Hence,

H (s) =V out

I in=

1

0.5s + 0.5 ! gm=

2

s +1! 2gm. Thus 1! 2gm > 0!"!gm < 0.5 .




12. What is vout

(t ) in sinusoidal steady state for the following circuit if vin(t ) = 10cos(2t ) V?

(1) 5 2 cos(2t!90o

)V (2) 5 2 cos(2t)V (3) 5 2 cos(2t+ 45o

)V

(4) 5 2 cos(2t! 45o

)V (5) 10cos(2t! 45o)V (6) 10cos(2t)V

(7) 10cos(2t+ 45o)V (8) None of these

ANSWER 12: (4) H (s) =4

4 + 3s +4

s

. Vout = 10 ! H ( j 2) =40

4 + j 6 " j 2=

10

1+ j = 5 2#" 45

o

13. The circuit shown below consists of a 100 Ω resistor in parallel with a real 10!µ F capacitor having

a Q = 10 at 1000 rad/sec. The input admittance of this combination (given in mhos or Siemens) is:

(1) 0.010 + (10µ )s (2) 0.011+ (10µ )s (3) 0.101+ (10µ )s (4) 0.110 + (10µ )s

(5) 0.01+1

(10µ )s (6) 100+

1

(10µ )s (7) 100 + (10µ )s (8) None of these

ANSWER 13: (2) 10 =! RC = 103"10

#5 R!$! R = 1000 Ω. Geq =

1

100+

1

1000= 0.011 mhos.




14. The coupled inductors shown below high-pass filter the input signal iin(t ) . At what frequency will

the –3dB point occur (i.e. at what frequency will the magnitude of the transfer function be 3 dB belowthe maximum value)?

(1)1

5rad/s (2)

1

3rad/s (3)

15

8rad/s (4)

17

7rad/s

(5) 3 rad/s (6) 5 rad/s 7) 7.5 rad/s (8) None of these

ANSWER 14: (6) V out (s) = !3sI in(s)+ 5sI 2(s) = !3sI in(s)! 5sV out (s)

25. Hence

1+ 0.2s( )V out (s) = !3sI in(s)!"! H (s) =!15s

s + 5. s = j 5 .

15. For the circuit below with i L (0!

) = 0 Aand V C (0!

) = 2V , the voltage, vC (t ) , in volts for t >0 is

given by:

(1) !e

!t + 3e

!2t

(2) 2 e

!t

!

e

!2t

( ) (3) 6e

!t ! 4e

!2t

(4)

1

9 3! (t )" 8e

"2

3t

#

$

%%

&

'

((

(5) !3e!t + 4e

!2t (6)

2

3!e

!t + 3e

!3t ( ) (7) 4e!t ! 2e

!2t (8) None of these

ANSWER 15: (3) M = k L1 L2 = 0.5 1 = 0.5 and Z in(s) = 3+2

s+ (2 !1)s =

s2+ 3s + 2

s. By voltage

division and the equivalent model of the capacitor, V C (s) =2

(s +1)(s + 2)+

2s + 6

(s +1)(s + 2)=

6

s +1!

4

s + 2.



EE-202, Ex 1 Sp 06 page 10

16. For the op-amp filter shown below determine the Q of the transfer function, H (s) .

(1)R1C1+Rf Cf

R1C1Rf Cf

(2) R1C1Rf Cf

R1C1+Rf Cf

(3) R1C1Rf Cf

R1C1+Rf Cf

(4)R1C1Rf Cf

R1 C1+Cf ( )

(5) R1C1Rf Cf

C1 R1+Rf ( ) (6)

R1C1Rf Cf

R1Cf +Rf C1

(7) R1C1Rf Cf

Cf R1+Rf ( ) (8) None of above

17. The step response of a certain system is given by 4 1!2 e!2t ( ) u(t ) . Determine the impulse

response h(t ) .

(1) 4 1!2e!2t

( )"(t ) (2) 4 1+ 4e!2t

( )"(t ) (3) 4!(t )+16e"2t

u(t )

(4) !4"(t )!16e!2t

u(t ) (5) 8e!2t

u(t ) (6) 16e!2t

u(t )

(7) !4"(t )+16e!2t

u(t ) (8) None of these

ANSWER 17: (7)d

dt 4 1!2e

!2t ( ) u(t )"#

$%= 16e

!2t u(t ) ! 4& (t )



EE-202, Ex 1 Sp 06 page 11

18. Determine the output impedance Z out (s) for the circuit shown below.

(1) 0 (2) aR (3) a2R (4)

R

a2

(5) 2 1-a( )R (6) 1-a( )2

R (7)1-a

a

!

"####

$

%

&&&&

2

R (8) None of these

ANSWER 18: (7) Short circuit V-source. V out = V secondary +V R , V

secondary = !V R

a,

V out = 1!1

a

" # $

% & ' V R =

a !1a

" # $

% & ' V R . Now V R = R I out + I prime( ) = R

a !1a

" # $

% & '

I out . Therefore

Z out =V out

I out

= Ra !1a

" # $

% & ' 2

.

19. Shown below is the small-signal high-frequency model for an FET. For this circuit determine the y-parameter yos , given that:

ig = yisvgs + yrsvds

id = y fsvgs + yosvdsor equivalently

ig

id

!

"#

$

%& =

yis yrs

y fs yos

!

"#

$

%&

vgs

vds

!

"#

$

%&

(1) j!Cgs

(2) j!(Cgs

+ Cgd

) (3) gm! j"C

gd

(4) gm+ j!(C

gs+C

gd) (5) g

m+

1

r d

+ j!Cds

(6)1

r d

+ j! Cgd+Cds( )

(7) gm+

1

r d

+ j! Cgd+C

ds( ) (8) None of these



EE-202, Ex 1 Sp 06 page 12



EE-202, Ex 1 Sp 06 page 13

20. In the circuit shown below determine the turns ratio, n1 : n2, that results in the maximum power

being delivered to the load resistor, RL.

(1) 1 : 400 (2) 1 : 40 (3) 1 : 20 (4) 3200 : 8(5) 20 : 1 (6) 40 : 1 (7) 400 : 1 (8) None of these

ANSWER 20: (5) 3200 =n1

n2

!

" #$

% &

2

8!!'!! n1

n2

!

" #$

% & = 20 .



EE-202, Ex 1 Sp 06 page 14

21. The circuit below is to realize a LP Butterworth filter having 3 dB down point at ! c= 3000 2

rad/s. The 3 dB NLP Butterworth transfer function is H 3dBNLP(s) =1

s2+ 2s +1

. Assuming no

magnitude scaling, the value of C in mF is:

(1) 1.333 (2) 2.333 (3) 0.333 (4) 4.333

(5) 12.000 (6) 1 12 (7) 0.666 (8) none of above

ANSWER 21: (3) H cir (s) =

1

LC

s2+

1

Ls +

1

LC

implies L =

1

2,!C = 2 . Hence,

C final =2

3000 2= 0.333 mF.

22. The NLP Butterworth circuit below, H cir (s) =1

s +1, is to realize a HP filter with 3 dB down point

of 5000 rad/s. The load resistance is to be R L,final = 100 Ω. The inductor becomes a capacitor of

value (in µ F):

(1) 100 (2) 2 (3) 20 (4) 200

(5) 5 (6) 50 (7) 101!105 (8) none of above

ANSWER 22: (2) L = 1!!!!!C = 1

100 " 5000= 2 "10

#6 F.



EE-202, Ex 1 Sp 06 page 15

23. For the circuit below, the admittance parameter matrix is (in mhos):

(1)4 !3

!gm ! 3 5

"

#$

%

&' (2)

1 2

!gm ! 3 5

"

#$

%

&' (3)

4 !gm ! 3

!3 5

"

#$

%

&'

(4)5 gm ! 3

!3 4

"

#$

%

&' (5)

1 !gm + 3

3 5

"

#$

%

&' (6)

5 gm + 3

3 4

!

"#

$

%&

(7)4 !3

gm ! 3 5

"

#$

%

&' (8) none of above

ANSWER 23: (1) See text example 19.6. Arrow now points up here.

Y 1 + Y 3 !Y 3!gm ! Y 3 Y 2 + Y 3

"

#$

%

&' =

4 !3

!gm ! 3 5

"

#$

%

&'



EE-202, Ex 1 Sp 06 page 17

26. In the circuit below, the h-parameters, h11 and h22 , are respectively (in standard units):

(1)1

b,!!1b

(2)1

b,!0 (3) 0,! R

b(4)

R2

b2,!0 (5)

R

b2,!1b

(6) R

b

2,!0 (7)

R

b

2,!!1

b(8) none of above

ANSWER 26: (See text example.) (6) R

b2,!0

27. Three two ports are connected in cascade as shown below. Standard labeling is assumed. The t-

parameter matrices of the first two two-ports are in standard units T 1 =2 1

1 1

!

"#

$

%& and T 2 =

1 !1

!1 2

"

#$

%

&'

respectively. The overall t-parameter matrix is:

(1)0.5 0

0 !2

"

#$

%

&' (2)

0.5 0

0 2

!

"#

$

%& (3)

3 3

2 3

!

"#

$

%& (4)

3 !1

1 0

"

#$

%

&'

(5)1 0

0 !1

"

#$

%

&' (6)

1 0

0 1

!

"#

$

%& (7)

!0.5 0

0 2

"

#$

%

&' (8) none of above

ANSWER 27: (2) T 1T 2 =

1 0

0 1

!

"

#$

%

& . Therefore the overall t-parameter matrix is: T =0.5 0

0 2

!

"

#$

%

&



EE-202, Ex 1 Sp 06 page 18

28. Consider the two-port interconnection below that is assumed to have the standard labeling. The y-

parameter matrix of the top two-port is y =3 5

2 4

!

"#

$

%& mhos. The z-parameter, z21 , of the

interconnected two port is (inΩ):

(1) 1 (2) 2 (3) 3 (4) 4 (5) –2

(6) –1 (7) 0 (8) none of above

ANSWER 28: (1) ztop = 0.54 !5

!2 3

"

#$

%

&' =

2 !2.5

!1 1.5

"

#$

%

&' . zbot =

2 2

2 2

!

"#

$

%& . Therefore z21 = 2 !1= 1 .



EE-202, Ex 1 Sp 06 page 19

29-30. (12 points) Sketch the convolution of the following two functions (i.e., y(t) = f(t) * g(t) ).

Be sure to carefully label the vertical axis in your sketch. Show work on attached page.

Integral of f(t)



EE-202, Ex 1 Sp 06 page 20

-1 0 1 2 3 4 5 6 7 80

1

2

3

4

5

6

Time in sec

I n t e g r a l o f f

TextEnd

dg

dt = ! (t +1)+! (t )" ! (t "1)"! (t " 2)

The convolution:

-1 0 1 2 3 4 5 6 7 80

1

2

3

4

5

6

7

C o n v o l u t i o n

o f f a n d

g

TextEnd

Time in sec



EE-202, Ex 1 Sp 06 page 21

202finalsolssp06

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