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

dt3! 2e!2t + e!t"#

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

the unit step function.

(1) 3s!

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+ 2y(t) = 4! (t) and 2x(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) 10s

(2) 2s

(3) 85

(4) 1.2s

(5) 8s

(6) 1.6s

(7) 65

(8) None of these

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

=2

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

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

.

EE-202, Ex 1 Sp 06 page 3

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

(1) 2 + 2s + 2s

(2) 2 + 2s + 2as

(3) 2 + 2s + 2 ! as

(4) 2 + 2s + 2 + as

(5) 2 + 2s + 2 + 2as

(6) 2 + 2s + 2 ! 2as

(7) 2 + 2s + 2a ! 2s

(8) None of these

ANSWER 3: (6) Vin (s) = (2 + 2s)IR(s) +2

s(IR(s) ! aIR(s)) = 2 + 2s +

2 ! 2as

"#$

%&'IR(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) Vin

1+1

Cs

= (s + 4) Vout !Vin( )!!"!!Vout

Vin

= 1+s

s +1

C

#$%

&'((s + 4)

!"!C = 0.5

5. The impulse response of the circuit below is (in V): (1) 0.25e!4tu(t) (2) 2.5e!0.25tu(t) (3) 0.25e!0.25tu(t) (4) 0.4e!4tu(t)

EE-202, Ex 1 Sp 06 page 4

(5) 0.025e!4tu(t) (6) 0.4e!0.25tu(t) (7) 0.025e!0.25tu(t) (8) None of above

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

Vprim (s) =

1

2s

2 +1

2s

Vin =1

4s +1Vin =

0.25

s + 0.25Vin . Hence Vout (s) = 10Vprim (s) =

2.5

s + 0.25Vin .

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!tu(t) (2) e!0.25tu(t) (3) e!10tu(t) (4) e!4tu(t)

(5) e!2tu(t) (6) e!0.5tu(t) (7) e!0.1tu(t) (8) None of above

ANSWER 6: (1) The equivalent resistance seen by the capacitor is 50100

= 0.5 Ω. Using the equivalent

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

2s +1

0.5

! 2 =1

s +1.

Hence, vout (t) = e!tu(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):

EE-202, Ex 1 Sp 06 page 5

(1) 1 (2) 2 (3) 3 (4) 4 (5) 5 (6) 6 (7) 20 (8) 50

ANSWER 7: (3) Initially, from left, V5Fl =20

10s=2

s. From right, V5Fr =

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.

Ceq = 20 F. Therefore V1 =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 ( j100) = 10 !(100 + j100)(100 " j100)

"1002+ j100 + (100)

2= 10 !

100 2( )2

100= 2000

EE-202, Ex 1 Sp 06 page 6

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

0 1 2 3 4 5 60

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

Frequency, rad/s

Magnitude o

f H

(jw

)

TextEnd

Then the best candidate for H(s) is:

(1) (s + 2)4

(s +1)2(s + 3)

2 (2) (s

2+1.8)(s

2+ 2.2)

(s + 2)4

(3) (s + j1.8)2(s ! j2.2)

2

(s +1)2(s + 4)

2 (4) (s

2+ 3.5)(s

2+ 4.6)

(s + j2)2(s ! j2)

2

(5) (s2+1.87

2)(s2+ 2.15

2)

(s + 0.375)2+1.5

2( ) (s + 0.62)2 + 2.52( ) (6) (s

2+ 4)(s

2+ 9)

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

2+ 8]

(7) (s2+ 2

2)(s2+ 2.5

2)

(s + 0.4)2+1.5

2( ) (s + 0.6)2 + 2.52( ) (8) none of above

ANSWER 9: (5)

EE-202, Ex 1 Sp 06 page 7

10. The low pass circuit shown below has a gain of 0.707 at ! = 1 rad/s. If the circuit is to be frequency 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 C2 is C2,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) KmK f

!4

3= 0.04 !"!Km =

3

4! 4 = 3 . C2,new =

1.5

3!100=0.5

100= 0.005 F.

11. Given that H (s) = VoutIin

in the circuit below, the COMPLETE rangegm 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) Iin = 0.5Vout ! gmVout + 0.5sVout = 0.5 ! gm + 0.5s( )Vout Hence,

H (s) =Vout

Iin=

1

0.5s + 0.5 ! gm=

2

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

EE-202, Ex 1 Sp 06 page 8

12. What is vout (t) in sinusoidal steady state for the following circuit if vin (t) = 10cos(2t) V? (1) 5 2 cos(2t!90

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

o)V

(4) 5 2 cos(2t! 45

o)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 ( j2) =40

4 + j6 " j2=10

1+ j= 5 2#" 45

o

13. The circuit shown below consists of a 100 Ω resistor in parallel with a real 10!µF 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

#5R!$!R = 1000 Ω. Geq =

1

100+

1

1000= 0.011 mhos.

EE-202, Ex 1 Sp 06 page 9

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 below the maximum value)?

(1) 15rad/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) Vout (s) = !3sIin (s) + 5sI2(s) = !3sIin (s) ! 5sVout (s)

25. Hence

1+ 0.2s( )Vout (s) = !3sIin (s)!"!H (s) =!15s

s + 5. s = j5 .

15. For the circuit below with iL (0

!) = 0A and VC (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

93! (t) " 8e

"2

3t#

$

%%

&

'

((

(5) !3e!t + 4e!2t (6) 23

!e!t

+ 3e!3t( ) (7) 4e!t ! 2e!2t (8) None of these

ANSWER 15: (3) M = k L1L2 = 0.5 1 = 0.5 and Zin (s) = 3+2

s+ (2 !1)s =

s2+ 3s + 2

s. By voltage

division and the equivalent model of the capacitor, VC (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+RfCfR1C1RfCf

(2) R1C1RfCf

R1C1+RfCf

(3) R1C1RfCfR1C1+RfCf

(4) R1C1RfCf

R1 C1+Cf( )

(5) R1C1RfCf

C1 R1+Rf( ) (6)

R1C1RfCf

R1Cf +RfC1

(7) R1C1RfCf

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

17. The step response of a certain system is given by 4 1!2e!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

!2tu(t) (8) None of these

ANSWER 17: (7) ddt

4 1!2e!2t( ) u(t)"#

$%= 16e

!2tu(t) ! 4& (t)

EE-202, Ex 1 Sp 06 page 11

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

(1) 0 (2) aR (3) a

2R (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. Vout = Vsecondary +VR , Vsecondary = !VR

a,

Vout = 1!1

a

"#$

%&'VR =

a !1a

"#$

%&'VR . Now VR = R Iout + I prime( ) = R

a !1a

"#$

%&'Iout . Therefore

Zout =Vout

Iout

= 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 + yosvds or equivalently ig

id

!

"#

$

%& =

yis yrs

y fs yos

!

"#

$

%&vgs

vds

!

"#

$

%&

(1)

j!C

gs (2)

j!(C

gs + C

gd) (3)

g

m! j"C

gd

(4) g

m+ j!(C

gs+C

gd) (5)

gm

+1

rd

+ j!Cds

(6)

1

rd

+ j! Cgd

+Cds( )

(7)

gm

+1

rd

+ j! Cgd

+Cds( ) (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 H3dBNLP (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) Hcir (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, Hcir (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 RL,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.

Y1 + Y3 !Y3!gm ! Y3 Y2 + Y3

"

#$

%

&' =

4 !3

!gm ! 3 5

"

#$

%

&'

EE-202, Ex 1 Sp 06 page 16

Consider the two port configuration for questions 24 and 25 below. The voltage gain GV1 =V1

Vs

=s +1

s + 2

and the z-parameter matrix is only partially known as: Z1 =z11 2

!2 2

"

#$

%

&' .

24. The parameter z11 = : (1) 1 (2) !s (3) s +1 (4) s + 2 (5) s (6) 0 (7) s !1 (8) none of above

ANSWER 24: Zin

Zin +1=s +1

s + 2!!!Zin = s +1 . Zin = s +1 = z11 !

z12z21

z22 + ZL

= z11 +1 z11 = s

25. The voltage gain VoutV1

= :

(1) !1

s +1 (2) !1

s (3) 1

s +1 (4) !1

s + 2 (5) 1

s

(6) 1

s + 2 (7) s + 2 (8) none of above

ANSWER 25: (1) Gv2 =V2

V1

=ZL

ZL + z22

z21

Zin

=1

2!

"2

s +1=

"1

s +1

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) 1b,!!1

b (2) 1

b,!0 (3) 0,!R

b (4) R

2

b2,!0 (5) R

b2,!1

b

(6) Rb2,!0 (7) R

b2,!!1

b (8) none of above

ANSWER 26: (See text example.) (6) Rb2,!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 T1 =2 1

1 1

!

"#

$

%& and T2 =

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) T1T2 =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 (in Ω): (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

Inte

gra

l of 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

Co

nvo

lutio

n o

f f

an

d g

TextEnd

Time in sec

EE-202, Ex 1 Sp 06 page 21