Final Exam Practice Problems - Purdue Engineeringee202/pastexams/pracfinal.pdf · Final Exam Sum 03...

Final Exam Practice Problems MULTIPLE CHOICE. *1. For a > 0, the Laplace transform of

f (t) = 2e(t+a)

!(t " a) + 2e(t"a)

!(t + a) is: (1)

2e2a (2)

2e2a

+ 2e!2a (3)

2e!2a

(4)

2e!a(s!2) (5)

2ea (s!2) (6)

2e!a(s!2)

+ 2ea(s!2)

(7) None of above ANSWER: (5) 2. The output of a circuit has Laplace transform

Vout (s) =64

s3(s+ 2)

With

vout (t) = Ae!2t

u(t) + Bu(t) + other ! terms the value of B is: (1) –8 (2) 2 (3) –4 (4) 4 (5) 8 (6) –16 (7) 32 ANSWER (5) 3. The Thevenin equivalent admittanace

Zin (s)of the circuit below is:

(1)

s+ 2 +3

s (2)

s+ 2 !1

s (3)

s+ 2 +1

s

(3)

1

s+ 0.5 ! s (4)

1

s+ 0.5 + 3s (5)

1

s+ 0.5 + s

(6)

s+ 2 !3

s (7) None of Above

ANSWER (2).

Final Exam Sum 03 202 page 2

4. The transfer function for the following op amp circuit is:

(1)

(s+ 2)

s(0.5s+1) (2)

s(0.5s+1)

s+ 2 (3)

0.5s+1

s(s+ 2)

(4)

!0.5s+1

s(s+ 2) (5)

!(s+ 2)

s(0.5s+1) (6)

s(s+ 2)

2s+1 (7)

s(s+ 2)

0.5s+1

ANSWER: (7)

*5. Suppose

L f (t) = !!ns

s+ 2

"

# $ %

& ' .

Then

L tf (t) is:

(1)

1

s+

1

s+ 2 (2)

1

s+ 2!

s

(s+ 2)2

(3)

!1

s+ 2+

s

(s+ 2)2

(4)

1

s!

1

s+ 2 (5)

!1

s+

s

(s+ 2)2

(6)

1

s!

s

(s+ 2)2

(7)

s+ 2

s (8) None of above

ANSWER (4)

6. The Laplace Transform of )(tf is given as

F(s) =1! e

!(s!a)

s! a. Then the Laplace Transform of

e!at

tf (t) is:

(1)

1! e!(s!a)

(s! a)2

!e!(s!a)

s! a (2)

1! e!s

s (3)

1! e!s

s2

!e!s

s (4)

1! e!s

s2

(5)

!1! e

!s

s2

(6)

1! e!(s!2a)

(s! 2a)2

!e!(s!2a)

s! 2a (7) None of above

ANSWER (3)


7. At t = 0-, all capacitors are relaxed. At t = 0, the switch moves to position A. At t = 1, the switch moves to position B. vout(1+) = (in volts): (1) 12 (2) 2 (3) 3 (4) 4 (5) 8 (6) 6 (7) none of above

ANSWER (4).

8. A linear circuit with a transfer function

H(s) =Vout (s)

Vin (s)=16

s+ 4

s2

+ 4s+16 has an input

vin (t) = 2 cos(4t) V. The output of the circuit in the steady state has a magnitude and phase (in degrees) equal to: (1)

4, 45o (2)

2, ! 45o (3)

4, ! 45o (4)

4 2, 45o

(5)

4 2, ! 45o (6)

8, ! 45o (7)

8, 45o (8) None of above

ANSWER: (6). 9. In the circuit below, vC(0-) = 2 V and iL(0-) = 0. The node equation at node A which accounts for all initial conditions is:

(1)

2 +5

s+ 0.5s

!

" #

$

% & Va '

5

sVb = 2Vin +1 (2)

2 +5

s+ 0.5s

!

" #

$

% & Va '

1

sVb = 2Vin +1

(3)

2 +5

s+ 0.5s

!

" #

$

% & Va +

4

sVb = 2Vin + 2 (4)

0.5 +5

s+ 0.5s

!

" #

$

% & Va '

1

sVb = 0.5Vin +1

(5)

0.5 +5

s+ 0.5s

!

" #

$

% & Va '

5

sVb = 0.5Vin +1 (6)

0.5 +5

s+ 0.5s

!

" #

$

% & Va +

4

sVb = 0.5Vin + 2

(7) None of above

ANSWER (2).


10. The impulse response of a circuit is h(t) shown below. Suppose the input is

vin (t) = !2" sin(2"t)u(t) . The value of the convolution at t = 4 seconds is:

(1) 1 (2) 2 (3) –1 (4) –2 (5) 3 (6) 0 (7) none of above

ANSWER (6). 11. The output of a linear and relaxed circuit with input x(t) and impulse response h(t) (shown below) at time t = 3 seconds is:

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

(5) 0 (6) 3.5 (7) None of the above

ANSWER (2) 12. The value of ω which makes circuit below resonant is (in rad/s):

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

(5) 5 (6) 16 (7) None of the above

ANSWER (4)


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

0 2 4 6 8 10 120

0.2

0.4

0.6

0.8

1

1.2

1.4

Frequency-rad/s

| H(jw

) |

TextEnd

Then the best candidate for H(s) is:

(1)

Ks

(s+ 0.5 + j5)(s+ 0.5 ! j5) (2)

K(s+ j2)(s! j2)

(s+ 0.5 + j5)(s+ 0.5 ! j5)

(3)

K(s+ j2)(s! j2)

s(s+ 0.5 + j5)(s+ 0.5 ! j5) (4)

K(s+ j2)(s! j2)

(s+ 0.3+ j4)(s+ 0.3! j4)(s+ 0.8 + j6)(s+ 0.8 ! j6)

(5)

Ks(s+ j2)(s! j2)

(s+ 0.3+ j4)(s+ 0.3! j4)(s+ 0.8 + j6)(s+ 0.8 ! j6)

(6)

K(s+ j2)(s! j2)

s(s+ 0.3+ j4)(s+ 0.3! j4)(s+ 0.8 + j6)(s+ 0.8 ! j6) (7)

K(s+ j2)(s! j2)

s(s+ 0.8 + j6)(s+ 0.8 ! j6)

ANSWER (5) 14. Referring again to the frequency response of problem 13, the type of frequency response is: (1) low pass (2) high pass (3) band pass (4) band reject (5) none of the above ANSWER (3)


15. Consider the circuit below. The approximate value of the higher half power frequency is: (1) 11.25 (2) 12.5 (3) 22.5 (4) 20.2 (5) 20.4 (6) 21.25 (7) none of above

ANSWER: (6) 16. Referring again to the circuit of problem 15, the maximum value of the frequency response is: (1) 10 (2) 16 (3) 0.8 (4) 4 (5) 50 (6) 8 (7) none of above ANSWER: (6) 17. Consider the RLC circuit below. Recall that the Q of a lossy capacitor near resonance is Qcap ≅ RLCω0. The approximate value of Qp of the RLC circuit is: (1) 1.667 (2) 2 (3) 1.5 (4) 4.333 (5) 10 (6) 2/3 (7) none of above

ANSWER (1). 18. The value of the turns ratio "a" for maximize the power transferred to the load is: (1) 0.2 (2) 25 (3)

12.5 (4) 20 (5) 5 (6)

20 (7) None of above

ANSWER: (5) Under the condition that the load is taken the circuit to right of A-B. ANSWER: (7) Under the condition that the load is taken as the resistor to the right of C-D:

Indeed, here 4a2 = 50 !10050 +100

= 33.34 !"!a = 2.8868


19. The input impedance Zin(s) for the circuit below is:

(1)

0.5 +1

3s (2) 3s + 2 (3)

1.5s

3s+ 0.5 (4)

4s

2s+ 2

(5)

6

3s+ 2 (6)

6s

3s+ 2 (7)

0.5s

3s+ 0.5

ANSWER: (6) 20. The t-parameter, t21, of the circuit below below is: (1) 1 (2) 2 (3) –0.25 (4) 0.25 (5) 0.5

(6) –0.5 (7) None of above

ANSWER (4). 21. Consider the 2-port below. The inner 2-port labeled N has z-parameters: z11=7 Ω, z12 =2 Ω, z21=10 Ω, z22= 3 Ω. Then the overall 2-port N* has y12 = ( in mhos): (1) s+3 (2) –s–2 (3) –2 (4) –s

(5) s+2 (6) s–2 (7) None of the above

ANSWER (2)


22. The impulse response, h(t), of the circuit below is: (1)

e!tu(t) (2)

4e!4 t

u(t) (3)

!e!tu(t) (4)

!e!4 t

u(t) (5)

!0.25e!tu(t)

(6)

0.25e!tu(t) (7) None of the above

ANSWER (5) 23. Consider the cascaded two port below.

The 2-port Na has y parameters and 2-port Nb has z parameters as given below (with standard port labeling and units):

ya[ ] =0.1 0.1

!0.2 0.1

"

# $

%

& ' ; zb[ ] =

25 1

20 2

"

# $

%

& '

Then

Zina =:

(1) 10 (2) 20 (3) ∞ (open circuit) (4) 30/7

(5) 5 (6) 0 (short circuit) (7) none of above

ANSWER (5).

24. Referring again to the circuit of problem 23, the voltage gain

V1b

V1a

is:

(1) 1 (2) 2 (3) –1 (4) –2

(5) 4/3 (6) –4/3 (7) None of above ANSWER (1).


25. The three terminal device below is represented by an indefinite admittance matrix (in mhos) a number of whose entries have been corrupted by a virus in Microsoft word.

I1

I2

I3

!

"

#

#

#

$

%

&

&

&

=

1 ? 1

'1 0 ?

? ? ?

!

"

#

#

#

$

%

&

&

&

V1

V2

V3

!

"

#

#

#

$

%

&

&

&

If the three terminal device is made into a 2-port with terminal 1 grounded and terminal 3 as the "input port" with terminal 2 taken as the secondary, then the admittance parameter y21 of the new configuration in mhos is: (1) 1 (2) 2 (3) –1 (4) –2 (5) 0.5 (6) 0 (7) none of above ANSWER (1). 26. Referring again to the indefinite admittance matrix of problem 25 above, if terminal 3 is grounded with terminal 1 the input port and terminal 2 the output port, then the z-parameter z21 of this new configuration is: (1) 1 (2) 2 (3) –1 (4) –2 (5) 0.5 (6) –0.5 (7) 0 ANSWER: (6)


27. The filter transfer function

HLPcir (s) =2

s2

+1

Qs+1

can be realized by the Sallen & Key

circuit below

If this circuit is to realize the transfer function

HLPcir (s) =4

s2

+ s+ 4, then C2 becomes C2new

whose value is: (1) 1 F (2) 2 F (3) 0.125 (4) 0.25 F (5) 0.5 F (6) 1/16 F (7) none of above ANSWER (4) 28. Suppose you have computed the values of the resistors and capacitors of the circuit of

problem 27 that actually realize the transfer function

HLPcir (s) =4

s2

+ s+ 4, If the circuit

is to be frequency scaled by Kf = 1000 and the value of R2 is to be 1000Q, then the new value of C1 is:

(1) 1 µF (2) 0.25 µF (3) 0.125 µF (4) 2 µF (5) 0.5 µF (6) 5 µF (7) none of above ANSWER (1) 29. Again consider the circuit of problem 27. The gain is to be adjusted using input attenuation in which R1 is replaced by a voltage divider of RA in series with RB as developed in class. The value of RB that realizes the proper gain is (in Ω): (1) 1 (2) 2 (3) 0.25 (4) 4 (5) 0.5 (6) 0.25 (7) none of above ANSWER:

RA = RB = 2 Ω.


30. The following LP circuit is to be used to realize a HP circuit with passband edge frequency of wp = 1000 rad/s. If the magnitude scale factor is Km = 1000, then the resistor R2 becomes a capacitor C2 of value: (1) 1 µF (2) 0.2 mF (3) 0.2 µF (4) 1 mF (5) 5 µF (6) 5 mF (7) 5 F

ANSWER (3) 31. For the circuit below, the values of R (in Ω) and C (in F) that make the input admittance

Yin (s) =4

2s+1+

4s

2s+1

are: (1) R = 1, C = 4 (2) R = 4, C = 0.25 (3) R = 4, C = 4

(4) R = 0.25, C = 0.25 (5) R = 0.5, C = 2 (6) R = 0.25, C = 4

(7) none of above


32. The transfer function for the following op amp circuit is:

(1)

!s+ 0.5

0.5s+1 (2)

!0.5s+1

s+ 0.5 (3)

s+ 0.5

0.5s+1 (4)

s+1

0.5s+1

(5)

!s+1

0.5s+1 (6) –2 (7) none of above

33. In a circuit shown below, the initial conditions are

iL (0!) =1amp and

vC (0!) = 2 volt. If

iL (t) = ae!0.5t

u(t) + be!dt

u(t) then a is: (1) 1 (2) 2 (3) 3 (4) 4 (5) –4 (6) –3 (7) –1

34. Let v(t) be the output and vi(t) the input. In order for the circuit below to be stable in the sense of BIBO, the range of m must be:

(1) m!1 (2) m > -1 (3) m!0 (4) 0 > m

(5) 1!"m (6) 1 > m (7) Something else


35. The transfer function of a system is given as

H(s) =2

s(s+ 2). When

vi (t) = 2 2 cos(2t + 45!)u(t) is applied as the input, the output becomes

vo(t) = Acos(2t + !)u(t) where ),( !A is:

(1)

(1,!90!) (2)

(1,!135!) (3)

(2,!45!) (4)

(1,!45!)

(5)

(2,!135!) (6)

(2,!90!) (7) Something else

36. The step response of a linear time invariant (RLC) circuit is given as 2r(t-1), a shifted ramp function. When

vi (t) = e!2t

u(t) is applied to this circuit as the input, the output )(tvo

is:

(1)

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

u(t) ! e!2t

u(t !1)

(3)

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

u(t !1) (4)

(1! e!2(t!1)

)u(t !1)

(5)

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

u(t !1) (6)

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

u(2t ! 2)

(7) Something else

37. The impulse response of a particular circuit is approximately measured on a scope as illustrated below.

The input to the circuit is

vin (t) = u(t) ! u(t !1) . Then the value of the response, y(t), at t = 2.5 seconds is: (1) 1 (2) 2 (3) 3 (4) 3.5 (5) 5

(6) 2.5 (7) none of above


38. For the band pass circuit shown below, the exact (not approximated) resonant frequency is:

(1) 1 (2) 2 (3) 2 (4) 1/ 2 (5) 4 (6) 0.25 (7) None of the above

39. A resonant circuit has the transfer function

H(s) =Vout (s)

Iin (s)=

cs

s2

+ as+ b

, with a = 0.1, b = 4

and c = -40. After both magnitude and frequency scale changes of 4=mK and 2=fK , the new c becomes: (1) –320 (2) –160 (3) –120 (4) –80

(5) –40 (6) –20 (7) None of the above

40. A coupling circuit is connected to a voltage source and a load as shown below. 2u(t) is applied as the input voltage. Then the Laplace transform I1(s) is:

(1)

s+1

s(4s+ 2) (2)

2s+ 2

s(s+ 2) (3)

4s+ 2

s(s+1) (4)

s+1

s(2s+1)

(5)

2s+1

s(s+1) (6)

2s+1

s+ 2 (7) None of the above

41. Refering again to the circuit of problem 10, the voltage vout(t) =: (1)

0.5!(t) " 0.5e"0.5t

u(t) (2)

0.5!(t) + 0.5e"0.5t

u(t)

(3)

u(t) ! e!0.5t

u(t) (4)

2u(t) ! 2e!0.5t

u(t)

(5)

e!0.5t

u(t) (6)

2e!0.5t

u(t)

(7) none of above


42. In order for an active circuit shown below (units are Ω and F) to have a transfer function with two poles, p1 = –0.6 and p2 = –0.4, R must be: (1) 1.44 (2) 1.20 (3) 0.96 (4) 0.72

(5) 0.48 (6) 0.24 (7) None of the above

43. In the circuit shown below, )()( tuaeti

bt!= where b is:

(1) 2 (2) 0.5 (3) 3 (4) 2.5

(5) 1.5 (6) 1 (7) None of the above

44. Consider the RLC circuit below. Recall that the Q of a lossy capacitor near resonance is Qcap ≅ RpCω0. The approximate value of the bandwidth (in rad/s) of the RLC circuit is: (1) 4.333 (2) 2 (3) 2.5 (4) 4

(5) 8 (6) 6.4 (7) none of above


45. In a circuit with an ideal transformer shown below, outv is:

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

(5) –1 (6) –6 (7) None of the above

46. Three circuits shown below are equivalent: (a) a non-ideal coupling circuit, (b) an equivalent modeled with an ideal transformer and (c) an equivalent without the ideal transformer. In order for (c) to be equivalent to the others, R must be:

(1) 50 (2) 10 (3) 20 (4) 4

(5) 100 (6) 400 (7) None of the above

47. For the network shown below, the output impedance is bs

aZout

+=1

where b is:

(1) 1/2 (2) 2 (3) 1 (4) 0

(5) 1/4 (6) 4 (7) None of the above

48. The Laplace transform of the Thevenin equivalent voltage seen at the output terminals of the circuit of problem 17 is:

(1)

2s

2s+1 (2)

10s

2s+1 (3)

5

s!4s+ 2

4s+1 (4)

10

2s+1 (5)

10

s(s+ 2)

(6)

5

s!4s+1

4s+ 2 (7) Something else


49. From the text, the gain of a standard two port terminated at the secondary by a conductance

YL(s) is

Gv2=

!y21

y22 +YL. Suppose a two-port network with !

"

#$%

&=

11

11][Y is connected as shown

below. Then, 2v is: (1) –5 (2) 2 (3) –2 (4) 2.5

(5) 5 (6) –2.5 (7) None of the above

50. The circuit shown below is a non-common-ground series connection of two identical two-port networks. Then, the 21z (in Ω) for the entire circuit boxed by a dotted line is:

(1) 1 (2) –3 (3) 3 (4) 4

(5) 5 (6) –4 (7) None of the above

51. Two two-port networks are connected in cascade as shown below. Then, the 12t of the entire network is:

(1) 14 (2) 2 (3) –2 (4) 4

(5) 10 (6 12 (7) 7


52. Three terminals of a transistor (shown as a circle) are connected as shown below, where the

transistor has !"

#$%

&=

020

01][Y mhos. Then, the 21y (in mhos) of the entire network is:

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

(5) 21 (6) –1 (7) 19

53. Recall the h-parameter relationship:

V1

I2

!

" #

$

% & =

h11 h12

h21 h22

!

" #

$

% &

I1

V2

!

" #

$

% & . Consider the circuit below.

Assume standard 2-port labels.

The parameter h21(s) =:

(1)

4s

s+1 (2)

!4s

s+1 (3)

!4

s+1

(4) 4 (5)

s

s+1 (6)

4s

s+ 4

(7) none of above

54. A 2-port has the following h parameters:

h11= 20 Ω, h12 =1, h21= 30, h22= 3 mho If the 2-port is terminated at port 1 with a 10 Ω resistance, then the resistance measured at port 2 is Rout = ( in ohms):


(1) 10 (2) 2 (3) 3 (4) 4

(5) 0.5 (6)

27 +9

13 (7) None of above

55. Consider the cascaded two port below.

The 2-port Na has y parameters and 2-port Nb has z parameters as given below (with standard port labeling and units):

ya[ ] =0.15 0.1

0.2 0.1

!

" #

$

% & ; zb[ ] =

25 1

20 2

!

" #

$

% &

Then

Zina =:

(1) 10 (2) 20 (3) 25

(4) 0.04 (5) 0.05 (6) –0.05

(7) none of above

56. The circuit below has a tranfer function of the form

Hcir (s) =

1

LC

s2

+ ??? s+ ???.

The circuit is to realize the 2nd order Normalized Butterworth transfer function

HNLP (s) =1

s2

+ 2 s+ 1

Then L (in henries) =:

(1)

2 (2)

1

2 (3) 1 (4) 0.25

(5) 2

2 (6)

1

2 2 (7) none of above


57. The NLP filter of problem 26 is to have a capacitor of value

1

2 nF in a final design in

which the 3 dB down frequency is to be 105 rad/s. The new value of the resistor is:

(1) 10

2 (2)

2 102 (3) 103 (4) 104 (5) 105 (6)

104

2

58. The normalized LP 2nd order Butterworth transfer function can be realized by the circuit below. If this is to be the basis for a HP filter with 3dB down point at

2 1000 rad/s the 1 Ω resistor is to be 1 kΩ, then the capacitor in the HP ciruit is: (1) 1 µF (2)

2 µF (3) 1 mF (4) 10 nF

(5) 0.5 µF (6)

1

2 µF (7) none of above

59. The filter transfer function of the circuit below is

HLPcir (s) =

4

3

s2

+1

Qs+1

. If the circuit was

to realize the transfer function

HLP (s) =

4

3

s2

+ 3s+1, then the value of C1 is:

(1) 1 (2)

3 (3) 3 (4) 9

(5)

1

3 (6)

1

2 µF (7) none of above

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