Final practice

Final practice - Final Exam Practice Problems MULTIPLE...

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Final Exam Practice Problems MULTIPLE CHOICE. *1. For a > 0, the Laplace transform of f ( t ) = 2 e ( t + a ) ! ( t " a ) + 2 e ( t " a ) ( t + a ) is: (1) 2 e 2 a (2) 2 e 2 a + 2 e ! 2 a (3) 2 e ! 2 a (4) 2 e ! a ( s ! 2) (5) 2 e a ( s ! 2) (6) 2 e ! a ( s ! 2) + 2 e a ( s ! 2) (7) None of above ANSWER: (5) 2. The output of a circuit has Laplace transform V out ( s ) = 64 s 3 ( s + 2) With v out ( t ) = Ae ! 2 t 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 Z in ( 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 + 3 s (5) 1 s + 0.5 + s (6) s + 2 ! 3 s (7) None of Above ANSWER (2).
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Final Exam Sum 03 202 page 2 4. The transfer function for the following op amp circuit is: (1) ( s + 2) s (0.5 s + 1) (2) s (0.5 s + s + 2 (3) 0.5 s + 1 s ( s + 2) (4) ! 0.5 s + 1 s ( s + 2) (5) ! ( s + 2) s (0.5 s + (6) s ( s + 2) 2 s + 1 (7) s ( s + 2) 0.5 s + 1 ANSWER: (7) *5. Suppose L f ( t ) { } = ! ! n s 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 ) ( t f 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 s 2 ! e ! s s (4) 1 ! e ! s s 2 (5) ! 1 ! e ! s s 2 (6) 1 ! e ! ( s ! 2 a ) ( s ! 2 a ) 2 ! e ! ( s ! 2 a ) s ! 2 a (7) None of above ANSWER (3)
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Final Exam Sum 03 202 page 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. v out (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 ) = V out ( s ) V in ( s ) = 16 s + 4 s 2 + 4 s + 16 has an input v in ( t ) = 2 cos(4 t ) V. The output of the circuit in the steady state has a magnitude and phase (in degrees) equal to: (1) 4, 45 o (2) , ! 45 o (3) 4, ! 45 o (4) 4 , 45 o (5) 4 ! 45 o (6) 8, ! 45 o (7) 8, 45 o (8) None of above ANSWER: (6). 9. In the circuit below, v C (0 - ) = 2 V and i L (0 - ) = 0. The node equation at node A which accounts for all initial conditions is: (1) 2 + 5 s + 0.5 s ! " # $ % & V a 5 s V b = 2 V in + 1 (2) 2 + 5 s + 0.5 s ! " # $ % & V a 1 s V b = 2 V in + 1 (3) 2 + 5 s + 0.5 s ! " # $ % & V a + 4 s V b = 2 V in + 2 (4) 0.5 + 5 s + 0.5 s ! " # $ % & V a 1 s V b = 0.5 V in + 1 (5) 0.5 + 5 s + 0.5 s ! " # $ % & V a 5 s V b = 0.5 V in + 1 (6) 0.5 + 5 s + 0.5 s ! " # $ % & V a + 4 s V b = 0.5 V in + 2 (7) None of above ANSWER (2).
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Final Exam Sum 03 202 page 4 10. The impulse response of a circuit is h(t) shown below. Suppose the input is v in ( 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)
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