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EE202/ Quiz 4 Jan 22, 2020

Section Morning 1. (4 pts) The Laplace transform of a signal f (t), f (0 )  0 , is F(s)  d f (t) is: dt

of e2t

Solution: (2 pts for each part) F(s) 

2 2

s 4

 sF(s) 

2s 2

s 4

2 s2  4

. The Laplace transform

 (s  2)F(s  2) 

2(s  2) (s  2)2  4

2. (3 pts) An integro-differential equation for a parallel RLC circuit driven by a current source, iin (t) , is given by

vC

dvC

1 C  R dt L

t

 vC (q)dq  iin (t)



with C = 0.5 F, L = 0.25 H, R = 0.5 Ω, iL (0 )  2 A, vC (0 )  0 , and iin (t)  0 . Then vC (t)  (in V):

(1) 4e2t sin(t)u(t)

(2) 2e2t sin(t)u(t)

(3) e2t sin(2t)u(t)

(4) e2t cos(2t)u(t)

(5) 2e2t sin(2t)u(t)

(6) 2e2t sin(2t)u(t)

(7) e2t cos(2t)u(t)

(8) None of above

Solution: VC (s)

 CsVC (s)  CvC (0 ) 

R implies VC (s)

 CsVC (s) 

VC (s)

VC (s) Ls

 CvC (0 ) 

1 L

0

 vC (q)dq



s

iL (0 )

2 s

 I in (s)

R Ls s implies  4 2 2 V 2  0.5s  (s)   V (s)  2 ANSWER: (6) C  s  C s (s  2)2  22


Section Afternoon 1. (4 pts) The Laplace transform of a signal f (t), f (0 )  0 is F(s)  transform of eat

4 (s  a)2   2

. The Laplace

d f (t) is: dt

Solution: (2 pts for each part) 4 4s 4(s  a) F(s)   sF(s)   (s  a)F(s  a)  (s  a)2   2 (s  a)2   2 s2   2

2. (3 pts) An integro-differential equation for a series RLC ( iL  iC  iR ) circuit driven by a voltage source, vin (t) , is given by t

diL

1  RiL  L dt C

 iC (q)dq  vin (t)



with C = 0.25 F, L = 0.5 H, R = 2 Ω, vC (0 )  20 A, iL (0 )  0 , and vin (t)  0 . Note that in a series RLC iC (t)  iL (t) . Then iL (t)  (in A): (1) 10e2t sin(2t)u(t)

(2) 20e2t sin(2t)u(t)

(4) 20e2t sin(2t)u(t)

(5) 20e2t sin(t)u(t)

(7) 10e2t sin(t)u(t)

(8) None of above

(3) 10e2t sin(2t)u(t) (6) 20e2t sin(t)u(t)

Solution:

RI L (s)  LsI L (s)  LiL (0 )  implies RI L (s)  LsI L (s) 

I L (s)

I L (s) Cs

 LiL (0 )  Cs

1 C

0

 iC (q)dq



vC (0 ) s

 Vin (s)

s 

20 s

implies  4 20 2 ANSWER: (4)  2  0.5s  s  I L (s)  s  IL (s)  20 (s  2)2  22

quiz4  

Solution: (2 pts for each part) F(s) 2 s 2 4 sF(s) 2s s 2 4 (s2)F(s2) 2(s2) (s2) 2 4 of e 2t d dtf(t) is: V C (s) R CsV C (s)...

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