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EE-202, Ex 2,

EE-202

Exam II

March 8, 2005

Name: __________________________________

Student ID: _________________

CIRCLE YOUR DIVISION

Division: 201-1 (morning) 201-2 (afternoon)

INSTRUCTIONS

There are 11 multiple choice worth 5 points each; there is 1

workout problem worth 40 points.

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

or calculators are permitted. A transform table and properties

table are attached at the back of the exam.

All students are expected to abide by the usual 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.

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EE-202, Ex 2,

1. For the circuit shown below, the Quality factor Q and BW are:

(1) Q=0.5, BW=1 rad/sec (2) Q=2, BW=o.25 rad/sec

(3) Q=1, BW=1 rad/sec (4) Q=1, BW=0.5 rad/sec

(5) Q=0.5, BW=0.5 rad/sec (6) None of the above

Since Req seen by the C and L in parallel = [(2//2) + 1 ] // 2 = 1 , BW = 1/Req C =

!p=

1

2*2

= 0.5"Q =!p

BW= 1

2. The output voltage of a circuit has the following transfer function, v(0+) and v() are:Vo =

s+1

s(s2+3s+1)

(1)v(0+)=0, v()=1 (2) v(0+)=1, v()=0(3) v(0

+)=, v()=0 (4) v(0

+)=1, v()=1

(5) v(0+)=0, v()= (6) v(0

+)=0, v()=0

V 0+( ) = lim

s!"

sV s( ) =s +1

s2+ 3s +1

= 0

V !( ) = lims"0sV s( ) =s +1

s2+ 3s +1

= 1

1

2F 2HVi 2

2

2

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EE-202, Ex 2,

3. For the pole-zero diagram shown below assuming that H(0)=2, H(s) is(1) H(s) = !8s

2+24s+32

s2+16

(2) H(s) = !8s2!24s+32

s2+16

(3) H(s) = !8s2!24s+32

s2!16

(4) H(s) = !8s2!24s!32

s2!16

(5) H(s) = s2+16

!8s2!24s+32(6) H(s) = 8s

2!24s!32

s2+16

By inspection H s( ) =Ks2+ 3s ! 4

s2+16

.

K= !8 , thus, H s( ) =!8s

2+ 24s + 32

s2+16

-4

4j

-4j

1

j

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EE-202, Ex 2,

4. Circuit shown below has a transfer function H(s) and is ??

(1) H(s) = ! 4s2+5s+1

2s,band reject (2) H(s) = ! 2s

4s2+5s+1, band stop

(3) H(s) =!

2s4s2+5s+1

, low pass (4) H(s) =!

4s2+5s+12s

,band pass

(5) H(s) = ! 4s+12s+1

, low pass (6) H(s) = ! 2s

4s2+5s+1

, band pass

H s( ) = !

1

1

s+1

2s + 0.5= !

2s

4 s +1( ) s + 0.25( ), a bandpass filter.

Vo

-

+

2F

1F 1

2

Vs

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EE-202, Ex 2,

7. What is the output at t=2 sec, y(2), for an input step {x(t)=4u(t)} response of a circuit whoseimpulse response h(t) is shown below?

(1) 6 (2) 24 (3) 32 (4) 14 (5) 46 (6) none of the above

y t( ) = ! t( )* 2 t+ 4( )2u t+ 4( )" 4 t( )

2u t( )+ 2 t" 4( )

2u t" 4( ) + 4r(t" 4) " 4r(t" 8)#

$%

&

'(

y(2) = 2 6( )2! 4 2( )

2+ 0

"

#$%

&'= 72 !16 = 56

8. For what value of gm the following circuit is stable?? H(s)=V1/I1

(1) gm>1 (2) gm 0 (4) gm>2 (5) gm>-2 (6) gm>-1(7) none of the above

V1s

( )=V s

( )+I

1s

( )and I

1

s

( )= g

m

V1

s

( )+

V s( )

5+ 2s

! I1

s

( )= g

m

V1

s

( )+

s V1

s( )" I1(s)( )5s + 2

!

1+s

5s + 2

!"#

$%&I1s( ) = gm +

s

5s + 2

!"#

$%&V1s( ) 'H s( ) =

6s + 2

5gm+1( )s + 2gm

,

H s( ) is BIBO ! Re"2g

m

5gm+1

#

$%

&

'(< 0 . !gm > 0 and gm < "0.2 . gm in the range of answers

(1) (3) and (4) allresult in the stable circuit.

V

0.5F

gmV1

1V1

5

I1

x(t)h(t)

44

4-4 8

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EE-202, Ex 2,

9. The circuit shown below is excited with a voltage source of Vs. If the circuit is to be scaledsuch that the output responds 1000 times faster and the new value of C2 is to be 1F. The new

values of R1, and L, then become.

(1) Lnew

=3H, R1new

=1k (2) Lnew

=3mH, R1new

=1k

(3) L

new

=3H, R1new

=1M (4) L

new

=3mH, R1new

=1M(5) Lnew

=3H, R1new

=1k (6) Lnew

=3H, R1new

=1M (7) none of the above

Kf = 1000 , Km =Cold

CnewKf=

1

1!1000= 1000 . Lnew =

LoldKm

Kf= 3H, R

new=R

oldK

m= 1k!

L=3H

C2=1F

R1=1 R2=2

C1=2F

Vs

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EE-202, Ex 2,

10. An input of v(t)=2(t)- (t-1)+3 (t-2) is applied to a circuit with the following impulseresponse. The circuit output at 2.5 seconds is.

(1) v(2.5)=3 (2) v(2.5)=4 (3) v(2.5)= 0(4) v(2.5)= 2 (5) v(2.5)=6 (6) v(2.5)= 1 (7) none of the above

Apply sifting property, yield the following results:

v t( ) = 2h t( )! h t!1( ) + 3h t! 2( )" v 2.5( ) = 2h 2.5( ) ! h 1.5( )+ 3h 0.5( )v(2.5) = 0 ! 0.5+ 3= 2.5

11. For h(t) and f(t) sketched below, the convolution y(t) = h(t)*f(t) is given by which picture:

(1) (2)

t (sec)

h(t)

1 2

1

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EE-202, Ex 2,

(3) (4)

(5) (6)

(7) none of the above

MC Sols

1. (4)

2. (1)3. (2)

4. (6)5. (2)

6. (1)7. (6)

8. (1), (3), and (4)9. (1)

10. (7)

11. Differentiating h(t) we have, h '(t) = !(t+1)+!(t"1) . Whereas f(q)dq

!"

t

# is given by the

picture

Summing a unit left shift and a unit right shift of this picture yields (6).

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EE-202, Ex 2,

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EE-202, Ex 2,

Workout Problem (45 points): Consider the circuit below in which iL(0! ) = 0 .

PART 1: (34 points) Switch in Position A.

(a) (6 pts) Find the transfer function H1(s) =VC(s)

Vin(s)

;

(b) (7 pts) (i) (2 pts) Find the impulse response.

(ii) (5 pts) Find the step response.(c) (10 pts) If the input to the system is given by the signal below, find vin(t) as a sum of scaled

and/or shifted step functions. Then find the zero-state response vC,zs(t) . Hint: what principles

can you apply to obtain the answer very easily.

(d) (6 pts) Find the zero-input response vC,zi(t) ifvC(0!

) = 4 V.(e) (5 pts) Find the complete response and identify the transient and steady state parts of it.

PART 2: (11 points) Unexpectedly, the MATRIX moves the switch to Position B at t = 5 sec.(f) (3 pts) Determine, APPROXIMATELY, vC(t) at t = 5 sec.

(g) (8 pts) For 5 sec t, compute VC

(s) . Do not computevC(t) . Hint: Draw the equivalent

circuit valid for 5 sec t.

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EE-202, Ex 2,

SOLUTION WORKOUT.

(a) (6 pts) The impedance of the parallel RC combination is

2

0.5s

2 +1

0.5s

=2

s +1. Hence

H1(s) =

VC(s)

Vin(s)

=

2

s +1

2 +2

s +1

=1

s + 2.

(b) (i) (2 pts) By inspection h(t) = e!2t

u(t) .

(ii) (5 pts)1

s(s + 2)=

0.5

s

!

0.5

s + 2Hence, the step response is: 0.5 1! e!2t( )u(t) .

(c) (10 pts) By inspection vin(t) = 10 u(t)!u(t

!10)( ) . From linearity and time shifting or time

invariance property of the circuit,

vC,zs

(t) = 10 Step Response(t) ! Step Response(t!10)"# $%

= 5 1! e!2t( )u(t) ! 5 1! e!2(t!10)( )u(t!10)

(d) (6 pts) The equivalent circuit is a current source of value CvC(0! ) = 2 in parallel with a 1

resistor and 0.5 F cap. Thus, vC,zi(t) =L

!1 2 "2

s + 2

#

$%

&

'( = 4e

!2tu(t) .

(e) (5 pts) The complete response is:

vC(t) = v

C,zs(t) + vC,zi(t) = 5 u(t) ! u(t!10)( )! e!2t

u(t) + 5e!2(t!10)

u(t!10)

Steady state is: 5u(t) ! 5u(t!10) . Transient is the rest.

(f) (3 pts) 5 V

(g) (8 pts) 2.5e!5s

= 0.5s +4

s+ 2

"

#$

%

&'VC =

0.5s2+ 2s + 4

s

"

#

$$

%

&

''VC

in which case

VC= 2.5se

!5s

0.5s2 + 2s + 4= 5se

!5s

s2+ 4s + 8

= 5se!5s

(s + 2)2 + 4