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Unformatted text preview: EE 350
Exam 3 23 November, 1998 Name: (Print) ID #: Section: DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO SO Test Form A Instructions: 1. This test consists of {our problems and each problem consists of either three or
four parts. Solve each part only in the space following each question marked as
Solution; if you need more space, continue on the reverse side and write the
question number; for example, Question (4.2) gogg'nued..; N0 credit will be
given to a solution that does not meet this requirement. 2. Box your answers wherever appropriate; failure to do so will result in reduced
credit for the question. 3. Do NOT remove the pages under any circumstance. Loose papers will NOT be accepted as valid exam; a grade of ZERO will be assigned. Time for the exam is 75 minutes. You are allowed to use only a 8 112"x11" note as reference. GOOD LUCK .UIP Have a Nice Thanksgiving Problem 1: (25 points)
For this problem assume that the operational ampliﬁer is ideal and vc(0')=2V
R=SQ
C=lOpF Figure 1. Circuit for problem 1. (1.1) (5 points) Redraw the circuit, representing the sources and components with their complex
frequency domain equivalents. Solution: (Do not box your answer) EE BSOIExam 3I'Page l of 11 (1.2) (10 points)
Find the zero—state response, stam. using Laplace analysis of the same circuit. Solution: EE BSOIExam 3IPage 2 of 11 (1.3) (10 points)
Find the zeroinput re5ponse, Yzmﬂ). using Laplace analysis of the same circuit. ﬁt) a sin(lﬂ,ﬂﬂcl t) u(t) C 3:0) _
v: (0 ) 2 v
' RSQ C = 10 ”F
Solution: EE 350fExam 3I'Page 3 or 11 mm: (25 Points} Consider the system: f (t) + y (t) Proportional Gain [deal Integrator Plant
Ampliﬁer Sensor
(2.1) (5 points)
Express the closedloop transfer function, H(s)  i—Esiin terms of II,(s). H2(s), H3(s),
s
and Gc(s).
Solution: EE 350lExam BlPage 4 of 11 (2.2) (5 points)
Suppose the transfer function of the plant is Is the plant BIBO stable? Justify your answer. Solution: (2.3) (5 points)
Write the transfer function of the ideal integrator. Solution: BE BSD/Exam BIPage 5 of 11 (2.4) (10 points)
Suppose the overall transfer function was found to be K H(s) 2
s +s+K—2 determine the range of K for the system to have an overdamped response to a unit
step input. Solution: EE350lExam SlPage 6 of 11 Problem 3: (25 points) NOTE: 1. u(t) is unit step function 2. Express the required transforms only as the ratio of two poiynomiais.
as2 + bs + c Exam Ie: —————
p dss+esz+fs+g (3.1) (8 points)
By direct integration find the Laplace transform of the function f (t) deﬁned as f(t)=cosm 0st50.5
" = 0 elsewhere and specify the region of convergence. Solution: EE350!Exnm3/Page 7 of If (3.2) (8 points)
Find the Laplace lmnsform of a signal g(t) = sin t u(t) — sin I u(tJr)
using standard table of transforms and properties. Solution: EEBSOIExam 3fPage 8 of 11 (3.3) (9 points)
The impulse response of a linear timeinvariant system is h(t) = exp(t) u(t). Find the response y(t) of the system when an input {(t) = u(tl) is applied to it
using Laplace transforms. Solution: EE 350fﬁxam 3fPage 9 of l 1 mm: (25 points)
This problem focuses on the concept of orthogonality. (4.1) (5 points)
For an arbitrary set of signals, x,(t), x2(t). , xN(t) which exist in the interval [[0, to+T], state and explain the criteria for determining if the signals form an orthonormal set. Solution: (4.2) (5 points)
In your own words, state the purpose of orthogonality. Solution: EE 350lExam 3/Page 10 of 11 (4.3) (15 points)
Suppose you have the functions xl(t) and f (t) as shown below. Fmd the bestﬁt approximation of f(t) in terms of x,(t). i.e., ﬁnd {0) 9‘ 0 X1“) x,(t) = 2 sin(ut) Solution: EE350IExam3lPagc 11 of 11 indefinite Integrals judv=uv—fvdu [name's = 1mm) ~ [Rama / . .1 f 1 .
sm ax dz = —— cos n: cos a: d: = — sm a:
a a
.7: sin 2a.: 2 1 sin 20:
sin azdx= §_ 4:: [cos urns—2+ 4a
1
zsin ax d1: = ——(sin ax  ax cos ax)
1 .
1 cos ardr: 2(cos a: + ax sm as)
a
2 1 2 2
a: sin a: d: = 1(211: sin a: + 2cos ax — a :1: cos ax)
2 1 2
2: cos a: d: = —3(an cos a:  2sin a: + a 2: sin ax)
sin (a  b): sin (a + b): 2 2
sin b  ———————  —————~
azsin zdz— 201—” 201+” a #6
cos (a — b): cos (a + b): 2 2
sin cos b a’ = — ~——————— ————— b
” ‘ ’ [201—5) + 2(a+b) ° 5'5
. _ b .
sm (a )1: sm (0 + b): a2 95 b2 2(a — b) + 2(a +b) 1:“ dx—  —e° re" dx = —(a:c — l)
172:“ d1: _ {Ma — 201: + 2)
ecu:
e J“sin b.1 dx—  a2 + (12 (a sin bx  bcos bx)
e” .
cos b1: dz— .. a2 + b2(a cos b: + bsm bx)
2d l tan If
12+a a a
1 2
=—in(:.:2 +a ) /
f
f
f
f
f
/
[cos a: cos b: d; =
f
f
f
f
f"
f
H: A Short Table of (Unilateral) Laplace Transforms m) PM
1 6(t) 1
2 u(t) l
S
. 3 in“) :15
n n!
4 t u(t) 3"“
5 ("a“) a l A
M 1
6 ts u(t) (3 A):
r
7 mung) (s 1')“,
s
80 cos bf 11(1) 3’ + b2
. b
Sb sm 6! u(t) 2 + b:
5
9a 3““ cos bl 11(1) ﬁa—p
gg . 6
9b C 51“ bt u(t) W
_ rcosﬁa+ arcosﬂwbrsina
1°“ " “”0" + 9’ “(0 W
0.511” 0.511.”
105 "" (at 9 —.._. —.
re cos( + )u(l:) a+a—jb+s+a+jb
10c 1'6““ cos (bt + 9) u(t) ﬁ—ifi‘ESB—H
T = ’A'lccl‘i'LZABa’ 9 = tan—l Arlee—pi:
b= c—a
, B — An .43 + B
_¢¢ .
10d e [.4 cos bt + b sm 6!] u(t) ——_32 + 2“ + c a ‘ ..1».14.x:l‘.}.‘:'1'.J.;'$i'.i.L'.""'29h "JAM" It'dP »«LALV‘u' ‘ ‘ ' lumi’...w._, The Laplace Transform Properties % Operation Ht) F(s) Addition f1(t)+ f2(t) . F1(a) + 5(3)
Scalar multiplication kf(t) kF(.s)
Time differentiation % _, sF(s) — f(0‘)
d2 
(“if We) — afar) — no")
(13)“ 3 2 _ _ . _ .. _
3 .., F(s)—s 1(0) sum1(0)
‘ 1
Time integration f(r) dr :Fh)
o wn: iron é ]_ ma: Time shift f(t — to)u(t  to) F(s)e"“° to 2 0
Frequency shift f(t)e"°' F(s — so)
dF
Frequency utf (t) %s}
differentiation
Frequency integration I—E—Q f F (z) d:
S
. 1 a
Scaling . ﬂat), at 2 0 :F (E)
Time convolution 1'10) 1 f2(t) F1(s)F2(a)
1
Frequency convolution f1(t)f2(t) EaFﬁs) : F2(a)
Initial value f(0+) .1352; 317(5) (n > 111)
Final value f(oo) limbsﬂs) (poles of 317(5) in LHP) m
W ...
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 SCHIANO,JEFFREYLDAS,ARNAB

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