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Unformatted text preview: Assignment #12 — Solutions
ECSE~24 I (j) Sigaals & Systems  Spring 2007 Due Tue 03/16/07 1+ja) 1(20); A linear, timeinvariant system has a frequency response given by H = ———————_~— .
. 1w Ja) 1+~—— [ 10 j The system respanse to the general input x(t) = M 008(6001‘ + 6) is
320} Mifﬂwg )1 cos(a)ot + 6 + AH(a)O )). 1+ j0.2 _ 1.0198eJ'0‘974 _5'0986_ﬂ..3934
j0.2 {a} Fain“ input x6 :10005(O.2t), H (0.2): 10.2[1 + ~16.) 0.2g”?1.0002e1'002 {amgieml s3€334 $79.8 °) [331611 jag: {z} 50.98 cos(0.21——79.8° WWMMMWWV 7%mewa ~ ‘ j1,3734 ’
{in} 1%? input 32;}(r)m10c:os(5t), H(5) = H 155 = :0 e = 0912261066“)
15(14 E) 56131.118ew‘4636 imngass2:41:36: 10 £37.88 °) i , iii‘2}? izzgzui m lOsin(5t)= 10008[5t —%], then since the frequency is the same, H (5) does not ms 11(5) = 0.9122e—10‘6610
' ' err«£3,636,310 $37.88") _\ {g} as; 9.122cos(51~37.88° —90°) = 9.122 cos(51137«88{J  r 7 11.4711 m: <r>=wcosaor>, H(lo)=~—1i—Ji9;5m339i3§99 ”
j10[1 + 117;] we; J27? = 0.7106e'1‘08851 43,3255} m—50.7159°) ' 7.106cos(IOt50.72°) ,_ (NMV Assign #12—Solutions. Spﬂng Big}? 9.1 Assignment #12  Solutions ease—24 ; a Signals & Systems  Spring 2007 Due Tue 03/1 6/07
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1. Continued. Matlab COdC. nummﬁ i}; den‘ﬁOJ 1 0]; Boddnumﬁen, {0. 1 , 10}) grici W2:11[O.2 :3 19]; {mag phasd‘bOddnumadenaw) Bode Diagram {:3 \_\ V \IL
g a? ‘ System sys ‘ y Frequency (rad/sec): 0203 System: sys
g} ‘ magmmdews): 14 ' Frequency (rad/sec): 4.92 System sys »
é ‘ ‘ \ Magnitude (dB): ~0.767 rad/sec>:10 . , . . wraguuuugl‘dBk ‘
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% ' i / Frequency (rad/sec):10 PhaSe (deg): 50.7
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‘ _ Frequency (rad/sec): 0.203 ' ' L ’ ’ 1 Phase (deg): 79.7 3 10 Frequency (rad/sec) iiéeigmi ih::r§1 the following portion of the program:
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lv‘jw 7. 1w Assignment #12— Solutions —— p. é ECSE~24 18 Signals & Systems — Spring 2007 /%V The signal x(z‘) is sampled by multiplying it by an impulse train. The sampled signal is
£190) = COMO t Whereggﬁ) is the sampled version of x(t) and C(t) is an impulse train described by it“; {:th z 5(1 ~ kT) . If x(t) = 1 +100c0321+0.01 sin4t, what should be the requirement on ismwe the peried, T, to permit full recovery of x(t) from x: (t) ? ~ ﬂ? ,5 cc.
Requirement on T: 0 L r L g Assignment #12— Solutions — p. 7 iiiCSEw2410 Signals; 85 Systems  Spring 2007 5 (20). Two signals are multiplied as follows Aft) +®——> y(t) 4‘ C(13)
27: where {:{g} is the: impulse train, C(t) = 250 — kT) , with T = 3000 k=~oa sec.
{2;} {1:3} the spectrum of x(t) be X (‘0)
A lOOO 1000 Skeieh ihe spectrum of y(t) , Y (60). Can x(t) be recovered from y(t) using lowpass
fiiiez’éng‘? Why or why not? U to = ﬂL—e 1 3,060 M/peg T ﬂlrgago
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MO) Wﬂﬁwuj k3 WW (444v Mgu‘ufm/e. Assignment #12— Solutions — p. X 3353324 16 Signals & Systems — Spring 2007 ﬂMMa‘ufé W
(0) {1(3) Repeat part (a) if the spectrum of x(t) is
X (‘0 )
A,
2000 2000 ‘0 and {:(z) remains the same. Can x(t) be recovered from y(t) using lowpass
fiiteying? Why or why not? /‘/\7/\\—7
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we; mmg ) J M (slaw/7} away. Assignment #12— Solutions — p. 7 ECSEQMO Signals & Systems  Spring 2007 of 3:333 in sampled signal xp (t) is generated by sampling the original signal
xéz’) @3622?) every 0.1 seconds as shown in the figure, where C(t) : 2 5(1 — kT),
k=—°o
with (3.1.
X(t) —>®——> xp(t) t C(t)
{3){43 ESketch x p (t) . {133(3} What is the sampling frequency in radians/sec. ? 
{eke} What is the highest frequency (rad/sec) in the on'ginal signal x(t)? (eke) Is it possible to recover the original signal x(t) from the sampled signal xp(t) ? Answer Yes or No and state why? a?) Kym; 605(5176, “U/ icon, 2/ ,
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 Spring '07
 WOZNY
 Signal Processing, impulse train, ECSEQMO Signals & Systems, Sigaals & Systems

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