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Unformatted text preview: University of Illinois Spring 2006
ECE 210: Third MidSemester Exam Thursday April 207 2006
7:00 pm. 7 8:15 pm. Section: B 9 am 3 E 1 pm E F 2 pm University ID Number: Instructions This exam is closed book and closed notes. Electronic equipment such as
calculators, laptop computers, PDAs, iPods, cellphones, email pagers, head—
phones, etc. may not be used. The last page of your exam booklet has tables of formulas (on both sides)
that you may use in solving problems. You may tear 01f this sheet if you like;
it need not be turned in with your exam. Write your answers in the spaces provided. SHOW YOUR WORK. Answers
without appropriate justiﬁcation will receive very little credit. If you need
extra space7 use the back of the previous page. Grading 1. 24 points
2. 25 points
3. 26 points 4. 25 points Total (100 points) _ W . [24 points] (a) [4 points] Find the energyrE' in the signal f (t) = sinc(t) 2 smog) “WT W=g~w _ Iletaw
‘5‘; m lat—maﬁrn‘iw
= 5%“ 'rr‘otw
= “Q ClCDJ 7’ l
(b) [4 points] An LTI system has impulse response h(t) = sinc(t).
If the input is f (t) = sinc(25t), what is the resulting output y(t)?r ya) 2 ESMLCE‘)
s‘mcﬁm—u “racecéﬁmuo Sim. 1.74 “W “J _
*6 Dewmecﬂ)‘, MN) {‘0 YL ; L w ‘ .
w) E meek) H36+3,ﬁl_s‘.mcc3
(c) [4 points] The Fourier transform of g(t) is C(w) 2 1 :jwe_2jw Find g(t). gm : M CtiD c. (d) [6 points] Express y1(t) and y2(t) below in the form A150: — 151) + A26<t — t2) + A3605 — t3) where A1, A2, A3 and t1, t2,t3 are constants (possibly with value 0).
a 31105) : [6(t) —— 26(t — 271') — 486(t + 487T)] sin(t) A1: 0131: 0 A2: 0 2:221? A3=O tgzelrS‘TT . 1,205) : [6(t + 3) — 26(2 ~ t) + 26(t — 2)]t2 ’Alzﬁ 13:3 A2=~9t2=9~ 143:8“:3‘ l (e) [6 points] For each of the following systems with input f (t) and
output y(t), determine whether the system is zero—state linear (L)
or nonlinear (N), timeinvariant (I) or timevarying (V), causal (C)
or noncausal (N) and circle the appropriate letters. 2. [25 points] In the sketches below, be sure to mark the salient points: locations and values of maxima and minima, points Where the function goes to zero, etc. Consider a signal g(t) : [sinc(t)]2 cos(207rt).
(a) [10 points] Sketch ]G(w), the magnitude of the Fourier transform 0f g(t).
10W)!
“Yr/L /\
‘20q’L —2:o'1‘ —20WOL zoﬁL—ZEf IOWHL $ w (b) [15 points] The signal g(t) is the input to the demodulator circuit
shown below, which produces outputs $(t) and y(t). cos(207rt) Sketch X and Y(w)[, the magnitudes of the Fourier transforms
of x(t) and IX(w)
"Tr/L
.4, I \/ > LL)
1, 1/
WM] 3. [26 points] (a) [10 points] Consider the circuit below with input voltage f (t) and
output voltage y(t): Find the frequency response H (w) = Y(w)/F(w) and the impulse
response h(t). PM) = r—— ha) = 2 mt (b) points] The input to a iowpass ﬁlter with frequency response H (w) = rect(w) exp(—j2w) is the signal f (t) = sinc(2t) + sinc(4t).
What is the output y(t)? W) = .527 sine (3&2’9] j E“ W chbrajmt<wue‘~3”
(t.
(a) a 366) :_'5_ 5:“ [g (92)]
[ Tool 8 4. [25 points] The signal f (t) Shown below is the input to a linear time
invariant system with impulse response 11(2)) as shown below. The output signal is y(t). f (t) (a) [5 points] Sketch f (3 — 75), indicating clearly the points Where the
function becomes zero. f(3t)
‘ > t
Lt 5'
(b) [20 points] Find y(t) for all t, —00 < t < oo.
olt<l 4; 1—t+', l5£<L
y(t) 2 46+ 4t —3 2 st<3 0/ {333
Colgw.\, 3/5! {,Bg—Lat<l
MI
£~eﬁrl ’
(mg; n} a «t < I
(4341 claiﬂt’zcﬁi—z c3 — t1 b r
7f\ 3 ) ' <— 3 I» —2 +
M —2
71+,2 ,14trtw—l a 1 <t<3 Case : _
g ’ﬂat 3L€)23 l2CT+1>&t:—tl+l_nJC—3 194" {’4 CageLtr ALT £L+7~\ ——9 t>3 6 ...
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