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Unformatted text preview: EE  350 Exam 3 1 April 2002 Last Name Sol V'Elg 05 First Name
Student # Section DO NOT TURN THIS PAGE UNTIL YOU ARE TOLD TO DO so
Problem 1
2
3
Total .EI Test Form B Instructions You have 2 hours to complete the exam. Calculators are not allowed. This exam is closed book. You may use one 8.5" x 11" note sheet. Solve each part of the problem in the space following the question. If you need more space, continue your solution on the reverse side labeling the page with the question number, for example, "Problem 1.2 Continued". NO credit will be given to solutions that do not meet
this requirement. 5. DO NOT REMOVE ANY PAGES FROM THIS EXAM. Loose papers will not be accepted
and a grade of ZERO will be assigned. 6. If you introduce a voltage or current in the analysis of a circuit, you must clearly label the
new variable in the circuit diagram and indicate the voltage polarity or current direction. If
you fail to clearly deﬁne the voltages and currents used in your analysis, you will receive
ZERO credit. 7. The quality of your analysis and evaluation is as important as your answers. Your reasoning must be precise and clear; your complete English sentences should convey what you are doing. To receive credit, you must show your work. PWNI" Problem 1 (25 points)
Deﬁne g(t) =eM [u(t+1)—u(t—1)] and let f(t)= 2 302") "3—90 1.1) (5 points) Determine any existing symmetry of the signal f(t) that implies that any
trigonometric Fourier series coefﬁcients are zero. 1.2) (10 points) Compute the trigonometric Fourier series coefﬁcients of the signal f(t). 1:" even 19’
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o n a 2., H, 6,... 1.3) (10 points) Consider a periodic signal x(t) that has the trigonometric Fourier series x(t) = a0 +21%l cos(n%t).
X n=l
Suppose ﬁt? is passed through a causal low — pass filter with impulse response h(t) = 3e'2'u (I).
Find the compact trigonometric Fourier coefﬁcients of the (zero — state) output signal, y(t), in
temis of the trigonometric Fourier series coefﬁcients of x(t) and the frequency response function of the ﬁlter. 710“ crave“: ares/Jame 'Flnc'Li'O/I “(ftQ 3! the C'HSGV‘ 7‘
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= .._..——3“~ mom — w" $95)) V“! +(ﬂwo " 1‘": ‘C‘ o“w5 vi Problem 2 (25 points)
2.1) (7 points) Consider a periodic signal f(t) where f(t)=e’2’, —1$t<1 and is found elsewhere by periodic extension. Determine the complex exponential Fourier series
of f(t). Tye math CH» 7: ItalHue» even
0” 0001, 13:7. —> "‘"°= "‘—“§= “7" 2.2) (8 points) Consider a periodic signal
f (t) = Z D.e’"”"’ where Dn as a function of n is shown in the ﬁgure below. Determine whether the signal f(t) is
even, odd or neither and whether f(t) is real valued, imaginary valued or complex valued. Justify
your answer using the symmetry properties of the Fourier series. Dn 1 ' Observe. .thmt (I) by, Is {tomb mag anno C23 Dn =bn is
0‘" oolOQ— cmp‘blw} % ﬂ. ' 95 L; 4: I2. 7:.
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so that ff”) 35 he va.alvee£. ’ 2.3) (10 points) A real periodic signal is applied to an ideal high pass ﬁlter with the frequency response function H(w)={1, lw21.57r 0, otherwise. Given that the power of f(t) is Pf = g , determine the power of the ﬁlter’s output y(t). ‘ From the, axpoﬂaﬂé Lag FM\€Y‘ [e vies perks20 nwo :1 run ) anoQ. So (Do: 11". o kcwun the ‘cmbuEﬂCy. % ﬁkQ 35 L5" 11', the. erter searum ComeﬂQﬂ‘AS a. 41*) a’é n: vi) 0 \ ourQ. y the. Qléer, 1L"; (:oi‘owi note. «Lth the cum/1’th and pit“!
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P—F = Z nanO [1: —09 D: 0th? rwl'SC) gs E o n =1,0, I Problem 3 (25 points)
3.1) (13 points) Consider the signal f<r)=jrect[;). a) (6 points) Determine, without computation of the spectra, whether F(0)) is even function, odd function, or neither and whether F( (0) is real valued, imaginary valued, or complex valued.
Justify your answer using the properties of the Fourier transform. b) (7 points) Compute the spectrum of f(t) using direct integration of the Fourier transform
integral. (a) .FGQ Fit) "5 f‘reia. Madmen; anti qn
. Qvnn aniswn ﬁ/ 4: «.5 ‘F’lb3= ¥C"+) ‘ BCCAUSQ. ‘5 4n even Quickly“ % ﬂung the $6le
MiG/rag) f':f'(t3$mw‘kJ&J Vantskei. Eeccwsg +L~kb is pure} (Imaginary)
“’50 PAH) ‘5 15 Uan‘
I au 0 p I I43 red: 6? CSmC ,t 'Cououﬁ a 2/ SMCW 3.2) (7 points) Given cos(a)0t) <—> 7r[6(a)— 04,)+ 6(w+ 6%)] , use the time shift property of the Fourier transform to determine 3 {cos (wot + 6)} . Simplify your expression. 1 Let 301:): (oS(%t)) qni abSevve, £110.14 cos[w.t +6] = Cos[‘*b(t+§;)] = a(*=+ gt.) . Usiy the ewe s)“ng property («Cyom Tabb ~11)
W'Lo
a_(l:+h5 e—a Soﬁe?" 3 to= G/wa 3.3) (5 points) Given f (t) <—> F (m) , show that dfdgt) (—) ij '9‘bar‘l: UH'U Itkq, gigCm, m % thQ, mver’se. FbvnQr
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gas3 é—é scab, ,t $6M“; we éif a éiw PCW) art 10 Problem 4 (25 points) Consider a realvalued signal f(t), with Fourier transform F( (0), that is band — limited as shown in
the ﬁgure below. F(w)
A "Dr 0% 4.1) (5 points) Find and sketch the Fourier transform of the modulated signal
m(t) = f (t)cos(a)ct)
for me >> 20);. Q) { m («:33 = 3'; (KNEW 3’ {cos wc'ir] .
‘ 51} Flu) a: [If HurtMA +17 SCw—waj Mtuh = J; F(w+w°) + "z‘ F(w— wc.) 11 Assume that m(t) = f (t)cos(a)ct) from part 1 is used as the input to the system shown below
where cos (coatF6) 4.2) (8 points) Assume that 9 = 0, sketch 0(0)), the spectrum of g(t), and ﬁnd an expression for
y(t) in terms of f(t). xcm (76+) = x (+3 C05 (wot)
.. ,L, 4%
% é: 6cm  szwm + ,_ m» )
—w¢, '00; We M. “’ (m3 a,qu cm. W4: m)
+w$ u»;
50*” Ym = “Lth 6.0»)
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W to; w W
'2“ 13'; ""4 “4 «5‘33; 2”“ no.5. we 12 4.3) (6 points) Assume that 6? %, determine the phase of G(co) in terms of the spectrum F(c0). 30:) = Me) (.05 (wt: + 1,5) = —x(t> sin (Wat) {Eye} = “ a}? fixmlx JisiMwLﬂB : g; mm a‘ 0:117‘E8Cwch) —~ sow—web] 2' XCw— 000) A j: X (W + Wc.) 6 (M M/v 13 4.4) (6 points) Assume 9 = 0 and Find the phase of Y(co). Fv‘om Part Ll3 #210060») ‘1an 50 2Hwa s 2wth + 2mm 0 ); gt“) : 0 199.60wa Gav) 2C) ‘C‘IY‘ 992 W, 1) H’L‘W3 7 30%
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