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Unformatted text preview: S‘O\UL'l‘\rovx {Qﬁ/ EE301 Signals and Systems In—Class Exam
Exam 3 Thursday, Apr. 21, 2011 Cover Sheet Test Duration: 75 minutes.
Coverage: Chaps. 5,7 Open Book but Closed Notes. One 8.5 in. x 11 in. crib sheet
Calculators NOT allowed. All work should be done on the sheets provided.
You must show all work for each problem to receive full credit.
Plot your answers on the graphs provided. VIP Note Regarding DTFT Plots: The abscissa in each plot is the frequency axis.
Ignore the numbers at the bottom. For each plot, the abscissa goes from —27r to +27? with tic marks every 77/ 8. There is a dashed vertical line at w = —7r and another dashed vertical
line at w = +7r. You only have to plot over —7r < w < 7r. Multiple Choice Question 1. Circle EVERY answer that is correct: Primary differences
between the CTFT and the DTFT include .Fhe time—domain variable for the DTFT is a discrete (integer—valued) variable for the
DTFT whereas it is continuous valued for the CTFT ® he DTFT is always periodic with period 271' whereas the CTFT is generally not
periodic except under special conditions @The DTFT is a summation over time whereas the CTFT is an integration over time (d) The inverse DTFT is a summation over frequency whereas the inverse CTFT is an integration over frequency Tke in v ev 39 D T FT {5 6m 1' n ’79 Z‘ ra'i‘icv.
M (e) The highest frequency for the DTFT is effectively 7r whereas the highest frequency for
the CTFT is 00 Multiple Choice Question 2. Circle EVERY answer that is correct: Properties of the
CTPT that am very different for the DTF T a1e . T 1me Scaling Property )Convolution Property .b Differentiation—in—Time Property
’ 's‘kg , a a s. 0.9 ¢ {2, / l” k1? G We
((1) Frequency—Shift Property} U“ \\ a CUP?" 9‘ )Time—Shift Property (( f))Duality Property Multiple Choice Question 3. Circle EVERY answer that is correct. (a) When you pass an analog signal through an LT 1 ﬁlter, the Nyquist rate for the output
signal is greater than the Nyquist rate for the input signal. hen you form the product of two analog signals, the Nyquist rate for the product
signal is the sum of the respective Nyquist rates for the two individual signals. he Nyquist rate for the square of a signal is twice the Nyquist rate of the original
signal. (d) The Nyquist rate for the derivative of a signal is greater than the Nyquist rate for the
original signal. When you multiply a signal by a realvalued sinewave, the Nyquist rate of the resulting
modulated signal is greater than the Nyquist rate for the original signal. Problem 2 (a). Consider an analog signal xa(t) with a bandwidth (maximum frequency) of
W in rads/sec. The sampling rate is twice the Nyquist rate: ws 2 4W (where ws 2 27r/Ts.)
Determine the range for acceptable values for the cutoff frequency, we, of the Ideal Lowpass
Filter that allows ma(t) to be reconstructed perfectly according to the formula below. @109 = f: $a(7’st)h(t — nTS) Where: h“) : TSSln(th) TL=*OO 7rt W< WC < (u;— \A/ w324W W<WC< 1W Problem 2 (b). Consider a CT signal ma(t) with bandwidth (maximum frequency) W in
rads/ sec. The sampling rate is chosen to be above the Nyquist rate at ws 2 3W, where
ws 2 27T/T3. xa(t) is reconstructed perfectly according to the formula below. Let H (to) be
the CTFT of h(t). Determine the respective values of cal and £02, both in terms of W, so
that the CTFT H (w) is ﬂat up to the bandwidth W and then rolls off to zero at ws — W. wad) = f: $a(nTs)h(t ~ nTs) where h(t) = TSlSiMwlt) sin(w2t) W1 7ft ’ITt and cos 2 3W n=—oo W‘AyW1: CUS~W1 gW~W12W ><< ) ”l0 ~— w+l0
.(: LU 4:
><(§m T yeC < D  ls VQC ( > Problem 2 (c). Consider the continuous—time signal Mt) below A discrete—time signal is 2
created by sampling 56(15) according to x[n] = 17(nTs) for T S = 3%. Plot the DTFT of LEM, X (w), over —7r < w < 7r in the space provided below. $05) = j2Ts {sin(10t) } sin(10t) 3M m «09,4 ’C m W)
T: % 5 11% Ny ‘NM—S‘f— Yale “1 46 1 7,0 7\ 3.: .. 1“
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6.§8IBB%MWW53M.WEEIE BEE13539203921554 851! MEHZSEWQMWZQQSMQBBZ Problem 2 (d). Consider the continuous—time signal :c(t) below A discrete—time signal is 2
created by sampling x(t) according to m[n] = 33(nTs) for T8 : 3—7; Plot the DTFT of mm], X (w), over —7r < w < 7r in the space provided below. 93“) = jQTs {8111(1015) } sin(10t)
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—6.55wu§m5mm53mmmmmwemznsenmasvmmmmmsamgnmsmm32 Problem 2 (e). Consider a DT LTI system with impulse response Mn] : 8 {sin (%n)} 7TH 2 Determine the output y[n] for the input $[n] given by 27. Z’Fr‘ 3/7? FLQCQM’Y 0+
+(s‘mngie 7‘ 2 “‘7: H (0» Mw e> so i H (23 C?) +H<¥>Cos<¥© + HUT} “5 @V‘B H ("it 1 H<T :\ Hi? :0 Uv‘keve3 HMS—'3 {’5 _. Tint/L517 9Ch>z ; + Zecg <§n>+ C“ (I?) Workout Problem 3 Let H Lp(w) be the DiscreteTime Fourier Transform (DTFT) of the
impulse response th[nl deﬁned below. ' 1 ‘ E71
s1n (8n) sm ( 8 n) ’ITTL 7F” (1) thln] = 8 (a) Note: th[n] is real—valued and even—symmetric as a function of discretetime. Thus,
H L p(w) is both real—valued and symmetric as a function of frequency. Plot H Lp(w) in
the space provided over —7r < w < 7T showing as much detail as possible. (b) hln] is deﬁned in terms of thln] as:
_ 7r
h[n] = 2 thln] s1n<§n> (2) Note that h[n] is odd—symmetric as a function of time. Thus, H (w) is purely imaginary
for all frequencies. Plot H (w) in the space provided. Note that the vertical axis values
have the multiplicative scalar j = v«l factored into them. (0) Consider the input signal $[n] below. :t[n] = 16 {W} cos (gm) Determine and plot the DTFT X (to) of the signal xln] in the space provided. (d) Denote the output y[n] when the signal in part (0) directly above is the input to the
UN system with impulse response h[n] in part (b) defined by eqn. (2). Determine and
plot Y(w) in the space provided over —7r < w < 7r showing as much detail as possible. (e) Create a complex—valued signal as
Zlnl = min] + jylnl Determine and plot the DTFT Z (w) in the space provided over —7r < w < 7T showing
as much detail as possible. VIP Note Regarding DTFT Plots: The abscissa in each plot is the frequency axis.
Ignore the numbers at the bottom. For each plot, the abscissa goes from —27r t0 +27r with
tic marks every 7r/ 8. There is a dashed vertical line at a) = —7r and another dashed vertical
line at w = +7r. You only have to plot over —7r < w < 7r. Plot your answer to Problem 3 (a) here. Show work above. ~’ZI“S‘ 2 _ I I  
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53920.39 I I I 23549.51! I  I I gazemsmmsz H (WI: ’5 HUD (WILE> I”) H“? (w+{£> Plot your answer to Problem 3 (b) here. Show work above. I I
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 Spring '06
 V."Ragu"Balakrishnan

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