FinalExamSP14Soln

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Col© Tioe N Name: - Final Exam ECE301 mignals and Systems Tuesday, May 6, 2014 Cover Sheet Write your name on this page and every page to be safe. Test Duration: 120 minutes. | Coverage: Comprehensive Open Book but Closed Notes. Three two-sided handwritten sheets. Calculators NOT allowed. This test contains four problems, each with multiple parts. You have to draw your own plots. You must show all work for each problem to receive full credit. Good luck! It was great having you in class this semester! Have a great Summer!

Problem 1. : (a) (b) (e) (£) For the signal s(t) below, plot the Fourier Transform S(w) which is purely real-valued: 7 {sin(2.5t) S(t) - é—g i }22 cos(5t) (1) An AM signal r(t) is formed from s(¢) above as prescribed below, where k£ = 0.1 and ¢ r(£) = [1 4+ k s(t)] cos (35t + o) ' (2) The signal r(t) above is applied to a square-law device, followed by amplification, to form z(t) = 1072(¢t) p(t) = 10 {[1+k s(t)] cos (35t + ¢)}> 10 [14+k s(t)]z‘cos2 (35t + ¢) (3) Plot the magnitude of the Fourier Transform of z(t) denoted |X(w)|, IGNORE as negligible any term which is scaled by &% = 0.01. Recall cos*(0) = £ + £ cos(26). Consider an LTI system with impulse response h(t) = {_ZT_ sin(15¢) Sin(25t)} 15 7t 7t (4) Determine and plot the frequency response, H(w), the Fourier Transform of h(%). For the LTI system with this impulse response, determine the output y(t) for the input z(t) above. Plot |Y(w)|. Again, ignore any term which is amplitude-scaled by k* = 0.01 as negligible. | y(t) = z(t) * h(t) Would your answer to part (d) change if the phase shift in the sinewave was changed from ¢ = 7 to ¢ = Z7 Does the value of the phase shift ¢ have any impact on the recovery of s(t) via the method above? Explain your answer. In the scheme above, a strong DC component is added before the signal is multiplied by the cosine wave. Determine the final output y(¢) via the same sequence of steps with the DC component removed: (1) Form r(t) = 2s(¢) cos (35t + ¢), (2) Multiply by - cosine at the same frequency z(t) = r(t) cos (35¢t), and (3) y(t) = z(¢) * h(t). Recall: 2 cos(0) cos(p) = cos(0 + ¢) + cos(@ — ¢). How does the output y(¢) depend on the value of gb7 What is the output When the phase is gb =T / 2‘7 N Prast

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P R I I 21 . I P I R -10 0 10 20 30 40 50 60 70 80 1. l -80-70-60-50-40-30-20 —2 gy Mm 4 ;@

Show all work and plot for Problem 1 (c.). R T T “ * 2 4 % e e s s e & w e @ » > ® s ® s ® e s e+ s 4 & s e & w s v e 8 = s s s s e s e e s s » . . . . . . . . . . . . . . . . . : . . . . . . . . . . . . . . . . . . . . . . D . ‘ . . . . . . . . . . . . . . . . . . . . . . . . . . a . . . . . . . . . . . . . . . . . . . N . . 0 . . . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . D . . . . . . . . . 0 . . . . 0 . . . . . . . . . . . . . ! . . . . . . . . . . . . 0 . . . . ; . . . . . , . . . B . . . . . . . 0 . . . * 0 . . . . . . . . . e S L T I L T S T T D T S T T O T I I TP SR AP PP S . . . . . . 0 . . . . . . . . . . . . 0 s . 0 . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; . . B . . . . { . . . . . . . . . . . . B . . . . . . . . D . . o . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B . . . . . . [ . . . . . . . . . . . . . . . . . . . . . . . . . . . e . . . ' . . . e s P L T T T N T T e P R . . . . . B . . . . i . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D . . . . . . . . . . . . i . 3 ( . . . . . . . . . i . . . . . . . . . . . . . . . . . H . . . . . . . ‘ . . . . . . . . . i . . . . . . . . . . . . . 0 . . . : [ . . . . . . . . . . . - . . ‘. . . . . . . . B . . . . . . . . . B . . « . . . . . . . . . . . . 0 . . . . . 0 . s . . * . . . ‘. . . . . . . . . . D . . . . . . . . . . . . . . . . . . . . . . . . . . . . ; . . . . . . . . . . . . . . . . . . B . . ’ . . . . . . . . . . e ¢ & s 4 e e s e e e v e+ 4 s s e s e e e s s e e e a s s s s e s e % e e e e e e e s e - e e e . B T T T S P S - e s . D T T S — . . . . D . O ’ . . . . D y . . . . . . . 0 . B . . . 0 . . . D . . . . . . . . . . a . . . . . . . . . . . + ’ . D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D . . . . . . . . ’ . . . . . . . . . . D . . . . . . . . . . . . . 0 . . . . . . . . . . . . » . . . . . . . . . . . . . . . [ . . . . . . . » . . . . . ’ . . . . . . . . . . . . . . . » . 0 . . fass & ® s a2 e = v a2 4 e 8 s s e = s s e s s s v s s e s s oa s s s s e s s o e s e s e s e s e v s+ e & 2 s s s e & s e s o m s = s s e e s o a s r s e s e s s e a w s e s e e e e e s e s e S—— . . . . . . . 0 . . . . . . . . . + D . . . . B . . . . . . . . . . . . . . . . . . . D . . . . ‘e . . . . . . . 0 . . . . . . . . . . . . . . . . B . . . . . . . . . . . . D . . . . . . . . . . . . . . . . B . 0 . . . . . . . . - ¢ B . . : . ’ . . . . . . . . . . . 0 . + . . . . . . . . 0 . . . . . . . . . . . . . 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . « l l l | I | l l | l l I I | | l | 3

e S [ T e N\ o~ 2y - . . e e e b s e e e e PO T TP s 4 s N Q1 S P W | T W L P N L & N L P —L_ o samede = - &L o m__. f— =+ mamnde = N L o N L Ul W L O QL | a1 = | o SO0 o1 . . . . . . . . ’ v . . e . . . . » . o , D . . . .. . . * P . 0 . . . . B . . . . . . . . . . . . . . o * . . . . . . " ' . . . . e . . . ’ g . . . . .., . . . . o . . . . . . . / 9 (P) T W[qOoIJ 0] j0[d pue I0M [[8 MOYS ) [

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Problem 2. The rectangular pulse z;,(t) = {u(t) — u(t — 2)} of duration 2 secs is input to the following integrator The output z4(t) is sampled every Ts = 1/2 seconds to form z[n] = z,(nT;). The sampling rate is fs = 2 samples/sec. Note: Part (b) can be done independently of Part (a). Show work. Clearly label and write your final answer in the space provided on the next few pages. (a) Do a stem plot of the DT signal x[n], or you can simply write the numbers that comprise z[n] (indicate with an arrow where the n = 0 value is.) (b) Determine a closed-form expression for the DTFT X (w). Show work and clearly label and write your final answer in the space provided on the next few pages. VIP: List the values of w within the range from —7 < w < 7 for which X (w) = 0. (c) The Discrete-Time (DT) signal z|n], created as described above, is input to the DT system described by the difference equation below: y[n] = —2y[n — 1] + z[n] 4 8z[n — 3] (i) First, determine and plot the impulse response h|n| for this system. Do the stem- plot for h[n| on the graph provided on the next page. (ii) Determine and plot the output y[n| by convolving the input z|n| defined above with the impulse response h[n|. Show all work in the space provided. Do the stem-plot for y[n| on the graph provided on the page after next.

Show your work and plots for Problem 2 here. R e, e, R s

Show your work and plots for Problem 2 here.

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Problem 3 (a). Consider an analog signal z,(t) with maximum frequency (bandwidth) wyr = 20 rads/sec. That is, the Fourier Transform of the analog signal z,(t) is exactly zero for |w| > 20 rads/sec. This signal is sampled at a rate ws; = 60 rads/sec., where w, = 27 /T, such the time between samples is T, = 22 sec. This yields the discrete-time sequence 60 | 60) (sin(Zn) sin(Zn 2 zn| = x,(nTs) = {—é—ﬂ—_} { ;; ) + fmi )} where: Ty = —é—g A reconstructed signal is formed from the samples above according to the formula below. oo 21 7 sin(10¢) sin(30¢) B - here: Tg= — t) ="Ts— CEr(t) n;mm[n]h (t —nTs) where 0 and h(t) T - / [N 20409 Problem 3 (b). Consider the SAME analog signal z,(¢) with maximum frequency (band- width) wys = 20 rads/sec. This signal is sampled at the same rate w, = 60 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does z,(t) = x,(¢)? For this part, you do not need to determine z,(¢), just need to explain whether z,(t) = z,(¢) or not. o N _ 2m ., ™ sin(15¢) sin(25¢) z.(t) = > z[nlh(t —nTs) where: Ty= a0 and h(t) = S _— nN=—0oo Problem 3 (c). Consider the SAME analog signal z,(¢) with maximum frequency (band- width) wys = 20 rads/sec. This signal is sampled at the same rate w, = 60 rads/sec., but is reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does z,(t) = z,(¢)? For this part, you do not need to determine z,(t), just need to explain whether z.(t) = z,(¢) or not. s 27 7 sin(5t) sin(35t) = - 1 " : S — = TS — z,(t) n:E_OO zinlh (t —nTs) where: T, w0 and (1) g - Show all your work for Prob. 3, parts (a)-(b)-(c) on next page.

o PIEER, i \g,;rw’ & (Eon— g B an 'mopaq (9)-(q)-(e) syred ‘¢ -qoiJ 10J 3I0M INOA MOYS

Problem 3 (d). Consider an analog signal z,(¢) with maximum frequency (bandwidth) wyr = 20 rads/sec. That is, the Fourier Transform of the analog signal z,(¢) is exactly zero for |w| > 20 rads/sec. This signal is sampled at a rate ws; = 50 rads/sec., where wy = 27 /T, such the time between samples is T, = 2% sec. This yields the discrete-time sequence 50 ' 50 (sin(Zn) sin(Zn 2 zn] = z,(nTs) = {ZT-} { frfl ) 4 ;Z >} where: T, = B—g A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal z,(¢). Show work. o0 2 . 2 z,(t) = Z znlh (t —nTs) where: T, = —5—73 and h(t) =T, sin(20t) n=-—00 0 i Problem 3 (e). Consider the SAME analog signal z,(¢) with maximum frequency (band- width) wys = 20 rads/sec. This signal is sampled at the same rate w; = 50 rads/sec., but reconstructed with a different lowpass interpolating filter according to the formula below. Does this achieve perfect reconstruction, that is, does z.(t) = z,(¢)? For this part, you do not need to determine z.(¢), just need to explain whether z,(t) = z,(¢) or not. 0. 0] 2 . z-(t) = Y z[n]h(t —nTs) where: T, = _5_% and h(t) = T, sin(35¢) Tt Problem 3 (f). Consider the SAME analog signal z,(t) with maximum frequency (band- width) wys = 20 rads/sec. This signal is sampled at the same rate w\s_::ﬂfi/(hrads_@c., but reconstructed with a different lowpass interpolating filter according ¥o the formula below. Does this achieve perfect reconstruction, that is, does z.(t) = z,(t)? For this part, you do not need to determine z.(¢), just need to explain whether z,(t) = z,(¢) or not. 6. @) 2 z-(t) = Y z[nlh(t—nTs) where: T = -5—7?— and h(t) =T, n=—00 0 sin(15t) t Show all your work for Prob. 3, parts (d)-(e)-(f) on next page.

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Iy ‘Show your work for Prob. 3, parts (d)-(e)-(f) below. e = PRt s

Show all your work for Prob. 3, parts (g)-(h) on this page. s =30

Problem 3 (g). Consider an analog signal z,(¢) with maximum frequency (bandwidth) wy = 20 rads/sec. This signal is sampled at a rate w, = 30 rads/sec., where wy = 27/T} such the time between samples is T = -25% sec, yielding the following discrete-time sequence: (30 (sin(®n) sin(%En) 27 = x,(nTy) = — 3 3 where: T, = — zln] = ,S) {271'} { TN i TTY > 30 A reconstructed signal is formed from the samples above according to the formula below. Determine a closed-form expression for the reconstructed signal x.(t). Show all work. oo 2 . 1 z.(t) = > z[nlh(t—nTs) where: Ty = —B—g— and h(t) = T, sin(15t) Nn=—00 Tt Problem 3 (h). Consider the SAME analog signal z,(¢) with maximum frequency (band- width) wy = 20 rads/sec. This signal is sampled at the same rate w; = 30 rads/sec., where ws = 2 /Ts and the time between samples is T, = —2—7—05 sec, but at a different starting point. This yields the Discrete-Time z|n]| signal below: 30 (sin(Z(n+0.5)) sin(*(n+0.5)) 27 [n| = z,(nT,+0.5T,) = { — 3. ! here: Ts=— Zeln] = Ta(nds+0.5T;) {zw}{ T(n+05) | a(n+05) WHEEE A reconstructed signal is formed from the samples above according to the formula below. Determine a simple, closed-form expression for the reconstructed signal z,(¢). 0.8 2 | . ()= Y, znlh(t— (n+0.5)T;) where: T = _é_g and h(t) =T, sin(15¢) oo e Show your. work for Prob. 3, parts (g)-(h) on next page.

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Show all y ur work for P rob. 3, parts (g)-(h) on this page. i e, wzn 3 S =5 T T mal R L | (a1 o 2 s e, S . s s — i P e SRR e 2 s — N S i S T

A DDENDOM Lot consvnsd o mulbipiping (6 C e P(t)‘: Z g(t’V\Ts) “f‘c P\'ck O‘F(: \/a{k@;> {,9) gé\m(;o‘e) Q’F_ r\‘:. -—OO Na(£> eve% xa ('E‘,) QVPV&- —Ts S@gcv\.e\S> CO*\S‘\‘QSQV ,%ZW:E N 2y pb-T)ers™E > e T (v * T - | | . T4 w72 wM> . - _)PLQWGKCw,p&:F Nno overlap of rep’ C9> Z: 6 | > onS Ty IS ev o’ O"\\y /PL—:'@ = Fevar which 15 original v hal = (PQ(F—ec—,\ * - YeConsTrucTioh T TP wS<ZwM> VPP\\'Q% \f\aung, ’Y\J{’J—e«en‘\- P\'Sm.se wetfk5~8wgg ansd : recensy vucred s anel will depend on