ECE 350 Signals and Systems Spring 2011 Final Exam - Solutions. Three 8 ½ x 11 sheets of notes, and a calculator are allowed during the exam.

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1 ECE 35 Spring - Final Exam 9 May ECE 35 Signals and Systems Spring Final Exam - Solutions Three 8 ½ x sheets of notes, and a calculator are allowed during the exam Write all answers neatly and show your work to get full credit Write neatly and WATCH YOUR ALGEBRA! Problem Grade /7 / 3 /46 4 /5 Individual Total (6%) / Group Total Exam Total 6%Ind+4%max(Ind,Grp) Course grade: ECE 35 Final Exam, Spring page /9

2 ECE 35 Spring - Final Exam 9 May ) (7 points) Consider the following system: x(t) X a(t) H (j) y(t) k p ( t) ( t kt) Suppose the spectrum for x(t) is X(j) ½ - - and p(t) is an impulse train of frequency s =/T H (j) is an ideal lowpass filter whose gain is G and cutoff frequency is c a) (9 points)the Nyquist Theorem says that T must be what What must G and c be to perfectly reconstruct x(t)? Why? X p (j) /T s- - s- s s s s s- s- s s if s > s- > s > (Nyquist) s = /T /T > / > T P(j) = ( ks ) T k X p (j) has magnitude of (/)()(/T) = /T T _< c = G = T ECE 35 Final Exam, Spring page /9

3 ECE 35 Spring - Final Exam 9 May b) (8 points) For this signal x(t) there is a large part of the spectrum between and + that is unused If it is known to the system developers that only the part of the spectrum between and and and - is used for the signal information (ie, it is know that X(j)= if < ), and > /, design a system that violates the Nyquist Theorem but samples the signal x(t) to produce a (t), and then processes a (t) to produce a y (t) that exactly matches x(t) Define T and draw the system diagram that shows what should go in the box labeled with a question mark Explain all the steps that show why this works X(j) x(t) X a (t)? y (t) ½ - - ( t) ( t kt k p ) There is space between the signal bands from to + Use this bandwidth As you decrease s, at first there is aliasing, X p (j) /T but - s- - s- - - s s s s- s- s s s X p (j) /T - s- - s- - - s s s then the empty bandwidth can be used > - means gap around zero is large enough for the signal to fit in so they don t overlap, need s < s < and s > s > T = / /, / so / < T < / = T max a =, b =, A = T x(t) X a (t) y (t) ( t) ( t kt k p ) The filter is a band pass filter with pass bands from to, and gain of T ECE 35 Final Exam, Spring page 3/9

4 ECE 35 Spring - Final Exam 9 May ( points) Match the S-domain pole/zero patterns (a)-(f) with the Bode plots ()-(6) shown below ( points each) ) ) ) 4) ) 6) - - ECE 35 Final Exam, Spring page 4/9

5 ECE 35 Spring - Final Exam 9 May Problem (continued) 6 Im Im (a) (b) is associated with bode plot #_4 This is a low pass filter 3 Im is associated with bode plot # This is a band pass filter There is a zero on the j axis So as, the magnitude approaches This is - db 3 Im (c) (d) is associated with bode plot #_5 This is a band pass filter is associated with bode plot #_6 There are more zeros than poles The magnitude as 3 Im 5 Im (e) (f) is associated with bode plot #_3 This is a low pass filter There are 3 poles and zero Thus more poles than zeros As, the plot will drop at -4 db/decade is associated with bode plot # There are the same number of poles and zeros, so as w the magnitude is a nonzero constant ECE 35 Final Exam, Spring page 5/9

6 ECE 35 Spring - Final Exam 9 May ) (46 points) Consider the causal discrete-time system characterized by the following inputoutput relationship: y[n ] + y[n - ] - 8y[n] = 3x[n-] x[n] a) ( points) Assuming the system is causal, use iterative methods from Chapter to determine the first three non-zeros terms of the impulse response -8 y[n] = 3 x[n-] x[n] y[n-] y[n-] y[n] = 3 / 8 x[n-] + / 8 x[n] + ¼ y[n-] + / 8 y[n-] y[- ] = y[-] = 3 / 8 x[-] + / 8 x[-] + ¼ y[-] + / 8 y[-3] = y[] = 3 / 8 x[-] + / 8 x[] + ¼ y[-] + / 8 y[-] = + / 8 () + + = / 8 y[] = 3 / 8 x[] + / 8 x[] + ¼ y[] + / 8 y[-] = - 3 / 8 () + + ¼ ( / 8 ) + = - / 38 y[] = 3 / 8 x[] + / 8 x[] + ¼ y[] + / 8 y[] = + + ¼ ( - / 38 ) + / 8 ( / 8 ) = - 9 / 8 (picture not needed for credcit) ECE 35 Final Exam, Spring page 6/9

7 ECE 35 Spring - Final Exam 9 May b) (5 points) What is the transfer function for this system? y[n ] + y[n - ] - 8y[n] = 3x[n-] x[n] z - Y(z) + z - Y(z) 8Y(z) = 3z - X(z) X(z) Y(z) = X(z) H(z) = H(z)= c) (5 points) Sketch the pole-zero plot for this transfer function Zeros: 3z - - = z - = /3 z = 3 Poles: z - + z - -8 = (z - -4)(z - +) z - -4 = z - + = z - = 4 z - = - z = ¼ z = - ½ d) (5 points) What would the ROC have to be for this system to be causal? Why? -½ > ¼ z > ½ The ROC must include + Note: This is z, not {z} or {s} e) (5 points) If it were causal, would it then also be stable? Why/why not? Yes, z > ½ also includes the unit circle ECE 35 Final Exam, Spring page 7/9

8 ECE 35 Spring - Final Exam 9 May f) (8 points) For the ROC you determined in (d), determine the first three non-zero terms of the inverse transform through use of the Power Function method Power Function Long division h[n] = / 8 δ[n] / 3 δ[n-] 9 / 8 δ[n-] + Note: g) (8 points) For the ROC you determined in (d), invert the transform found in (b) using tables to determine the impulse response for this system ECE 35 Final Exam, Spring page 8/9

9 ECE 35 Spring - Final Exam 9 May 3) (5 points) If x(t), a periodic signal with period T=, is passed through a real filter with H(j) = 4 j The non-zero Fourier Series coefficients for the resulting y(t) are b = -, b = b * - = e j/4, b 5 = b * -5 = What was the real signal x(t) that was input to this system that produced this y(t)? Explain clearly how you arrived at this answer b k = a k H(j) a k = b k / H(j) = k ω = π /T = π /π = k b k ω H(jω) a k - -/ e jπ / e -jπ / x(t) = - ½ + 8 (e j(()()t+5+π /4) + e j(()(-)t-5-π /4) ) + 4 (e j(()(5)t+55) + e j(()(-5)t-55) ) = - ½ 64 cos(4t+5 + π/4) + 8 cos(t + 55) (this step is necessary The question said real function x(t)) ECE 35 Final Exam, Spring page 9/9