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EECS 451 EXAM #1 Winter 2009PRINT YOUR NAME HERE:HONOR CODE PLEDGE: ”I have neither given nor received aid on this exam, nor have Iconcealed any violations of the honor code.” Closed book; 2 sides of 8.5×11 ”cheat sheet.”SIGN YOUR NAME HERE:CIRCLE ONE: Undergraduate GraduateWrite your answer to each question in the answer space to the right of that question.Problems #1-15 are multiple choice (same as fill-in-the-blank) worth 5 points each.1. The period of the discrete-time sinusoid 4 cos(0.56πn +0.7) is:(a) 12.5 (b) 25 (c) 40 (d) 50 (e) Not periodic2. sin(40πt) + 2 sin(160πt) and which of these are identical after sampling at 100Hz:(a) 0 (b) − sin(40πt) (c) 3sin(40πt) (d) − sin(160πt) (e) 3 sin(160πt)3. The convolution {1, 2}∗{3, 4, 5} =: (a) {3, 6, 4, 10} (b) {3, 10, 13, 10}(c) {3, 10, 14, 10} (d) {3, 11, 13, 10} (e) {3, 11, 14, 10}4. The system (transfer) function if 2nu[n] → |LT I| → u[n]+2nu[n]is:(a)1+2zz+2(b) 1+1z+2(c)2z−3z−1(d)1z2−3z+2(e) 1+1z−15. The system (transfer) function if {1, 2, 3}→|LT I| →{4, 5, 6} is:(a)3z2+2z+16z2+5z+4(b)6z2+5z+43z2+2z+1(c)z2+2z+34z2+5z+6(d)4z2+5z+6z2+2z+3(e) 1+z + z26. The z-transform of (2)nu[n]+(−4)nu[n]is:(a)1z2+2z−8(b)z2+zz2+2z−8(c) 2z2+zz2+2z−8(d)z2−zz2+2z−8(e) 2z2−zz2+2z−87. The inverse z-transform of2zz2+1is:(a) (−1)n(b) 2cos(πn)u[n] (c) 2sin(πn)u[n] (d) 2cos(π2n)u[n] (e) 2sin(π2n)u[n]8. The inverse z-transform ofz2−5z+6z2(z−1)is:(a) { 1, −5, 6}∗u[n] (b) {1, −6} (c) {1, −5, 6}∗u[n − 1] (d) {0, 1, −6}9. The impulse response if 2nu[n]+4nu[n] → |LT I| → δ[n]is:(a) (12)nu[n]+(14)nu[n] (b) {1, 2, 4} (c) {1, −2, −4} (d) C3nfor n>3forsomeC10. The zero-input response for y(n)-2y(n − 1)=2x(n)+3x(n − 1) with y(−1)=1 is:(a) 2n+1u(n)(b)2nu(n)(c)2n−1u(n − 1) (d) 2nu(n)+2n−1u(n − 1) (e) 0For problems #11-15: An LTI system has transfer function H(z)=z(z−1)(z−2).11. The poles and zeros are, respectively, (a) {0}; {1, 2} (b) {0}; {−1, −2} (c) {1, 2}; {0} (d) {−1, −2}; {0}12. The difference equation is: (a) y[n] − 3y[n − 1] + 2y[n − 2] = x[n − 2](b) y[n ] − 3y[n − 1] + 2y[n − 2] = x[n − 1] (c) y[n] − 3y[n − 1] + 2y[n − 2] = x[n](d) y[n ]+3y[n − 1] + 2y[n − 2] = x[n − 1] (e) y[n]+3y[n − 1] + 2y[n − 2] = x[n]13. The impulse response h[n]is:(a) 2nu[n](b)2nu[n]+u[n](c)2nu[n]−u[n] (d) 3(2)nu[n]+2u[n] (e) 3(2)nu[n]−2u[n]14. The response to x[n]={2, −6, 4} is:(a) 2nu(n)+u(n )(b)2nu[n] − u[n](c)2n+1u[n]+3u[n](d)2δ[n +1](e)2δ[n − 1]15. The response to x[n]=δ[n] − δ[n − 1] is:(a) 2n+1u[n +1] (b) 2nu[n](c)2n−1u[n −1] (d) 2nu[n]+u[n](e)2nu[n] −2n−1u[n −1][24] 16. X(z)=zz−0.2+zz−2has 3 different inverse z-transforms. For each one, compute:[3@3] (a) x[n ] [3@3] (b) ROCs [3@1] (c) if stable [3@1] (d) if causal. Write below.✫✪✬✩✫✪✬✩✫✪✬✩STABLE CAUSAL STABLE CAUSAL STABLE CAUSAL[1] 17. Did you: (a) PRINT your name; (b) SIGN your name; (c) CIRCLE grad or


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U-M EECS 451 - EXAM #1

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