EECS 451 EXAM #1 Fall 2008PRINT 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 (here same as fill-in-the-blank) worth 5 points each.1. The period of x[n]=3cos(2π0.075n +1)forinteger n is:(a) 1 (b) 1/0.075 (c) 40 (d) 75 (e) not periodic2. sin(32πt)+sin(48πt) is sampled at 40 Hz, then ideally interpolated. The result is:(a) 0 (b) sin(16πt) (c) 2sin(16πt) (d) sin(32πt) (e) 2sin(32πt)3. The convolution {1, 2, 3}∗{4, 5, 6} =: (a) { 7, 8, 9} (b) {4, 13, 27, 18}(c) { 4, 13, 28, 27, 18} (d) {4, 14, 32, 28, 18} (e) { 5, 11, 20, 23, 9}4. The system (transfer) function of a LTI system described by the difference equationy[n]+2y[n − 1] + 3y[n − 2] = 4x[n]+5x[n − 1] + 6x[n − 2] is:(a)z2+2z+34z2+5z+6(b)3z2+2z+16z2+5z+4(c)4z2+5z+6z2+2z+3(d)6z2+5z+43z2+2z+1(e) z2+ z +15. The system function of a LTI system with impulse response h[n]=u[n]+2nu[n]is:(a)z+2z(b)z+2z−1(c)z+2z2−3z+2(d)z2−3zz2−3z+2(e)2z2−3zz2−3z+26. Z{2nu[n]+3nu[n]} =: (a)z2−5z+61(b)1z2+5z+6(c)5z2+5z+6(d)2z−5z2−5z+6(e)2z2−5zz2−5z+67. If H(z)=6/[(z +1)(z − 2)], then h[n]=Z−1{H(z)} =:(a) 2nu[n]+(−1)nu[n] (b) 2nu[n] − (−1)nu[n] (c) 2(2n)u[n]+2(−1)nu[n](d) 2(2n)u[n] − 2(−1)nu[n] (e) 2nu[n]+2(−1)nu[n] − 3δ[n].8. The z-transform of {1, −3, 2}∗u[n]is:(a)z−2z(b)2z3−4z2+5z−2z3−z2(c) 1+3z−1+2z−2+zz−1(d) 1 − 2z (e) z − 29. The impulse response if δ[n]+2nu[n] → |LT I| →{2, −2} is: (a) δ[n] − 2nu[n](b) (12)nu[n] (c) 2(12)nu[n] − 2(12)n−1u[n − 1] (d) {1, −2} (e) 2δ[n] − 2(2n−1)u[n − 1]10. The zero-input response for y(n)-2y(n − 1)=x(n)+x(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−1)(z−6)(z−2)(z−3).11. The zeros, poles, and BIBO stability of the system are: (a) {1, 6}; {2, 3};stable(b) {1, 6}; {2, 3};unstable (c) {2, 3}; {1, 6};stable (d) {2, 3}; {1, 6};unstable12. The difference equation for the system is:(a) y[n] − 7y[n − 1] + 6y[n − 2] = x[n] − 5x[n − 1] + 6x[n − 2](b) 6y[n] − 7y[n − 1] + y[n − 2] = 6x[n] − 5x[n − 1] + x[n − 2](c) y[n] − 5y[n − 1] + 6y[n − 2] = x[n] − 7x[n − 1] + 6x[n − 2](d) 6y[n] − 5y[n − 1] + y[n − 2] = 6x[n] − 7x[n − 1] + x[n − 2]13. The response of the system to x[n]={1, −5, 6} is y[n]=:(a) {1, −7, 6} (b) (2)n+1u[n]−2(3)nu[n] (c) {6, −7, 1} (d) δ[n]+(2)n+1u[n]−2(3)nu[n]14. The response of the system to x[n] = 7 for all n is y[n]=:(a) 0 (b)76cos(πn) (c) ∞ (d) 1.22 cos(πn − 0.165) (e) 1.22 cos(πn +0.165)15. The impulse response of the system is h[n]=:(a) {1, −7, 6} (b) (2)n+1u[n]−2(3)nu[n] (c) {6, −7, 1} (d) δ[n]+(2)n+1u[n]−2(3)nu[n][24] 16. X(z)=1z−0.5−1z−3has 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