ECGR4124 Digital Signal Processing Final Spring 2009
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1 ECGR4124 Digital Signal Processing Final Spring 2009 Name: LAST 4 NUMBERS of Student Number: Do NOT begin until told to do so Make sure that you have all pages before starting Open book, 2 sheet front/back notes, NO CALCULATOR, NO CELL PHONES/WIRELESS DEVICES DO ALL WORK IN THE SPACE GIVEN Do NOT use the back of the pages, do NOT turn in extra sheets of work/paper Multiple-choice answers should be within 5% of correct value Show ALL work, even for multiple choice ACADEMIC INTEGRITY: Students have the responsibility to know and observe the requirements of The UNCC Code of Student Academic Integrity. This code forbids cheating, fabrication or falsification of information, multiple submission of academic work, plagiarism, abuse of academic materials, and complicity in academic dishonesty. Unless otherwise noted: F{} denotes Discrete timefourier transform {DTFT, DFT, or Continuous, as implied in problem} F -1 {} denotes inverse Fourier transform ω denotes frequency in rad/sample, Ω denotes frequency in rad/second denotes linear convolution, N denotes circular convolution x*(t) denotes the conjugate of x(t) Useful constants, etc: e 2.72 π 3.14 e e e e /e e e ln( 2 ) 0.69 ln( 4 ) 1.38 log 10 ( 2 ) 0.30 log 10 ( 3 ) 0.48 log 10 ( 10 ) 1.0 log 10 ( 0.1 ) -1 1/π cos(π / 4) 0.71 cos( A ) cos ( B ) = 0.5 cos(a - B) cos(a + B) e jθ = cos(θ) + j sin(θ) 1/12
2 5 Points Each, Circle the Best Answer 1. The signal x[n] = sin(n) has a discrete-time frequency ω = a) 1 b) π/2 c) π d) none above 2. If x[n] = 2u[n] - 3 ( u[n-3] ) 2 then, x[3] equals a) 0 b) 1 c) -1 d) none above 3. The first two points of x[n] (at n=0,1) in sampling x(t)= cos(2πt) at 4 samples/second are: a) {1,0} b) {1,-1} c) {1,1} d) none above 4. A continuous time signal is defined as x(t)=cos(ω t), with Ω = 200π rad/second. To prevent aliasing, x(t) should be sampled at a rate greater than a) 2π samples/s b) 100 samples/s c) 200 samples/s d) none above 5. ((9)) 8 = a) 1 b) 7 c) 17 d) none above 2/12
3 5 Points Each, Circle the Best Answer 6. The circular convolution of the two 4-point sequences x[n] = {1,1,0,0} and y[n] = {1,0,0,1} a) {2,1,0,1} b) {1,2,1,0} c) {2,0,1,2} d) none above 7. Implementation of convolution using the FFT results in circular convolution. 8. Circle the BIBO stable impulse response below. a) h[n]=(0.1) -n u[n] b) h[n]=(2) -n u[n] c) h[n]= (2) n u[n] d) none above 9. The system with z-transform z 1.3 H ( z) = and ROC z >0.1 is BIBO stable. z A cube root of -1 is a) e jπ/3 b) j c) e jπ/6 d) none above 3/12
4 5 Points Each, Circle the Best Answer 11. The difference equation for a system is y[n] = 2x[n] + 3x[n-2]. The impulse response of the system is h[n] = a) 2δ [n] + 3δ [n-2] b) 2u[n] + 3u[n-2] c) 3δ [n] + 2δ [n-2] d) none above 12. The difference equation of a system with output y[n] is y[n] = 3x[n] 2x[n-4]. The z- transform of the system is H(z) = a) 1/(3+ 2z -4 ) b) 3-2z -4 c) 3+2z -4 d) none above 13. The DTFT of (0.1) n u[n] is a) (1/10)e -jωn/10 b) 1/(1-10e -jωn ) c) 1/(1-0.1 e -jω ) d) none above 14. The dc frequency response (H(ω) at ω=0) of a system with impulse response h[n] = δ[n-1] + 3δ[n-2] - 4δ[n-3] is: a) 0 b) z z -2-4z -3 c) 7 d) none above 15. The DFT of x[n] is X [ k] = X ( z) jk / N z = e 4/12
5 5 Points Each, Circle the Best Answer 16. The z-transform of h[n]= (1/4) n-1 u[n-1] is H(z) = a) c) 2 z z z 0.25 ; z <0.25 b) z z ; z >0.25 d) none above ; z >0.25 z 17. If a filter has H ( z) = and ROC z >0.5 then, the dc response of the filter is z 0.5 a) 0 b) 2 c) -2 d) none above 18. The system with z-transform z H ( z) = and ROC z >0.9 is BIBO stable. z The system with z-transform z H ( z) = and ROC 0.2< z <1.2 is BIBO stable. ( z 0.2)( z 1.2) 20. The 2-point Fourier matrix W === 2 is 1 1 a) 1 j 1 1 b) c) 0 1 D) none above 5/12
6 5 Points Each, Circle the Best Answer 21. A 2 sample/second filter with impulse response h[n] is constructed using the impulse invariance method for h(t) = 2-2t. Given this filter, h[2]= a) 1 b) 1 / 4 c) 1 / 8 d) none above 22. Pre-warped bilinear filter designs suffer from frequency aliasing. 23. An FIR filter Hamming window has a lower peak side-lobe than a Bartlett window. 24. The system with impulse response h[n] = 4 δ[n-1] + 2δ[n-2] - 1δ[n-3] could be implemented as an FIR filter. z 25. If a filter has H ( z) = and ROC z >0.5 then, the filter is z 0.5 a) FIR b) IIR c) none above 6/12
7 Assume the following system is initially at rest, and let h[n] be the impulse response. x[n] Σ y[n] z-1 z 1.5 1/2-1 z Points Each, Circle the Best Answer 26. h[1] = a) 1/2 b) 1 c) 2 d)none above 27. h[2] = a) 0 b) 1 c) 2 d)none above 28. The system is represented by difference equation y[n]= 0.5y[n-1] + x[n] + 1.5x[n-1] + x[n-2]. 29. The DC response of the system is a) 7/5 b) 5 c) 7 d)none above 30. The causal system is BIBO stable. 7/12
8 The following questions are for a filter with H(z) = 2 z j 1.4( z + 0.5)( z 0.5) and ROC z >1/2. F j P Q -1 A B C D E 1 R G S -j 5 Points Each, Circle the Best Answer 31. The proper locations of the zeroes of H(z) are at locations a) A and E b) ) O and R c) O and S d) none above 32. The proper locations of the poles of H(z) are at locations a) B and D b) C and D c) B and C d) none above 33. At frequency ω=0 rad/sample, the frequency response H(ω) most nearly equals a) 0 b) 1/3 c) 4/3 d) 5 8/12
9 In the following questions, a discrete-time filter is to be designed using the impulse invariance method. The sampling rate of the digital system is 2 samples/second. The causal continuoustime filter is given as: H ( s) 1 = s Points Each, Circle the Best Answer 34. The z-transform of the discrete-time filter is H(z)= a) 1 e z 1 ; z > e b) 2z z + 4 ; z >4 c) 1 e 2 2 z 1 ; z >e 2 d) none above 35. For the discrete-time filter, the dc response is H(ω) ω=0 is H(0)= a) 2 1 e 2 b) e c) 2/5 d) none above 36. The resulting discrete-time filter is BIBO stable. 9/12
10 A discrete-time filter is to be designed using the bilinear transform method. The sampling rate of the digital system is 2 samples/second. The continuous-time filter is given as: 1 H ( s) = s Points Each, Circle the Best Answer 37. H(z)= a) c) z + 1 2z + 1 z + 1 5z 3 z 1 ; z >0.5 b) ; z >5 z + 5 ; z >3/5 d) none above 38. At analog frequency Ω=0 rad/second, the dc response of the continuous-time filter H(Ω) most nearly equals a) 0 b) 0.2 c) 1 d) none above 39. At frequency ω=0 rad/sample, the magnitude of the frequency response of the discrete-time filter H(ω) = a) 0.25 b) 0.5 c) 1 d) none above 40. At frequency ω=π rad/sample, the magnitude of the frequency response of the discrete-time filter H(ω) = a) 0 b) 0.75 c)1.75 d) none above 10/12
11 The following questions refer to the design of a moving-average filter with H(ω) shown below with a sampling rate of 8000 samples/second. H(ω) 5 Points Each, Circle the Best Answer 41. For the discrete-time filter, the DTFT H(ω) is (IGNORING PHASE TERMS) ω a) sin( ω) sin( ω / 2) ; z >0 b) 1 sin(4ω ) 8 sin( ω / 2) ; z >0 c) 1 sin(2ω ) ; z >0 d) none above 8 sin( ω / 2) 42. For the discrete-time filter, the first null is at a continuous-time frequency of a) 250Hz b) 500 Hz c) 1000 Hz d)none above 43. For the discrete-time moving average filter, the number of points being averaged is a) 2 b) 4 c) 8 d)none above 11/12
12 5 Points Each (Circle the best answer) The following questions refer to the Java class below and the program main(). public class Red { private int a; private int b; public Red(int aa, int bb) { a=aa; b=bb; } public void qq(red c) { this.a= c.a; this.b=c.b - c.a; } public void fn(red c) { this.a= c.a - this.a; this.b = c.b - c.a; } } public static void main(string[] args) { Red x = new Red (2,2); Red y = new Red (3,6); Red z = new Red (2,4); int xx=1,yy=2,zz=3; x.fn(y); z.qq(x); } 44. At the end of the main program, x.a = a) 1 b) 2 c) 3 d) None above 45. At the end of the main program, x.b = a) 1 b) 2 c) 3 d) None above 46. At the end of the main program, z.b = a) 1 b) 2 c) 3 d) None above 12/12
ECGR4124 Digital Signal Processing Midterm Spring 2010
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