ECGR4124 Digital Signal Processing Midterm Spring 2010

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ECGR4124 Digital Signal Processing Midterm Spring 2010 Name: LAST 4 DIGITS of Student Number: Do NOT begin until told to do so Make sure that you have all pages before starting Open book, 1 sheet front/back notes, NO CALCULATOR 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 1/e 0.37 2 1.41 3 1.73 5 2.22 7 2.64 10 3.16 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 log 10 ( e ) 0.43 cos(π / 4) 0.71 cos( A ) cos ( B ) = 0.5 cos(a - B) + 0.5 cos(a + B) e j θ = cos(θ) + j sin(θ) 1/11

1. If x[n] = 5u[n-1] - u[n-4] + 3 δ[n-1] then, x[4] equals a) -1 b) 2 c) 4 d) none above 2. The signal x[n] = sin[2n] has a discrete-time frequency ω = a) 2 rad/sample b) 2π rad/sample c) 4π rad/sample d) none above 3. A continuous time signal is defined as x(t)=cos( 10πt) + cos(5πt ). To prevent aliasing, x(t) should be sampled at a rate greater than a) 10 samples/second b) 20 samples/second c) 40π samples/second d) none above 4. A continuous-time signal cos(1000πt) sampled at 1000 samples/second would correspond to a discrete-time frequency of ω = a) π/4 rad/sample b) π rad/sample c) 500 rad/sample d) none above 5. The DTFT of x[n] = ( 0.1 ) n u[n] is X(ω) = a) 1/( 1 10 e -jω ) b) 1/( 1 0.1 e -jω ) c) 1/( 1 0.1 e -jω ) d) none above 2/11

6. The system with input x[n] and output y[n] = x[n] + x 2 [n-1] is linear. 7. If a linear time invariant system has impulse response h[n]=2δ[n-2], the output response to an input signal of u[n-2] is. a) u[n-1] b) 2u[n-2] δ[n-2] c) 2u[n-4] d) none above 8. If x[n] = δ[n-1], h[n] = δ[n] - 2δ[n-1], and y[n] = x[n] h[n], where denotes convolution, then, y[1] equals a) 1 b) 2 c) -2 d) none above 9. The difference equation for a system is y[n] = 3x[n] + x[n-2]. The impulse response of the system is h[n] = a) 3u[n] + u[n-2] b) 3δ [n] + δ [n-2] c) 3 + e -2jt d) none above 10. The circular convolution of the two 4-point sequences x[n] = {1,0,0,1} and y[n] = {1,2,3,4 } is a) {2,2,3,5} b) {4,5,3,7} c) {3,5,7,5} d) none above 3/11

11. The ideal reconstruction filter for reconstructing a signal from its samples is an ideal lowpass filter with amplitude T s and cutoff frequency Ω = 1/T s where T s is the sampling period of the system. 12. Circle the BIBO stable impulse response below. a) h[n]=(0.01) -n u[n] b) h[n]=(1.11) -n u[n] c) h[n]= (1.01) n u[n] d) none above 13. The FFT is a fast version of the DFT. 14. (( - 4 )) 4 = a) -4 b) 4 c) 0 d) none above 15. If the four point sequence x[n] = {1,2,3,4}, then x[ ((n+1)) 4 ] = a) {4,3,2,1} b) {4,1,2,3} c) {1,4,3,2} d) none above 4/11

16. The dc response of a system with impulse response h[n] = 3δ [n-1] -2 δ [n-2] is H(ω) ω=0 = a) e j ω - e j3ω b) 2 c) 3 d) none above 17. The DTFT of x[n] = δ[n-3] is X(ω) = a) 1 b) 1/( 1 δ[ω-3]e -jω ) c) e -j3ω d) none above 18. A system has an impulse response h[n] = δ[n] + δ [n-2]. The frequency response of the system H(ω), at ω=π, is H(π)= a) 0 b) 1 c) 2 d) none above 19. The DFT of the four point sequence x[n] ={1,0,-1,0} is X[k] = a) {1, 1, 0, -1} b) {0,2,0,2} c) {1,0,-1,0} d) none above 20. An LTI system with y[n] = x[k] is an accumulator. n k= 5/11

21. The circular convolution of the two 8-point sequences x[n] = {1,1,0,0, 0,0,0,0} and y[n] = {1,1,1,1, 0,0,0,0} is a) {1,2,3, 2, 1,0,0,0} b) {1,2,2,2, 1,0,0,0} c) {1,2,3,3, 2,1,0,0} d) none above 22. The system with impulse response h[n]= 4 -n u[n] is BIBO stable. 23. A discrete-time sinusoid with ω = π/4 rad/sample (sampled at 8000 samples/second) has a corresponding continuous-time frequency of f = a) 500 Hz b) 500π Hz c) 1000 Hz d) none above 24. The difference equation for a system is y[n] + 5y[n-1]= x[n] + 2x[n-1]. The frequency response of the system is H(ω) = a) 1+ 2e jω 1 5e jω b) 1+ 2e jω jω 1+ 5e c) jω 1+ 5e 1 2e jω d) none above 25. The DFT of the 4-point sequence x[ ((n-1)) 4 ] is X[k] = a) X[k] δ[ω-1] b) ( j) k X[k] c) e -jkω X[k] d) none above 6/11

26. The first four points of x[n] (at n=0,1,2,3) in sampling x(t)=e jπ t at 2 sample/second are: a) {1,0,1,0} b) { e j0, e jπω, e j2πω, e j3πω } c) {1, j,-1,-j} d) none above 27. The signal x[n] = ( -1 ) n is periodic with discrete-time frequency ω = π/2. === 28. The 2-point Fourier matrix W 2 is a) b) c) 1 1 1 0 b) none above 29. If (12) 0.5 3.5, then the rms quantization noise of a 3-bit ADC with 0.7 volt step size is a) 0.2 V rms b) 0.7 V rms c) 3.5 V rms d) none above 30. The frequency response of the system below is H(ω) = a) 1+ 2e jω 1 5e jω b) 1+ 2e jω jω 1+ 5e c) jω 1+ 5e 1 2e jω d) none above 7/11

The system below is represented by difference equation y[n] = Ay[n-1] + Bx[n] + Cx[n-1] + Dx[n-2]. 31. In the difference equation, the coefficient A should be A = a) 1 b) 3 c) 5 d) 7 e) none above 32. In the difference equation, the coefficient B should be B = a) 1 b) 3 c) 5 d) 7 e) none above 33. In the difference equation, the coefficient C should be C = a) 1 b) 3 c) 5 d) 7 e) none above 34. The first 3 points of the impulse response of the system above is a) {1, 3, 5} b) {1, 7, 5} c) {1, 5, 15} d) none above 35. The system above is BIBO stable. 8/11

The frequency spectrum X(Ω) of continuous time signal x(t) is given below. 1 0-40π 0 40π Ω 5 Points Each (Circle the best answer) 36. Highest frequency component in the time signal x(t) is a) 5 Hz b) 10 Hz c) 15 Hz d) 20 Hz 37. If the continuous time signal is sampled at 20 samples per second, the following plot is the magnitude of the discrete- time frequency spectrum X(ω). 20 0-2π 0 2π ω 38. If the continuous time signal x(t) is sampled at 15 samples per second, there will be aliasing 9/11

The frequency spectrum X(ω) of discrete time signal x[n] is given below. X(ω) 8 0 -π 0 π ω 5 Points Each (Circle the best answer) 39. The 6 db bandwidth of the system X(ω) above is a) π/8 rad/sample b) π/4 rad/sample c) π/2 rad/sample d) none above 40. The 8-point DFT, X[k], corresponding to the signal with spectrum X(ω) shown above is X[k] = {8,6,4,2,0,2,4,6} as shown in the plot below. X[k] 8 0 0 2 4 6 k 10/11

The following questions refer to the Java class below and the program main(). public class Green { private int a; private int b; public Green(int aa, int bb) { a=aa; b=bb; } public void equals(green c) { this.a= c.a; this.b=c.b; } public void fn(green c) { this.a= c.a + this.a; this.b = c.b - this.b; } } public static void main(string[] args) { Green x = new Green (2,4); Green y = new Green (3,5); Green z = new Green (7,8); int xx=1,yy=2,zz=3; x.fn(y); z.equals(y); } 41. At the end of the main program, x.a= a) 1 b) 2 c) 3 d) None above 42. At the end of the main program, x.b= a) 1 b) 2 c) 3 d) None above 43. At the end of the main program, y.a= a) 1 b) 2 c) 3 d) None above 44. At the end of the main program, z.a= a) 1 b) 2 c) 3 d) None above 11/11