Reference Solution of DSP Test 2 2005 1

Reference Solution of DSP Test 2 2005 1

Problem 1 2

Problem 1 - Solution • (a) The Nyquist rate is 2 times the highest frequency. –→ T=1/(5 k. Hz x 2) = 1/10000 sec. • (b) 3

Problem 2 4

Problem 2 - Solution • The output xr[n] = x[n] if no aliasing occurs as result of downsampling. That is, X(ejw) = 0 for pi/3 ≦|w|≦pi. • (a) X(ejw) has impulses at w = ± pi/4, so there is no aliasing. xr[n] = x[n]. • (b) X(ejw) has impulses at w = ± pi/2, so there is aliasing. xr[n] ≠ x[n]. 5

Problem 3 6

Problem 3 - Solution • (a) A random process that satisfies: – 1) its mean is constant. – 2) its autocovariance is a function that depends only on the distance in placement. • (b) A random process for which time averages equal ensemble averages. • (c) A function that describes the average power of a signal over a particular frequency band. • (d) A random process with a flat power spectral density. • (e) A bit reduction technique to minimize quantization error. 7

Problem 4 8

Problem 4 – Solution (1/2) x[n] H(z) y[n] ↓M w[n] = y[n. M] • Observation 1: The goal of the system is to sample y[n] every M data points. – Apply H(z) only on the 1/M signal portions will effectively reduce the needed computational operations. • Observation 2: The two systems below are equivalent. H(z) ↓M x[n] y[n] ↓M y’ [n] w[n] H(z) w’ [n] 9

Problem 4 – Solution (2/2) x[n] Z 0 (z. M) x[n] ↓M z-1 Z 0(z. M) ↓M Z 1(z. M) z-1 Z 1(z. M) ↓M + z-1 ↓M ZM+1 (z. M) ↓M original y[n] + z-1 ↓M y[n] ZM+1(z. M) answer 10

Problem 5 11

Problem 5 - Solution • (a) Yes. The poles z = ±j(0. 9) are inside the unit circle so the system is stable. • (b) minimum phase poles & zeros outside unit circle 12

Problem 6 13

Problem 6 - Solution x[n] y[n] z-1 1/2 5/6 z-1 1/2 1/6 14