Reference Solution of DSP Test 2 2005

Problem 1

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

Problem 2

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].

Problem 3

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.

Problem 4

Problem 4 – Solution (1/2) 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] w[n] = y[nM] H(z) ↓M x[n] y[n] w[n] ↓M H(z) x[n] y’ [n] w’ [n]

Problem 4 – Solution (2/2) + + answer original Z0(zM) ↓M Z1(zM) ZM+1(zM) + x[n] y[n] z-1 Z0(zM) ↓M Z1(zM) ZM+1(zM) + x[n] y[n] z-1 answer original

Problem 5

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

Problem 6

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