1. Reference Solution of DSP Test 2
2005

2. Problem 1

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

4. Problem 2

5. 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].

6. Problem 3

7. 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.

8. Problem 4

9. 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]

10. 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

11. Problem 5

12. 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

13. Problem 6

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