Reference Solution of DSP Test 2 -...
Transcript of Reference Solution of DSP Test 2 -...
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)
sec/ 1250625210000
18
101
rad
Tw
kHzT
c
c
ππ
π
=⋅=Ω
Ω=
Ω=
=
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) ↓Mx[n] y[n] w[n] = y[nM]
H(z) ↓Mx[n] y[n] w[n]
↓M H(z)x[n] y’ [n] w’ [n]
10
Problem 4 – Solution (2/2)
Z0(zM) ↓M
Z1(zM) ↓M
ZM+1(zM) ↓M
+
x[n]
y[n]z-1
z-1
Z0(zM)↓M
Z1(zM)↓M
ZM+1(zM)↓M
+
x[n]
y[n]z-1
z-1
original answer
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)
191
81.012.01)(
2
2
1 −
−
− −⋅
++
=z
zzzH
minimum phase poles & zerosoutside unit circle
)()( 911
9181.01
)911()2.01(
911
911
191
81.012.01)(
1
2
2
2
21
2
22
2
1
zHzH
z
zz
zz
z
zzzzzH
ap=
−
−⋅
+
−⋅+=
−
−⋅
−⋅
++
=
−
−
−
−−
−
−−
−
−
13
Problem 6
14
Problem 6 - Solution
x[n] y[n]
1/2
1/2
5/6
1/6
z-1
z-1