Reference Solution of DSP Test 2 -...

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1 Reference Solution of DSP Test 2 2005

Transcript of Reference Solution of DSP Test 2 -...

Page 1: Reference Solution of DSP Test 2 - cmlab.csie.ntu.edu.twcmlab.csie.ntu.edu.tw/DSPCourse/...Solution_of_DSP... · Reference Solution of DSP Test 2 2005. 2 Problem 1. 3 Problem 1 -

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Reference Solution of DSP Test 2

2005

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Problem 1

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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=

=

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Problem 2

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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].

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Problem 3

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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.

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Problem 4

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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]

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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

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Problem 5

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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=

−⋅

+

−⋅+=

−⋅

−⋅

++

=

−−

−−

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Problem 6

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Problem 6 - Solution

x[n] y[n]

1/2

1/2

5/6

1/6

z-1

z-1