adsp ws1415 solutions

7/25/2019 Adsp Ws1415 Solutions

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TECHNICAL FACULTY,CHRISTIAN-ALBRECHTS-UNIVERSITYOF KIEL

DIGITALSIGNAL PROCESSING AND

SYSTEM THEORY

Problem 1 (relationship between continuous and discrete signals)

(a) Keeping in mind that after sampling , =T, the Fourier transform ofv(n) is

Va(j)

0 1 2 0 12

V(ej)

1

T

(b) A straight-forward application of the Nyquist criterion would lead to an incorrectconclusion that the sampling rate is at least twice the maximum frequency ofva(t),or 22. However, since the spectrum is bandpass, we only need to ensure that thereplications in frequency which occur as a result of sampling do not overlap withthe original. (See figure below ofVs(j ).) Therefore, we only need to ensure

2 2/T < 1 T 22/T

Digital Signal Processing and System Theory, Prof. Dr.-Ing. Gerhard Schmidt, www.dss.tf.uni-kiel.deAdvanced Digital Signal Processing, Exercise Solutions WS 2014/2015

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SYSTEM THEORY

Problem 2 (overall system for filtering a continuous-time signal in digitaldomain)

(a) vi(t) =va(t) p(t)

Vi(j)

Vi(j)

2/T 2/T

V(ej)

2 2

1/T

1/T

(b) Since H(ej) in an ideal lowpass filter with c = /4, we dont care about anysignal aliasing that occurs in the region /4 . We require:

2/T 2 10000 Hz /(4T)1/T 8

7 10000 Hz

T 78 104s

Also, once all of the signal lies in the range || /4, the filter will be ineffective,i.e., /4 T(2 104 Hz). So, T 12.5s.

Digital Signal Processing and System Theory, Prof. Dr.-Ing. Gerhard Schmidt, www.dss.tf.uni-kiel.de

Advanced Digital Signal Processing, Exercise Solutions WS 2014/2015

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SYSTEM THEORY

(c) Sketch ofwc as a function of 1/T

c

Slope =/4

8/7104 8104 1/T



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SYSTEM THEORY

Problem 3 (quantization)

(a) Number of quantization levelsL

L= 2b; b= 4

L= 24

= 16

Quantization step

= R/L

= 10/16 V

= 0.625 V

(b) Input-output characteristic of themidtreatquantizer(See lecture slide Slide II-22)

Differences between a midriseand a midtreatquantizer:

(i) Amidrisequantizer has no zero level.

(ii) Amidrisequantizer is odd symmetric along the y-axis.

(c) Forn = 1250

v(n) = 2.925

vq(n) = Q[v(n)]= 3.125

eq(n) =v(n) vq(n)= 0.2

Bipolar code (sign and magnitude representation) of the quantized value

Code for 5 = 0101

(d) Real system

v(n)Q[ ]

vq(n)

Mathematical model



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SYSTEM THEORY

v(n)

eq(n)

vq(n)

(e) Power of the quantization noise PN

PN = 2/12

= (0.625)2

12= 0.03255

(f) SNR/dB = 6.02b + 10.8

20log10

(R/v

)R is the range of the quantizerv is the RMS amplitude of the input signalR= 10 V, v=

52

= 3.5355, b = 4

SNR/dB = 25.8477 dBLinear value = 384.388

(g) New v= 1

2= 0.707

New SNR/dB = 11.8683 dB

(h) Compute b from the equation 6.02b+ 10.8 20log10(R/v) by equating it to 45dB.For 1V input:b 10 ( Because b has to be an integer )

For 5V input:b 8 ( Because b has to be an integer )



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SYSTEM THEORY

Problem 4 (DFT and convolution)

M= 8

(a) H8() =7

k=0h(n) W8n = 1 + W8

Y8() = 1 + W8 + W8

2 + W83

(b) y(n) =v(n) h(n) =7

k=0v(k)h((k n))M

Y8() =V8() H8() V8() = 1 + W82v(k) = {1, 0, 1, 0, 0, 0, 0, 0}

(c) v(n) has length 3 = M1h(n) has length 2= M2 M1 + M2 - 1 < M

cyclic convolution equals linear convolution

y(n) = v(n) h(n)



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SYSTEM THEORY

n

v(n)

763210 54

76543210

0 1 2 3 4 5 6 7

h((0 n))M

h((1 n))M

0 1 2 3 4 5 6 7

h((2n))M

h((3 n))M

h((4 n))M76543210

0 1 2 3 4 5 6 7

0 1 2 3 4 5 6 7

h((7 n))M

n

n

n

n

n

n



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SYSTEM THEORY

Problem 5 (DFT)

(a) LetTbe the period of the time-limited signal, v0(t).T = 2

w0t= 4w0 = 2T (two periods)

1

t

1

vo(t)

T T

(b) v(n) =v0(t)

t=nTA

= v(nTA) = sin(w0 n 4w0 ) = sin(4 n)

whereTA is the sampling periodv(n) = sin(4 n) n = 0, 1, 2, . . . , 15TA =

4w0

= T8

wA= 8w0

n

-1

2T A 8T ATA

v(n)1



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SYSTEM THEORY

(c) V() =DF T16{v(n)} =15n=0

v(n)Wn16 =Z{v(n)}z=W

16

z= ej

WM =ej 2M z = WM here, M= 16

V() =V(z)

z=e

j2M

,

=1

V() = (1 + W816) I

(1 W416) II

( W162

+ W216 +W316

2)

II I

I) (1 + W816) = 2 if is even0 if is odd

II) (1 W416) =

0 = 0, 4, 8, 12

1 (j) otherwise

V() equals zero for all except = 2, 6, 10, 14 (III) needs to be computedonly for = 2, 6, 10, 14, however due to the symmetric property of the DFT (III)is computed only for = 2, 6

III) W162

+ W216 +W316

2

for = 2: 12

( 12j 1

2) + (j) + 1

2( 1

2j 1

2) = 2j

for = 6: 12

( 12j 1

2) + (j) + 1

2( 1

2j 1

2) = 0

Now, using the results of (I), (II) and (III) we getV() as

V() = (1 + W8

16)

I

(1

W4

16)

II

(W16

2+ W2

16 +

W316

2)

II I

V(2) = (2)(2)(2j) = (8j)V(6) = (2)(2)(0) = 0

for real valued sequences,

V() =V(M )



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SYSTEM THEORY

V(10) =V(16 10)=V(6)

= 0

V(14) =V(16 14)=V(2)

= 8j

V() =

8j = 2

8j = 14

0 otherwise

(1)

(d) The discrete sequencev(n) is also time-limited (finite-length sequence)

M= length of sequence v(n)v(n) = sin(4 n)

1(n) 1(n M)

fM(n)

D TFT{fM(n)} =M1n=0

1 ejn

=1 ejM

1 ej =ej

M2 (ej

M2 ejM2 )

ej2(ej

2 ej2)

=ej(M12 ) sin(

M2 )

sin( 2 )

v(n) =sin(

4n) fM(n)

F{v(n)} =V(ej)

V(ej) = 1

2 DTFT{v(n)}

ej

2(M1) sin

M2

sin 2

= 1

2

j

0( +

4) 0(

4)

ej(M1)2 sin

M2

sin 2

= j

2ej (M1)2 (+4 ) sin[( +

4

)M

2

]

sin[+4

2 ] ej

(M1)

2 (

4 ) sin[(

4

)M

2

]

sin[4

2 ]

(e) V(ej) =

n=v(n) fM(n).ejn =

15n=0

v(n) ejn

DFT{v(n)} = V(ej)

= 2M

= 0, 1, 2, . . . , 15



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SYSTEM THEORY

V(ej)

= 216 . = 8

=V()

= j

2

ej

152 (

8 +

4 ) sin[(

8 +

4 )8]

sin[8 +

4

2 ] ej 152 (8 4 ) sin[(

8 4 )8]

sin[8

4

2 ]

(2)

(1) and (2) should be equal.



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SYSTEM THEORY

Problem 6 (DFT, zero padding, leakage)

Letva(t) be a time-continuous periodic signal

va(t) = 1 + cos(240t) + 3 cos(2120t).

The signal is sampled (s = 2280s1) to produce the sequence v(n). For practical

purposes (delay, complexity) the sequence is limited to L samples. M is the length ofthe DFT.

0 0.005 0.01 0.015 0.02 0.025 0.033

2

1

0

1

2

3

4

5

a) one period L=7, DFTlength M=7, sampling frequency s=2280 s

1

Time [s]

Amplitude

0 1 2 3 4 5 6 70

2

4

6

8

10

12

DFT VM

() and Fourier transform V(ej

)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2



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SYSTEM THEORY

0 0.005 0.01 0.015 0.02 0.025 0.033

2

1

0

1

2

3

4

5

b) one period L=7, DFTlength M=14 (zero padding), sampling frequency s=2280 s

1

Zeit [s]

Amplitude

0 2 4 6 8 10 12 140

2

4

6

8

10

12

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

Problem 7 (FFT)

(a) Even indexed sequence: ve(n) = [v(0), v(2), v(4), v(6)],

Odd indexed sequency: vo(n) = [v(1), v(3), v(5), v(7)]



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SYSTEM THEORY

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.13

2

1

0

1

2

3

4

5

c) four periods L=28, DFTlength M=28, sampling frequency s=2280 s

1

Zeit [s]

Amplitude

0 5 10 15 20 250

10

20

30

40

50

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

DFT expressions of the sequences:

Ve,() =DF T{ve(n)} =4

n=0

veWn4

=4

n=0

veej 24 n

Vo,() =DF T{vo(n)} =4

n=0

voWn4

=4

n=0

voej 24 n

(b) DFT ofv(n) is given by

V8() =M/21n=0

v(2n) ej 28 2n +M/21n=0

v(2n + 1) ej 28 (2n+1)



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SYSTEM THEORY

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.13

2

1

0

1

2

3

4

5

d) four periods L=28, DFTlength M=56 (zero padding), sampling frequency s=2280 s

1

Zeit [s]

Amplitude

0 10 20 30 40 500

10

20

30

40

50

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

ej28 (2n) =ej

24 (n) =Wn4

V8() =

M/21n=0

v1(n) Wn4 Ve,()

+ W8

M/21n=0

v2(n) Wn4 Vo,()

(c)

Direct DFT:

Complexity: M2 = 64

Modified method:

Complexity: 2 (M/2)2 + M = 40

(d) Yes, the complextiy can be further reduced by applying the same principle. Weget,



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SYSTEM THEORY

0 0.01 0.02 0.03 0.04 0.053

2

1

0

1

2

3

4

5

e) two periods L=14, DFTlength M=15 (zero padding), sampling frequency s=2280 s

1

Time [s]

Amplitude

0 5 10 150

5

10

15

20

25

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

V8() =1

n=0v1,1(n)W

n2 +W

28

1n=0

v1,2(n)Wn2 +

1n=0

v2,1(n)Wn2 +W

28

1n=0

v2,2(n)Wn2



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SYSTEM THEORY

0 0.01 0.02 0.03 0.04 0.053

2

1

0

1

2

3

4

5

f) two periods L=14, DFTlength M=21 (zero padding), sampling frequency s=2280 s

1

Time [s]

Amplitude

0 2 4 6 8 10 12 14 16 18 200

5

10

15

20

25

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

Problem 8 (FFT)

Given x(n) =ej(/M)n2

WhenM is even, X() = M ej/4

ej(/M)2



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SYSTEM THEORY

0 0.02 0.04 0.06 0.08 0.13

2

1

0

1

2

3

4

5

g) L=30, DFTlength M=30, sampling frequency s=2280 s

1

Time [s]

Amplitude

0 5 10 15 20 25 300

10

20

30

40

50

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

y(n) =ej(/M)n2

Y() =2M1n=0

y(n)ej(2/2M)n

=M1

n=0 ej(/M)n2ej(2/2M) +

2M1

n=M ej(/M)n2ej(2/2M)n

=M1n=0

ej(/M)n2

ej(2/2M) +N1l=0

ej(/M)(l+M)2

ej(2/2M)(l+M)|n= l+ M

=M1n=0

ej(/M)n2

ej(2/2M) +N1l=0

ej(/M)(l+M)2

ej(2/2M)(l+M)

=M1n=0

ej(/M)n2

ej(2/2M) +N1l=0

ej(/M)(l+M)2

ej(2/2M)(l+M)

= (After simplification)

=

M

1n=0

ej(/M)n2ej(2/2M) + (1)M

1n=0

ej(/M)n2ej(2/2M)

= (1 + (1))M1n=0

ej(/M)n2

ej(2/M)/2



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SYSTEM THEORY

0 0.01 0.02 0.03 0.04 0.05 0.063

2

1

0

1

2

3

4

5

h) L=15, DFTlength M=30 (zero padding), sampling frequency s=2280 s

1

Time [s]

Amplitude

0 5 10 15 20 25 300

5

10

15

20

25

30

DFT VM


)

Amplitude

0 pi/2 pi 3pi/2 2pi

0 40 80 120 160 200 240 280

Frequency , , /2

Y() =

2X(/4) is even

0 is odd



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SYSTEM THEORY

v(0)v(0)

v(1)v(2)

v(3)

v(4)

v(5)

v(6)

v(7)

V8(0)

V8(1)V8(2)

V8(3)

V8(4)

V8(5)

V8(6)

V8(7)

DFT

Order 8

v(0)v(0)

v(2)v(4)

v(6)

v(1)

v(3)

v(5)

v(7)

V8(0)

V8(1)V8(2)

V8(3)

V8(4)

V8(5)

V8(6)

V8(7)

DFT

DFT

Order 4

Order 4

W08W18

W2

8W38

W48W58W68W

68

W78

Problem 9 (FFT)

(a) Since x(n) is real valued x(n) =x(n)

Compute theM-point DFT X()

X() =M1

n=0x(n)ej2/Mn

=M1

n=0

x(n)ej2/Mn

=M1

n=0

x(n)ej2/Mnej2/MM

=X(M )

XR() =XR(M ) & XI() =XI(M ) (3)



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SYSTEM THEORY

(b) Given real valued sequencesx1(n) X1() and x2(n) X2()

Complex sequence g(n) =x1(n) +jx2(n)G() =G

R() +jG

I()

G() =X1() + X2()

= (X1ER() +jX1OI()) +j(X2ER() +jX2OI())

= (X1ER() X2OI()) Real part GR()

+j(X1OI() + X2ER()) Imaginary part GI()

Even and odd parts ofG()

GER() = 1/2{

GR() + GR(M)}

=X1()

Similarly,

GOR() = X2OI()GEI() =X2ER()GOI() =X1OI()

And finally,

X1() =GER() +jGOI()X2() =GEI() jGOR()



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SYSTEM THEORY

Problem 10 (signal flow graph)

The signal flow graph in figure of the problem describes the input-output relationshipofv(k) and y(k).

X(z) = V(z) + 0.5 X(z)z1

X(z) = V(z)

1 0.5z1Y(z) = 2 (3V(z) + X(z)) + 2X(z)z1

Y(z) = 6V(z) + 2V(z)

1 0.5z1 + 2V(z)z1

1 0.5z1

Y(z) = 6V(z) 3V(z)z1 + 2V(z) + 2V(z)z1

1 0.5z1

Y(z) = 8 z11 0.5z1 V(z)

h(k) = 8 (0.5)k1(k) (0.5)k11(k 1)



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SYSTEM THEORY

Problem 11 (signal flow graph)Bottom figure in the question:

Y(z) =V(z)z1

+ 2rcos(0)Y(z)z1

r2

Y(z)z2

H(z) =Y(z)

V(z)=

z1

1 2rcos(0)z1 + r2z2

Top figure in the question:

X(z) = V(z) r2sin2(0)Y(z)z1W(z) = X(z) + rcos(0)W(z)z

1

W(z) = X(z)

1 rcos(0)z1 =V(z) r2sin2(0)Y(z)z1

1 rcos(0)z1

Y(z) = W(z)z1 + rcos(0)Y(z)z1 = W(z)z

11 rcos(0)z1

Y(z) = (V(z) r2sin2(0)Y(z)z1)z1

(1 rcos(0)z1)2

Y(z)((1 rcos(0)z1)2 + r2sin2(0)z2) =V(z)z1

H(z) =Y(z)

V(z)=

z1

1 2rcos(0)z1 + r2z2



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SYSTEM THEORY

Problem 12 (round-off effects in digital filters)

a)

Y(z) = V(z) +1

4Y(z)z1 =

V(z)

1 14 z1 =

V(z)z

z 1/4V(z) = .5

z

z 1Y(z) =

12 z

2

(z 1/4)(z 1) =12 z

2

z2 54 z+ 1/4= 1/2 +

58 z 1/8

(z 1/4)(z 1)

= 1/2 1/24z 1/4+

2/3

z 1invers transform

y(n) = 1/2 o(n) 1/24 (1/4)n11(n 1) + 2/3 (1)n11(n 1)y(k)|n = 2/3

b)unquantized case (working from the difference equation):

y(n) = v(n) + 1/4 y(n 1)y(0) = 1/2, y(1) = 1/2 + 1/4 1/2 = 5/8, y(2) = 1/2 + 1/4 5/8 = 21/32,y(3) = 85/128, y(4) = 341/512, y(5) = 1364/2048

quantized case (working from the difference equation):

y(n) = v(n) + Q[1/4 y(n 1)]y(0) = 1/2,

y(1) = 1/2 + Q[1/4 1/2] = 1/2 + Q[1/8] = 5/8y(2) = 1/2 + Q[1/4 5/8] = 1/2 + Q[5/32] truncation!= 1/2 + 1/8 = 5/8y(3) = 5/8 . . .

c)direct form II:

H(z) = 1 + z1

1 14 z1

V(z) = 1/2

1 + z1

Y(z) = H(z) V(z) = 1/21 14 z1

invers transform

y(n) = 1/2 (1/4)k



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SYSTEM THEORY

unquantized case:y(n)|n= 0

quantized case (working from the difference equation):

y(n) = v(n) + v(n 1) + Q[1/4 y(n 1)]y(0) = 1/2 + 0 + 0 = 1/2,

y(1) = 1/2 + 1/2 + Q[1/4 1/2] = 1/8,y(2) = 1/2 1/2 + Q[1/4 1/8] = 0 . . .



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SYSTEM THEORY

Problem 13 (round-off effects in digital filters)

Let h(n), h1(n), and h2(n) represent the unit sample responses corresponding to thesystem functions H(z), H1(z), and H2(z), respectively. It follows that

h1(n) = (1/2)n1(n)

h2(n) = (1/4)n1(n)

H(z) = H1(z) H2(z)=

1

1 0.5z1 1

1 0.25z1 = z2

z2 34 z+ 1/8polynom division

= 1 +34 z 1/8

z2

3

4 z+ 1/8

= 1 +34 z 1/8

(z 1/2)(z 1/4)partial fraction expansion

= 1 + A

(z 1/2)+ B

(z 1/4)

A = limz1/2

34 z 1/8(z 1/4) =

3/8 1/81/4

= 1

B = limz1/4

34 z 1/8(z 1/2) =

3/16 1/81/4 = 1/4

H(z) = 1 + 1

z 1/2 1/4

z 1/4

inverse transform :h(n) = 0(n) + ( 1/2)

n11(n 1) 1/4 (1/4)n11(n 1)= 0(n) + 2 (1/2)n1(n 1) (1/4)n1(n)= (2 (1/2)n (1/4)n)1(n)

first cascade realization:

v(n)

12

14

z1 z1

y(n)

e1(n) e2(n)

Figure 1: Cascade system realization 1



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SYSTEM THEORY

Qa1(z) lecture eq.: (4.39)

= 1/2 Qa1(z)z1 + E1(z) = E1(z)1

1

2z

1

Qa2(z) lecture eq.: (4.39)

= 1/4 Qa2(z)z1 + E2(z) + E1(z)1 12 z1

= E2(z)

1 14 z1 corresponds to H2(z)

+ E1(z)

(1 12 z1)(1 14 z1) corresponds to H(z)

So the quantization noise variance 2qa at the output of the first cascade realization canbe calculated with help of the impulse responses h2(n) and h(n) as

2

qa =

2

e

n=0 h2(n) + 2

e

n=0 h22(n) , 2

e denoting the variance ofe1/2(n)

For the second cascade realization, the variance 2qb is calculated in the same way andgives

2qb = 2e

n=0

h2(n) +n=0

h21(n)

n=0h21(n) =

1

1 1/4 = 4/3n=0

h22(n) = 1

1 1/16 = 16/15n=0

h2(n) = 4

1 1/4 4

1 1/8+ 1

1 1/16= 1.83

Therefore,

2qa = 2.90 2e2qb = 3.16 2e

and the ratio of noise variances is

2qb2qa

= 1.09

Consequently, the noise power in the second cascade realization is 9% larger than in thefirst realization.



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SYSTEM THEORY

Problem 14 (digital filter design)

Determine the unit sample response hi of a linear-phase FIR filter of length L = 4 forwhich the amplitude frequency responseH0() at = 0 and = /2 is specified as

H0(0) = 1, H0(/2) = 1/2.

Even lengthL type-2 or type-4 linear phase system.H0(0) = 1 no type-4 linear phase system (H0(0) = 0!).

Type-2 linear phase system:

H0() = 2

L/21i=0

hi cos

L 1

2 i

= 21

i=0

hi cos

3

2 i

.

At = 0,

1 = 21

i=0hi cos(0)

1/2 = h0+ h1,

at =/2,

1/2 = 21

i=0

hi cos

3

2 i

/2

= 2 (h0 cos( 3

4 ) + h1 cos(

4))

1/4 = 1

2h1 1

2h0.

Solving these equations, we get

h0 = 0.073

h1 = 0.427

and with symmetry

h2 = h1

h3 = h0



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SYSTEM THEORY

Problem 15 (digital filter design)(a)

Hd(ej) = (1

2

1())ej(/2) for


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SYSTEM THEORY

(d) The delay is L12 = 211

2 = 10 samples. L is odd, it is therefore a type III system.

hi

i

0.5

0.5

0

0 5 10 15 20

|H(ej

)|

0.5

00

1

Figure 2: Scetches to problem 15 part d)

(e) The delay is L12 = 201

2 = 9.5 samples. L is even, it is therefore a type IV system.

hi

i

0.5

0.5

0

0 5 10 15

|H(ej)|

0.5

00

1

Figure 3: Scetches to problem 15 part e)



30


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SYSTEM THEORY


From the given equations we havec(k) = 2 h(Sk), 1 k S. For type III linear-phasefilters h(S) = 0 andL is odd.

H03( ) =Si=1

ci sin(( )i)

= Si=1

ci sin(k)cos(i) =Si=1

(1)i+1ci sin(i).

Thus H03() =H03( ) impliesSi=1

ci sin(i) =Si=1

(1)i+1ci sin(i),

or equivalently,Si=1

(1 (1)i+1)ci sin(i) = 0 which in turn implies that ci = 0 fori= 2, 4, 6, . . ..

ci= 2 h(S i), 1 i Scan be rewritten as hi= 1/2 cSi. As S is evenhi= 0 for ieven.



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SYSTEM THEORY


(a) From the given equations, we get

1 = 1 101/20 = 1 100.005 = 0.01144602 = 10

2/20 = 101.75 = 0.01778279

N = 10log10(12) 132.324

=10log10(0.00020355796) 13

2.324 2( 2kHz1.8kHz12kHz )= 98.2730

Since the number must be an integer, we round up the value yielding N = 99.As the length L = N+ 1 is even, a type II FIR filter can be designed to meetthe specifications. A type I filter can be designed by increasing the order by 1 to

N= 100.(b) Note that the width of the transition bands are not equal. We therefore use the

width of the smallest transition band to compute the order N.

1 = 2

FP1 FS1

FT

= 0, 031416

2 = 2

FS2 FP2

FT

= 0, 062832

N = 10log10(0.002 0.001) 132.324 2( 0.35kHz0.3kHz10kHz )

= 602.51

The order of the required FIR filter is N= 603. As the length L = N+ 1 is even,a type II FIR filter can be designed to meet the specifications.



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SYSTEM THEORY


(a) In the impulse invariance design, the poles transform as zi= esiT and we have the

relationship1

s si 1

1 esiTz1Therefore,

Ha(s) = 2

s + 0.1 1

s + 0.2

In this case the solution is unique, since ha(t) is real, and the poles are both on the-axis in the s-plane. Due to the periodicity of z = ej a more general answer for a

complex impulse response ha(t) would be

Ha(s) = 2

s + (0.1 +j 2nT ) 1

s + (0.2 +j 2mT )

where n and m are integers.(b) Using the inverse relationship for the bilinear transform

z =1 + (T /2)s

1 (T /2)swe get

Ha(s) = 2

1 e0.2 1(T/2)s1+(T/2)s 1

1 e0.4 1(T/2)s1+(T/2)sT=2=

2

1 e0.2 1s1+s 1

1 e0.4 1s1+s=

2(s + 1)

s(1 + e0.2) + (1 e0.2) s + 1

s(1 + e0.4) + ( 1 e0.4)=

2

1 + e0.2 s + 1

s + 1e0.2

1+e0.2

11 + e0.4

s + 1s + 1e

0.4

1+e0.4

Since the bilinear transform does not introduce any ambiguity, the representation isunique.



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SYSTEM THEORY


(a) Recall that =T, Tdenoting the sampling period. So the specifications for thecontinuous-time signal are

0.89125 |H(ej)| 1, 0 || 0.2/T,|H(ej)| 0.17783, 0.3/T || /T.

(b) At the passband edge, we have

|H(j0.2/T)|2 = 11 + ( 0.2cT)

2N

!= 0.891252. (4)

At the stopband edge, we have

|H(j0.3/T|2 = 11 + ( 0.3cT)

2N = 0.177832. (5)

We can solve these two equations for N and cTand we get

N = 5.8858

cT = 0.70474

to solve the equations above exactly. Rounding up to the next integer N= 6 andinsertingNin (1) we getcT = 0.7032 to meet the specifications of the continuous-

time filter for the passband edge exactly. Then the stopband edge specificationsof the continuous-time filter are exceeded.



34


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SYSTEM THEORY


Given:

- Filter C: Continous-time IIR-Filter with System fincrionH(s)

- Filter B: Stable discrete-time filterH(z) derived through bilinear transform fromFilter C

Question: Can Filter B be an FIR-Filter?

-

Excurs: IIR-Filter(System function):

H(z)IIR = Y(z)X(z)

= m=0 bzn=0 az

relation to the Laplace-domain: zi= es,i/0,i contains poles and zeros

FIR-Filter(System function):

H(z)FIR=m=0

bz

contains m zeros andm-th order pole at z = 0

Filter is stable, if the poles are located within the unit circle (z-domain) or within theleftLaplace-(s)-plane.

-

In this case, filter C is an IIR-Filter, this means, it will have poles, that are nonzero,because otherwise it woukd be an FIR-Filter. Therefore, and due to the fact, that thebilinear transform is unique in converting from s-domain to z-domain, also filter B willhabe poles. Thus, filter B cannot be a FIR-Filter.



35


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SYSTEM THEORY


First we determine the values for 1 and 2. They are used in the magnitude represen-tation of filter specifications.

1 = 1 101/20 =.1087490618662 = 10

40/20 =.01

1 is related to by

(1 1)2 = 11 + 2

2 = 1

(1

1)2

1

= 0.508847139905

and is given by

=

1

22 1 = 99, 995

The normalized frequencies for the passband edge and stopband edge in the digitaldomain are given by

p = 2 40/240s = 2 60/240

As the bilinear transform warps the frequency scale, we have to determine the passband

edge and stopband edge in the analog domain by inverse transformation of the valuesfor the digital domain:

p = 2/T tan(p/2)s = 2/T tan(s/2)

c/s = tan(p/2)/tan(s/2) = 1.73205080758

This leads to the following filter length:

Butterworth filter: Nmin log10(/)log10(s/p)

= 9.613 10

Chebyshev filter: Nmin log10((1 22+ 1 22(1 + 2))/(2))

log10(s/c+

(s/p)2 1) 5.212 N= 6

Elliptic filter: Nmin K(p/s)K(

1 (/)2)K(/)K(

1 (p/s)2)

= 3.2 N= 4



36


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SYSTEM THEORY

0 0.2 0.4 0.6 0.8 1100

80

60

40

20

0

20 log10

(H(ej

))

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

|H(ej

)|

0.1 0.2 0.3

4

3

2

1

0

Passband detail

0.5 0.6 0.7 0.8 0.9

120

100

80

60

40

Stopband detail

Figure 4: Butterworth MATLAB: [b_b,a_b]=butter(10,4/12);

0 0.2 0.4 0.6 0.8 1100

80

60

40

20

0

20 log10

(H(ej

))

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

|H(ej

)|

0 0.2 0.4

8

6

4

2

0

Passband detail

0.4 0.6 0.8 1

60

50

40

30

20

Stopband detail

Figure 5: Chebyshev 1 MATLAB: [b_c1,a_c1]=cheby1(6,1,4/12);



37


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SYSTEM THEORY

0 0.2 0.4 0.6 0.8 1100

80

60

40

20

0

20 log10

(H(ej

))

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

|H(ej

)|

0.1 0.2 0.3 0.4

1

0.5

0

0.5

Passband detail

0.4 0.6 0.8 1

70

60

50

40

Stopband detail

Figure 6: Chebyshev 2 MATLAB: [b_c2,a_c2]=cheby2(6,40,6/12);

0 0.2 0.4 0.6 0.8 1100

80

60

40

20

0

20 log10

(H(ej

))

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

|H(ej

)|

0.1 0.2 0.3 0.4

10

8

6

4

2

0

Passband detail

0.4 0.6 0.8 1

55

50

45

40

35

30

25

20

Stopband detail

Figure 7: Elliptic MATLAB: [b_e,a_e]=ellip(4,1,40,4/12);



38


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SYSTEM THEORY

Problem 22 (multirate digital signal processing)

The output y(n) = x(n) if no aliasing occurs as result of downsampling. That is,X(ej) = 0 for /3 || .

(a) x(n) = cos(n/4). X(ej) has impulses at =/4, so there is no aliasing.y(k) =x(k).

(b) x(n) = cos(n/2). X(ej) has impulses at = /2, so there is aliasing.y(n) =x(n).

(c) x(n) = ( sin(n/8)n )2 = 1/64 ( sin(n/8)n/8 )2 = 1/64 (sinc(n/8))2. The spectrum

of sinc(n/8) is a rectangular in the range of/8 /8. The squaredfunction leads to a triangular spectrum with double bandwidth (multiplication intime domain convolution in frequency domain). So the highest signal frequencyis|max| =/4< /3 and no aliasing will occur.



39


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SYSTEM THEORY

Problem 23 (multirate digital signal processing)

We can analyze the system in the frequency domain:

X(ej) Y(ej)

X(e2j)

2 2X(e2j)H1(e

j)

H1(ej)

Y(ej) is X(e2j) H1(ej) downsampled by 2:

Y(ej) = 1/2

X(e2j/2)H1(ej/2) + X(e2j(2)/2)H1(e

j(2)/2)

= 1/2 X(ej)H1(ej/2) + X(ej(2))H1(ej(2))= 1/2

H1(e

j/2) + H1(ej(

2))

X(ej)

= H2(ej)X(ej)

H2(ej) = 1/2

H1(e

j/2) + H1(ej(2))



40


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SYSTEM THEORY

Solution to Problem 24 (multirate digital signal processing)

(a) h(n) = 0 for|n| > (RL 1). For a causal system we have a delay by RL 1samples.

(b) General interpolation condition:

h(0) = 1

h(nL) = 0, n= 1,2, . . .

(c)

y(k) =

RL1k=(RL1)

h(k)v(n k) =h(0)v(n) +RL1k=1

h(k)(v(n k) + v(n + k))

This requires only RL 1 multiplications (assuming h(0) = 1).(d) Show, that only 2R Multiplications per output sample are required.

y(n) =n+RL1

k=n(RL1)v(k)h(n k)

Due to part b) it has been shown, that only h(0) = 1, and h(Ln) = 0 for n =+

1,+

2,... This is a general condition related to the interpolation (include (L 1)zeros).

Ifn = mL (man integer), then we dont have any multiplications since h(0) = 1and the other non-zero samples of v(n) hit at the zeros h(n). Otherwise theimpulse response spans 2RL 1 samples of v(n), but only 2R of these are non-zero. Therefore, we have 2R multiplications in total.



41


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SYSTEM THEORY

10 5 0 5 101

0.5

0

0.5

1

n

Amplitude v(n)

10 5 0 5 101

0.5

0

0.5

1

n

Amplitude h(0n)

10 5 0 5 101

0.5

0

0.5

1

n

Amplitude h(1n)

Solution to Problem 25 (multirate digital signal processing)

Steps to do:

a)b)[c) Polyphase decomposition of decimation filters -> moving of all decimation factors

through alls branches before the filter decomposition (efficient structure)

d)e) M= 2, L = 3 -> decimator and interpolator can be changed/turned

f) Polyphase decomposition of interpolator filter -> moving of interpolator factorthrough all branches (efficient structure) (z3 > z)



42


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SYSTEM THEORY

1

Y(z) X(z)

3

3

3

3

3

3

3

3

3

3

3

3

3 2

2

2

2

2

2

2

2

2

2

2

G(z)

G0(z)

G0(z)

G0(z)

G0(z)

G1(z)

G1(z)

G1(z)

G1(z)

G1(z)

G00(z)

G01(z)

G02(z)

G10(z)

G12(z)

z1

z1

z1

z1

z1z1

z1z1

z1

z1

z3

z2

z

a)

b)

c)

d)

e)

f)

adsp ws1415 solutions

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