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Signal Processing

AssignmentDonald

CarrMay 18, 2005

Question

1Common

factsMy -3 dB frequency was : 2 kHz

= 2 p 2000c

The circuits were taken from the Electronic Filter Design Handbook (Williams,

1981). Standard values were acquired for the respective configurations and de-

normalised accordingly using the FSF and Z adjustment.

CLCR = 600

s

C = 0.132 F1

L = 95.5

mL2

C = 0.132 F3

R = 600o

R Ls 2

CC 21

Ro

Figure 1: The p CLC circuit

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T

LCLR = 600

s

L = 0.048mL

1

Standard values were acquired C = 0.265 F2

L = 0.048mL

3

R = 600o

L L1 3

Rs

C R2 o

Figure 2: The T LCL circuit

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4/39

- 10 = I + ( R + sL + 1 )I

1 o 3 2sC sC2 2

R + sL + I U1 - 1

=s 1 1sC

sCR + sL + I 0- 1 1

o 3 2sC

sC

- 1I R + sL + U1 - 1

1 s 1= sC

sCI R + sL + 0- 1 1

2 o 3sC

sC

UI =

2sC R R + s CL R + s CL R + s CL L + sL + sL + R + R2 2 3

2 s o 3 s 1 o 1 3 1 3 s o

But : y = i R2 o

Y = I R2 o

URoY =

sC R R + s CL R + s CL R + s CL L + sL + sL + R + R2 2 32 s o 3 s 1 o 1 3 1 3 s o

Since : y = h * u

Y = HU

YH =U

RoH =

sC R R + s CL R + s CL R + s CL L + sL + sL + R + R2 2 32 s o 3 s 1 o 1 3 1 3 s o

Since L = L = L, R = R = R, C1 3 s o

R

H = CL

2

s + s + ( + )s +2R 2 R 2R3 2 2L L

CL L C2 2

After substituting in component values and reducing in terms ofc

3

c

H(s) = 2s + 2 s + 2 s +3 2 2 3

c c c

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state variable

technique

di1 = a i + a v + a i + b u

1 1 1 1 2 2 1 3 3 1 1dt

dv 2= a i + a v + a i + b u

2 1 1 2 2 2 2 3 3 2 1dt

di3 = a i + a v + a i + b u

3 1 1 3 2 2 3 3 3 3 1dt

y = c i + c v + c i + d u1 1 1 1 2 2 1 3 3 1 1

dx= [ A ]x + [ B ]u

dt

sX = [ A ]X + [B ]U

( s [I] - [A ])X = [ B ]U

X = ( s [I] - [A ]) [B ]U- 1

and y = [ C]x + [ D ]u

Y = [ C]X + [ D ]UC]adj (s [I ] - [A ]) [B ]UT

Y = [ D ]Udet(s [I] - [A ]) + [

Y C]adj (s [I ] - [A ]) [B ]TBut H = H = [ D ]

U det (s [I ] - [A ]) + [

Where :

s + 0R 1 1LL L 1

(s[I]- [A ]) = s , [C] = 0 0 R , [B ] = , [D ] = 0- 1 1 0oC C

0 s +- 1 R 0L L

R R R R R + Ro s o s o s3 2det(s [I] - [A ]) = s + ( + )s + ( 1 + 1 + )s +

L L L C L C L L L L C3 1 3 2 1 2 3 1 1 3 2

Roand [ C]adj (s [I] - [A ]) [B ] =TCL L

1 3

Ro

CL

LH(s ) = 1 3s + ( + )s + ( + + )s +R R 1 1 R R R + R3 2o s o s o s

L L L C L C L L L L C3 1 3 2 1 2 3 1 1 3 2


3

c

H(s ) = 2s + 2 s + 2 s +3 2 2 3

c c c

It was very reassuring that both the matrix loop equation and state variable

technique produced the same transfer function for their common circuit.

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R Ls 2

1 2

CC 21 Ro

Figure 4: The p CLC circuit

Question

3Matrix node

equationNode 1

i = 0 = i + i + iRs L 2 C 1

v dv tRs0 = + C + 1 vdt + i (0)1R dt L

s 2 0

u - v d (0 - v ) t1 10 = + C + 1 ( v - v ) dt + i(0)

1 2 1R dt Ls 2 0

-du d v d ( v )2110 = - C + 1 (v - v )dt dt

1 2 1R dt Ls 2

s( U - V )1= > 0 = - s C V + 1 ( V - V ) (Laplace)2

1 1 2 1R L

s 2

U - V10 = - sC V + 1 (V - V )

1 1 2 1R sL

s 2

Rs s

U = (1 + sC R + ) V - R V 1 s 1 2sL sL

2 2

Node 2

i = 0 = i + i + iL 2 C 2 Ro

tv dvRo0 = + C + 1 vdt + i(0)

2R dt Lo 2 0

v d( v ) 1 t2 20 = + C (v - v ) dt - i (0)

2 1 2R dt - L

o 2 0

d ( v ) 1dv 2220 = + C ( v - v )dt

2 1 2R dt - Lo 2

sV 12= > 0 = + s C V - (V - V ) (Laplace)2

2 2 1 2R Lo 2

V 120 = + sC V - (V - V )

2 2 1 2R sL

o 2

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- 10 = V + ( 1 + sC + 1 )V

1 3 2sL sL R2 2 o

(1 + sC R + )R -R V Us s1 s 1 =sL

sL

2 2

V 0( + sC + )- 1 1 13 2s

LsL

R2 2 o

- 1sC R + )R -RV Us s

1 1 s= (1 + sL

sL

2 2

V ( + sC + ) 0- 1 1 12 3s

LsL

R2 2 o

But : y = v2

Y = V2

UY = 1

det sL2

where : sL det =2

c R C L (s + ( + )s + ( + + )s + +3 1 1 2 1 1 1 11 s 3 2 C R C R C R C R C L C L R C C L

1 s 3 o 1 s 3 o 3 2 1 2 o 1 3 2

)1R C C L

s 1 3 2

But : Y = H U

1

c R C LH = 1 s 3 2(s + ( + )s + ( + + )s + + )1 1 1 1 1 1 13 2

C R C R C R C R C L C L R C C L R C C L1 s 3 o 1 s 3 o 3 2 1 2 o 1 3 2 s 1 3 2


3

c

H(s ) = 2s + 2 s + 2 s +3 2 2 3

c c c

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state variable

technique

di1 = a i + a v + a i + b u

1 1 1 1 2 2 1 3 3 1 1dt

dv 2= a i + a v + a i + b u

2 1 1 2 2 2 2 3 3 2 1dt

di3 = a i + a v + a i + b u

3 1 1 3 2 2 3 3 3 3 1dt

y = c i + c v + c i + d u1 1 1 1 2 2 1 3 3 1 1

Identical algebra to that covered in Question 2 : state variable

C]adj (s [I ] - [A ]) [B ]TH = [ D ]

det(s [I ] - [A ]) + [

s + 01 1 1R C C R C

s 1 1 s 1

(s[I]- [A ]) = s , [C] = 0 0

1

, [B ] = , [D ] = 0- 1 1 0L L

2 20 s + 0- 1 1C Ro

C3 3

det(s [I] - [A ]) =

s +( + )s + ( + + )s + + +3 1 1 2 1 1 1 1 1 1R C R C C L C L R R C C R R R R C L R R C L

o 1 s 3 1 2 3 2 o s 1 3 o s s o 1 2 s o 3 2

and [ C]adj (s[I] - [A ]) [B ] = 1TR C L C

s 1 2 3

H (s ) =

1

Rs C L C1 2 3

s +( + )s +( + + )s + + +3 1 1 2 1 1 1 1 1 1Ro C Rs C C L C L Ro Rs C C Ro Rs Rs Ro C L Rs Ro C L1 3 1 2 3 2 1 3 1 2 3 2

After substituting in component values and reducing in terms of c3

c

H(s) = 2s + 2 s + 2 s +3 2 2 3

c c c

Again the separate derivations converged on the same transfer function, and

corroborated the earlier transfer function garnered from the T configuration

topology. This consistency verified the inherent characteristics of the filter which

are independent of the topology we choose to adopt.

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Question

4Characteristic polynomial : s + s 2 + s2 +3 2 2 3

c c c3

d3 y d2 y d ySystem di erential equation : + 2 + 2 + y = U2 3 cc c cdt dt dt 2

Question

5There are no zeros since there are no s terms in the numerator. We need to dis-

cover the poles of the transfer function, so we simply factorise the denominator.

1

s + s 2 + s 2 +3 2 2 3c c c

1

(s + )(s + s + )2 2c c c

1v v

(s + )(s + + j )(s + -j

)3 3c cc c c2 2 2 2

eigenfrequencies : s = -c

orv

3c

s = - jc2 + 2

orv

3cs = - -

jc2 2

j

c

s

Figure 5: Pole Zero diagram

Since the poles are all on the left hand side of the imaginary axis, we know

from our notes that the filter is stable.

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Question

6

3

c

H(s) = 2

s + 2 s + 2 s +3 2 2 3

c c c

3

c

H(s) = 2v v(s + )(s + + j )(s + -

j

)3 3c cc c c2 2 2 2

3

c

H(j ) = 2v v(j + )(j + - j )(j + + j )3 3c c

c c c2 2 2 2

Magnitude is given by :

3

| |c|H (j ) | = 2v v

|(j + ) || (j + - j ) || (j + + j ) |3 3c cc c c2 2 2 2

Phase is given by :v v

3 3c cH(j ) = - ( j + ) - ( j + - j ) - ( j + j )

c c c2 2 2 + 2

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3rd o rd e r B u tt er wo rt h h ig h p as s f ilt er wc = 2 k h z

1.2

1

0.8

0.6

0.4

0.2

0

-2 5000 -2 0000 -1 5000 -10 000 -50 00 0 500 0 1000 0 1500 0 2000 0 2500 0

Fr eq uen cy [r ad s]

Figure 6: excel : Normalised gain vs frequency

25 0

20 0

15 0

10 0

5 0

0

-2 500 0- 200 00-1 5000 - 10 0 00 -5 000 0 50 0 010 000 150 00 2 0 00 0 25 0 00

-5 0

- 10 0

- 15 0

- 20 0

- 25 0

fre que nc y [r a ds]

Figure 7: excel : phase vs frequency

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1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0

- 2 - 1.5 - 1 -0 .5 0 0.5 1 1.5 2

freq uenc y [r ads ] x 10 4

Figure 8: matlab : gain vs frequency

200

150

100

50

0

- 50

- 100

- 150

- 200-2 -1. 5 - 1 -0.5 0 0.5 1 1. 5 2

fr equenc y [r ads ]x 10 4

Figure 9: matlab : phase vs frequency

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Question 8 wc 12566.37061

h = 100wc/2 = 6283.185307

root(3)/2*jwc = 10882.7961854053i

p1 p2 p3

wc -wc/2 - j(root(3)/2*wc) -wc/2 + j(root(3)/2*wc)

-12566.37061 -6283.18530717959-10882.7961854053i -6283.18530717959+10882.7961854053i

alpha 9.92201E+11

residues

k1 6283.18530717961

k2 -3141.5926535898+1813.79936423422i

k3 -3141.5926535898-1813.79936423422i

Figure 10: Excel calculated residues

4/23/05 2:10 PM MATLAB Command Window 1 of 1

alpha =

9.9220e+011

K1 =

6.2832e+003

K2 =

-3.1416e+003 +1.8138e+003i

K3 =

-3.1416e+003 -1.8138e+003i

Figure 11: Matlab calculated residues

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Question

9The Laplace transform of

K e + K e + K e = h ( t)a p t p t p t 1 2 3v v 1 2 3(s + )( s + c + j )( s + c -j )3 3

c c c2 2 2 2

Where :

K =

6283.185307179611

K =

-3141.5926535898+1813.79936423422i2

K = -3141.5926535898-1813.79936423422i

3

p =-12566.37061

1

p = -6283.18530717959-10882.7961854053i

2

p =-6283.18530717959+10882.7961854053i

3

h (t) = K e + K e + K ep t p t p t1 2 31 2 3

h (t) = K e + |K|e ( e e + e e )1 2 5 6 6 .3 7 0 6 1t - 62 8 3 . 1 8 5 3 0 7 1 7 959

t i K -i 1 0 8 8 2.7 9 6 1 8 5 4 0 53

t i K i 1 0 8 8 2 .7 9 6 1 8 5 4 0 53

t2 3

1

Since K = K (Conjugate pair)2 3

h (t) = K e + |K|e ( e + e )1 2 5 6 6 .3 7 0 6 1t - 62 8 3 . 1 8 5 3 0 7 1 7 959

t - i ( K +1 0 8 8 2 .7 9 6 1 8 5 4 0 53

t) i( K +1 0 8 8 2 .7 9 6 1 8 5 4 0 53

t)3 3

1

h (t) = K e + 2 |K|e cos ( K +10882

.7961854053

t)1 2 5 6 6 .3 7 0 6 1t - 62 8 3 . 1 8 5 3 0 7 1 7 959

t

1 3

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3 0 0 0

2 5 0 0

2 0 0 0

1 5 0 0

1 0 0 0

50 0

0

0 0. 000 2 0. 0004 0. 00 0 60. 00 080 . 00 10.00 120 . 001 40 .0 0 160 . 001 80 .0 0 2

- 50 0

ti m e (s )

Figure 12: Excel impulse response

3000

2500

2000

1500

1000

500

0

-5000 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2

ti me [s ec onds ]x 10 - 3

Figure 13: MatLab impulse response

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Question

10Using the the bilinear transform within Matlab :

1 + T PP = i2i 1 - PT

i2

f [hz] P P Ps ample

1 2 3

48000 0.7685 0.8561 - 0.1975i 0.8561 +0.1975i8000 0.1202 0.1595 - 0.5663i 0.1595 +

0.5663i

Table 1: Matlab generated poles for the digital filters

This bilinear transformation shifts the frequency response of the resulting

filters. The higher the sampling frequency, the milder the resulting shift in

frequency response. The filter with f = 48kHz therefore basically retainss ampleits initial frequency characteristics, while the filter with f = 8kHz has its

s amplef dropped to 1700Hz.

c

Question11

(1 + z ) (1 -Z

z )- 1 N -M M - 1iH(z) = i =1

N (1 - P z )- 1ii =1

Where there are M = 0 zeros and N = 3 poles

(1 + z )- 1 3H(z) =

3 (1 - P z )- 1ii=1

Where the values of the poles are given in table 1 (above) and

MT (1 - Z )Ti

= K ( N -M i =1 22 ) N (1 - P )T

ii= 1 2

T 13= c 3

2 ( 2 ) 3 (1 - P )Tii =1 2

f requency

80000.056548000

0.000864

Table 2: values

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for f =48000Hz

sample

(1 + z )- 1 3H(z) =

4 8 0 0 0 (1 - 0 .7685 z )(1 - (0 .8561 - 0 .1975 i )z )(1 - (0 .8561 +

0

.1975 i)z )- 1 - 1 - 1

for f =8000Hz

sample

(1 + z )- 1 3H(z) =

8 0 0 0 (1 - 0 .1202 z )(1 - (0 .1595 - 0 .5663 i )z )(1 - (0 .1595 +0

.5663 i)z )- 1 - 1 - 1

Question

12There are three zeros at -1, divulged from the numerator term. The denominator

reveals three poles, demarcated by Xes on the graph

I m (z)

-11

R e (z)

Figure 14: 48000 Hz filter Pole Zero diagram

I m (z)

-11

R e (z)

Figure 15: 8000 Hz filter Pole Zero diagram

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Question

13

Similarly to question 6

Magnitude :

(1 + z ) |- 1 3|H (z) | = |

|(1 - P z ) || (1 - P z ) || (1 - P z ) |- 1 - 1 - 11 2 3

Where is taken from table 2 and P values are taken from table 1

Phase :

H(z) = S ( zeros ) - S ( poles )

H(z) = 3 (1 + z ) - (1 - P z ) - (1 - P z ) - (1 - P z )- 1 - 1 - 1 - 11 2 3

Both Matlab and excel were used in order to establish corroborative evidence

for the frequency response of the filter. I felt this was necessary since the shifting

inherent in the bilinear transform was not immediate obvious, and I was initially

concerned with my results.

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1. 2

1

0. 8

0. 6

0. 4

0. 2

0

-4 -3 -2 -10 1 23 4

the ta [r a ds]

Figure 16: Excel: 8000 Hz digital filter normalised gain

30 0

20 0

10 0

0

-4 -3 -2 -10 1 23 4

- 10 0

- 20 0

- 30 0

the ta [r a ds]

Figure 17: Excel: 8000 Hz digital filter phase response

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1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0-4 - 3 -2 -1 0 1 2 3 4

theta

Figure 20: Matlab : 8000 Hz digital filter normalised gain

300

200

100

0

-100

-200

-300-4 -3 - 2 - 1 0 1 2 3 4

theta

Figure 21: Matlab : 8000 Hz digital filter phase response

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1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0-4 - 3 -2 -1 0 1 2 3 4

theta

Figure 22: Matlab : 48000 Hz digital filter normalised gain

300

200

100

0

-100

-200

-300-4 -3 - 2 - 1 0 1 2 3 4

theta

Figure 23: Matlab : 48000 Hz digital filter phase response

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Question

14

(1 + z )- 1 3H(z) =

(1 - P z )(1 - P z )(1 - P z )- 1 - 1 - 1

1 2 3Y (z)

But : H(z) =X (z)

Y (z) (1 + z )- 1 3

X (z) = (1 - P z )(1 - P z )(1 - P z )- 1 - 1 - 11 2 3

Y (z) (1 - P z )(1 - P z )(1 - P z ) = X (z) (1 + z )- 1 - 1 - 1 - 1 31 2 3

Y (z) (1 - P z - P z - P z + P P z + P P z + P P z - P P P z ) =- 1 - 1 - 1 - 2 - 2 - 2 - 31 2 3 1 2 2 3 1 3 1 2 3

X (z) (1 + 3 z + 3 z + z )- 1 - 2 - 3

Using the inverse Z transform properties given on pg. 25 of the digital section

y - P y - P y - P y + P P y + P P y + P P y - P P P y ) =n 1 n- 1 2 n - 1 3 n - 1 1 2 n - 2 2 3 n - 2 1 3 n - 2 1 2 3 n - 3

(x + 3 x + 3 x + x )n n - 1 n - 2 n - 3

y = ( P + P + P )y - (P P + P P + P P )y + P P P y + (x + 3 x + 3 x + x )n 1 2 3 n - 1 1 2 2 3 1 3 n - 2 1 2 3 n - 3 n n - 1 n - 2 n - 3

y = ay - y + y + (x + 3 x + 3 x + x )n n - 1 n - 2 n - 3 n n - 1 n - 2 n - 3

This recurrence relation is common to both filters, with the coe cients of

the terms assuming di erent values depending on the sampling frequency.

1 1 1

x 1 ynnz z- 1 - 1

3 a

1z z- 1 - 1

-

3

1z z- 1 - 1

1

Figure 24: Signal ow diagram for both filters

The following coe cients were calculated in Matlab :

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f [Hz ] a s ample 8000 0.4392 0.3845

0.041648000 2.4808 2.08780.5933

Table 3: Matlab generated values for a , and

Question

15The unit sample responses of both digital filters corresponded strongly to the

original impulse response calculated for the analogue filter, with the impulse

response being clearly recognisable. The lower the initial sampling frequency,

the faster the emergence of the recognisable response was when observing the

discrete impulse response.

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0 .06

0 .05

0 .04

0 .03

0 .02

0 .01

0

0 10 2 030 405 0 60 70

-0 .01

n

Figure 25: Matlab : 48000 Hz filter impulse response

0 .25

0.2

0 .15

0.1

0 .05

0

0 10 2 030 405 0 60 70

-0 .05

-0.1

n

Figure 26: Matlab : 8000 Hz filter impulse response

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Question

16In designing an ideal low pass filter, we want a top-hat function in the frequency

domain, ranging from - toc c

1

-c c

Figure 27: Ideal low pass filter frequency response

This transforms, via the inverse Fourier transform, into a sinc function in the

time domain. We sample the signal at 41 di erent points in order to discover

the magnitude of the signal at that point and the sampled impulse response can

subsequently be digitally recreated by summing Dirac functions multiplied by

these magnitudes.

The function is non-causal, so it must be shifted in the positive direction,

and incomplete since the sinc function continues on to infinity in reality and

we have discarded all the unsampled points. This impacts on the accuracy of

recreation, and there are many di erent windowing metho ds devoted to easing

this cut-o point.As is clearly seen from graph 29, the di erent windowing functions attenuate

the function to di ering degrees and in di erent ways, as it approaches its cuto

frequency.

Our coe cients ( a )are calculated by multiplying the shifted magnitudes ofi

the sampled points by an appropriate windowing coe cient.

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1

0.8

0.6

0.4

0.2

0

-0.2

-0.40 5 10 15 20 25 30 35 40 45

n

Figure 28: Recreated impulse response

Fi na l c oeffi c i ents

1

0

-10 5 10 1 5 20 2 5 30 35 4 0 45

r ec ta ngul ar w i ndow

1

0

-10 5 10 1 5 20 2 5 30 35 4 0 45

H amm i ng w i ndo w

1

0

-10 5 10 1 5 20 2 5 30 35 4 0 45

Bla c k man wi nd ow

Figure 29: Di erent windowing methods

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no Rectangular Hamming Blackman

1 0 002 -0.0335 -0.0029-0.00013 0 004 0.0374 0.00490.00085 0 006 -0.0424 -0.0091-0.00287 0 008 0.049 0.0162

0.00719 0 0010 -0.0579 -0.0271

-0.015411 0 0

012 0.0707 0.0433

0.029913 0 0014 -0.0909 -0.0681

-0.054615 0 0016 0.1273 0.11020.098517 0 0

018 -0.2122 -0.2016-0.193619 0 0020 0.6366 0.633

0.630221 1 1122 0.6366 0.6330.630223 0 0

024 -0.2122 -0.2016-0.193625 0 0026 0.1273 0.1102

0.098527 0 0028 -0.0909 -0.0681-0.054629 0 0

030 0.0707 0.0433

0.029931 0 0

032 -0.0579 -0.0271-0.015433 0 0034 0.049 0.01620.007135 0 0

036 -0.0424 -0.0091-0.002837 0 0038 0.0374 0.00490.000839 0 0040 -0.0335 -0.0029-0.000141 0 00

Table 4: Table of coe cients

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Question

17The recreated signal is therefore represented by :

40h ( n ) = S a d(n - i )ii=0

The recurrence relation is there given by :

y ( n ) = h ( n ) * x ( n )

y ( n ) = S a d(n - i ) * x ( n )40ii=0

y ( n ) = S a x ( n - i )(unit impulse convolution)40ii=0

The transfer function can be deduced as :

y ( n ) = S a x ( n - i )4 0ii =0

Y (z) = S a X (z)z (Z transform)4 0 -iii =0

Y (z)a z4 0 -iii =0X (z) = S

H(z) = S a z4 0 -iii =0

Question

18

r ectan gula r wi ndow

1.5

1

0.5

0

-0.5

-1

-1.5-1.5 -1 - 0.5 0 0.5 1 1.5

Re(z )

Figure 30: Rectangular window, pole zero diagram

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Hammi ng w indow

1.5

1

0.5

0

-0.5

-1

-1.5-1.5 -1 - 0.5 0 0.5 1 1.5

Re(z )

Figure 31: Hamming window, pole zero diagram

For some reason unbeknownst to me, the matlab roots function malfunc-

tioned at 2000 kHz, leaving me with a severely unsatisfactory Blackmans win-

dow pole zero diagram. At 1995 kHz, the function stabilised, and gave me a

pole zero diagram that was more readily believable in the context of the other

plots.

There are three poles in each pole zero diagram, though they are clustered

at the origin of each of the respective pole zero diagrams.

The zeros radiating beyond the unit circle were a point of some concern,

before it was revealed that the lo cation of the zeros was unrestricted, and that

the presence of the poles within the unit circle was the only point of concern

regarding filter stability.

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B l ac kma n wi ndow

1.5

1

0.5

0

-0.5

-1

-1.5

-1.5 -1 - 0.5 0 0.5 1 1.5

Re(z )

Figure 32: Blackman window, pole zero diagram

B l ac kma n wi ndow

1.5

1

0.5

0

-0.5

-1

-1.5

-1.5 -1 - 0.5 0 0.5 1 1.5

Re(z )

Figure 33: Blackman window, pole zero diagram ( f = 1995)c

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Question

19I used Matlab to calculate the frequency response of the three fir digital filters,

which simplified matters until I tried to calculate the phase of the filters. This

returned a sharply discontinuous phase response graph, with Matlab automat-

ically cycling values within a range of [ -p . . . p ]. The unwrap function, which

was previously used to successfully maintain phase progression information un-

der the iir filters, could not cope with the steep phase shift inherent to the filter.

I had to write my own very rudimentary unwrap script (donunroll3), which was

limited (very) to use with linearly varying phases. (Which the initial plots of

the discontinuous phase revealed it to be.)

The gain and phase are plotted around the unit circle in the s plane. This

unit circle in the s plane is related to the frequency plane by :

=T

s ample2pf = f

s amplefs ample

f =2p

Where = [ -p . . . p ]

f = [ - 4000 . ..

4000]

All the FIR filters were discovered to have linear phase response. Gibbs

phenomenon is normally associated with sudden cuto s in the frequency domain

resulting in a ripple in the time domain. The sudden cuto s applied by our

di erent windowing functions, in sampling the impulse response, induce similar

rippling in the frequency domain. This e ect is incredibly pronounced in the

rectangular window filter, and results in a strong ripple at the edge of the

passband. It results in a barely perceivable waver in the passband when using

the Hamming window and in absent in the passband when using the Blackman

window.

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Rec tangu lar wi ndo w, 2k hz LP F ,s ampl e fr equenc y 8k hz

1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0-4 - 3 -2 -1 0 1 2 3 4

th eta [r ads ]

Figure 34: Rectangular window, digital filter normalised gain

Rec tangul ar w indow, 2khz LP F,s ampl e fre quency 8k hz

4000

3000

2000

1000

0

-1000

-2000

-3000

-4000

-4 -3 -2 - 1 0 1 2 3 4

the ta [rads ]

Figure 35: Rectangular window, digital filter phase response

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Ham min g wi ndow , 2k hz LP F ,sa mpl e fre quenc y 8k hz

1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0-4 - 3 -2 -1 0 1 2 3 4

th eta [r ads ]

Figure 36: Hamming window, digital filter normalised gain

Ham ming w indow , 2khz LP F,s ampl e fr equency 8k hz

4000

3000

2000

1000

0

-1000

-2000

-3000

-4000

-4 -3 -2 - 1 0 1 2 3 4

the ta [rads ]

Figure 37: Hamming window, digital filter phase response

35


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B l ac k mans wi ndo w, 2k hz LP F ,s ampl e fr equenc y 8k hz

1

0 .9

0 .8

0 .7

0 .6

0 .5

0 .4

0 .3

0 .2

0 .1

0-4 - 3 -2 -1 0 1 2 3 4

th eta [r ads ]

Figure 38: Blackman window, digital filter normalised gain

B lac k man wi ndow, 2k hz LP F ,sa mple f requenc y 8k hz

4000

3000

2000

1000

0

-1000

-2000

-3000

-4000

-4 -3 -2 - 1 0 1 2 3 4

the ta [rads ]

Figure 39: Blackman window, digital filter phase response

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It became apparent in the later comparison of implementation vs theory, that

the graphs o er far more information when the gain was displayed in decibels.

The graphs below clearly show the ripple at the edge of the passband, moving

away from the passband as we progress from the rectangular window to the

Blackman window. Gibbs phenomena can not avoided, but the more advanced

windowing methods shift the oscillation away from the passband, where they

have a greatly diminished e ect on the filtering process.

Rec tangul ar wi ndow, 2k hz LP F, sam ple fr equ enc y 8k hz

0

-1 0

-2 0

-3 0

-4 0

-5 0

-6 0

-7 0

-8 0

-9 00 0.5 1 1.5 2 2.5 3 3.5

theta [rads ]

Figure 40: Rectangular window, digital filter normalised gain in dB

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Hamm ing w indow , 2kh z LP F,s am ple fr equenc y 8k hz

0

- 50

- 100

- 150

0 0.5 1 1.5 2 2.5 3 3.5

th eta [rads ]

Figure 41: Hamming window, digital filter normalised gain in dB

B l ac km an wi ndow, 2k hz LPF ,s ampl e fr equen cy 8k hz

0

- 20

- 40

- 60

- 80

- 100

- 120

- 140

- 160

0 0.5 1 1.5 2 2.5 3 3.5

th eta [rads ]

Figure 42: Blackman window, digital filter normalised gain in dB

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