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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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- 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
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
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
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
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
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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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