chapter 7 finite impulse response(fir) filter design
DESCRIPTION
Chapter 7 Finite Impulse Response(FIR) Filter Design. 1. Features of FIR filter. Characteristic of FIR filter FIR filter is always stable FIR filter can have an exactly linear phase response - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 7Chapter 7 Finite Impulse Response(FIR) Finite Impulse Response(FIR)
Filter DesignFilter Design
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Characteristic of FIR filter
– FIR filter is always stable– FIR filter can have an exactly linear phase response– FIR filter are very simple to implement. Nonrecursive FIR
filters suffer less from the effect of finite wordlength than IIR filters
1. Features of FIR filter1. Features of FIR filter
0
( ) ( ) ( )N
k
y n h k x n k
0
( ) ( )N
k
k
H z h k z
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Phase response of FIR filter
– Phase delay and group delay
2. Linear phase response2. Linear phase response
0
( )
( ) ( )
= ( )
Nj T j kT
k
j T j
H e h k e
H e e
where ( ) arg ( )j TH e
p g( )
p
( )g
dd
(1)
(2)
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– Condition of linear phase response
( )
( )
Where and is constant Constant group delay and phase delay response
(3)
(4)
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• If a filter satisfies the condition given in equation (3)– From equation (1) and (2)
1 0
0
( )
( )sin = tan
( )cos
N
nN
n
h n nT
h n nT
thus
0
0
( )sintan =
( ) cos
N
nN
n
h n nT
h n nT
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– It is represented in Fig 7.1 (a),(b)
0
( ) cos sin sin cos 0N
n
h n nT nT
0
( )sin( ) 0N
n
h n nT
( ) ( 1), 0h n h N n n N
/ 2NT
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• When the condition given in equation (4) only– The filter will have a constant group delay only– It is represented in Fig 7.1 (c),(d)
( ) ( )h n h N n
/ 2NT
/ 2
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Center of symmetry
Fig. 7-1.
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Table 7.1 A summary of the key point about the four types of linear phase FIR filters
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Example 7-1(1) Symmetric impulse response for linear phase response.
No phase distortion
(2)
( ) ( ) or ( ) ( )h n h N n h n h N n
( ) ( ), 0 , 10h n h N n n N N
(0) (10)(1) (9)(2) (8)(3) (7)(4) (6)
h hh hh hh hh h
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• Frequency response ( )H
10
0
2 3 4 5
6 7 8 9 10
5
( ) ( )
= ( )
= (0) (1) (2) (3) (4) (5)
+ (6) (7) (8) (9) (10)
= [ (0)
j T
jk T
k
j T j T j T j T j T
j T j T j T j T j T
j T
H H e
h k e
h h e h e h e h e h e
h e h e h e h e h e
e h
5 4 3 2
2 3 4 5
(1) (2) (3) (4) (5)
+ (6) (7) (8) (9) (10) ]
j T j T j T j T j T
j T j T j T j T j T
e h e h e h e h e h
h e h e h e h e h e
5 5 5 4 4 3 3
2 2
5
( )= [ (0)( ) (1)( ) (2)( )
(3)( ) (4)( ) (5)]
[2 (0)cos(5 ) 2 (1)cos(4 ) 2 (2)cos(3 ) 2 (3)c
j T j T j T j T j T j T j T
j T j T j T j T
j T
H e h e e h e e h e e
h e e h e e h
e h T h T h Th
os(2 ) 2 (4)cos( ) (5)]T h T h 5
5 ( )
0
( )= ( )cos( ) ( )j T j
k
H a k k T e H e
where 5
0
( ) = ( )cos( )k
H a k k T
( ) 5 T
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(3) 9N (0) (9)(1) (8)(2) (7)(3) (6)(4) (5)
h hh hh hh hh h
9 /2 9 9 7 7 5 3
3 3
9 /2
( )= [ (0)( ) (1)( ) (2)( )
(3)( ) (4)( )]
[2 (0)cos(9 / 2) 2 (1)cos(7 / 2) 2 (2)cos(5 / 2) 2
j T j T j T j T j T j T j T
j T j T j T j T
j T
H e h e e h e e h e e
h e e h e e
e h T h T h T
( )
(3) cos(3 / 2) 2 (4)cos( / 2)]
= ( ) j
h T h T
H e
5
1
( )= ( ) cos[ ( 1/ 2) ]k
H b k k T
where
( ) (9 / 2) T 1 1( ) 2 ( ), 1,2, ,
2 2N Nb k h k k
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Transfer function for FIR filter
3. Zero distribution of FIR filters3. Zero distribution of FIR filters
0
( ) ( )N
k
k
H z h k z
0
0
1
( ) ( )
= ( )
= ( )
Nk
k
k N
k N
N
H z h N k z
h k z z
z H z
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Four types of linear phase FIR filters– have zero at ( )H z 0z z
0jz re
1 10
jz r e
is real and is imaginary( )h n 0z*
0jz re
* 1 10( ) jz r e
1 1 1 1 1 1(1 )(1 )(1 )(1 )j j j jre z re z r e z r e z
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– If zero on unit circle
– If zero not exist on the unit circle
– If zeros on
1, 1/ 1r r
0jz e
1 *0 0
jz e z
1 1(1 )(1 )j je z e z
1 1 1(1 )(1 )rz r z
1z 1(1 )z
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Necessary zero
Necessary zero
Necessary zero
Necessary zero
Fig. 7-2.
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Filter specifications
4. FIR filter specifications4. FIR filter specifications
peak passband deviation (or ripples)
stopband deviation
passband edge frequency
stopband edge frequency
sampling frequency
p
r
p
r
s
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ILPF
Satisfies spec’s
Fig. 7-3.
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Characterization of FIR filter
– Most commonly methods for obtaining • Window, optimal and frequency sampling methods
0
( ) ( ) ( )N
k
y n h k x n k
0
( ) ( )N
k
k
H z h k z
( )h k
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FIR filter– Frequency response of filter– Corresponding impulse response
– Ideal lowpass response
5. Window method5. Window method
( )IH
( )Ih n
1( ) ( )2
j nI Ih n H e d
1 1( ) 12 2
2 sin( ) , 0, - =
2 , 0
c
c
j n j nI
c c
c
c
h n e d e d
f n n nn
f n
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Fig. 7-4.
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Truncation to FIR
– Rectangular Window
( ) ( ) ( ) "windowing"Ih n h n w n
12
( )12
) ) )
) )
j j jI
j jI
H(e H (e W(e
H (e W(e d
1 , 0,1,...,( )
0 ,n N
w nelsewhere
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Fig. 7-5.
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Fig. 7-6.
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Fig. 7-7.
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Table 7.2 summary of ideal impulse responses for standard frequency selective filters
and are the normalized passband or stopband edge frequencies; N is the length of filter
1, cf f 2 f
( )Ih n
( )Ih n
( )Ih n
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Common window function types– Hamming window
0.54 0.46cos 2 / , 00 ,
n N n Nw n
elsewhere
3.32 /F N
where N is filter length and
is normalized transition widthF
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– Characteristics of common window functions
Fig. 7-8.
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Table 7.3 summary of important features of common window functions
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– Kaiser window
2
0
0
21
, 0
0 ,
n NIN
w n n NI
elsewhere
2
01
( / 2)( ) 1!
kL
k
xI xk
where typically
where is the zero-order modified Bessel function of the first kind0 ( )I x
25L
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• Kaiser Formulas – for LPF design
10
0.4
20log ( )min( , )
0, if 21dB
0.5842( 21) 0.07886( 21), if 21dB 50dB0.1102( 8.7) if
p s
A
A
A A AA A
50dB
7.9514.36AN
F
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Example 7-2– Obtain coefficients of FIR lowpass using hamming
window
• Lowpass filter
Passband cutoff frequency
Transition width
Stopband attenuation
Sampling frequency
: 1.5pf kHz
: 0.5f kHz
: 8sf kHz
50dB
sin
2 , 0
2 , 0
cc
I c
c
nf n
h n nf n
/ 0.5 / 8 0.0625sF f f
3.32 / 3.32 / 0.0625 53.12N F
54N
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• Using Hamming window( ) ( ) ( )Ih n h n w n 0 54n
( ) 0.54 0.46cos(2 / 54)w n n 0 54n ' / 2 (1.5 0.25) 1.75[ ]c pf f f kHz
' / 1.75 / 8 0.21875c c sf f f
2 sin ( )2 , , 0
2( ) ( )2
2 ,2
c c
I c
c
Nf nNn n N
Nh n n
Nf n
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2 0.218750 : (0) sin( 27 2 0.21875)27 2 0.218750.00655
(0) 0.54 0.46cos(0) 0.08(0) (0) (0) 0.00052398
I
I
n h
wh h w
2 0.218751 : (1) sin( 26 2 0.21875)26 2 0.218750.011311
(1) 0.54 0.46cos(2 / 54)0.08311
(1) (1) (1) 0.00094054
I
I
n h
w
h h w
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2 0.218752 : (2) sin( 25 2 0.21875)25 2 0.218750.00248397
(2) 0.54 0.46cos(2 2 / 54)0.092399
(2) (2) (2) 0.000229516
2 0.2187526 : (26) sin( 1 2 0.21875)1 2 0.21875
0.312936
I
I
I
n h
w
h h w
n h
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(26) 0.54 0.46cos(2 26 / 54)0.9968896
(26) (26) (26) 0.3112226I
w
h h w
27 : (27) 2 2 0.218750.4375
(27) 0.54 0.46cos(2 27 / 54)1
(27) (27) (27) 0.4375
I c
I
n h f
w
h h w
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Fig. 7-9.
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Example 7-3– Obtain coefficients using Kaiser or Blackman window
Passband cutoff frequency
Transition region
Sampling frequency
: 1200pf Hz
: 500f Hz
: 10sf kHz
Stopband attenuation : 40dB
passband attenuation : 0.01dB
20log(1 ) 0.01 , 0.00115p pdB
20log( ) 40 , 0.01r rdB
0.00115r p
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• Using Kaiser window
7.95 58.8 7.95 70.8214.36 14.36(500 /10000)AN
F
71N
0.1102(58.8 8.7) 5.52 ' / 1,450 /10,000 0.145c c sf f f
2 sin2( ) , 0
2
c c
I
c
Nf nh n n N
Nn
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( ) ( ) ( )Ih n h n w n
2 0.1450 : (0) sin( 35.5 2 0.145)35.5 2 0.145
0.00717
In h
2
0
0
0 0
21(0)
(0) 0.023( ) (5.52)
n NIN I
wI I
(0) (0) (0) 0.000164935Ih h w
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2 0.1451: (1) sin( 34.5 2 0.145)34.5 2 0.145
0.0001449
In h
2
0
0
0 0
695.52 171 (1.3)
(1) 0.0337975(5.52) (5.52)
II
wI I
(1) ( 1) (1) (1) 0.000004897Ih h h w
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2 0.1452 : (2) sin( 34.5 2 0.145)33.5 2 0.145
0.007415484
In h
2
0
0
0 0
695.52 171 (1.8266)(2) 0.04657999
(5.52) (5.52)
II
wI I
(2) ( 2) (2) (2) 0.000345413Ih h h w
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2 0.14535 : (35) sin( 0.5 2 0.145)0.5 2 0.145
0.280073974
In h
2
0
0
0 0
15.52 171 (5.51945)(35) 0.999503146
(5.52) (5.52)
IIw
I I
(35) ( 35) (35) (35) 0.279934818Ih h h w
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Fig. 7-10.
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Summary of window method1. Specify the ‘ideal’ or desired frequency response of filter,2. Obtain the impulse response, , of the desired filter by evaluating the inverse Fourier transform3. Select a window function that satisfies the passband or attenuation specifications and then determine the number of filter
coefficients4. Obtain values of for the chosen window function and the
values of the actual FIR coefficients, , by multiplying by
( )IH
( )Ih n
( )h n
( )w n
( )w n
( )Ih n( ) ( ) ( )Ih n h n w n
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Advantages and disadvantages– Simplicity– Lack of flexibility– The passband and stopband edge frequencies cannot be
precisely specified– For a given window(except the Kaiser), the maximum
ripple amplitude in filter response is fixed regardless of how large we make N
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Basic concepts– Equiripple passband and stopband
• For linear phase lowpass filters– m+1 or m+2 extrema(minima and maxima)
6. The optimal method6. The optimal method
( ) ( )[ ( ) ( )]IE W H H
min[max ( ) ]E
where m=(N+1)/2 (for type1 filters) or m =N/2 (for type2 filters)
WeightedApprox. error
Weightingfunction
Ideal desiredresponse
Practical response
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Ideal responsePractical response
Fig. 7-11.
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Fig. 7-12.
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– Optimal method involves the following steps• Use the Remez exchange algorithm to find the optimum set of
extremal frequencies• Determine the frequency response using the extremal
frequencies• Obtain the impulse response coefficients
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Optimal FIR filer design
0
( ) ( )N
k
k
H z h k z
where ( ) ( )h n h n
/2 /2
1 0
( ) (0) 2 ( )cos ( )cosN N
k k
H h h k k T a k k T
where and , (0) (0)a h ( ) 2 ( )a k h k 1, 2, , / 2k N
Let / 2 / 2sf f T
1
1 , 00 , 0.5
p
r
f fH f
f f
1 , 0
1 , 0.5
p
r
f fW f k
f f
This weighting function permits different peak error in the two band
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0
( ) ( ) ( ) cos 2N
j f
k
H f H e a k k f
min[max ( ) ]E f
Find ( )a k
( ) ( )[ ( ) ( )]IE f W f H f H f
where are and 0, pf f ,0.5rf
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– Alternation theorem
0, ,0.5p rF f f Let
If has equiripple inside bands and more than m+2 extremal point ( )E f F
then
1( ) ( ) , 0,1, , 1i iE f E f e i l
0 1 ( 1)lf f f l m
,0.50, &max ( )
p rf f fe E f
where
( ) ( )IH f H f
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• From equation (7-33) and (7-34)
• Equation (7-35) is substituted equation (7-32)
• Matrix form
( ) ( ) ( ) ( 1) , 0,1,2, , 1ii i I iW f H f H f e i m
0
( ) cos 2 ( ) ( 1) , 0,1,2, , 1( )
mi
i I ik i
ea k kf H f i mW f
0 0 0 0 0
1 1 1 1
!1 1 1 1
1 cos 2 cos 4 cos 2 1/ ( ) ((0)1 cos 2 cos 4 cos 2 1/ ( ) (1)
1 cos 2 cos 4 cos 2 ( 1) / ( ) ( )1 cos 2 cos 4 cos 2 ( 1) / ( )
I
mm m m m
mm m m m
f f mf W f H faf f mf W f a
f f mf W f a mf f mf W f e
1
1
)( )
( )( )
I
I m
I m
H f
H fH f
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– SummaryStep 1. Select filter length as 2m+1 Step 2. Select m+2 point in FStep 3. Calculate and e using equation (7)Step 4. Calculate using equation (5). If in some of f , go to step 5, otherwise go to step 6Step 5. Determine m local minma or maxma pointsStep 6. Calculate when
if
( )a k
( )E f ( )E f e
(0) (0), ( ) ( ) / 2h a h k a k 1, 2, ,k m
where 0
( ) ( )N
k
k
H z h k z
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– Example 7-4• Specification of desired filter
– Ideal low pass filter– Filter length : 3– –
• Normalized frequency
1[rad],pT 1.2[rad]rT
2k
0 1 20.5 / 2 , 1/ 2 , 1.2 / 2f f f
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• From
• Cutoff frequency
1 0.8776 2 (0) 11 0.5403 2 (1) 11 0.3624 1 0
aa
e
0 1 2 0 1( ) ( ) 1, ( ) 0, ( ) ( ) 1/ 2I I IH f H f H f W f W f 2and ( ) 1W f
( ) 0.645 2.32cos 2H f f : not the optimal filter
0 1 21 1.2 0.5
2 2f f f
0 1 2 0( ) 1, ( ) ( ) 0, ( ) 0.5,I I IH f H f H f W f 1 2and ( ) ( ) 1W f W f
(0) 0.645, (1) 2.32,a a 0.196e
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1 0.5403 2 (0) 11 0.3624 1 (1) 01 1 1 0
aa
e
(0) 0.144, (1) 0.45,a a 0.306e
( ) 0.144 0.45cos 2H f f : has the minimum
(N=3)
max ( )E f
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Fig. 7-13.
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Optimization using MATLAB– Park-McClellan– Remez
remez( )b N,F, M
remez( )b N,F, M, WT
where N is the filter order (N+1 is the filter length)
F is the normalized frequency of border of pass band
M is the magnitude of frequency response
WT is the weight between ripples
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– Example 7-5• Specification of desired filter
– Band pass region : 0 – 1000Hz– Transition region : 500Hz– Filter length : 45– Sampling frequency : 10,000Hz
• Normalized frequency of border of passband
• Magnitude of frequency response
[0, 0.2, 0.3, 1]F
[11 0 0]M
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Table 7-4.
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Fig.7-14.
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– Example 7-6• Specification of desired filter
– Band pass region : 3kHz – 4kHz– Transition region : 500Hz– Pass band ripple : 1dB– Rejection region : 25dB– Sampling frequency : 20kHz
• Frequency of border of passband
[2500, 3000, 4000, 4500]F
[0 1 0]M
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• Transform dB to normal value
• Filter length– Remezord (MATLAB command)
20
20
10 1
10 1
p
p
A
p A
2010
rA
r
where and ripple value(dB) of pass band and rejection band
pArA
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Table 7-5.
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Fig. 7-15.
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Frequency sampling filters– Taking N samples of the frequency response at intervals
of– Filter coefficients
7. Frequency sampling method7. Frequency sampling method
/skF N 0,1, , 1k N
( )h n
1(2 / )
0
1( ) ( )N
j N nk
k
h n H k eN
where are samples of the ideal or target frequency response( ), 0,1, , 1H k k N
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– For linear phase filters (for N even)
– For N odd• Upper limit in summation is
12
1
1( ) 2 ( ) cos[2 ( ) / ] (0)
N
k
h n H k k n N HN
where ( 1) / 2N
( 1) / 2N
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Fig. 7-16.
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– Example 7-7(1) Show the
• Expanding the equation
/2 1
1
1( ) 2 ( ) cos[2 ( ) / ] (0)N
k
h n H k k n N HN
12 /
0
12 / 2 /
0
12 ( ) /
0
1
0
1
0
1( ) ( )
1 ( )
1 ( )
1 ( ) cos[2 ( ) / ] sin[2 ( ) / ]
1 ( ) cos[2 ( ) / ]
Nj nk N
k
Nj k N j kn N
k
Nj n k N
k
N
k
N
k
h n H k eN
H k e eN
H k eN
H k k n N j k n NN
H k k n NN
( )h n is real value
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(2) Design of FIR filter– Band pass region : 0 – 5kHz– Sampling frequency : 18kHz– Filter length : 9
Fig. 7-17.
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Table 7-6.
1, 0,1,2( )
0, 3,4k
H kk
• Samples of magnitude in frequency
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– Window method• The easiest, but lacks flexibility especially when passband and
stopband ripples are different
– Frequency sampling method• Well suited to recursive implementation of FIR filters
– Optimal method• Most powerful and flexible
8. Comparison of most commonly 8. Comparison of most commonly methodmethod