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7/29/2019 fwhomfbrhi27435FWHOMFBRHI

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Dr. M. Venu Gopala Rao, Professor, KL University, Vijayawada

1

Finite Impulse Response Filters

Introduction: In the design of frequency selective filters, the desired filter characteristics are specified in the

frequency domain in terms of the desired magnitude and phase response of the filter. In the filter design

process, we determine the coefficients of a causal FIR or IIR filter that closely approximates the desired

frequency response specifications. The issue of which type of filter to design, FIR or IIR depends on the

nature of the problem and on the specifications of the desired frequency response.

In practice, FIR filters are employed in filtering problems where there is a requirement for a linear phase

characteristic within the pass band of the filter. If there is no requirement for a linear phase characteristic,

either an IIR or an FIR filter may be employed. However, as a general rule, an IIR filter has lower side lobes

in the stop band than an FIR filter having the same number of parameters. For this reason, if some phase

distortion is either tolerable or unimportant, an IIR filter is preferable, primarily because its implementation

involves fewer parameters, requires less memory and has lower computational complexity.

FIR filters: If the impulse response of the digital filter is determined by some finite number of impulse

sequences then these filters are known as FIR Filters. In other words it can be defined as the unit sample

response of an LTI system is of finite duration, then the system is said to be FIR system or filter. FIR

filters are recursive filters. The present output sample depends only on present and past inputs, but not on

past outputs. Since FIR filters have no feedback, they have no poles and are therefore always stable. Theadvantages and disadvantages of FIR filters over IIR filters are described below.

Advantages:

1, FIR filters are always stable.

2. Fir filters with exactly linear phase can easily be designed.

3. Fir filters can be realized in both recursive and non-recursive structures.

4. FIR filters are free limit cycle oscillations, when implemented on a finite word length digital system.

5. Excellent methods are available for various kinds of FIR filters.

Disadvantages:

1. The implementation of narrow transition band FIR filters are very costly, as it requires considerably more

arithmetic operations and hardware components such as multipliers, adders and delay elements.

2. Memory requirement and execution time are very high.


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2

Linear Phase FIR Filters and its properties:

An FIR filter of length N with input [ ]x n and output [ ]y n is described by the difference equation

0 1 1

1

0

[ ] [ ] [ 1] . [ 1]

[ ]

. . . NN

kk

y n b x n b x n b x n N

b x n k

=

= + + +

=

+

(1)

where kb is the set of linear coefficients. Alternatively we can express the output sequence as the

convolution of the unit sample response [ ]h n of the system with the input signal. Thus we have

1

0

[ ] [ ] [ ]N

k

y n h k x n k

== (2)

where the lower limit and upper limits on the convolution sum reflect the causality and finite duration

characteristics of the filter.

The filter can also be characterized by its system function

1

0

[ ] [ ]N

n

n

H h nz z

== (3)

which we view as a polynomial of degree 1N in the variable 1z . The roots of this polynomial constitute

the zeros of the filter.

The frequency response of the filter is obtained by substituting jz e= in equation (3), thus we get

1

0

[ ] [ ]N

j jn

n

H h ne e

== (4)

Which is periodic with period 2 . The frequency response is represented by its magnitude and phase

responses as

[ ][ ] | [ ] |j j j H H ee e = (5)

where | [ ] |jH e is magnitude response and [ ] is the phase response.

Phase delay: The phase delay is defined as the negative ratio of phase [ ] and frequency of a filter.

i.e.,[ ]

p

= . (6)

Group delay: The group delay is defined as the negative differentiation of [ ] with respect to . In

other words it also can be defined as the rate of phase response with respect to frequency i.e.,

[ ]d

gd

= . (7)

Linear Phase Characteristics:

For FIR filters with linear phase we can define [ ] = (8)

where is a constant phase delay in samples. Substituting in equations (6) and (7), we get


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3

[ ] p

= == and [ ]

d d g

d d = == . Therefore p g = = which means

independent of frequency. Therefore to obtain a linear phase characteristics, the FIR digital filter both phase

delay and group delay should be constant.

Proof:

From equations (4) and (5) , we can write1

[ ]

0

[ ] | [ ] |N

jn j j

n

h n H ee e

== or

1

0

[ ]{cos sin } | [ ] |{cos [ ] sin [ ]}N

j

n

h n n j n H j e

= = + (9)

which gives us1

0

[ ] cos | [ ] | cos [ ]N

j

n

h n n H e

== and

1

0

[ ] sin | [ ] | sin [ ]N

j

n

h n n H e

= =

By taking their ratio

1

01

0

[ ]sin| [ ] | sin [ ] sin [ ] sin( ) sin( )

cos [ ] cos( ) cos( )| [ ] | cos [ ][ ]cos

N

jn

jN

n

h n nH

H h n n

e

e

=

=

= = = =

After cross multiplying and simplifying we get1

0

[ ] sin ( ) 0N

n

h n n

= = (10)

The above equation is zero when [ ] [ 1 ]h n h N n= and 12

N = (11)

Refer Fig 1(a) and (b)

In the case of only constant group delay is required, and not phase delay, we define [ ] k = . If

2 = then [ ] [ 1 ]h n h N n= and

1

2

N

= . Refer Fig 1(c) and (d).


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4

Frequency Response of Linear Phase FIR Filters:

Case 1: Symmetrical Impulse Response and N is odd:

1

| [ | [ ] 2 [ ]cos ,k

jk

n

H e h h k n n=

= + and [ jH e k = , where 12Nk = .

1

| [ ] | [ ] 2 [ ]cos ,k

jk

n

H e h h k n n=

= + and [ ]jH e k = , where 12Nk = .

Applications: L.P.F., H.P.H., B.P.F., and B.R.F

Case 2: Symmetrical Impulse Response and N is even:

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12

| [ ] | 2 [ ]cos )N

j

n

NH e h n ( n=

= and ( )12[ ]j NH e =

Applications: L.P.F., and B.P.F

Case 3:Anti-symmetrical Impulse Response and N is odd:

12

1

12

| [ ] | 2 [ ]sin

N

j

n

NH e h n n

=

= and ( )12 2[ ]j NH e =

Applications: Differentiator and Hilbert Transform.


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5

Case 4:Anti-symmetrical Impulse Response and N is even:

2

)12 2

1

| [ ] | 2 [ ]sin

N

j N

n

H e h n (n - n=

= and ( )12 2[ ]j NH e =

Applications: Differentiator and Hilbert Transform.

Designing FIR Filters:

Fourier Series Method: The frequency response [ ]jH e of a system is periodic in period 2 . By

Fourier series method, any periodic function can be expressed as a linear combination of complex

exponentials. Therefore the desired frequency response of an FIR filter can be represented by Fourier Series.

That is [ ] [ ] ,j jndn

H e h n e

== where 12[ ] [ ]

j jndh n H e e d

= .

The Z.T. of sequence is, [ ] [ ] ndn

H z h n z

== which is an infinite duration sequence. To get FIR filter

transfer function the series can be truncated by assigning [ ] [ ]dh n h n= for1 1

2 2N Nn . That is

1 12 2[ ], for[ ]0, Otherwise

d N Nnh nh n

=

. Then

( ) ( )

[ ]

11 12

2 2

12

12

1 1

1

1 12 2

[ ] [ ] . . . . ( 1) (0) (1) . . .

(0) [ ] [ ]

NN N

N

N

n

n

n n

n

N NH z h n z h z h z h h z h z

h h n z h n z

=

=

= = + + + + + +

= + +

For a symmetrical impulse response having symmetry at 0n = . That is [ ] [ ]h n h n = .

Therefore [ ]

12

1

[ ] (0) [ ]

N

n n

n

H z h h n z z

== + + . This transfer function is not physically realizable. Realizability

can be brought by multiplying by1

2N

z

, where 12

N is the delay in samples.


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Therefore the physically realizable filter transfer function can be written as

[ ]

12

11 2

2

1

[ ] [ ]

(0) [ ]

N

NN

n n

n

H z z H z

z h h n z z

=

=

= + +

Ex1:Design an ideal LPF with a frequency response| |

2 2

| |2

1, for[ ]

0, for

jd

H e

=

. Find the values of

[ ]h n for 11N= . Find [ ]H z and plot its response. Use Fourier series method.

Solution:

1 12 2

21 12 2

2

[ ] [ ] [ ]

sinsin1.

c

c

c

c

j jn j jn

d

cjn

h n H e e d = H e e d

n n= e d = n

n n

=

=

Given that 11N = . Therefore 1 11 12 2

5Nn = = = .

21

22

sin, for 5 5

[ ]

0, Otherwise

d

nn

nh n

=

21 12 20 0

2

sin[0] lim [ ] limd dn n

nh h n

n

= = =

or

01 1 122 2 2

2

[0] [ ] 1c

c

j jd

h H e e d = d =

=

2 12

2 13

2 15

sin[ 1] [1] 0.3183

sin[ 2] [2] 0

2

sin3[ 3] [3] 0.106

3

sin2[ 4] [4] 0

4

sin5[ 5] [5] 0.063665

d d

d d

d d

d d

d d

h h

h h

h h

h h

h h

= = = =

= = =

= = = =

= = =

= = = =

The transfer function of the filter

[ ] [ ]

12

1 1

1 3 3 5 5

5[ ] (0) [ ] [ ] 0.5 [ ]

0.5 0.3183( ) 0.106( ) 0.06366( )

N

n n n n

n n

H z h h n z z H z h n z z

z z z z z z

= =

= + + = = + +

= + + + + +


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The transfer function of realizable filter1

2[ ] [ ]N

H z z H z

=

152

5

5

1 3 3 5 5

2 4 6 8 10

[ ] [ ] [ ]

0.5 0.3183( ) 0.106( ) 0.06366( )

0.06366 0.106 0.3183 0.5 0.3183 0.106 0.06366

N

H z z H z z H z

z z z z z z z

z z z z z z

= =

= + + + + +

= + + + +

Frequency Response: To find the frequency response substitute jz e= in [ ]H z .

3 3 5 5[ ] 0.5 0.3183( ) 0.106( ) 0.06366( )

0.5 0.6366 cos 0.212cos3 0.127cos5

j j j j j j j dH e e e e e e e

= + + + + +

= + +

The magnitude response is shown in Figure.

Designing FIR filters using Windows: The main limitations of Fourier series method is oscillations in

the pass band and stop band due to the slow convergence. This effect is known as Gibb;s phenomenon. To

reduce these oscillations, the Fourier series coefficients are modified by multiplying a weighting sequence

called window [ ]w n ,

where1

2

12

0,for | |[ ] [ ]

0,for | |

N

N

nw n w n

n

= =

= >

After multiplication of window sequence [ ]w n with [ ]dh n , we get finite duration sequence [ ]h n , that

satisfies the desired magnitude response.1

2

12

[ ] [ ], for | |[ ]

0, for | |

N

N

dh n w n nh nn

=

>

The frequency response [ ] [ ] [ ]j j jdH e H e W e= .

The window must have the following Characteristics:i. The central lobe of the frequency response of the window should contain most of the energy and should

be narrow.

ii. The highest side lobe level of the frequency response should be small

iii. The side lobes of the frequency response should decrease in energy rapidly as tends to .

Various windowing techniques are illustrated below


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8

(a)Rectangular Window:

The rectangular sequence is given by

1 12 2

1,for[ ]

0, elsewhere

N Nnw n

=

12

12

sin / 2[ ]

sin / 2

N

Nn

j jR

NW e e n

=

= =

i. For the rectangular window, the amplitude of the side lobes is unaffected by the length of the

window. So increase in length will not reduce the ripple but increase its frequency.

ii. J.W. Gibbs showed that a finite length LPF an 8.9% maximum ripple, no matter how long the filter is

made.

iii. The effect where maximum ripple occurs just before and after the transition band is known as Gibbs

phenomenon.

(b)Bartlett Window (Triangular Window): The N-point triangular window is given by

1 12 2

( 1) / 2

| |1 , for

[ ]0, elsewhere

N N

N

nn

w n

=

( )142

sin[ ]

sin / 2

Nj

T

W e

=

Advantage: The triangular window produces a smooth

magnitude response in both pass band and stop band.

Disadvantage: When compared to rectangular window

i. The transition region is more

ii. The attenuation in stop band is lessHence it is not a good choice.

(c)Hanning Window: The Hanning window is defined as

i. The main lobe width is twice of the rectangular window.

ii. The magnitude of the side lobe level is -31 dB, which is 18 dB lower that of rectangular window. That

is the first side lobe of Hanning window spectrum is 1/10 th of rectangular window.

iii. Smaller ripples in both pass band and stop band.

(d)Hamming Window: The Hamming window is defined as

1 12 2

( 1) / 20.54 0.46cos , for

[ ]

0, elsewhere

N N

N

nn

w n

+

=

1 12 2( 1) / 2

0.5 0.5cos , for[ ]

0, elsewhere

N N

N

nn

w n

+

=


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9

i. The peak side lobe level is down about 41 dB from the main lobe peak an improvement 10dB relative

to the Hanning window.

ii. At higher frequencies the stop band attenuation is low when compared to that of Hanning window.

Because less oscillations in the side lobes than Hanning window the Hamming window is generally

preferred.

(e)Blackman Window: The Blackman window is defined as

1 12 2

( 1) / 2 ( 1) / 2

20.42 0.5cos 0.08cos ,for

[ ]

0, elsewhere

N N

N NB

n nn

w n

+ + =

The additional Cosine term reduces side lobes, but increases the

main lobe width to 12N

.

The peak side lobe level down about 57 dB from the main lobe peak, animprovement of 16dB relative to the Hamming window

The comparison of various windowing techniques are illustrated below

Comparison between FIR filters and IIR filters:

S.No FIR Filter IIR Filter

1These filters can be easily designed to have

perfectly linear phase,These filters dont have linear phase,

2

FIR filters can be realized recursively and non-

recursively, FIR filters are easily realized recursively ,

3Greater flexibility to control the shape of their

magnitude response,

Less flexibility, usually limited to specific kind

of filters.

4Errors due to round of noise are less severe in FIR

filters, mainly because feedback is not used.The round of noise in IIR filters is more.


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Design Procedure for FIR Filters using Windows

Step1: From the given specifications draw the graph for [ ]jdH e in the range ( , ) .

Step2: Determine is impulse response using Inverse DTFT formula 12

[ ] [ ]j jnd dh n H e e d

= .

Step3: For the given order N

, determine ,[0] [1], [2]d d dh h h etc., using the following formulas.

For lim0

[0] [ ]d dnh h n

= or 1

2[0] [ ]jd dh H e d

= .

For [1], [2]d dh h . . . use the formula defined in Srep2.

Step4: Choose the appropriate window for the problem and write its equation for the given order N.

(a) Rectangular Window:1 1

2 21,

[ ]0,

N N

R

for nw n

elsewhere

=

(b) Bartlett Window:

1 1

2 2

| |

1 ,( 1) / 2[ ]

0,

N N

T

n

for nNw n

elsewhere

=

(c) Hanning Window:

1 12 2( 1)/2

0.5 0.5cos ,[ ]

0,

N NN

Hn

n for nw n

elsewhere

+ =

(d) Hamming Window:

1 12 2( 1)/2

0.54 0.46cos ,[ ]

0,

N NN

Hm

n for nw n

elsewhere

+ =

(e) Blackman window:

1 12 2( 1)/2 ( 1)/2

20.42 0.5 cos 0.08cos ,[ ]

0,

N NN N

B

n n for nw n

elsewhere

+ + =

Step5: After choosing appropriate window, determine the window coefficients [0], [1], [2]w w w for the order N.

Step6: Determine the filter coefficients according to the given window by using the formula

[ ] [ ] [ ]dh n h n w n= for Nvalues. i.e., [0], [ 1] [1], [ 2] [2]h h h h h = = etc.

Step7: Determine the transfer function using the formula { }

12

1

[ ] [0] [ ]

N

n

n nH z h h n z z

=

= + + .

Step8: Detrmine the realizable filter transfer function1

2[ ] [ ]N

H z H zz

= .

Step9: Determine the frequency response by substitutingj

ez = in [ ]H z calculated in step7.Step10: Formulate a table for a table for and 20log [ ]|j| H e and draw its magnitude response.

34

12

14

0 14 12

34

[ ]|j| H e

20log [ ]|j| H e


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Ex2:Design an ideal LPF with a frequency response| |

2 2

| |2

1, for[ ]

0, for

jd

H e

=

. Find the values of

[ ]h n for 11N= . Find [ ]H z . Use various windows.

Ans: From ex1,

21

2

2

sin, for 5 5

[ ]0, Otherwise

d

nn

nh n

=

21 12 20 0

2

sin[0] lim [ ] limd dn n

nh h n

n

= = =

or 01 1 12

2 2 22

[0] [ ] 1c

c

j j

d h H e e d = d =

=

2 12

2 13

2 15

sin[ 1] [1] 0.3183

sin[ 2] [2] 0

2

sin3

[ 3] [3] 0.1063

sin2[ 4] [4] 0

4

sin5[ 5] [5] 0.06366

5

d d

d d

d d

d d

d d

h h

h h

h h

h h

h h

= = = =

= = =

= = = =

= = =

= = = =

(a) Rectangular Window:1 1

2 21,

[ ]0,

N N

R

for nw n

elsewhere

=

12

[ ] [ ] [ ]

[0] 1 [0] [0] [0]

[ 1] [1] 1 [ 1] [1] [1] [1] 0.3183

[ 2] [2] 1 [ 2] [2] [2] [2] 0

[ 3] [3] 1 [ 3] [3] [3] [3] 0.106

[ 4] [4] 1 [ 4] [4] [4] [4] 0

Rd

R Rd

R R Rd

R R Rd

R R Rd

R R Rd

R

h n h n w n

w h h w

w w h h h w

w w h h h w

w w h h h w

w w h h h w

w

=

= = = = = = = =

= = = = =

= = = = =

= = = = =

[ 5] [5] 1 [ 5] [5] [5] [5] 0.06366R Rdw h h h w = = = = =

(b) Bartlett Window:

1 12 2

1,[ ]

0,

N N

R

for nw n

elsewhere

=

| | | |[ ] 1 1 [ ] [ ] [ ]( 1) / 2 5

[0] 1 [0] [0] [0] 0.5

[ 1] [1] 0.8 [ 1] [1] [1] [1] 0.3183 0.8 0.25464

[ 2] [2] 0.6 [ 2] [2] [2] [2] 0 0.6 0

[ 3] [3] 0.4 [ 3] [3] [3]

T Td

T Td

T T Td

T T Td

T T d

n nw n h n h n w nN

w h h w

w w h h h w

w w h h h w

w w h h h w

= = == = =

= = = = = =

= = = = = =

= = = =

X

X

[3] 0.106 0.4 0.0424

[ 4] [4] 0.2 [ 4] [4] [4] [4] 0 0.2 0

[ 5] [5] 0 [ 5] [5] [5] [5] 0.06366 0 0

T

T T Td

T T Td

w w h h h w

w w h h h w

= =

= = = = = =

= = = = = =

X

X

X


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(c) Hanning Window:

1 12 2( 1)/2

0.5 0.5cos ,[ ]

0,

N NN

Hn

n for nw n

elsewhere

+ =

5[ ] 0.5 0.5cos [ ] [ ] [ ]

[0] 1 [0] [0] [0] 1 0.5

[ 1] [1] 0.9045 [ 1] [1] [1] [1] 0.3183 0.9045 0.2879

[ 2] [2] 0.6545 [ 2] [2] [2] [2] 0 0.6545 0[ 3] [3] 0.

Hn Hn

Hn

Hn Hn Hn

Hn Hn Hn

Hn Hn

d

Td

d

d

nw n h n h n w n

w h h w

w w h h h w

w w h h h w

w w

= + =

= = =

= = = = = =

= = = = = = = =

X

X

X

3455 [ 3] [3] [3] [3] 0.106 0.3455 0.0366

[ 4] [4] 0.0955 [ 4] [4] [4] [4] 0 0.0955 0

[ 5] [5] 0 [ 5] [5] [5] [5] 0.06366 0 0

Hn

Hn Hn Hn

Hn Hn Hn

d

d

d

h h h w

w w h h h w

w w h h h w

= = = =

= = = = = =

= = = = = =

X

X

X

(d) Hamming Window:

1 12 2( 1)/2

0.54 0.46cos ,[ ]

0,

N NN

Hm

n for nw n

elsewhere

+ =

5[ ] 0.54 0.46cos [ ] [ ] [ ]

[0] 1 [0] [0] [0] 1 0.5

[ 1] [1] 0.9121 [ 1] [1] [1] [1] 0.3183 0.9121 0.2862[ 2] [2] 0.6821 [ 2] [2] [2] [2] 0 0.6821 0

[ 3] [3]

Hm

Hm

Hm Hm

Hm Hm

Hm Hm

Hmd

Hmd

Hmd

Hmd

nw n h n h n w n

w h h w

w w h h h ww w h h h w

w w

= + =

= = =

= = = = = = = = = = = =

=

X

XX

0.3979 [ 3] [3] [3] [3] 0.106 0.3979 0.0422

[ 4] [4] 0.1679 [ 4] [4] [4] [4] 0 0.1679 0

[ 5] [5] 0.080 [ 5] [5] [5] [5] 0.06366 0.080 0.0051Hm Hm

Hm Hm

Hmd

Hmd

Hmd

h h h w

w w h h h w

w w h h h w

= = = = =

= = = = = =

= = = = = =

X

X

X

(e) Blackman window:

1 12 2( 1)/2 ( 1)/2

20.42 0.5cos 0.08cos ,[ ]

0,

N NN N

B

n n for nw n

elsewhere

+ + =

5 52[ ] 0.42 0.5cos 0.08cos [ ] [ ] [ ][0] 1 [0] [0] [0] 1 0.5

[ 1] [1] 0.8942 [ 1] [1] [1] [1] 0.3183 0.8942 0.2846

[ 2] [2] 0.5098 [ 2] [2] [2] [2] 0 0.5098 0

[ 3] [3]

B B

B B

B B B

B B B

B B

d

d

d

d

n nw n h n h n w nw h h w

w w h h h w

w w h h h w

w w

= + + == = =

= = = = = =

= = = = = =

= =

X

X

X

0.2008 [ 3] [3] [3] [3] 0.106 0.2008 0.0213

[ 4] [4] 0.0402 [ 4] [4] [4] [4] 0 0.0402 0

[ 5] [5] 0 [ 5] [5] [5] [5] 0.06366 0 0

B

B B B

B B B

d

d

d

h h h w

w w h h h w

w w h h h w

= = = =

= = = = = =

= = = = = =

X

X

X

Ex3:Design a high pass filter to meet the following specifications: Cutoff frequency = 250 Hz, Sampling

frequency 1KHz. Filter length 7.

22

c

S

F

Fc

= = rad/sec

| |

2

1, for[ ]

0,jd

H e

else

=

2

2

1[ ] [ ]

21

1 12

sinsin 1

2

c

c

j jnd

jn jn-

h n H e e d

e d e d

n nc

n n

=

= +

= =


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13

[0] 0.5; [ 1] [1] 0.3183; [ 2] [2] 0; [ 3] [3] 0.1061;d d d d d d d h h h h h h h= = = = = = =

{ }

12

1

[ ] [0] [ ]

N

n

n nH z h h n z z

=

= + + . The realizable filter transfer function1

2[ ] [ ]N

H z H zz

= .

Ex4: Design a Band Pass Filter with the following specifications:1 2

100Hz: 200Hzc cF F= = and

1KHzSF = . Filter length = 9.

Ans: 11

1002 2 0.2

1000SF

cc

F = = = and

22

2002 2 0.4

1000SF

cc

F = = =

1 2

2 1

1 1[ ] [ ] 1 1

2 2sin(0.4 ) sin(0.2 )

0.4 0.20.4 0.2

c c

c c

j jn jn jn

d

-

-h n H e e d e d e d

n n

n n

= = +

=

A suitable window function [ ]w n for 4 4n is computed and can be used to determine [ ] [ ] [ ]dh n h n w n= .

{ }

12

1

[ ] [0] [ ]

N

n

n nH z h h n z z

=

= + + . The realizable filter transfer function1

2[ ] [ ]N

H z H zz

= .

Ex5: Design a Band Stop Filter with the following specifications:1 2

100Hz: 200Hzc cF F= = and

1KHzSF = . Filter length = 9.

Ans: 11

1002 2 0.2

1000SF

cc

F = = = and

2

2

2002 2 0.4

1000SF

cc

F = = =

2 1

1 1

1[ ] [ ]

21

1 1 12

sin(0.2 ) sin(0.4 )[ ] 0.2 0.4

0.2 0.4

c c

c c

j jnd

jn jn jn

d

-

- -

h n H e e d

e d e d e d

n nh n

n n

=

= + +

=


14/19


14

Kaiser window: In previous sections, we discussed four different types of windows. The values of samples

of this windows depend only upon N, i.e., length of window. In other words, the width of the main lobe and

attenuation of side lobes depends only upon length N of window. They cannot controlled independently.

However, Kaiser window allows separate control of width of the main lobe and attenuation of side lobes. It

provides a large main lobe width for the given stop band attenuation, which implies the sharpest transition

width.

The Kaiser window is defined as

( )

[ ]

22

0 11 1

2 20

1

[ ] ,

0,

NN N

k

nI

w n for nI

elsewhere

=

where is an adjustable parameter, and [ ]0I x is the modified zeroth order Bessel function of the first

kind of order zero given by

[ ]2 2 2 2 3

0 2 2 21

21 0.25 (0.25 ) (0.25 )

1 1 . . .! 2 (1!) (2!) (3!)k

kx x x x

I xk=

= + = + + + +

Design: Consider the LPF specification shown in figure. Let the pass band

ripple and minimum stop band attenuation in decibels are given by

120log

1p

+=

and 20logS = respectively.

The transition width S pB = rad/sec and cutoff frequency

2

S pc

+= ., where p and S are pass band and stop band frequencies.

Design Steps:

Step1: Determine [ ]dh n using Inverse DTFT formula1

2[ ] [ ]j jnd dh n H e e d

= .

Use2

S pc

+=

Step2: Choose 1 2min( , ) = , where0.05

1 10 S = and

0.05

2 0.05

10 1

10 1

p

p

=

+.

Step3: Determine new stop band attenuation 20logS = .Step4: Determine the parameter from the following equation

0.4

0, 21

0.5842( 21) 0.07886( 21), 21 50

0.1102( 8.7), 50

S

S S S

S S

for

for

for

= + >


15/19


15

Step5: Choose the parameterD using equation

0.9222, 21

7.95, 21

14.36

S

SS

for

Dfor

= >

Step6: Choose the filter order for the lowest odd value ofNusing 1Sf D

NB

+ where Sf is

sampling frequency.Step7:Compute the window sequence

( )

[ ]

22

0 11 1

2 20

1

[ ] ,

0,

NN N

k

nI

w n for nI

elsewhere

=

Step8: Compute the modified impulse response using [ ] [ ] [ ]d kh n h n w n=

Step9: Determine the transfer function

12[ ] [ ]NH z H zz = , where { }

1

2

1

[ ] [0] [ ]

N

n

n nH z h h n z z

== + +

Ex6:Design an FIR filter satisfying the following specifications 0.1dBp ; 44.0dBS

20p = rad/sec; 30S = rad/sec; 100Sf = rad/sec;

Ans: From the given specifications 10S pB = = rad/sec; 252

S pc

+= = rad/sec;

2(Radians) 25

100 2c

= = ;

Step1: 1, | 20,

[ ] forjdelse

|H e

=

Then 2

2

sin1[ ]2

dnh n

n

= .

Step2: 0.05 0.05(44) 3X1 10 10 6.309 10S = = = , and

0.053X2 0.05

10 15.7563 10

10 1

p

p

= =+

3X1 2min( , ) 5.7563 10 = =

Step3: New 20log 44.797dBS = = .

Step4: For 44.797dBS = , 0.40.5842( 21) 0.07886( 21) 3.9524S S = + = .

Step5: For 44.797dBS = ,7.95

2.56614.36

SD

= = .

Step6: 100(2.566)1 1 26.6610

Sf DNB

+ = + = . 27N = .

Step7:

( )

[ ]

22

0 1

0

1

[ ] , 13 13N

k

nI

w n for nI

=

For 3.9524 = , [ ] 10.73790I =


16/19


16

[ ][ ]

[ ]

[ ]

[ ][ ]

0[0] 1

03.9406 10.73790[ 1] [1] 0.9899

10.8468 10.84683.9073 10.41630[ 2] [2] 0.9603

10.8468 10.84683.8457

9.89640[ 3] [3] 0.912410.8468 10.8468

3.7606 9.20180[ 4] [4]10.8468 10

Iwk I

Iw wk k

Iw wk k

Iw wk k

Iw wk k

= =

= = = =

= = = =

= = = =

= = =

[ ]

[ ]

[ ]

[ ]

0.84835.8468

3.623 8.18560[ 5] [5] 0.7546510.8468 10.8468

3.5062 7.41680[ 6] [6] 0.6837810.8468 10.8468

3.3305 6.40250[ 7] [7] 0.5902710.8468 10.8468

3.1154 5.36130[ 8] [8]

10.8468 10.

Iw wk k

Iw wk k

Iw wk k

Iw w

k k

=

= = = =

= = = =

= = = =

= = =

[ ]

[ ]

[ ]

0.49428

84682.852 4.33310[ 9] [9] 0.399510.8468 10.8468

2.5257 3.35530[ 10] [10] 0.3093410.8468 10.8468

2.1063 2.45740[ 11] [11] 0.226510.8468 10.8468

[1.52] 1.66660[ 12] [12]10.8468

Iw wk k

Iw wk k

Iw wk k

Iw wk k

=

= = = =

= = = =

= = = =

= = =

[ ]

0.153610.8468

0 10[ 13] [13] 0.092210.8468 10.8468

Iw wk k

=

= = = =

The impulse response are given belown [ ]dh n [ ] [ ] [ ]dh n h n w nk=

0 0.5 0.5

1 0.318 0.31479

2 0 0

3 -0.106 -0.0967

4 0 0

5 0.06366 0.4804

6 0 0

7 -0.0454 -0.0268

8 0 0

9 0.03536 0.01412610 0 0

11 -0.0289 -0.006546

12 0 0

13 0.02448 0.002267

The Transfer function is given by1

2[ ] [ ]N

H z H zz

= , where { }

12

1

[ ] [0] [ ]

N

n

n nH z h h n z z

=

= + +


17/19


17

Solved Problems:

Ex7:Designa a LPF with the following specifications2 , |

4

0,[ ]

j forjd

else

e |H e

=

.

Ans: For the given specifications of a LPF with4c

= , 2j e indicates the delay of 2 units. This implies

12

2

N = or 5N= (Filter length).

Method 1: The filter length of 5 has been reinforced by defining rectangular window function for 0 4n .It is to be noted that [ ] [ ] [ ]dh n h n w n= and in this case both [ ]dh n and [ ]w n are defined with the view of

obtaining a causal filter directly, unlike in the previous examples where we first obtained [ ]h n

for 1 12 2

N Nn and then causilised [ ]h n by right shift so that [ ] 0h n = for 0n < .

Now

2

4

4

1 1[ ] [ ]

2 2sin ( 2)1

24 ( 2)

c

c

j jn j jn

d d h n H e e d e e d

nn

n

= =

=

1[2] 0.25

4dh = = ,

[0] [4] 0.15923d dh h= = and

[1] [3] 0 .22519d dh h= =

For rectangular window,

[0] [1] [2] [3] [5] 1R R R R Rw w w w w= = = = =

and [ ] [ ] [ ]dh n h n w n=

[0] [4] [0] [0] 0.15923

[1] [3] [1] [1] 0.22519[ ] [2] [2] 0.25

Rd

RdRd

h h h w

h h h wh n h w

= = =

= = =

= =

1 4

0 0

[ ] [ ] [ ]N

n n

n nH z h n z h n z

= =

= =

1 2 3 4

1 2 3 4

[0] [1] [2] [3] [4]

0.15923 0.22519 0.25 0.22519 0.15923

h h z h z h z h z

z z z z

= + + + +

= + + + +

For Hamming window: defined for 1 12 2

N Nn

1 12 2( 1)/2

2

[ ] 0.54 0.46 cos ,

0.54 0.46 cos , 2 2

N NHm N

nw n for n

n for n

= +

= +

[0] 1; [ 1] [1] 0.5400; [ 2] [2] 0.0800Hm Hm Hm Hm Hmw w w w w= = = = =

Causal Hamming window coefficients are given by' ' ' ' '[2] 1; [0] [4] 0.0800; [1] [3] 0.5400Hm Hm Hm Hm Hmw w w w w= = = = =

Then '[ ] [ ] [ ]Hmdh n h n w n=

'

'

'

[2] [2] [2] 1 2.5 0.25

[0] [4] [0] [0] 0.0800 0.15923 0.01274

[1] [3] [1] [1] 0.22519 0.5400 0.1216

Hmd

Hmd

Hmd

h h w

h h h w

h h h w

= = =

= = = =

= = = =

X

X

X


18/19


18

1 41 2 3 4

0 01 2 3 4

[ ] [ ] [ ] [0] [1] [2] [3] [4]

0.01274 0.1216 0.25 0.1216 0.01274

N

n n

n nH z h n z h n z h h z h z h z h z

z z z z

= =

= = = + + + +

= + + + +

Method 2: Use Causal Hamming window defined as

( 1)/2

2

00.54 0.46cos

0.54 0.46cos

[ ] ,

4,

1

0NHm

w n for

f

n n Nn or n

=

=

[2] 1;[0] [4] 0.0800;[1] [3] 0.5400

Hm

Hm Hm

Hm Hm

ww ww w

== == =

which are equivalent to ' [ ]Hmw n calculated in method1.

1 41 2 3 4

0 0

[2] [2] [2] 1 2.5 0.25

[0] [4] [0] [0] 0.0800 0.15923 0.01274

[1] [3] [1] [1] 0.22519 0.5400 0.1216

[ ] [ ] [ ] [0] [1] [2] [3] [4]

0.01274 0.1216

Hm

Hm

Hm

d

d

dN

n n

h h w

h h h w

h h h w

n nH z h n z h n z h h z h z h z h z

z

= =

= = =

= = = =

= = = =

= = = + + + +

= +

X

X

X

1 2 3 40.25 0.1216 0.01274z z z + + +

Method 3: Consider the magnitude function as , |4

| [ ] 1|j fordH e |= and design the filter in usual

procedure. The phase information is useful in determining the filter length.

4

4

1 1[ ] [ ] 1

2 2sin1

4

c

c

j jn jn

d d h n H e e d e d

n

n

= =

=

[0] 0.25[ 1] [1] 0.2251[ 2] [2] 0.1592

d

d d

d d

hh hh h

= = = = =

For Hamming window: defined for 1 12 2

N Nn

1 12 2( 1)/2

2

[ ] 0.54 0.46 cos ,

0.54 0.46 cos , 2 2

N NHm N

nw n for nn for n

= +

= +

[0] 1; [ 1] [1] 0.5400; [ 2] [2] 0.0800Hm Hm Hm Hm Hmw w w w w= = = = =

Then [ ] [ ] [ ]Hmdh n h n w n=

122

1 01 2 2

[0] [0] [0] 1 2.5 0.25

[ 1] [1] [1] [1] 0.22519 0.5400 0.1216

[ 2] [2] [1] [1] 0.0800 0.15923 0.01274

[ ] [0] [ ]{ } [ ]{ }

[0] [1]( ) [2]( )

0.25

Hm

Hm

Hm

d

d

dN

n n

h h w

h h h w

h h h w

n n n nH z h h n z z h n z z

h h z z h z z

= =

= = =

= = = =

= = = =

= + + = +

= + + + += +

X

X

X

1 2 20.1216 0.1216 0.01274 0.01274z z z z + + +

The realizable filter1

2 2[ ] [ ] [ ]N

H z H z H zz z = = .


19/19


19

( )1 2 21 2 3 4

2 2[ ] [ ] 0.25 0.1216 0.1216 0.01274 0.01274

0.01274 0.1216 0.25 0.1216 0.01274

H z H z z z z z

z z z z

z z

= = + + + += + + + +

Observe that All three methods gives the same results.

The Causal window function are given below

Hanning:( 1)/2

0.5 0.5 co 0 1,s[ ]n NHnw n f n Nor= .

Hamming:( 1)/2

0.54 0.46c[ ] ,os 0 1H Nm nn Nnw for = .

Bartlett:( 1)/2 ( 1

1) 2/

122

20.42 0.5 cos 0.08[ ] ,cosN

NB N

Nw n fon nn r + = .

Ex8: Design a linear phase FIR filter using Hamming and Hanning windows for the following desired

frequency response3 , |

4

0,[ ]

j forjd

else

e |H e

=

.

Ans: This is a HPF with cut-off frequency4

c

= and1

3 72

NN

= = .

3 3

4

1 1[ ] [ ]2 2

sin ( 3)

( 3)

cc

j jn j jn j jnd d

h n H e e d e e d e e d

n

n n

= =

=

[3] 0.75

[0] [6] 0.07506

[1] [5] 0.15923

[2] [4] 0.22586

d

d d

d d

d d

h

h h

h h

h h

=

= =

= =

= =

For Hanning window:3

0.5 0.5c[ ] 0 6os ,Hn

nw n for n=

[3] 1

[0] [6] 0

[1] [5] 0.25

[2] [4] 0.75

Hn

Hn Hn

Hn Hn

Hn Hn

w

w w

w w

w w

=

= =

= =

= =

Now [ ] [ ] [ ] { 0, 0.0398, 0.16939,0.75, 0.16939, 0.0398,0 }Hmdh n h n w n

= =

Causal 1 2 3 4 5[ ] 0.0398 0.16939 0.75 0.16939 0.0398H z z z z z z = +

For Hamming window:3

0.54 0.46[ ] 0 6os ,cHmnw n for n=

[3] 1

[0] [6] 0.08

[1] [5] 0.31

[2] [4] 0.77

Hm

Hm Hm

Hm Hm

Hm Hm

w

w w

w w

w w

=

= =

= =

= =

Now [ ] [ ] [ ] { 0.006, 0.04936, 0.17391,0.75, 0.17396, 0.04936, 0.006 }Hndh n h n w n

= =

Causal 1 2 3 4 5 6[ ] 0.006 0.04936 0.17391 0.75 0.17396 0.04936 0.006H z z z z z z z = +

fwhomfbrhi27435fwhomfbrhi

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