Pulse Characteristics of Linear Phase Fir Digital Filter

must satisfy condition  1  1 1

j 

6. The undulation in the stop band  2 . In the stop band the frequency amplitude response

j 

must satisfy the conditions:

H bp ( e )

 2

Actual filter has

  p ,  1 and  2The smaller the frequency amplitude characteristic, the larger the frequency amplitude characteristic.

almost rectangular in shape, so the signal selectivity is better.

4.2. Pulse characteristics of linear phase FIR digital filter

FIR digital filters have a finite impulse response h(n) , so the system transfer function

To be :

N  1

H  z  h ( n ) z  nn  0

Since the impulse response h(n) is finite, the FIR filter is always stable, which means that all

The poles of the system function H(z) lie within the digital unit circle of the FIR digital filter:

z  1 . Frequency response



N  1

H ( z )  h ( n ) e j  n  A ( e j ) e j ( )

n  0

For linear phase FIR digital filters:

     

(4.12)



In which and are constants , and is the propagation time of the signal through the filter:

d 



d 

(4.13)

From this we can see that all frequency components of the signal passing through the linear phase FIR digital filter are delayed equally, so the signal is not distorted.

Because H ( e j )

cyclical

2 So we only study the boundary response.

frequency

H ( e j ) and phase  (  ) when

(    )

or (0   2 ) .

On the other hand, if the digital filter has an impulse response h(n) that is a real sequence, then according to the properties of the Fourier transform we have:

 H( e j ) 



H( e j  )

          

So

H ( e j )

is an even and symmetric function, and

(  )

is odd and antithetical

symmetric. Therefore, when the impulse response h(n) is a real sequence, it is only necessary to study the digital filter in

interval (0   ) .

According to (4.12), there are 2 cases of linear phase FIR filter:

1.  0  (  )   

2.  0  (  )   

a. Case  0  (  )   

Expanding Euler's formula, the frequency response is expressed as:

On the other hand there are:

H(e j ) A(e j )e j A(e j )  cos(  ) j sin(  ) 

H(e j ) A(e j )cos(  ) jA(e j ) sin(  ) (4.14)

  

N  1 N  1

H(e j ) h(n)e j  n h(n). cos(  n) j sin(  n)

n  0 n  0

N  1 N  1

H(e j )  h(n).cos(  n) j  h(n) sin(  n)

n  0 n  0

From (4.14) and (4.15) we have:

(4.15)



N  1

A ( e j ) cos(  )  h ( n ) cos(  n )

n  0



N  1

A ( e j ) sin(  )  h ( n ) sin(  n )

n  0

Infer:

N  1



h ( n ) sin( n  )

N  1

tg (  ) n  0

 h ( n ) cos( n  )

n  0

Since sin(0) = 0 and cos(0) = 1 , we can rewrite the above expression as:



N  1

h ( n ) sin( n  )

N  1

tg (  ) n  0

h (0)   h ( n ) cos( n  )

n  1

From here there are 2 cases  0 is a zero-phase filter, and  0



N  1

h ( n ) sin( n  )

N  1

 With  0 : tg (0  ) n  0 0

h (0)   h ( n ) cos( n  )

n  1

That is :

ì 0 h( n ) = ì

yes yes

ïî ¹

" n ¹ 0

.

0 n = 0

Such a filter is impractical and impossible to implement, because the signal transmitted through the filter is always delayed, even if the delay time is minimal.

 With  0

sin(  )

N  1



h(n).sin(n  )

tg( ) n  00



cos(  )

N  1

N  1

h(n).cos(n  )

n  0

N  1

Or:

tg (  )  sin(  )  h ( n ) cos( n  )  cos(  )  h ( n ) sin( n  )

n  1 n  0

N  1 N  1

So:

tg (  )  sin(  )  h ( n ) cos( n  )  cos(  )  h ( n ) sin( n  )  0

n  1 n  0

Continuing to transform trigonometry will get the equation:



N  1

h ( n ) sin  ( n  )

n  0

N  1

2

(4.16)

(4.17)

And: h(n) = h(N-1-n) with n  (0, N-1) (4.18)

According to (4.18) the impulse response h(n) of the linear phase FIR digital filter when the array is symmetric.

 0 is

- When  0

- When  0

Example 1 :

and N is odd, called a linear phase FIR digital filter of the first type.

and N is even is called a type 2 linear phase FIR digital filter.

Linear phase FIR digital filter has

h(2) =2 .

(  )  , with N =5 and h(0) = -1, h(1) = 1,

Find and plot the impulse response h(n ) of the filter.

Prize :

Because  0 and N is odd so this is a linear phase FIR digital filter of the first type. According to (4.17) we have:

 N  1  5  1  2

2 2

According to (4.18) we have: h(n) = h(5-1-n) = h(4-n)

So :

Impulse response graph h(n) :

h(4) = h(0) = -1

h(3) = h(1) = 1 h(2) = 2

h ( n )

2

1

-1 1 2 3 n

Figure 4.10. h(n) of a linear phase FIR filter of the first type.

Example 2 :

Linear phase FIR filter has (  )   

, with N = 4 and h(0) = -1, h(1) = 1.

Find and plot the impulse response h(n) of the filter.

Prize:

Because  0 and N is even so this is a type 2 linear phase FIR filter.

According to (4.17) we have:

N  1 4  1  1.5

2 2

According to (4.18) we have: h(n) = h(4-1-n) = h(3-n)

So :

Impulse response graph h(n):

h(3) = h(0) = -1

h(2) = h(1) = 1

1 1

-1

0 1 2 3

n

h ( n )

Figure 4.11. h(n) of a type 2 linear phase FIR filter.

Comment :

- Type 1 and type 2 linear phase FIR digital filters have symmetric impulse responses h(n) like ideal filters.

- The center of symmetry of h(n) coincides with the sample at n = .

+ If N is odd then is an integer and the axis of symmetry h(n) coincides with the sample at

N - 1

n = .

2

N - 1

+ If N is even then is a decimal and the axis of symmetry lies between the two denominators.

N

in

n = and

2

n = .

2

b. Case  0  (  )   

By the same transformation as in the above case, we get.

 h ( n ) sin   (  )  0 

The above Fourier equation has only one solution at:

(4.19)

N  1 ;

2



2

(4.20)

h(n) = -h(N-1-n) with n  (0, N  1) (4.21)

According to (4.21), the impulse response h ( n ) of the linear phase FIR filter in the case

 0 is an antisymmetric sequence.

- When  0

- When  0

Example 3:

and N is odd, called a type 3 linear phase FIR digital filter. and N is even, called a type 4 linear phase FIR digital filter.

Given a linear phase FIR filter with (  )   , with N=7 and h (0) = -1,

h (1) = -0.5 and h (2) = 1.5 .

Find and plot the impulse response of the filter.

Prize:

Because  0 and N is odd so this is a type 3 linear phase FIR filter.

According to (4.20) we have:

 N  1  7  1  3

2 2

According to (4.21) we have: h(n) = -h(7-1-n) = -h(6-n)

So :

Impulse response graph h(n):

h(6) = -h(0) = 1

h(5) = -h(1) = 0.5

h(4) = -h(2) = 1.5 h(3) = 0

h ( n )

1.5

1

0 1

-0.5

-1

2

3

4

-0.5

5

6

n

-1

Figure 4.12. h(n) of a type 3 FIR filter.

Example 4 :

Given a linear phase FIR filter with (  )   with N=4 and h(0) = -1, h(1) = 1 Find and plot the impulse response of the filter.

Prize:

Because  0 and N are even so this is a type 4 linear phase FIR filter

According to (4.20) we have:

N  1 4  1  1.5

2 2

According to (4.21) we have: h(n) = -h(4-1-n) = -h(3-n)

So :

Impulse response graph h(n):

1

1

0 1 2

-1

3

-1

n

h ( n )

h(3) = -h(0) = 1

h(2) = -h(1) = -1

Figure 4.13. h(n) of a type 4 FIR filter.

Comment :

worthy

-The antisymmetric center of h(n) is at n = .

+ If N is odd then is an integer and the antisymmetric center h(n) coincides with the denominator at

N - 1

n = and there h(n) = 0 .

2

+ If N is even then is a decimal and the anti-symmetric center lies between the two samples at

N - 1 N

n = and

2

n = .

2

Thus, there are four types of linear phase FIR filters with (  )   :

- Type 1 filter:

- Type 1 filter:

- Type 1 filter:

- Type 1 filter:

 0 , N is odd , impulse response h(n) is symmetric.

 0 , N is even, the impulse response h(n) is symmetric.

 / 2 , N odd , impulse response h(n) is antisymmetric.

 / 2 , N is even , the impulse response h(n) is antisymmetric.

4.3. Frequency characteristics of linear phase FIR digital filter

4.3.1. Frequency characteristics of type 1 linear phase FIR filter

Linear phase FIR filter has (  )   and N is odd, the frequency response is:

N  1

H ( e j )   h ( n ) e j  n  A ( e j ) e j n  0

Because N is odd, expand the above expression into the sum of 3 components:

H ( e j )  

N  1

h ( n ) e j  n    h  N  1  e j 

N  1



2 

N  1



h ( n ) e j  n 

 

 n  0

  



   2 

  

 n  N  1 

2 

Change the variable to the third part, let m = (N-1-n) => n = (N-1-n),

When

n N  1 

then

m   N  1  

1 

2



1 

2



When n = (N-1) then m = 0 :

N  1  1 

N  1

H ( e j )   2

h ( n ) e j  n    h  N  1  e j 2

  

h ( N  1  m ) e j ( N  1  m ) 

0



  2 

  

n  0

   

 m N  1  1 

2 

Reverse the index and change the variable of the 3rd component according to n :

N  1  1 

N  1

N  1  1 

H ( e j )   2

h ( n ) e j  n    h  N  1  e j 2

   2

h ( N  1  n ) e j ( N  1  n ) 





 n  0

  

    2 

  

   n  0  

Since the linear phase FIR filter of type 1 has h(n) =h(N-1-n), then:

N  1

N  1  1 

H ( e j ) 

N  1 

j 

2



2 

h ( n )  e j  n e j  ( N  1  n )  

(4.22)

h  e

2 



 n  0

 j  N  1  

j  N  1  n 



 

 j  N  1  n  

In which:  e j  n e j  ( N  1  n )  e



2  e 2

 



e 2 

 

  j N  1

  N  1  

Or:

e j  n

e j  ( N  1  n )

 2 e



2 cos   

n  

Therefore (4.22) is reduced to the form:

N  1

N  1  1

  2

 

 N  1 

j N  1 

 j 2

  N  1

 j 2 

H ( e

) h 

 e 2

 2 h ( n ) cos   

n   e 

Or:

2 

n  0

  2  

N  1  1

N  1 

N  1 

j  

N  1 2

  N  1

 j  

2 

j 

2

H ( e

)   h 

  

2 h ( n ) cos   

n   e 

 e 

   2 

n  0

  2    