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Notice from [link] for N even, A(ω) is an even function around ω=0 , an odd function around ω=π , and is periodic with period . This requires A(π)=0 .

For the case in [link] where K 1 = π / 2 , an odd symmetry is required of the form

h ( n ) = - h ( N - n - 1 )

which, for N odd, gives

H ( ω ) = j A ( ω ) e j M ω

with

A ( ω ) = n = 0 M - 1 2 h ( n ) sin ω ( M - n )

and for N even

A ( ω ) = n = 0 N / 2 - 1 2 h ( n ) sin ω ( M - n )

To calculate the frequency or amplitude response numerically, one must consider samples of the continuous frequency response function above. L samples of the general complex frequency response H ( ω ) in [link] are calculated from

H ( ω k ) = n = 0 N - 1 h ( n ) e - j ω k n .

for k = 0 , 1 , 2 , , L - 1 . This can be written with matrix notation as

H = F h

where H is an L by 1 vector of the samples of the complex frequency response, F is the L by N matrix of complex exponentials from [link] , and h is the N by 1 vector of real filter coefficients.

These equations are possibly redundant for equally spaced samples since A ( ω ) is an even function and, if the phase response is linear, h ( n ) is symmetric. These redundancies are removed by sampling [link] over 0 ω k π and by using a defined in [link] rather than h . This can be written

A = C a

where A is an L by 1 vector of the samples of the real valued amplitude frequency response, C is the L by M real matrix of cosines from [link] , and a is the M by 1 vector of filter coefficients related to the impulse response by [link] . A similar set of equations can be written from [link] for N odd or from [link] for N even.

This formulation becomes a filter design method by giving the samples of a desired amplitude response as A d ( k ) and solving [link] for the filter coefficients a ( n ) . If the number of independent frequency samples is equal to the number of independent filter coefficients and if C is not singular, this is the frequency sampling filter design method and the frequency response of the designed filter will interpolate thespecified samples. If the number of frequency samples L is larger than the number of filter coefficients N , [link] may be solved approximately by minimizing the norm A ( ω ) - A d ( ω ) .

The discrete time fourier transform with normalization

The discrete time Fourier transform of the impulse response of a digital filter is its frequency response, therefore, it is an important tool.When the symmetry conditions of linear phase are incorporated into the DTFT, it becomes similar to the discrete cosine or sine transform(DCT or DST). It also has an arbitrary normalization possible for the odd length that needs to be understood.

The discrete time Fourier transform (DTFT) is defined in [link] which, with the conditions of an odd length-N symmetrical signal, becomes

A ( ω ) = n = 1 M a ( n ) cos ( ω n ) + K a ( 0 )

where M = ( N - 1 ) / 2 . Its inverse as

a ( n ) = 2 π 0 π A ( ω ) cos ( ω n ) d ω

for n = 1 , 2 , , M and

a ( 0 ) = 1 K π 0 π A ( ω ) d ω

where K is a parameter of normalization for the a ( 0 ) term with 0 < K < . If K = 1 , the expansion equation [link] is one summation and doesn't have to have the separate term for a ( 0 ) . If K = 1 / 2 , the equation for the coefficients [link] will also calculate the a ( 0 ) term and the separate equation [link] is not needed. If K = 1 / 2 , a symmetry results which simplifies equations later in the notes.

Four types of linear-phase fir filters

From the previous discussion, it is seen that there are four possible types of FIR filters [link] that lead to the linear phase of [link] . These are summarized in [link] .

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Source:  OpenStax, Digital signal processing and digital filter design (draft). OpenStax CNX. Nov 17, 2012 Download for free at http://cnx.org/content/col10598/1.6
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