newbie: fir filter design

aine_canby@yahoo.com wrote:
> Thanks Jerry.
>
> As part of my class project, I have to produce the coefs myself.
>
> I'm happy enough now with what I have, just wondering though about the
> implementation of a 256 tap filter. Should I simply calculate for 255
> taps and then leave the last tap empty, or is there a version the
> following equation for even numbered taps -
>
> h[n] = (1/R)sinc(n/R) n =-16 to 16

Here you hit n/R = 0 for n = 127 or so. For even numbers of taps,
displace the sinc slightly:

h[n]= 1/R sinc((n-0.5)/R)

(it could be that the sign should be "+" instead of "-"...)
This should give you an even symmetric sinc pulse where

h[n] = h[N-n],  n=0,...,N/2.

Rune

Hi Andor!
Uff, I guess you are right. To be honest, I have copied + pasted this one 
sentence from the web page http://www.dspguru.com/info/faqs/fir/props.htm 
:-( I should be more carefull doing that!
Of course an FIR is also linear phase if it satisfies :  . Then it is called 
antisymmetric, isn't it? But in the end it looks also like a kind of 
symmetry to me ;-)

Regards,
Karin

"Andor" <an2or@mailcircuit.com> schrieb im Newsbeitrag 
news:1120036884.916167.293210@g44g2000cwa.googlegroups.com...
> Karin wrote:
>
>> A FIR filter is linear-phase if (and only if) its coefficients are 
>> symmetrical
>> around the center coefficient, that is, the first coefficient is the same 
>> as
>> the last; the second is the same as the next-to-last, etc.
>
> Hello Karin, I'm not sure about that second "if". I think hermitian
> symmetric FIRs are also linear-phase ...
>
> Regards,
> Andor
> 


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Uff, I guess you are right. To be honest, I have copied + pasted this one 
sentence from the web page http://www.dspguru.com/info/faqs/fir/props.htm 
:-( I should be more carefull doing that!
Of course an FIR is also linear phase if it satisfies : h[n] = - h[L-n-1] 
for n = 0...L-1  Then it is called antisymmetric, isn't it? But in the end 
it looks also like a kind of symmetry to me ;-)

Karin

"Andor" <an2or@mailcircuit.com> schrieb im Newsbeitrag 
news:1120036884.916167.293210@g44g2000cwa.googlegroups.com...
> Karin wrote:
>
>> A FIR filter is linear-phase if (and only if) its coefficients are 
>> symmetrical
>> around the center coefficient, that is, the first coefficient is the same 
>> as
>> the last; the second is the same as the next-to-last, etc.
>
> Hello Karin, I'm not sure about that second "if". I think hermitian
> symmetric FIRs are also linear-phase ...
>
> Regards,
> Andor
> 

Cheers everyone,

What about a high pass filter? For an odd number of taps I worked out
that you could just change the sign of every second coef.The first
being even the second odd etc. But what about an even number of taps?
You end up with -

b[n] = -b[ L-n-1] rather than b[n] = b[ L-n-1]

Does this give linear phase? The remez plot suggests it does, but I'm
still not sure. Here's my code for the lowpass and thanks again.

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

double sinc(double t);

#define PI ( 3.141592653589793 )
#define L 1024
#define R 5

int main()
{
   FILE *Fp;

   Fp = fopen("coef.dat", "w");

	for(int i=0;i<L;i++)
	{

		fprintf(Fp,"%lf\t",
1/double(R)*sinc((i-((L-1)/2.0))/double(R))*(0.54-0.46*cos(2*PI*i/double(L-1)))
   );
	}
	return EXIT_SUCCESS;
}

double sinc(double t)
{
	if(t)
		return sin(PI*t)/(PI*t);
	else
		return 1;
}

Karin,

you are right, antisymmetric FIR filters are also linear phase. I
looked it up [1]: the characterizing condition (if and only if)  for
linear phase response of an FIR filter with impulse response h[n] of
order N, with h[0] =/= 0 =/= h[N], is given by

h[n] = d h*[N-n],                    (1)

where d is a complex number of length 1 ( |d| = 1 ). If you want real
coefficients, then condition (1) reduces to

h[n] = +/- h[N-n],

ie. symmetric or antisymmetric impulse response.

Regards,
Andor

[1] Vaidyanathan, "Multirate Systems and Filter Banks",  see chapter
"Linear Phase Filters"

Yes, this is also a linear phase condition like Andor pointed out correctly.

Sorry, if I put you on the wrong way in my first post... but I corrected it 
here:

"Karin Dressler" <muncho@web.de> schrieb im Newsbeitrag 
news:d9ttqs$cg4$1@online.de...
> Uff, I guess you are right. To be honest, I have copied + pasted this one
> sentence from the web page http://www.dspguru.com/info/faqs/fir/props.htm
> :-( I should be more carefull doing that!
> Of course an FIR is also linear phase if it satisfies : h[n] = - h[L-n-1]
> for n = 0...L-1  Then it is called antisymmetric, isn't it? But in the end
> it looks also like a kind of symmetry to me ;-)
>
> Karin

PS: regarding your code, it is always safe to convert i to float or double 
before using it in your formula. Don't know about compiler behaviour in this 
special case, but if you put ( double ) i it is clear.

Karin

<aine_canby@yahoo.com> schrieb im Newsbeitrag 
news:1120045207.647633.311640@g47g2000cwa.googlegroups.com...
> Cheers everyone,
>
> What about a high pass filter? For an odd number of taps I worked out
> that you could just change the sign of every second coef.The first
> being even the second odd etc. But what about an even number of taps?
> You end up with -
>
> b[n] = -b[ L-n-1] rather than b[n] = b[ L-n-1]
>
> Does this give linear phase? The remez plot suggests it does, but I'm
> still not sure. Here's my code for the lowpass and thanks again.
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <math.h>
>
> double sinc(double t);
>
> #define PI ( 3.141592653589793 )
> #define L 1024
> #define R 5
>
> int main()
> {
>   FILE *Fp;
>
>   Fp = fopen("coef.dat", "w");
>
> for(int i=0;i<L;i++)
> {
>
> fprintf(Fp,"%lf\t",
> 1/double(R)*sinc((i-((L-1)/2.0))/double(R))*(0.54-0.46*cos(2*PI*i/double(L-1)))
>   );
> }
> return EXIT_SUCCESS;
> }
>
> double sinc(double t)
> {
> if(t)
> return sin(PI*t)/(PI*t);
> else
> return 1;
> }
> 

Wow, thank you! Again I've learned something!
Karin

"Andor" <an2or@mailcircuit.com> schrieb im Newsbeitrag 
news:1120046220.795848.195950@g44g2000cwa.googlegroups.com...
> Karin,
>
> you are right, antisymmetric FIR filters are also linear phase. I
> looked it up [1]: the characterizing condition (if and only if)  for
> linear phase response of an FIR filter with impulse response h[n] of
> order N, with h[0] =/= 0 =/= h[N], is given by
>
> h[n] = d h*[N-n],                    (1)
>
> where d is a complex number of length 1 ( |d| = 1 ). If you want real
> coefficients, then condition (1) reduces to
>
> h[n] = +/- h[N-n],
>
> ie. symmetric or antisymmetric impulse response.
>
> Regards,
> Andor
>
> [1] Vaidyanathan, "Multirate Systems and Filter Banks",  see chapter
> "Linear Phase Filters"
> 

Andor wrote:
> 
> Karin,
> 
> you are right, antisymmetric FIR filters are also linear phase. I
> looked it up [1]: the characterizing condition (if and only if)  for
> linear phase response of an FIR filter with impulse response h[n] of
> order N, with h[0] =/= 0 =/= h[N], is given by
> 
> h[n] = d h*[N-n],                    (1)
> 
> where d is a complex number of length 1 ( |d| = 1 ). If you want real
> coefficients, then condition (1) reduces to
> 
> h[n] = +/- h[N-n],

well, not quite, that would be sloppy and could be misleading. It still
would be

h[n] = d * h[N-n] where d = 1 or -1  
	
d has to be the same value for all n.


-jim

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