Hi,
I have learnt some amount of DSP recently by reading a few artickes
(and using this newsgroup). I had a question regarding cut-off
frequency.
Given an FIR Filter with Transfer function:
H(z,N) = J . sigma_k={0...N-1} (z^{k-N}[k -1 ])
where J is a constant, is it possible to get a relation for the cutoff
frequency in terms of N (in closed form)? In general, how does one
determine the cutoff frequency of a FIR filter?
Thank you
Dinkar.
cutoff frequency of a FIR Filter
Started by ●September 23, 2004
Reply by ●September 23, 20042004-09-23
dnb wrote:> Hi, > > I have learnt some amount of DSP recently by reading a few artickes > (and using this newsgroup). I had a question regarding cut-off > frequency. > > Given an FIR Filter with Transfer function: > > H(z,N) = J . sigma_k={0...N-1} (z^{k-N}[k -1 ]) > > where J is a constant, is it possible to get a relation for the cutoff > frequency in terms of N (in closed form)? In general, how does one > determine the cutoff frequency of a FIR filter? > > Thank you > Dinkar.I can't quite make out your notation, but it should be possible. In general one determines the characteristics of a FIR filter by doing a DFT and plotting the result. As far as I know this is the only generally practical method since the taps on an FIR filter can be quite arbitrary. On the up side, one often designs one's FIR filter using the DFT, and the DFT (even for non-power of two tap counts) isn't very expensive on modern processors with modern math software. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 23, 20042004-09-23
Tim Wescott wrote: ...> In general one determines the characteristics of a FIR filter by doing a > DFT and plotting the result. ...Of course, you mean an FFT of the impulse response (which is the same as the list of coefficients). Jerry -- ... they proceeded on the sound principle that the magnitude of a lie always contains a certain factor of credibility, ... and that therefor ... they more easily fall victim to a big lie than to a little one ... A. H. �����������������������������������������������������������������������
Reply by ●September 23, 20042004-09-23
Jerry Avins wrote:> Tim Wescott wrote: > > ... > >> In general one determines the characteristics of a FIR filter by doing >> a DFT and plotting the result. ... > > > Of course, you mean an FFT of the impulse response (which is the same as > the list of coefficients). > > JerryAny DFT will do, it doesn't have to be fast. But yes, a DFT on the list of coefficients. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Reply by ●September 23, 20042004-09-23
"Tim Wescott" <tim@wescottnospamdesign.com> wrote in message news:10l63sf9lee040e@corp.supernews.com...> Jerry Avins wrote: > > Tim Wescott wrote: > > > >> In general one determines the characteristics of a FIR filter by doing > >> a DFT and plotting the result. ... > > > > > > Of course, you mean an FFT of the impulse response (which is the same as > > the list of coefficients). > > Any DFT will do, it doesn't have to be fast.It seems in modern common usage, FFT has become synonymous with the Fourier transform. Audio analysis products are "FFT-based" as opposed to working in the time domain. It is rarely pointed out that FFT is just a computationally-efficient algorithm for computing the DFT, not some special type of transform itself. Just as the bubble sort is a fast algorithm of sorting data, but there are other algorithms that can also generate the same sorted data.
Reply by ●September 23, 20042004-09-23
"dnb" <dbhat2@yahoo.com> wrote in message news:fa0deac8.0409230739.1b0b14dc@posting.google.com...> Hi, > > I have learnt some amount of DSP recently by reading a few artickes > (and using this newsgroup). I had a question regarding cut-off > frequency. > > Given an FIR Filter with Transfer function: > > H(z,N) = J . sigma_k={0...N-1} (z^{k-N}[k -1 ]) > > where J is a constant, is it possible to get a relation for the cutoff > frequency in terms of N (in closed form)? In general, how does one > determine the cutoff frequency of a FIR filter? >Dinkar, I'm not sure about your notation. Did you mean: N-1 H(z,N) = J*sum {(k-1)*z^(k-N)] k=0 so that H(z,4) would be: 3 J*sum{(k-1)*z^(k-4) = J*(-z^-4 + z^-2 + 2*z^-1) where J is a constant 0 ????? (note that z^-3 has a zero coefficient) Normally there is a z^0 term which only represents a unit delay difference from the above. Shifting one delay and reversing the order of the expression and without bothering with the scale factor the above becomes: H(z) = 2*z^0 + z^-1 - z^-3 This makes me think that you meant something slightly different: N J*sum{(k)*z^[k-N)]} = J*(-z^-3 + 2*z^-2 + 3*z^-1 + 4*z^0) where J is a constant k=1 So the filter coefficients are: [4 3 2 1] for N=4 and [5 4 3 2 1] for N=5, etc. Either way, these coefficient sequences aren't generally low pass filters - although some cases could look like a lowpass in some sense. One thing one can say is that if all the coefficients are positive, then the highest gain will occur at zero frequency - because the gain at zero frequency is the sum of the coefficients. The output of a filter with real coefficients can never be greater than the sum of the magnitudes of the coefficients - thus, if all the coefficients are positive, can never be greater than the sum of the coefficients which met at zero frequency (and possibly other frequencies depending on if the temporal sequence of coefficients has periodic zeros in it. You cannot state the "cutoff frequency" of a FIR filter because you cannot assume it is a lowpass filter. If you state "a priori" that it *is* a lowpass filter then you can estimate its transition band width being the reciprocal of the length of the filter. Because a lowpass filter impulse response resembles a sinc function, you might estimate the bandwith from the distance between zeros in the tails of the impulse response. Fred
Reply by ●September 23, 20042004-09-23
Jon Harris wrote:> "Tim Wescott" <tim@wescottnospamdesign.com> wrote in message > news:10l63sf9lee040e@corp.supernews.com... > >>Jerry Avins wrote: >> >>>Tim Wescott wrote: >>> >>> >>>>In general one determines the characteristics of a FIR filter by doing >>>>a DFT and plotting the result. ... >>> >>> >>>Of course, you mean an FFT of the impulse response (which is the same as >>>the list of coefficients). >> >>Any DFT will do, it doesn't have to be fast. > > > It seems in modern common usage, FFT has become synonymous with the Fourier > transform. Audio analysis products are "FFT-based" as opposed to working in the > time domain. It is rarely pointed out that FFT is just a > computationally-efficient algorithm for computing the DFT, not some special type > of transform itself. Just as the bubble sort is a fast algorithm of sorting > data, but there are other algorithms that can also generate the same sorted > data.That's how language degenerates, to be reborn in new form. But Tim is right and I was sloppy. Jerry -- ... they proceeded on the sound principle that the magnitude of a lie always contains a certain factor of credibility, ... and that therefor ... they more easily fall victim to a big lie than to a little one ... A. H. �����������������������������������������������������������������������
Reply by ●September 23, 20042004-09-23
"Jon Harris" <goldentully@hotmail.com> wrote in message news:2rgknpF195dcvU1@uni-berlin.de...> "Tim Wescott" <tim@wescottnospamdesign.com> wrote in message > news:10l63sf9lee040e@corp.supernews.com... > > Jerry Avins wrote: > > > Tim Wescott wrote: > > > > > >> In general one determines the characteristics of a FIR filter bydoing> > >> a DFT and plotting the result. ... > > > > > > > > > Of course, you mean an FFT of the impulse response (which is the sameas> > > the list of coefficients). > > > > Any DFT will do, it doesn't have to be fast. > > It seems in modern common usage, FFT has become synonymous with theFourier> transform. Audio analysis products are "FFT-based" as opposed to workingin the> time domain. It is rarely pointed out that FFT is just a > computationally-efficient algorithm for computing the DFT, not somespecial type> of transform itself. Just as the bubble sort is a fast algorithm ofsorting> data, but there are other algorithms that can also generate the samesorted> data.Bubble sort is O(n^2). There are quite a few faster sorts. See http://en.wikipedia.org/wiki/Sort_algorithm Best wishes, --Phil Martel> >
Reply by ●September 23, 20042004-09-23
Thanks for the reply!
I should restate my question - I hope it makes sense. I have a low
pass
FIR filter with transfer function whose coefficients are functions
of the sample size N. The function is:
H(z,N) = J. sigma_k{k=0,1,...(N-1)} [ (k-1). z^{k-N} ]
where J is a constant. Is it possible to determine/express the
bandwidth/cut-off frequency of the low-pass filter in terms of N?
Thanks,
Dinkar.
Tim Wescott <tim@wescottnospamdesign.com> wrote in message news:<10l5vrf5tl8as92@corp.supernews.com>...
> dnb wrote:
> > Hi,
> >
> > I have learnt some amount of DSP recently by reading a few artickes
> > (and using this newsgroup). I had a question regarding cut-off
> > frequency.
> >
> > Given an FIR Filter with Transfer function:
> >
> > H(z,N) = J . sigma_k={0...N-1} (z^{k-N}[k -1 ])
> >
> > where J is a constant, is it possible to get a relation for the cutoff
> > frequency in terms of N (in closed form)? In general, how does one
> > determine the cutoff frequency of a FIR filter?
> >
> > Thank you
> > Dinkar.
>
> I can't quite make out your notation, but it should be possible.
>
> In general one determines the characteristics of a FIR filter by doing a
> DFT and plotting the result. As far as I know this is the only
> generally practical method since the taps on an FIR filter can be quite
> arbitrary.
>
> On the up side, one often designs one's FIR filter using the DFT, and
> the DFT (even for non-power of two tap counts) isn't very expensive on
> modern processors with modern math software.
Reply by ●September 24, 20042004-09-24
On Thu, 23 Sep 2004 17:51:24 -0400, Jerry Avins <jya@ieee.org> wrote: (snipped)>> >> >> It seems in modern common usage, FFT has become synonymous with the Fourier >> transform. Audio analysis products are "FFT-based" as opposed to working in the >> time domain. It is rarely pointed out that FFT is just a >> computationally-efficient algorithm for computing the DFT, not some special type >> of transform itself. Just as the bubble sort is a fast algorithm of sorting >> data, but there are other algorithms that can also generate the same sorted >> data. > >That's how language degenerates, to be reborn in new form. But Tim is >right and I was sloppy. > >JerryHi, it's gets even worse. We use "FFT" to mean the popular "radix-2 FFT". But there many other forms of the FFT. Dinkar used the phrase "FIR filter". He should, rightly, have said "transversal FIR filter" or perhaps "convolutional FIR filter". I say that because there a "recursive" (using feedback) FIR filters. [-Rick-]
