On Sun, 03 May 2009 14:05:23 -0500, "dkumar" <joindk@gmail.com> wrote:

>>
>>The value for N depends on what you are trying to do.  Why are you
>>taking an FFT?
>>
>>If you are trying to analyze a signal, the size of your FFT will
>>dictate the frequency resolution and one's ability to distinguish
>>neighboring tones.
>>
>>Freq. Resolution = Fs/N
>>
>>If you are trying to filter in the frequency domain, the size of your
>>FFT will be dictated by the length of the filter.
>>If you are trying to process Multi-Carrier modulated signals, your FFT
>>will be some multiple of a fundamental length dictated by the
>>communication system.
>>
>>FFT has many purposes and the reasons for choosing N will depend on
>>why you are using an FFT in the first place.
>>
>>-V
>>
>>
> hi
>So how do i decide on N for a sampled data system, I am sampling a input
>signal whose input frequency is carefully chosen and sampling frequency is
>2GHz. How do i  decide on number of points of FFT i should take?

To do what?  N, the number of points of the FFT, will be dependent on
your resolution requirements.  The FFT resolution will be ( 2 GHz / N
), where N is the number of points of the FFT.  Unless you take a very
large FFT, the resolution will be poor (1024 FFT = 1.953 Mhz per FFT
bin).  You could always decimate down to a lower rate, after applying
an anti-alias filter, and then apply the FFT.  The resolution would
then change to ( 2 GHz / ( D * N ) ), where D is the decimation
factor.

On May 3, 3:05&#4294967295;pm, "dkumar" <joi...@gmail.com> wrote:
> >The value for N depends on what you are trying to do. &#4294967295;Why are you
> >taking an FFT?
>
> >If you are trying to analyze a signal, the size of your FFT will
> >dictate the frequency resolution and one's ability to distinguish
> >neighboring tones.
>
> >Freq. Resolution = Fs/N
>
> >If you are trying to filter in the frequency domain, the size of your
> >FFT will be dictated by the length of the filter.
> >If you are trying to process Multi-Carrier modulated signals, your FFT
> >will be some multiple of a fundamental length dictated by the
> >communication system.
>
> >FFT has many purposes and the reasons for choosing N will depend on
> >why you are using an FFT in the first place.
>
> >-V
>
> &#4294967295;hi
> So how do i decide on N for a sampled data system, I am sampling a input
> signal whose input frequency is carefully chosen and sampling frequency is
> 2GHz. How do i &#4294967295;decide on number of points of FFT i should take?
>
> thanks.

This is basic time sampled signal stuff.  You need to understand the
effects of sample rate and sample duration.  If you sample at a rate
of fs for N samples, your buffer duration will be N/fs.  This tells
you something very basic about the signal content you will be able to
see.  If you don't know the answer, you need to go back to your books
or ask your professor.

Rick

"dkumar" <joindk@gmail.com> wrote in message
news:YtGdnbvDHcNud2DUnZ2dnUVZ_uednZ2d@giganews.com...
> >
> >The value for N depends on what you are trying to do.  Why are you
> >taking an FFT?
> >
> >If you are trying to analyze a signal, the size of your FFT will
> >dictate the frequency resolution and one's ability to distinguish
> >neighboring tones.
> >
> >Freq. Resolution = Fs/N
> >
> >If you are trying to filter in the frequency domain, the size of your
> >FFT will be dictated by the length of the filter.
> >If you are trying to process Multi-Carrier modulated signals, your FFT
> >will be some multiple of a fundamental length dictated by the
> >communication system.
> >
> >FFT has many purposes and the reasons for choosing N will depend on
> >why you are using an FFT in the first place.
> >
> >-V
> >
> >
>  hi
> So how do i decide on N for a sampled data system, I am sampling a input
> signal whose input frequency is carefully chosen and sampling frequency is
> 2GHz. How do i  decide on number of points of FFT i should take?
>
> thanks.
>
>

On 3 Mai, 21:05, "dkumar" <joi...@gmail.com> wrote:
> >The value for N depends on what you are trying to do. &#4294967295;Why are you
> >taking an FFT?
>
> >If you are trying to analyze a signal, the size of your FFT will
> >dictate the frequency resolution and one's ability to distinguish
> >neighboring tones.
>
> >Freq. Resolution = Fs/N
>
> >If you are trying to filter in the frequency domain, the size of your
> >FFT will be dictated by the length of the filter.
> >If you are trying to process Multi-Carrier modulated signals, your FFT
> >will be some multiple of a fundamental length dictated by the
> >communication system.
>
> >FFT has many purposes and the reasons for choosing N will depend on
> >why you are using an FFT in the first place.
>
> >-V
>
> &#4294967295;hi
> So how do i decide on N for a sampled data system, I am sampling a input
> signal whose input frequency is carefully chosen and sampling frequency is
> 2GHz. How do i &#4294967295;decide on number of points of FFT i should take?

Read the post you quote, and then conemplate what you read.
After that, try and answer some of the questions that are
mentioned there.

Rune

>
>The value for N depends on what you are trying to do.  Why are you
>taking an FFT?
>
>If you are trying to analyze a signal, the size of your FFT will
>dictate the frequency resolution and one's ability to distinguish
>neighboring tones.
>
>Freq. Resolution = Fs/N
>
>If you are trying to filter in the frequency domain, the size of your
>FFT will be dictated by the length of the filter.
>If you are trying to process Multi-Carrier modulated signals, your FFT
>will be some multiple of a fundamental length dictated by the
>communication system.
>
>FFT has many purposes and the reasons for choosing N will depend on
>why you are using an FFT in the first place.
>
>-V
>
>
 hi
So how do i decide on N for a sampled data system, I am sampling a input
signal whose input frequency is carefully chosen and sampling frequency is
2GHz. How do i  decide on number of points of FFT i should take?

thanks.

dbell wrote:
> Just to save me the time of trying to figure this out, how many bands?

Up to about 450.

> How many taps are you executing per input sample?

I'm not sure to understand that question.

> Are you getting one output from each band per input sample?

To some point, yes. Then I downsample it to the user-specified desired
sample rate.

> What are your stopband attenuation requirements?

Well I don't have any precisely defined requirements, and right now
it's whatever it is when you have a sinc function windowed with a
blackman in double precision float arithmetic.

> What are your transition band requirements?

Well like I said before overlap each other their bandwidth has to be
calculated so that filters overlap smoothly and do not have any flat
top, meaning that for example for let's say band #10 the low-pass
filter part of the bandpass filter should have its rolloff bandwidth
to equal the distance between the central frequency of band #10 and of
band #12. The un-zero-tapped windowed sinc function is considered to
have a rolloff width of 4 taps, so it's calculated accordingly.

> How are you doing the envelope detection?

I think I do it in an original way (since I've never read about it
anywhere). I simply perform an FFT, and add zero taps *before* the
existing frequency domain signal at the bass end, without cutting
anything at the treble end. The number of zeroes to add is determined
by the specified desired output sampling rate, so that the lowest
frequency in the original signal becomes above half the sampling
frequency. For example, if I choose an output sampling rate of 120 Hz,
then signals originally at 20 Hz will move to at least 80 Hz, signals
at 20000 Hz will move to at least 20060 Hz and the nyquist frequency
will move from 22,050 Hz to at least 22,110 Hz. Then in the time
domain I turn all negative samples to positive, and I perform a low-
pass filter, and there I got a very accurate envelope.

> If your filters ae different lengths, do you have to time align them?

Kind of. I keep track of their length, then I simply chop off the
"tails" in the output signals.

On Feb 6, 12:36 am, "Michel Rouzic" <Michel0...@yahoo.fr> wrote:
> dbell wrote:
> > Not knowing any more than you have said about your requirements so
> > far, a multirate implementation is a very efficient way to do a very
> > narrow bandpass filter, or a lot of them.  Depending on what you are
> > doing with your bank of filters, and performance required, there might
> > be a better way.
>
> > More details=> better answer.
>
> > Dirk
>
> OK, my program is a spectrographer with a log base 2 frequency scale.
> This is based on a filter bank and envelope detection. My FIRs that
> constitute my filter bank are defined in a way so that their central
> frequency matches to a logarithmically calculated frequency, and the
> two FIRs that constitute them have their central frequency and roll-
> off bandwidth calculated so that the BP FIRs overlap each other
> smoothly and do not have any flat top (so they're pretty much a
> slightly assymetrical bell curve, slight assymetrical due to the
> logarithmic spacing in frequency of each bandpass kernel)
>
> Usually I do not need more than 48 bands/octave (that's already quite
> a lot) and I never really need anything under 27.5 Hz, and my sampling
> frequency is almost always 44,100 Hz, so my largest BP FIR is never
> much bigger than about 400,000 taps.
>
> In case it matters, it's fine with me if the output is shifted by a
> certain frequency in the output, since all I use my filtered output
> for is for envelope detection.
>
> Thanks in advance :-)


Just to save me the time of trying to figure this out, how many bands?

How many taps are you executing per input sample?
Are you getting one output from each band per input sample?
What are your stopband attenuation requirements?
What are your transition band requirements?
How are you doing the envelope detection?
If your filters ae different lengths, do you have to time align them?


Dirk

dbell wrote:
> Not knowing any more than you have said about your requirements so
> far, a multirate implementation is a very efficient way to do a very
> narrow bandpass filter, or a lot of them.  Depending on what you are
> doing with your bank of filters, and performance required, there might
> be a better way.
>
> More details=> better answer.
>
> Dirk

OK, my program is a spectrographer with a log base 2 frequency scale.
This is based on a filter bank and envelope detection. My FIRs that
constitute my filter bank are defined in a way so that their central
frequency matches to a logarithmically calculated frequency, and the
two FIRs that constitute them have their central frequency and roll-
off bandwidth calculated so that the BP FIRs overlap each other
smoothly and do not have any flat top (so they're pretty much a
slightly assymetrical bell curve, slight assymetrical due to the
logarithmic spacing in frequency of each bandpass kernel)

Usually I do not need more than 48 bands/octave (that's already quite
a lot) and I never really need anything under 27.5 Hz, and my sampling
frequency is almost always 44,100 Hz, so my largest BP FIR is never
much bigger than about 400,000 taps.

In case it matters, it's fine with me if the output is shifted by a
certain frequency in the output, since all I use my filtered output
for is for envelope detection.

Thanks in advance :-)

On Feb 5, 2:57 am, "Michel Rouzic" <Michel0...@yahoo.fr> wrote:
> dbell wrote:
> > Is your goal to avoid the computational load or to reduce the length
> > of the reponse?
>
> To avoid the computational load.

Not knowing any more than you have said about your requirements so
far, a multirate implementation is a very efficient way to do a very
narrow bandpass filter, or a lot of them.  Depending on what you are
doing with your bank of filters, and performance required, there might
be a better way.

More details=> better answer.

Dirk

dbell wrote:
> Is your goal to avoid the computational load or to reduce the length
> of the reponse?

To avoid the computational load.