>On Apr 23, 11:11 am, "anando" <meet_ana...@yahoo.com> wrote:

>> Thanks for your reply. I do realize that R_b(f) can be written as a
>> product of R(f) and a boxcar function in the frequency space. As
mentioned
>> before, I know the inverse FT of R(f) [as r(t) is known] and also the
>> boxcar function [some combination of sinc functions - let us call it
s(t)
>> ].
>>
>> So I guess my query is how would I write down the inverse FT of R_b(f)
>> using r(t) and s(t) in a computationally efficient manner.
>
>Given that s(t) has infinite support, the most computationally
>efficient manner might well be to just (re)do a finite length
>FFT/boxcar/IFFT (or pruned FFT/IDFT), or to accept some form
>of approximation to your desired boxcar filtered r(t), using
>something such as FIR or IIR time domain filters, over a
>circular buffer if needed, for instance.
>
>
>IMHO. YMMV.
>--
>rhn A.T nicholson d.0.t C-o-M
>

Hi 

Many thanks for your suggestions. Any links to time-domain FIR/IIR
filtering will be greatly appreciated (second part in your reply).

Also, could wavelets be used for my problem ? For example, could I wavelet
transform r(t) to give me the boxcar filtered r(t) ?

Cheers,
Anando.

On Apr 23, 11:11 am, "anando" <meet_ana...@yahoo.com> wrote:
> Hi
>
>
>
>
>
> >Another way to express your band-limited signal is  R_b(f) = R(f) H(f)
> >Where H(f)= 1     f_1<= |f| <=f_2
> >           = 0     otherwise
>
> >There are some variations between the versions of the Fourier transform,
> >but there is properties that relates multiplication in Fourier space to
> >convolution in time.  In your case (FFT), it goes by two names:
> >"circular convolution" property and "cyclic convolution" property.
>
> >Circular convolution has some subtleties of which you need to be aware.
> >It's not the convolution your mother taught you. So if you are not
> >familiar with it do some web searches.
>
> >Cheers,
> >J. Elms
>
> Thanks for your reply. I do realize that R_b(f) can be written as a
> product of R(f) and a boxcar function in the frequency space. As mentioned
> before, I know the inverse FT of R(f) [as r(t) is known] and also the
> boxcar function [some combination of sinc functions - let us call it s(t)
> ].
>
> So I guess my query is how would I write down the inverse FT of R_b(f)
> using r(t) and s(t) in a computationally efficient manner.

Given that s(t) has infinite support, the most computationally
efficient manner might well be to just (re)do a finite length
FFT/boxcar/IFFT (or pruned FFT/IDFT), or to accept some form
of approximation to your desired boxcar filtered r(t), using
something such as FIR or IIR time domain filters, over a
circular buffer if needed, for instance.


IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

On 25 Apr., 07:27, "anando" <meet_ana...@yahoo.com> wrote:
> >>Thanks for your reply. I do realize that R_b(f) can be written as a
> >>product of R(f) and a boxcar function in the frequency space. As
> >mentioned
> >>before, I know the inverse FT of R(f) [as r(t) is known] and also the
> >>boxcar function [some combination of sinc functions - let us call it
> >s(t)
> >>].
>
> >>So I guess my query is how would I write down the inverse FT of R_b(f)
> >>using r(t) and s(t) in a computationally efficient manner.
>
> >>Again, many thanks for your help.
>
> >>Cheers,
> >>Anando.
>
> >Hi Anando,
> >I agree with what the others have said, but I'm not sure they answered
> >what you were asking. &#4294967295;When you say "write down...", do you mean how to
> >digitally calculate the result, or how to write down the equation? &#4294967295;As
> an
> >equation, the Inverse FT of R_b(f) is equal to the convolution of r(t)
> and
> >s(t). As a previous post pointed out, if this is for digital signals you
> >need to worry about circular convolution. &#4294967295;Here's some links:
>
> >http://www.dspguide.com/ch9/2.htm
> >http://www.dspguide.com/ch9/3.htm
>
> >Steve
>
> Hi Steve
>
> Thanks for your mail.
>
> I was hoping to calculate the result I guess.
>
> Just to clarify my question, in a slightly abstract form, my query is,
> would it be possible to recreate the inverse FT of a band-pass signal from
> the inverse FT of the full signal ? I'd be very grateful if you have
> pointers for that.

Sure. Let's say that y[n] is the desired bandpass signal, and x[n] is
the full band signal. You want to calculate y[n], given x[n]. That's
easy:

y[n] = IDFT[ rect[k] DFT[x[n] ]

where rect[k] is the suitably chosen rectangular bandpass.

Regards,
Andor

"anando" <meet_anando@yahoo.com> wrote in message 
news:v5CdndW3OpOF94zVnZ2dnUVZ_v2pnZ2d@giganews.com...
>
>>
>>For an IFFT of length N, if the number of non-zero coefficients
>>K is much less than log N, then one could just compute a naive
>>inverse DFT pruned to just calculating for K bins.
>>
>>If you actually want something different from the entire
>>and precise IFFT result vector, then you could note that
>>this complete IFFT looks very similar to a modulated
>>and upsampled result for some narrow baseband signal.
>>You might be able to translate your non-zero bins to
>>baseband, un-zero-pad, and perform a much shorter IFFT.
>>But this is unlikely to be more efficient unless you don't
>>need, or have some other (analog?) solution for, upsampling
>>interpolation and modulation later in your desired process.
>>
>
> Thanks a ton to all of you who have replied. This idea is one that appeals
> to me a lot and I might try it.
>
> So if I understand you correctly, I could
> * heterodyne the signal such that my zero frequency lies at
>     f_0 = 0.5*(f_1 + f_2) - then
> * I could take an IFFT of a much much smaller vector.
>
> Of course the time samples will be much coarser than the original IFFT -
> but I guess I can always interpolate ?
>
> Please correct me if I am mistaken. What did you mean by un-zero pad ?
>
> Cheers,
> Anando.

After decimating you don't need to interpolate any more than you did before.

Consider this:

An AM signal.  The carrier is of no real interest.  The interesting part is 
all in the waveform envelope.  So, the decimation still gets you a good 
representation of the envelope.

Fred 

>
>>Thanks for your reply. I do realize that R_b(f) can be written as a
>>product of R(f) and a boxcar function in the frequency space. As
>mentioned
>>before, I know the inverse FT of R(f) [as r(t) is known] and also the
>>boxcar function [some combination of sinc functions - let us call it
>s(t)
>>]. 
>>
>>So I guess my query is how would I write down the inverse FT of R_b(f)
>>using r(t) and s(t) in a computationally efficient manner.
>>
>>Again, many thanks for your help.
>>
>>Cheers,
>>Anando.
>>
>
>Hi Anando,
>I agree with what the others have said, but I'm not sure they answered
>what you were asking.  When you say "write down...", do you mean how to
>digitally calculate the result, or how to write down the equation?  As
an
>equation, the Inverse FT of R_b(f) is equal to the convolution of r(t)
and
>s(t). As a previous post pointed out, if this is for digital signals you
>need to worry about circular convolution.  Here's some links: 
>
>http://www.dspguide.com/ch9/2.htm
>http://www.dspguide.com/ch9/3.htm
>
>Steve
>


Hi Steve

Thanks for your mail. 

I was hoping to calculate the result I guess. 

Just to clarify my question, in a slightly abstract form, my query is,
would it be possible to recreate the inverse FT of a band-pass signal from
the inverse FT of the full signal ? I'd be very grateful if you have
pointers for that.


Cheers,
Anando.

>
>For an IFFT of length N, if the number of non-zero coefficients
>K is much less than log N, then one could just compute a naive
>inverse DFT pruned to just calculating for K bins.
>
>If you actually want something different from the entire
>and precise IFFT result vector, then you could note that
>this complete IFFT looks very similar to a modulated
>and upsampled result for some narrow baseband signal.
>You might be able to translate your non-zero bins to
>baseband, un-zero-pad, and perform a much shorter IFFT.
>But this is unlikely to be more efficient unless you don't
>need, or have some other (analog?) solution for, upsampling
>interpolation and modulation later in your desired process.
>

Thanks a ton to all of you who have replied. This idea is one that appeals
to me a lot and I might try it. 

So if I understand you correctly, I could 
* heterodyne the signal such that my zero frequency lies at 
     f_0 = 0.5*(f_1 + f_2) - then   
* I could take an IFFT of a much much smaller vector. 

Of course the time samples will be much coarser than the original IFFT -
but I guess I can always interpolate ?

Please correct me if I am mistaken. What did you mean by un-zero pad ?

Cheers,
Anando.

>Thanks for your reply. I do realize that R_b(f) can be written as a
>product of R(f) and a boxcar function in the frequency space. As
mentioned
>before, I know the inverse FT of R(f) [as r(t) is known] and also the
>boxcar function [some combination of sinc functions - let us call it
s(t)
>]. 
>
>So I guess my query is how would I write down the inverse FT of R_b(f)
>using r(t) and s(t) in a computationally efficient manner.
>
>Again, many thanks for your help.
>
>Cheers,
>Anando.
>

Hi Anando,
I agree with what the others have said, but I'm not sure they answered
what you were asking.  When you say "write down...", do you mean how to
digitally calculate the result, or how to write down the equation?  As an
equation, the Inverse FT of R_b(f) is equal to the convolution of r(t) and
s(t). As a previous post pointed out, if this is for digital signals you
need to worry about circular convolution.  Here's some links: 

http://www.dspguide.com/ch9/2.htm
http://www.dspguide.com/ch9/3.htm

Steve

  


       

I am guessing you can take advantage of the fact that for a bandpass
signal it is sufficent to sample it by (f_2-f_1), but the method that
you describe will yield samples that are sampled at the original Fs..

This means that you can save not only at the first stage, but also the
last stage of the FFT.. i.e. you will get [Fs/(f_2-f1)] periods in the
last stage

If (f_2-f_1) is known or not variable, you can try some tricks to
simplify.. Hope that helps

On Apr 23, 3:20 am, "anando" <meet_ana...@yahoo.com> wrote:
> Hi
>
> I'd be grateful if somebody could point me to the right literature for the
> following problem:
>
> I have a function r(t) which is known as well as its Fourier transform
> R(f).  Thus R(f) --> IFFT --> r(t). Both of them are known.
>
> Now suppose I have a bandpass version of R(f) [let us call it R_b(f)] such
> that
>    R_b(f) = R(f)  for  f_1 <= f <= f_2
>           = 0     otherwise
>
> Is there a way of calculating the inverse fourier transform of R_b(f) in
> terms of the known quantities ?

For an IFFT of length N, if the number of non-zero coefficients
K is much less than log N, then one could just compute a naive
inverse DFT pruned to just calculating for K bins.

If you actually want something different from the entire
and precise IFFT result vector, then you could note that
this complete IFFT looks very similar to a modulated
and upsampled result for some narrow baseband signal.
You might be able to translate your non-zero bins to
baseband, un-zero-pad, and perform a much shorter IFFT.
But this is unlikely to be more efficient unless you don't
need, or have some other (analog?) solution for, upsampling
interpolation and modulation later in your desired process.



IMHO. YMMV.
--
rhn A.T nicholson d.0.t C-o-M

"anando" <meet_anando@yahoo.com> wrote in message 
news:1MedncKzcPL2hpLVnZ2dnUVZ_jmdnZ2d@giganews.com...
> Hi
>
> I'd be grateful if somebody could point me to the right literature for the
> following problem:
>
> I have a function r(t) which is known as well as its Fourier transform
> R(f).  Thus R(f) --> IFFT --> r(t). Both of them are known.
>
> Now suppose I have a bandpass version of R(f) [let us call it R_b(f)] such
> that
>   R_b(f) = R(f)  for  f_1 <= f <= f_2
>          = 0     otherwise
>
> Is there a way of calculating the inverse fourier transform of R_b(f) in
> terms of the known quantities ?
>
> Many thanks,
> Anand.

This is a good question and one that I pondered a couple of years ago. 
Unfortunately, there's no improvement possible.

An arm-waving explanation is that after the first step in the FFT/IFFT none 
of the values are any longer zero.  Thus, most of the computations can't be 
eliminated.

Fred