Thread Subject:

IDTFT

Hi,

I have a question about IDTFT.

If I have X(e^(jw/K)) where K is an integer greater than 1. How can I recover the time series x(n)?

I have performed the IDTFT and obtained x(p) = sum_{n=0}^{N-1}x_n sinc(p-n/K) where sinc(x) = sin(pi*x)/pi/x.

But then I cannot obtain x(n).

Can anyone help?

Thanks

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <hvn745$jgp$1@fred.mathworks.com>...
> Hi,
>
> I have a question about IDTFT.
>
> If I have X(e^(jw/K)) where K is an integer greater than 1. How can I recover the time series x(n)?
>
> I have performed the IDTFT and obtained x(p) = sum_{n=0}^{N-1}x_n sinc(p-n/K) where sinc(x) = sin(pi*x)/pi/x.
>
> But then I cannot obtain x(n).
>
> Can anyone help?
>
> Thank

Hi, see the documentation for ifft

>>doc ifft


Wayne

Hi Wayne,

Thanks for your advice. In fact, I want to ask a conceptual question.

DTFT:
X(e^(j w/K)) = sum_{n=0}^{N-1}x(n) e^{-jwn/K}......................(1)

IDTFT:
x(p) = integrate_{- pi}^{pi} X(e^(j w/K)) e^{jwp}dw/2/pi
       = sum_{n=0}^{N-1}x(n) sinc(p-n/K);......................................(2)

However, when I compute the FFT of (2), it is different from (1).

Can you solve this problem?

Thanks a lot.

Frank

"Frank " <allinone_2003@yahoo.com.hk> wrote in message <hvo3p5$ar2$1@fred.mathworks.com>...
> Hi Wayne,
>
> Thanks for your advice. In fact, I want to ask a conceptual question.
>
> DTFT:
> X(e^(j w/K)) = sum_{n=0}^{N-1}x(n) e^{-jwn/K}......................(1)
>
> IDTFT:
> x(p) = integrate_{- pi}^{pi} X(e^(j w/K)) e^{jwp}dw/2/pi
> = sum_{n=0}^{N-1}x(n) sinc(p-n/K);......................................(2)
>
> However, when I compute the FFT of (2), it is different from (1).
>
> Can you solve this problem?
>
> Thanks a lot.
>
> Frank

Frank, we you say you have the DTFT at X(e^(j w/k)), this is continuous function of w. All you're doing is dilating the frequency axis with w/K. Are you trying to sample the continuous frequency axis to arrive at the DFT?? If so, then you want X(e^j(2*pi*k/N)) k=0,.....N-1 The arriving at the inverse DFT would be a summation over k.

If you are working in a compute, you are not going to be computing the DTFT. Again, that is a complex-valued function of a continuous variable.

Wayne