How to get inverse discrete time Fourier transform (IDTFT) of an array?

Apparently, there is no function to get IDTFT of an array. Is there any?

I used 'freqz(array,1)' to find the DTFT of an array from '0' to 'pi-pi/512'.

I also developed a code for DTFT for ab array A of length N from '-pi' to 'pi'

i=1
for wval=-pi:pi/512:pi
    DFT(i)=sum(a.*exp(-1j*(0:N-1).*wval));
    i=i+1;
end

But I do not know how to find IDTFT, when applied on a DTFT giving the original array.

Answer 1

[Abderrahim. B]

Hi!

Are you trying to implement DFT and its IDFT based on their equations ?

There are optimized algorithms to calculate these equations, the very well known one is the Fast Fourier Transform(FFT) and its inverse (IFFT). You may need to refer to this link for more explanation .

Demo:

Array = [1 2 3 4 5] % Sequence

Array = 1×5

1 2 3 4 5

arrFFT = fft(Array)      % DFT based FFT

arrFFT = 
  15.0000 + 0.0000i  -2.5000 + 3.4410i  -2.5000 + 0.8123i  -2.5000 - 0.8123i  -2.5000 - 3.4410i

Array = ifft(arrFFT ) 

Array = 1×5

1 2 3 4 5

Hope this helps

[Palguna Gopireddy]

@Abderrahim BELISSAOUI

I am actually looking for IDTFT. I followed the link you quoted. It is good. I am able to get DTFT using 'fft' function.

But when I am trying to get IDTFT using 'ifft' function i am unable to get the original values.

Could you please help me on getting IDTFT.

[Abderrahim. B]

@Palguna Gopireddy please share the data you re testing on. As you can see, using the same function ( ifft ) was able to get the original array.

[Palguna Gopireddy]

I have taken a simple array A = [ 1 3 2]

I have calculaed the DTFT of 512 samples using fft(A, 512)

A= [1 3 2];
B= fft(A,512);

i verified it B is DTFT of A

I am unable to get A by applying ifft on B.

Help me with this

[Paul]

A = [1 3 2];
B = fft(A,512);
C = ifft(B,'symmetric');
C(1:3) % should be A to within roundoff

ans = 1×3

1 3 2

[Abderrahim. B]

@Palguna Gopireddy

A = [1 3 2] ;
% N = 512 will zero pad the sequence. Somthing like this: A =  [1 2 3 0 0 0
% 0 0 .................0] 
A_fft = fft(A, 512) ;
lenB = length(A_fft) ;

% Default reconstructed sequence length will be the same as B ( 512)
A_ifft = ifft(A_fft) 

A_ifft = 1×512

1.0000 3.0000 2.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 0.0000 0.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 -0.0000 0.0000 0.0000 0.0000 0.0000 -0.0000 0.0000 -0.0000 -0.0000 0.0000 -0.0000

lenA = length(A_ifft) ;
lenA == lenB

ans = logical
   1

% Original A
A = A_ifft(1:length(A))

A = 1×3

1 3 2

[Palguna Gopireddy]

Thanks, I thought of rounding it off, so that all zeros are padded to original sequence.

Let me try it for 2D case to see the results.

[Palguna Gopireddy]

Its working for 2D also. But it is done by using fft two times once row, once column.

A = [1,2,3;4,5,6];
A_fft = fft(A, 512,1);
A_fftxy=fft(A_fft,512,2);
A_ifftx = ifft(A_fftxy,512,2);
A_ifft=ifft(A_ifftx,512,1);
%or
A_iffty=ifft(A_fftxy,512,1);
A_iffto=ifft(A_iffty,512,2);

It works directly using fft2 also.

A_fft2=fft2(A,512,512);
A_ifft2=ifft2(A_fft2,512,512);

%A_ifft2,A_ifft,A_iffto all are same as A, but they are not equal outside
%the size of A, which is almost zero. So it's ok.

[Abderrahim. B]

Use fft2 and ifft2.

A = [1,2,3; 4,5,6] ;
[n_row, n_col ] = size(A) ;

A_2Dfft = fft2(A, 512, 512) ;
A_ifft2 = ifft2(A_2Dfft) ; 
reconst_A = A_ifft2(1:n_row, 1:n_col) ;
isequal(reconst_A , A)

ans = logical
   1

[Palguna Gopireddy]

Thanks @Abderrahim BELISSAOUI.

But I can't prove the convolution theorem for 2D.

I posted it as another question. Could you help me withis this.

https://in.mathworks.com/matlabcentral/answers/1768915-can-t-prove-the-convolution-theorem-of-fourier-theorem-for-two-dimensional-matrices