Thread Subject: DFT identity of derivative

Hi,
i am new to Mathworks and i am using matlab for FFT calculations.
when i take FFT of say a simple sequence [1 2 3 4] it turns out to be
[10 -2+2i -2 -2-2i]
and when i take FFT of gradient([1 2 3 4]) result comes up as
[4 0 0 0], everything is alright till now.
but when i try to verify the DFT identity related to gradient
   f'(n)=2*pi*i*k*F(K) where f'(n) is derivative of original sequence.
                                    and F(K) is FFT of original sequence.
then it does not work.

see 2*pi*i*k*[10 -2+2i -2 -2-2i] is not equal to [4 0 0 0]
what is wrong with this?
Thanks
                                    

On Nov 13, 9:38 pm, "wasim nawaz" <wasim.na...@yahoo.com> wrote:
> Hi,
> i am new to Mathworks and i am using matlab for FFT calculations.
> when i take FFT of say a simple sequence [1 2 3 4] it turns out to be
> [10  -2+2i  -2  -2-2i]
> and when i take FFT of gradient([1 2 3 4]) result comes up as
> [4 0 0 0], everything is alright till now.
> but when i try to verify the DFT identity related to gradient
>    f'(n)=2*pi*i*k*F(K)     where  f'(n) is derivative of original sequence.
>                                     and F(K) is FFT of original sequence.
> then it does not work.
>
> see  2*pi*i*k*[10  -2+2i  -2  -2-2i] is not equal to [4 0 0 0]
> what is wrong with this?
> Thanks

There are at least two issues.

1) The derivative identity of the continuous Fourier transform (FT)
can only be extended to the finite discrete Fourier transform (DFT)
for these functions for which the DFT of the samples equals the
samples of the FT.

2) For signals satisfying 1), the Matlab's gradient() does not
necessarily represent the derivative.
Consider samples of a cosine function:
>> x = [1 .707107 0 -.707107 -1 -.707107 0 .707107];
>> fft(x)
ans =
  Columns 1 through 7
         0 4.0000 0 -0.0000 0 -0.0000 0
  Column 8
    4.0000
>> gradient(x)
ans =
  Columns 1 through 7
   -0.2929 -0.5000 -0.7071 -0.5000 0 0.5000 0.7071
  Column 8
    0.7071

Dale B. Dalrymple

Note also that if DFT[x(n)]=X(k), then we have the identitity,

DFT[x(n-1)]= exp(j*2*pi*k)*X(k) %shift n-1 is modula N

This means that the DFT of a first difference approximation to the derivative will be

DFT[x(n)-x(n-1)]= ( exp(j*2*pi*k) - 1 ) *X(k)

and not something like k*X(k).