Re: Purely real result of IFFT

On Oct 4, 1:07 pm, "David Doria" <daviddo...@xxxxxxxxx> wrote:
Ok that tells me that I came to the correct conclusion, but
here is a bit further question

I thought the sum in the IDFT was from k = -N/2 to N/2, but
here it is from k=0 to N-1

if you assume that Xk is a complex number (a+bi), and use
Eulers formula to get cos(2*pi*j*k*n/N) + i*sin(2*pi*j*k*n/N)

you can show that if X(k) = X(-k)* then all the imaginary
terms cancel. This doesn't work if there are never negative
numbers in the exponent. How could you show this property
using k = 0 to N-1 ??

The DFT is periodic, with period N.

X(k) = sum( n=0,N-1 ) exp(j*2*pi*k*n/N) * x(n)

X(k+N) = sum(n=0,N-1) x(n)*exp[j*2*pi*(k+N)*n/N]

= sum(n=0,N-1) x(n)*exp[j*2*pi*k*n/N] exp[j*2*pi*n*N/N]

= sum(n=0,N-1) x(n)*exp[j*2*pi*k*n/N]

= X(k)

So X(-k) = X(N - k)

- Randy

.

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