# Fourier transform Discreet time-shift

1. May 21, 2006

### electronic engineer

suppose we have this discreet fucntion:

x(k)=rect(k+N/2)= 1 ; when -N/2=<K<=N/2-1
x(k)=0; otherwise

This is discreet function(not continuous) of k shifted forward by N/2, we need to find fourier transform for it ..

anyway let N=6 for simplicity, then:

x(k)=rect(k+3)= 1 ; when -3=<K<=2
x(k)=0; otherwise

i think there's a law for finding FT of shifted signals but i can't remember ,and i need a guidance to get the solution

thanks

2. May 21, 2006

### vsage

First you must be careful and assert that N is even, because the function would be nothing if N was odd and it was shifted over by a fractional amount, which you did specify.

To answer your question though, a positive phase shift of p corresponds to a shift of the transform by $$e^{\frac{2 \pi i k p}{N}}$$ where k is the counting variable, N is the period, so if x[n] $$\rightarrow X[k]$$ then $$x[n+\frac{p}{2}] \rightarrow X[k] e^{\frac{\pi i k p}{N}}$$. See http://en.wikipedia.org/wiki/Discrete_Fourier_transform

Well I shouldn't have used the same counting variables for both x and its transform. Let me clean this up some.

Last edited by a moderator: May 21, 2006
3. May 22, 2006

### electronic engineer

so the solution would be like this:

x(k)<<DFT...<< X(f)
x(K+N/2) <<DFT....<< X(f).e^{(j*2*pi*k*N/2)/N}=X(f)*e^(j*pi*k)

where k is counting variable,N number of samples, j: imaginary unit

so the next step is to find X(f) of non-shifted disctreet function x(k)