Circular Time Shift Property

Circular Time Shift

Statement:

Shifting the sequence in time domain by ‘l’ samples is equivalent to multiplying the sequence in frequency domain by the twiddle factor.

Proof:

We will be proving the property

x((n-l))N = x(n-l) ==> DFT[x(n-l)] = X(k)e^-2πjlk/N or X(k)W^kl_N

where W is the twiddle factor.

According to DFT,

DFT [x(n)] = ∑_n=0^N-1 x(n) e^-2πjkn/N

DFT [x(n-l)] = ∑_n=0^N-1 x(n-l) e^-2πjkn/N

Let n-l=p

n=p+l

DFT [x(p)] = ∑_p=0^N-1 x(p) e^{-2πjk(p+l)/N}

= e^-2πjkl/N ∑_p=0^N-1 x(p) e^-2πjkp/N

= X(k)e^-2πjkl/N

Hence, proved.

The step-by-step values for the provided input is shown at the bottom of the page:

Circular Time Shift

Statement:

Proof:

According to DFT,

DFT [x(n)] = ∑_n=0^N-1 x(n) e^-2πjkn/N

DFT [x(n-l)] = ∑_n=0^N-1 x(n-l) e^-2πjkn/N

Let n-l=p

n=p+l

DFT [x(p)] = ∑_p=0^N-1 x(p) e^{-2πjk(p+l)/N}

= e^-2πjkl/N ∑_p=0^N-1 x(p) e^-2πjkp/N

= X(k)e^-2πjkl/N

Hence, proved.

Enter the sequence :

Enter the time-shifting factor :

Input sequence :

Time-Shifted sequence :

Circular Time Shift

Statement:

Proof:

According to DFT,

DFT [x(n)] = ∑n=0N-1 x(n) e-2πjkn/N

DFT [x(n-l)] = ∑n=0N-1 x(n-l) e-2πjkn/N

Let n-l=p

n=p+l

DFT [x(p)] = ∑p=0N-1 x(p) e-2πjk(p+l)/N

= e-2πjkl/N ∑p=0N-1 x(p) e-2πjkp/N

= X(k)e-2πjkl/N

Hence, proved.

Enter the sequence :

Enter the time-shifting factor :

Input sequence :

Time-Shifted sequence :

DFT [x(n)] = ∑_n=0^N-1 x(n) e^-2πjkn/N

DFT [x(n-l)] = ∑_n=0^N-1 x(n-l) e^-2πjkn/N

DFT [x(p)] = ∑_p=0^N-1 x(p) e^{-2πjk(p+l)/N}

= e^-2πjkl/N ∑_p=0^N-1 x(p) e^-2πjkp/N

= X(k)e^-2πjkl/N