# Explain Shift property of DFT

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Here it is said that if you delay your input signal by D samples, then each complex value in the FFT of the signal is multiplied by the constant exp(−j2πkD/N).

My question is that if i have a DFT vector F(u)={11,-3+2j,-1,-3-2j} and D=2,N=4 then I will get {11,3-2j,-1,3+2j}. But if we see this new vector we can see that here no translation of the signal is done. Signal has not moved to the new location. Just the value of each coefficient has changed in it own location. According to the translation property the answer should be {-1,-3-2j,11,-3+2j}. What is the problem ? Please explain .

• The answer that I am getting is by using the fftshift function of MATLAB. – Navdeep Sep 28 '14 at 17:02

## 1 Answer

But if we see this new vector we can see that here no translation of the signal is done.

You have modified the data in the frequency domain, by multiplying by exp(−j2πkD/N). The translation thus occurs in the time domain. It would be indeed weird if multiplyig a vector by exp(−j2πkD/N) would shift this same vector!

The inverse DFT of [11,3-2j,-1,3+2j] is: [4, 4, 1, 2].

The inverse DFT of [11,-3+2j,-1,-3-2j] is: [1, 2, 4, 4].

The shift property is observed - the signal has been shifted by 2 (modulo 4).

• @ pichenettes Thanks from heart ...actually i was very confused and frustrated in this property. – Navdeep Sep 28 '14 at 17:57