application of DFT multiple number of times

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purnapragna

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hi what will happen if i apply DFT on a sequence x[n] for four times? Actually this was asked in one of our quizzes. can anybody give the exact answer to it.

thnx

purna
 

After applying fft to vector 'x' four times
you will obtain 'x' multiplied by N^2, where N is length of 'x'

It is because of F=conj(F^(-1)), where F is fft matrix, F^(-1) is IFFT matrix

F*F^(-1)=I*N, where I is unity matrix

Combain these two properties with your brain and you obtain answer
 

looking at definition of DFT and IDFT you can obtain duality property.
->DFT(x[n])=N.conj(IDFT(conj(x[n]))),N=length x[n].
so
DFT(x[n])=X[k]
DFT(X[k])=N.conj(IDFT(conj(X[k])))=N.x[((-n))N]=N.y[n]
DFT(N.y[n])=N.Y[k]
DFT(N.Y[k])=N.N.conj(IDFT(conj(Y[k])))=N^2.y[((-n))N]=N^2.x[n].
why y[n]=x[((-n))N]->y[((-n))N]=x[n].
 

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