application of DFT multiple number of times

[purnapragna]

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

 

[vadkudr]

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

 

[mehtesham]

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].