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Justify DFT as a linear transformation.
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If x1(n)(<>)DFTx1(k) And

x2(n)(<>)DFTx2(k) Then

a1x1(n)+a2x2(n)(<>)DFTa1x1(k)+a2x2(k)

Proof:- By the definition

x(k) = ∑ x(n) (Wn)Kn

here x(n) = a1x1(n) + a2x2(n)

=> X(K) = ∑[a1x1(n) + a2x2(n)](Wn)Kn

X(K) = a1X1(k) + a2X2(k)

Hence proved.

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