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Let X(K) = {20, 0, -4+4j, 0, -4} is the 8 point DFT of a real valued sequence x(n)

1. Find X(K) for K= 5, 6, 7.

2. Find the 8 point DFT P(K) such that

$p(n) = (-1)n \otimes x(n)$ Using DFT property.

Mumbai University > Computer Engineering > Sem 7 > Digital Signal Processing

Marks: 10 Marks

Year: Dec 2015

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  1. N=8

    For k=5, $X(5) = x*(8-5) \\ = x*(3) = 0$

    For k=6, $X(6) = x*(8-6) \\ = x*(2) = (-4-4j)$

    For k=7, $X(7) = x*(8-7) \\ = x*(1) = 0$

    $\therefore, X(5) = 0, X(6) = -4-4j, X(7) = 0$

    Result: X(K)={20, 0, -4+4j, 0, -4, 0, -4-4j, 0}

(i) Let $x_1(n)=(-1)n$

$\therefore$

enter image description here

$\therefore, x_1$ (n)={1, -1, 1, -1, 1, -1, 1, -1}

enter image description here

$\therefore, X_1$(k)={0, 0, 0, 0, 8, 0, 0, 0}

By Convolution Property of DFT, Convolution of two signals in time domain is equivalent to multiplication in frequency domain.

$\therefore, P(K)= X_1(k). X(k) \ \ \ \ as, p(n) = x_1(n) * x(n)$

$\therefore$, By Convolution Theorem,

P(K) = {0, 0, 0, 0, 8, 0, 0, 0}. {20, 0, -4+4j, 0, -4, 0, -4-4j, 0}

Result: P(K)={0, 0, 0, 0, -32, 0, 0, 0}

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