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Special Cases-2

The output yt...

Question

The output y(t) of a linear time invariant system is related to its input x(t) by the following equation :
$y (t) = 0.5 x (t - t_{d} + T) + x (t - t_{d}) + 0.5 x (t - t_{d} - T)$
The filter transfer function $H (ω)$ of such a system is given by

(1 - c o s ω T) e^{- j ω t_{d}}

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(1 - 0.5 c o s ω T) e^{- j ω t_{d}}

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(1 + 0.5 c o s ω T) e^{- j ω t_{d}}

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(1 + c o s ω T) e^{- j ω t_{d}}

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Solution

The correct option is D $(1 + c o s ω T) e^{- j ω t_{d}}$
$y (t) = 0.5 x (t - t_{d} + T) + 0.5 x (t - t_{d} - T) + x (t - t_{d})$

Taking Fourier transform

$Y (ω) = [0.5 e^{j ω (- t_{d} + T)} + 0.5 e^{j ω (- t_{d} - T)} + e^{- j ω t_{d}}] X (ω)$

$\frac{Y (ω)}{X (ω)} = e^{- j ω t_{d}} [0.5 e^{j ω T} + 0.5 e^{- j ω T} + 1]$

$\frac{Y (ω)}{X (ω)} = e^{- j ω t_{d}} [c o s ω T + 1]$

$H (ω) = (1 + c o s ω T) e^{- j ω t_{d}}$

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