The Fourier transform of a continuous time signal x(t) is given by x(\omega)=\frac{1}{(10+j \omega)^{2}}. -\infty

GATE Electrical Engineering (EE) 2018 Solved Paper

Question : 59 of 65

Marks: +1, -0

The Fourier transform of a continuous time signal

x(t)

is given by

x(ω)=\;{1}/{(10+j ω)^{2}}

-∞<ω<∞

, where

j=√{-1}

and

ω

denotes frequency. Then the value of

|\ln x(t)|

t=1

is (upto 1 decimal place). (In denotes the logarithm to base e).

Solution:

X(ω)=\;{1}/{(10+j ω)^{2}}

By taking inverse Fourier transform,

x(t)=t e^{-10 t} u(t)

Now,

\;\; x(t)_{t=1}=1 × e^{-10} × 1=e^{-10}

Thus,

|\ln x(t)| \;\; =|\ln e^{-10}|

\;\; =|-10|=10

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