Question

Let f(t) be a continuous-time signal and let F(ω) be its Fourier Transform defined by

\(F\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\)

Define g(t) by

\(g\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty F\left( u \right){e^{ - ju t}}du\)

What is the relationship between f(t) and g(t)?

1. g(t) would always be proportional to f(t)
2. g(t) would be proportional to f(t) if f(t) is an even function
3. g(t) would be proportional to f(t) only if f(t) is a sinusoidal function
4. g(t) would never be proportional to f(t)

Answer 1

Correct Answer - Option 2 : g(t) would be proportional to f(t) if f(t) is an even function

Given the Fourier transform pair,

\(\begin{array}{l} F\left( \omega \right) = \mathop \smallint \limits_{ - \infty }^\infty f\left( t \right){e^{ - j\omega t}}dt\\ g\left( t \right) = \mathop \smallint \limits_{ - \infty }^\infty F\left( u \right){e^{ - j\omega t}}du \end{array}\)

The Inverse Fourier transform of F (ω) is given as

\(f\left( t \right) = \frac{1}{2π}\mathop \smallint \limits_{ - \infty }^\infty F\left( \omega \right){e^{j\omega t}}d\omega\)

Replacing t by ‘ –t ‘ and ω by u in the above expression, we have,

\(f\left( { - t} \right) = \frac{1}{2π}\mathop \smallint \limits_{ - \infty }^\infty F\left( u \right){e^{-ju t}}du\)

g(t) =  2π f (-t)

If f (t) is even, then f (-t) = f (t)

∴ g (t) = 2π f (t)

⇒ g (t) will be proportional to f (t), if f (t) is an even function.