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.