Correct Answer - Option 2 : Frequency domain
The Fourier transform of a function is equal to its two-sided Laplace transform evaluated on the imaginary axis of the s-plane.
Explanation:
The Fourier transform of x(t) is, denoted by X(jω), is defined as:
\(X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)
The Laplace transform of x(t), denoted by X(s), is defined as:
\(X\left( s \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - st}}dt\)
Where s is a continuous complex variable.
We can also express s as: s = σ + jω
Where σ and ω are the real and imaginary parts of s, respectively
The Laplace transform can be written as:
\(X\left( {\sigma + j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - \left( {\sigma + j\omega } \right)t}}dt = \mathop \smallint \limits_{ - \infty }^\infty \left( {x\left( t \right){e^{ - \sigma t}}} \right){e^{ - j\omega t}}dt\)
By comparing the above Laplace and Fourier transform equations, it is clear that Laplace transform of x(t) is equal to the Fourier transform of \(x\left( t \right){e^{ - \sigma t}}\).
When σ = 0 or s = jω, both are identical.
\({\left. {X\left( s \right)} \right|_{s = j\omega }} = X\left( {j\omega } \right) = \mathop \smallint \limits_{ - \infty }^\infty x\left( t \right){e^{ - j\omega t}}dt\)
That is, Laplace transform generalizes Fourier transform.
