Question

Which of the following statement is not correct?

• A. Laplace transform is a complex Fourier transform
• B. Fourier transform of a function can be obtained from its Laplace transform by replacing s by j\(\omega\)
• C. Fourier transform is the Laplace transform evaluated along the imaginary axis of the s-plane
• D. Convolution integrals cannot be evaluated using Fourier transform

Hint:

The Fourier Transform is a special case of the Laplace Transform evaluated on the \(j\omega\)-axis, provided the ROC includes it.
Convolution Theorem is a fundamental property of Fourier Transforms: \(f_1(t) * f_2(t) \longleftrightarrow F_1(j\omega) F_2(j\omega)\).

Answer 1

The Correct Option is D

(a) "Laplace transform is a complex Fourier transform": The Laplace transform \(F(s) = \int_0^\infty f(t)e^{-st}dt\) with \(s=\sigma+j\omega\). It can be seen as the Fourier transform of \(f(t)e^{-\sigma t}u(t)\). This statement is generally considered true in the sense that they are closely related and Laplace is a generalization. (b) "Fourier transform of a function can be obtained from its Laplace transform by replacing s by j\(\omega\)": This is true if the Region of Convergence (ROC) of the Laplace transform includes the \(j\omega\)-axis. So, conditionally true. (c) "Fourier transform is the Laplace transform evaluated along the imaginary axis of the s-plane": Same as (b), conditionally true. (d) "Convolution integrals cannot be evaluated using Fourier transform": This is FALSE. The Convolution Theorem states that convolution in the time domain corresponds to multiplication in the frequency (Fourier) domain: \(\mathcal{F}\{f(t)*g(t)\} = F(j\omega)G(j\omega)\). This property is often used to simplify or evaluate convolutions. Since the question asks for the statement that is "not correct", option (d) is the incorrect statement. \[ \boxed{\text{Convolution integrals cannot be evaluated using Fourier transform}} \]