Fourier transform and Laplace transform are the significant mathematical tools used for signal and system analysis. The limitation of Fourier transform is that it is not defined for some of the signals. Therefore for such signals analysis Laplace transform is used to overcome the drawback of Fourier transform. Due to availability of convergence factor Laplace transform is completely integrable and hence defined for those signals for which Fourier transform is not absolutely integrable. The LTI is represented by differential equation. The system response is obtained by solving the given differential equation. Any differential equation is converted into simpler algebraic equation by Laplace transform.
For two signals \(x_1(t)\) and \(x_2(t)\). The correct Laplace transform expression for \(x_1(t)\cdot x_2(t)\) is :
\(\frac{1}{2\pi j}\left[X_1(s)*X_2(s)\right]\)
Start from the property you already know. Convolution in the time domain becomes ordinary multiplication in the s-domain:
\(x_1(t)*x_2(t)\ \longleftrightarrow\ X_1(s)X_2(s)\)
This is the property behind \(Y(s)=H(s)X(s)\) for LTI systems, and it is the reason transform methods are used at all — a messy convolution integral turns into a product.
Duality gives the reverse. The two operations swap roles when you go the other way, so multiplication in the time domain must become convolution in the complex-frequency domain — with a scaling factor coming from the inverse-transform integral:
\(x_1(t)\cdot x_2(t)\ \longleftrightarrow\ \dfrac{1}{2\pi j}\left[X_1(s)*X_2(s)\right]\)
Where the \(1/2\pi j\) comes from. Writing one signal as its inverse Laplace integral,
\(x_1(t)=\dfrac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}X_1(p)e^{pt}dp\)
and substituting into the transform of the product, the \(1/2\pi j\) carries through and the remaining integral is a convolution in the complex variable:
\(\dfrac{1}{2\pi j}\int_{\sigma-j\infty}^{\sigma+j\infty}X_1(p)\,X_2(s-p)\,dp\)
evaluated along a vertical line lying in the common region of convergence.
Why the other options are wrong. \(X_1(s)X_2(s)\) is the transform of the convolution, not of the product — the classic mix-up this question tests. \(X_1(s)+X_2(s)\) is the transform of the sum, by linearity. The third option is not a transform property at all.
The Fourier analogue is the familiar modulation theorem, \(x_1(t)x_2(t)\leftrightarrow\frac{1}{2\pi}X_1(\omega)*X_2(\omega)\) — the reason multiplying a signal by a carrier shifts and spreads its spectrum.
Hence, the correct expression is \(\dfrac{1}{2\pi j}\left[X_1(s)*X_2(s)\right]\).
Similar Questions
Fourier transform and Laplace transform are the significant mathematical tools used for signal and system analysis. The limitation of Fourier transform is that it is not defined for some of the signals. Therefore for such signals analysis Laplace transform is used to overcome the drawback of Fourier transform. Due to availability of convergence factor Laplace transform is completely integrable and hence defined for those signals for which Fourier transform is not absolutely integrable.
The LTI is represented by differential equation. The system response is obtained by solving the given differential equation. Any differential equation is converted into simpler algebraic equation by Laplace transform.
The Laplace transform converts integro-differential equation in _________ domain.
Laplace transform is equivalent to Fourier transform when, the real term \((\sigma)\) of complex frequency \(S=\sigma+j\omega\) ,should be
If \(x(t)=e^{-ct}-e^{-dt}\,u(-t)\), Laplace transform of x(t) is
Unit parabolic function is represented by its Laplace transform as:
A function f(t) is given by e–αt cos ωt. Its pole location in s-plane will be given by
Following statements are given for Laplace transforms :
(a) \(L[x(t)]=\displaystyle\int_{-\infty}^{\infty}x(t)e^{st}\,dt\)
(b) \(L[x(t)]=\displaystyle\int_{-\infty}^{\infty}x(t)e^{-st}\,dt\)
(c) \(L\left[te^{-at}u(t)\right]=\dfrac{1}{(s+a)^{2}}\)
(d) \(L\left[te^{-at}u(t)\right]=\dfrac{1}{(s-a)^{2}}\)
Out of the above, the following is the correct answer :
Match the following :
List - I List - II a) $(\dfrac{1}{s-\alpha})$ i) 
b) $(\dfrac{1}{s+\alpha})$ ii) 
c) $(\dfrac{1}{(s-\alpha)^{2}+\omega_{o}^{2}})$ iii) 
d) $(\dfrac{1}{(s+\alpha)^{2}+\omega_{o}^{2}})$ iv) 
Codes :
Assertion (A) : Laplace and Z transforms can be applied to the analysis of many unstable systems and consequently play an important role in the investigation of the stability or instability of the systems.
Reason (R) : The region of convergence of the Laplace transform of x(t) is given by the expression : \(\displaystyle\int_{-\infty}^{\infty}x(t)e^{-\sigma t}\,dt\ \ge\ \infty\)
Select your answer using the codes given below :
Laplace transform of e–at sin ωt is
Important Questions from Laplace Transform
Which of the following is the final value of the impulse response of the system whose transfer function is
(2s + 1)/(s 4 + 8s 3 + 16s 2 + s)
Find the Laplace transform for the following time domain.
y(t) = -2te -t + 4e -t - 4e -2t
Match List I with List II
List – I
List – II
f(t)
F(S)
A.
e -at
I.
\(\rm \frac{s}{s^2+ \omega^2}\)
B.
te at
II.
\(\rm \frac{\omega}{s^2+ \omega^2}\)
C.
sinωt
III.
\(\rm \frac{1}{(s- a)^2}\)
D.
cosωt
IV.
\(\rm \frac{1}{(s+ a)}\)
Choose the correct answer from the options given below:
The Laplace transform of sin h (at) is
The unilateral Laplace transform of f(t) is \(\frac{1}{s^2+s+1}\). The unilateral Laplace transform of t f(t) is
