The Laplace transform converts integro-differential equation in _________ domain.
S
The Laplace transform carries a problem from the time domain into the s domain — option 2.
\(F(s)=\int_{0}^{\infty}f(t)\,e^{-st}\,dt,\qquad s=\sigma+j\omega\)
Why this is worth doing. The transform replaces calculus with algebra, because differentiation and integration become multiplication and division:
| Time domain | s domain |
|---|---|
| \(\dfrac{df}{dt}\) | \(sF(s)-f(0)\) |
| \(\displaystyle\int_{0}^{t}f\,d\tau\) | \(\dfrac{F(s)}{s}\) |
| Convolution | Multiplication |
An integro-differential equation therefore becomes an ordinary algebraic equation in \(F(s)\), which is solved by rearrangement and taken back to the time domain by partial fractions and a table.
Note also that the initial conditions appear automatically in the differentiation rule — the \(f(0)\) term. There is no separate step of finding a complementary function and fitting constants: the transform handles the transient and the steady state together, which is exactly why circuit analysis uses it.
Why option 3 is a genuine near-miss. \(j\omega\) is the Fourier domain, and the two are closely related: setting \(\sigma=0\) in \(s=\sigma+j\omega\) turns the Laplace kernel into the Fourier kernel, so the Fourier transform is the Laplace transform evaluated on the imaginary axis. But that restriction discards the \(e^{-\sigma t}\) factor, and with it both the convergence of growing signals and the initial conditions. The full complex variable is \(s\).
The other two options are not domains at all. \(\sigma\) is merely the real part of \(s\), and \(\xi\) (or \(\zeta\)) is the damping ratio — a parameter of a second-order system's response, not a transform variable.
The reason the s plane repays study is that the pole positions in it read off directly as behaviour: the real part sets how fast a term decays, the imaginary part sets its frequency of oscillation, and the left half plane is exactly the region of stability.
Hence, the Laplace transform works in the s domain.
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.
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
For two signals \(x_1(t)\) and \(x_2(t)\). The correct Laplace transform expression for \(x_1(t)\cdot x_2(t)\) 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
