Laplace transform of e–at sin ωt is
\(\dfrac{\omega}{(s+a)^{2}+\omega^{2}}\)
This is the frequency-shifting property applied to a standard pair — no integration is needed.
Start from the known transform
\(\mathcal{L}\left\{\sin\omega t\right\}=\dfrac{\omega}{s^{2}+\omega^{2}}\)
The shifting theorem states that multiplying a signal by \(e^{-at}\) in time replaces \(s\) by \(s+a\) in the transform:
\(\mathcal{L}\left\{e^{-at}f(t)\right\}=F(s+a)\)
so
\(\mathcal{L}\left\{e^{-at}\sin\omega t\right\}=\dfrac{\omega}{(s+a)^{2}+\omega^{2}}\)
which is option 1.
Two traps are built into the distractors.
The sign of the shift. A decaying exponential \(e^{-at}\) shifts the poles to \(s=-a\pm j\omega\) — into the left half plane, where a decaying signal's poles belong. Writing \((s-a)^{2}\) as options 2 and 3 do would put them at \(+a\pm j\omega\), describing a signal that grows. The substitution is deliberately counter-intuitive: minus in the time domain becomes plus in the s domain.
The sign inside the denominator. A \(+\omega^{2}\) gives complex conjugate poles and hence a sinusoid; a \(-\omega^{2}\) factorises into two real poles and belongs to the hyperbolic pair:
\(\mathcal{L}\left\{e^{-at}\sinh\omega t\right\}=\dfrac{\omega}{(s+a)^{2}-\omega^{2}}\)
so option 4 is the transform of a damped sinh, not a damped sine.
| Signal | Transform |
|---|---|
| sin ωt | \(\dfrac{\omega}{s^{2}+\omega^{2}}\) |
| cos ωt | \(\dfrac{s}{s^{2}+\omega^{2}}\) |
| e−at sin ωt | \(\dfrac{\omega}{(s+a)^{2}+\omega^{2}}\) |
| e−at cos ωt | \(\dfrac{s+a}{(s+a)^{2}+\omega^{2}}\) |
This pair is the workhorse of transient analysis, since it is exactly the step response of an underdamped second-order system: \(a=\zeta\omega_{n}\) sets how fast the ringing dies away and \(\omega=\omega_{n}\sqrt{1-\zeta^{2}}\) sets its frequency.
Hence, the transform is ω/[(s + a)2 + ω2].
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.
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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 :
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
