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 :
(b) and (c) are correct
The statements come in two contradictory pairs, (a) against (b) and (c) against (d), so exactly one of each pair is right.
The definition — (b) is correct, (a) is not. The Laplace transform kernel carries a negative exponent:
\(L[x(t)]=\int_{-\infty}^{\infty}x(t)e^{-st}\,dt\)
The minus sign is not a convention that could go either way. It is what makes the integral converge: for \(s=\sigma+j\omega\) the factor \(e^{-\sigma t}\) decays for positive σ and tames signals that grow with time. Statement (a), with \(e^{+st}\), would diverge for every ordinary signal and no region of convergence would exist.
The transform pair — (c) is correct, (d) is not. Build it in two steps.
Start from the ramp.
\(L\left[t\,u(t)\right]=\dfrac{1}{s^{2}}\)
Apply frequency shifting. Multiplying by \(e^{-at}\) in time replaces s by \(s+a\) in the transform:
\(e^{-at}x(t)\ \longleftrightarrow\ X(s+a)\)
\(L\left[te^{-at}u(t)\right]=\dfrac{1}{(s+a)^{2}}\)
So (b) and (c) survive, which is option 2.
The pole-location check that settles the sign instantly. The signal \(te^{-at}u(t)\) decays for a > 0, and a decaying signal must have its pole in the left half plane. Statement (c) puts a double pole at \(s=-a\) ✓. Statement (d) puts it at \(s=+a\), the right half plane, which would describe the growing signal \(te^{+at}u(t)\) ✗. Reading off the sign of the pole is quicker and safer than re-deriving the pair.
Why the transform is squared. Multiplication by t in the time domain corresponds to differentiation in s:
\(t\,x(t)\ \longleftrightarrow\ -\dfrac{dX(s)}{ds}\)
Applying it to \(e^{-at}u(t)\leftrightarrow1/(s+a)\) gives \(-\dfrac{d}{ds}\dfrac{1}{s+a}=\dfrac{1}{(s+a)^{2}}\) — the same answer by a second route, and a reminder that each extra factor of t raises the pole order by one.
Hence, statements (b) and (c) are correct.
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
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
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
