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.
If \(x(t)=e^{-ct}-e^{-dt}\,u(-t)\), Laplace transform of x(t) is
\(X(s)=\frac{2s+c+d}{(s+c)(s+d)}\)
The two standard pairs needed here.
Right-sided (causal) exponential:
\(e^{-ct}u(t)\ \longleftrightarrow\ \dfrac{1}{s+c}, \qquad \text{ROC: } \operatorname{Re}(s) \gt -c\)
Left-sided (anticausal) exponential — note the sign, which is where marks are usually lost:
\(-e^{-dt}u(-t)\ \longleftrightarrow\ \dfrac{1}{s+d}, \qquad \text{ROC: } \operatorname{Re}(s) \lt -d\)
Deriving the second pair. For the left-sided term the integration runs only over negative time:
\(\int_{-\infty}^{0}-e^{-dt}e^{-st}dt=-\int_{-\infty}^{0}e^{-(s+d)t}dt=\dfrac{1}{s+d}\)
The minus sign in front of the signal cancels the minus that comes out of the integration limits, so the transform ends up with a plus sign — the same algebraic form as the causal term, but with the opposite ROC.
Add the two transforms.
\(X(s)=\dfrac{1}{s+c}+\dfrac{1}{s+d}\)
Combine over a common denominator.
\(X(s)=\dfrac{(s+d)+(s+c)}{(s+c)(s+d)}=\dfrac{2s+c+d}{(s+c)(s+d)}\)
Checking the wrong options. Options 2 and 3 place poles at +c or +d, which would require exponentials of the form \(e^{+ct}\); the given signal decays as \(e^{-ct}\), so its pole must be at −c. Option 4's numerator \(2s-c-d\) would result from subtracting the two transforms, i.e. from mishandling the sign of the anticausal term.
The ROC of the result. Being the intersection of \(\operatorname{Re}(s) \gt -c\) and \(\operatorname{Re}(s) \lt -d\), it is the vertical strip \(-c \lt \operatorname{Re}(s) \lt -d\) — the signature of a two-sided signal, and the reason the same X(s) can correspond to different time signals unless the ROC is stated.
Hence, \(X(s)=\dfrac{2s+c+d}{(s+c)(s+d)}\).
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
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
