A function f(t) is given by e–αt cos ωt. Its pole location in s-plane will be given by
Take the Laplace transform of the signal. The standard pair for a damped cosine is
\(e^{-\alpha t}\cos\omega t\ u(t) \ \longleftrightarrow\ \dfrac{s+\alpha}{(s+\alpha)^{2}+\omega^{2}}\)
Find the poles by setting the denominator to zero:
\((s+\alpha)^{2}+\omega^{2}=0 \Rightarrow (s+\alpha)^{2}=-\omega^{2}\)
\(s+\alpha=\pm j\omega \Rightarrow s=-\alpha \pm j\omega\)
So there is a complex conjugate pair sitting at real part −α and imaginary parts ±ω — in the left half of the s-plane.
Read the two parts of the signal separately — this is the quickest way to place any such pole pattern:
The factor \(e^{-\alpha t}\) is a decaying envelope, and decay always corresponds to a negative real part. That alone eliminates options 2 (undamped, poles on the axis) and 3 (growing, poles on the right).
The factor \(\cos\omega t\) is an oscillation, and oscillation always corresponds to an imaginary part ±jω. Poles must therefore appear as a conjugate pair, symmetric about the real axis — which is why option 4, with poles at −α and +α on the real axis, cannot describe an oscillating signal.
The general dictionary worth carrying away:
| Pole location | Time behaviour |
|---|---|
| Left half plane (σ < 0) | decaying — stable |
| On the jω axis | sustained oscillation — marginal |
| Right half plane (σ > 0) | growing — unstable |
| Real poles | pure exponentials, no oscillation |
| Complex pair | oscillation at ω, envelope set by σ |
Hence, the poles lie at s = −α ± jω, the left-half-plane conjugate pair of option 1.
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:
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
