Previous Year Questions- Laplace Transform | Signals and Systems - Electrical Engineering (EE) PDF Download

Q1: If u(t) is the unit step function, then the region of convergence (ROC) of the Laplace transform of the signal is      (2024)
(a) $-\infty <\mathrm{Re}\left(𝑠\right)<\infty$−∞ < Re(s) < ∞
(b) Re(s) ≥ 10
(c) Re(s) ≤ 1
(d) $1\le \mathrm{Re}\left(𝑠\right)\le 10$1 ≤ Re(s) ≤ 10
Ans: (a)
Sol: Since x(t) is finite duration signal. So ROC will be −∞ < σ < ∞.

Q2: Which of the following statements is true about the two sided Laplace transform?     (2020)
(a) It exists for every signal that may or may not have a Fourier transform.
(b) It has no poles for any bounded signal that is non-zero only inside a finite time interval.
(c) The number of finite poles and finite zeroes must be equal.
(d) If a signal can be expressed as a weighted sum of shifted one sided exponentials, then its Laplace Transform will have no poles.
Ans: (b)
Sol: It has no poles for any bounded signal that is nonzero in a finite time interval. This is true as we know for finite amplitude finite width signal ROC is entire s plane and ROC never includes any pole.
It implies for such signals there is no poles. Hence the correct answer is option (B).

Q3: The output response of a system is denoted as y(t), and its Laplace transform is given by  
The steady state value of y(t) is      (2019)
(a)
(b) 10√2
(c)
(d) 100√2
Ans: (a)
Sol: Steady state value of y(t)

Q4: A system transfer function is  and all other coefficients are positive, the transfer function represents a        (2019)
(a) low pass filter
(b) high pass filter
(c) band pass filter
(d) notch filter
Ans: (a)
Sol: which represents second order low pass filter.

Q5: The inverse Laplace transform of      (2019)
(a) 3te^−t + e^−t 
(b) 3e^−t
(c) $2𝑡𝑒-𝑡+𝑒-𝑡$2te^−t + e^−t 
(d) 4te^−t + e^−t 
Ans: (c)
Sol: 
Q6: Consider a linear time-invariant system with transfer function If the input is cos(t) and the steady state output is Acos(t + α), then the value of A is _________.      (SET-2 (2016))
(a) 0.70
(b) 0.26
(c) 0.96
(d) 1.2
Ans: (a)
Sol: 
Q7: The transfer function of a system is  The steady state output y(t) is Acos(2t + φ) for the input cos(2t). The values of A and φ, respectively are      (SET-1 (2016))
(a)
(b)
(c)
(d)
Ans: (b)
Sol: 
Q8: The Laplace Transform of f(t) = e^2tsin(5t)u(t) is       (SET-1 (2016))
(a)
(b)
(c)
(d)
Ans: (a)
Sol: Laplace transform of sin5tu(t)
Q9: The Laplace transform of  The Laplace transform of g(t) =  is       (SET-2 (2015))
(a) $3𝑠-5/2/2$3s^−5/2/2
(b) s^−1/2
(c) $𝑠1/2$s^1/2 
(d) 𝑠^3/2
Ans: (b)
^{Sol: Given that,}
By using property of differentiation In time,

^{Q10: Consider an LTI system with transfer function} If the input to the system is cos(3t) and the steady state output is Asin(3t + α), then the value of A is       (SET-2 (2014))
(a) 1/30
(b) 1/15
(c) 3/4
(d) 4/3
Ans: (b)
Sol: Given,
A = 1/15.

Q11: Which one of the following statements is NOT TRUE for a continuous time causal and stable LTI system?      (2013)
(a) All the poles of the system must lie on the left side of the jω axis
(b) Zeros of the system can lie anywhere in the s-plane
(c) All the poles must lie within |s| = 1
(d) All the roots of the characteristic equation must be located on the left side of the jω axis.
Ans: (c)
Sol: All poles must lie within |Z| = 1

Q12: Assuming zero initial condition, the response y(t) of the system given below to a unit step input u(t) is      (2013)
(a) u(t)
(b) tu(t)
(c) (t²/2) (u(t))
(d) $𝑒-𝑡𝑢\left(𝑡\right)$e^−tu(t)
Ans: (b)
Sol: 
Q13: Consider the differential equation
The numerical value of  is,      (2012)
(a) -2
(b) -1
(c) 0
(d) 1
Ans: (d)
Sol: 
Q14: The unilateral Laplace transform of  The unilateral Laplace transform of  tf(t) is       (2012)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: 
Q15: Let the Laplace transform of a function F(t) which exists for t > 0 be F₁(s) and the Laplace transform of its delayed version f(t − τ) be F₂(s). Let  be the complex conjugate of F₁(s) with the Laplace variable set  s = σ + iω. If   then the inverse Laplace transform of G(s) is      (2011)
(a) An ideal impulse δ(t)
(b) An ideal delayed impulse δ(t − τ)
(c) An ideal step function u(t)
(d) An ideal delayed step function u (t − τ)
Ans: (b)
Sol: 
Q16: Given f(t) and g(t)as shown below:
The Laplace transform of g(t) is       (2010)
(a)
(b)
(c)
(d)
Ans: (c)
Sol: 
Q17: Given f(t) and g(t)as shown below:
g(t) can be expressed as      (2010)
(a) $𝑔\left(𝑡\right)=𝑓(2𝑡-3)$g(t) = f(2t − 3)
(b)
(c)
(d)  
Ans: (d)
Sol: Since, g(t) has width of 2-unit and f(t) has 1 unit, therefore we have to first expand f(t) by 2 unit and for this we have to scale f(t) by 1/2.
Now shift it by three unit to get g(t)

Q18: The running integration, given by     (2006)
(a) has no finite singularities in its double sided Laplace Transform Y(s)
(b) produces a bounded output for every causal bounded input
(c) produces a bounded output for every anticausal bounded input
(d) has no finite zeroes in its double sided Laplace Transform Y(s)
Ans: (d)

Q19: The Laplace transform of a function f(t) is  f(t) approaches      (2005)
(a) 3
(b) 5
(c) 17/2
(d) ∞
Ans: (a)
Sol:  As F(s) has only left hand poles.
By final value theorem,

Q20: Consider the function,  where F(s) is the Laplace transform of the of the function f(t). The initial value of f(t) is equal to      (2004)
(a) 5
(b) 5/2
(c) 5/3
(d) 0
Ans: (d)
Sol: By initial value theorem,

Q21: Let Y(s) be the Laplace transformation of the function y(t), then the final value of the function is       (2002)
(a) lim_s→0 Y(s)
(b) lim_s→∞ Y(s)
(c) lim_s→0 sY(s)
(d)  lim_x→∞ sY(s)  
Ans: (c)
Sol: Final value = lim_s→0 sY(s)

Q22: Given the relationship between the input u(t) and the output y(t) to be
the transfer function Y(s)/U(s) is      (2001)
(a)
(b)
(c)
(d)
Ans: (d)
Sol: