Signals & Systems Questions and Answers – Common Laplace Transforms – 1

This set of Signals & Systems Multiple Choice Questions & Answers (MCQs) focuses on “Common Laplace Transforms – 1”.

1. The Laplace transform of f(t) = (e-2t – 1)2 is ________________
a) $$\frac{2}{s+2} + \frac{1}{s}$$
b) $$\frac{1}{s+4}$$
c) $$\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}$$
d) –$$\frac{2}{s+2} + \frac{1}{s}$$

Explanation: (e-2t – 1)2 = e-4t – 2e-2t + 1
We know that, L {e-at} = $$\frac{1}{s+a}$$
L {1} = $$\frac{1}{s}$$
∴L {(e-2t – 1)2} = L {e-4t – 2e-2t + 1} = $$\frac{1}{s+4} – \frac{2}{s+2} + \frac{1}{s}$$.

2. Given f (t) = t2e-2x cos (3t). The value of L {f(t)} is __________________
a) $$\frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}$$
b) $$\frac{2(s-2)(s^2-4s-23)}{(s^2+4s+13)^3}$$
c) $$\frac{2(s+2)(s^2+4s+23)}{(s^2+4s+13)^3}$$
d) $$\frac{2(s-2)(s^2+4s-23)}{(s^2+4s-13)^3}$$

Explanation: Let g (t) = cos (3t); h (t) = e-2x cos (3t) = e-2x g (t)
Then, f (t) = t2h (t)
Let G(s) = L {g (t)}, H(s) = L {h (t)}, F(s) = L {f (t)}
So, G(s) = $$\frac{s}{s^2+9}$$
And H(s) = $$\frac{s+2}{(s+2)^2+9}$$
∴ F(s) = $$-\frac{d}{ds}[-\frac{d}{ds} H(s)] = \frac{2(s+2)(s^2+4s-23)}{(s^2+4s+13)^3}$$.

3. The inverse Laplace transform of F(s) = $$\frac{2}{s+c} e^{-bs}$$ is _________________
a) 2 e-k (t-b)
b) 2 e-k (t-b) u (t-b)
c) e-k (t-b) u (t-b)
d) 2 u (t-b)

Explanation: Let G(s) = $$\frac{2}{s+c}$$
Or, g (t) = L-1 {G(s)} = 2e-ct
Again, F(s) = L-1 {G(s) e-bs} = 2 e-k (t-b) u (t-b).

4. The inverse Laplace transform of F(s) = $$\frac{2}{s^2+3s+2}$$ is ______________
a) -2e-2t + 2e-t
b) 2e-2t + 2e-t
c) -2e-2t – 2e-t
d) 2e-t + e-2t

Explanation: s2 + 3s + 2 = (s+2) (s+1)
Now, F(s) = $$\frac{A}{(s+2)} + \frac{B}{(s+1)}$$
Hence, A = (s+2) F(s) |s=-2
= $$\frac{2}{s+1}$$|s=-2 = -2
And, B = (s+1) F(s) |s=-1
= $$\frac{2}{s+2}$$| s=-1 = 2
∴ F(s) = $$\frac{-2}{(s+2)} + \frac{2}{(s+1)}$$
∴ F (t) = L-1{F(s)}
= -2e-2t + 2e-t for t≥0

5. The Laplace transform of signal u(t-2) is ___________
a) $$\frac{-e^{-2s}}{s}$$
b) $$\frac{e^{-2s}}{s}$$
c) $$\frac{e^{-2s}}{1+s}$$
d) Zero

Explanation: X(s) = $$\int_0^∞ x(t) e^{-st} \,dt$$
Here, the given signal is u (t-2).
Hence, u (t-2) = 1 for all t>2 and =0 for all t<2.
So, limit is from 2 to ∞
∴ Laplace of signal= $$\int_2^∞ e^{-st} \,dt = \frac{e^{-2s}}{s}$$.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. The Laplace transform of the signal u (t+2) is _________
a) $$\frac{1}{s}$$
b) $$\frac{-1}{s}$$
c) $$\frac{e^{-2s}}{s}$$
d) $$\frac{-e^{-2s}}{s}$$

Explanation: X(s) = $$\int_0^∞ x(t) e^{-st} \,dt$$
Here, the given signal is u (t+2).
Hence, u (t+2) = 1 for all t>-2 and = 0 for all t<-2.
So, limit is from 0 to ∞
∴ Laplace of signal = $$\int_0^∞ e^{-st} \,dt = \frac{1}{s}$$.

7. The Laplace transform of the signal e-2tu(t+1) is ___________
a) $$\frac{1}{s+2}$$
b) $$\frac{e^{-s}}{s+2}$$
c) $$\frac{e^{-(s+2)}}{s+2}$$
d) $$\frac{-e^{-s}}{s+2}$$

Explanation: X(s) = $$\int_0^∞ x(t) e^{-st} \,dt$$
Here, the given signal is u (t+1).
Hence, u (t+1) = 1 for all t>-1 and = 0 for all t<-1.
So, limit is from 0 to ∞
∴ Laplace of signal = $$\int_0^∞ e^{-2t} e^{-st} \,dt = \frac{1}{s+2}$$.

8. The Laplace transform of the signal e2tu(-t+2) is ___________
a) $$\frac{e^{2(s-2)-1}}{s-2}$$
b) $$\frac{e^{-2s}}{s+2}$$
c) $$\frac{1-e^{-2(s-2)}}{s-2}$$
d) $$\frac{e^{-2s}}{s-2}$$

Explanation: X(s) = $$\int_0^∞ x(t) e^{-st} \,dt$$
= $$\int_0^∞ e^{2t} u(-t+2) e^{-st} \,dt$$
= $$\int_0^2 e^{t(2-s)} \,dt$$
= $$\frac{e^{2(2-s)-1}}{2-s}$$
= $$\frac{1-e^{-2(s-2)}}{s-2}$$.

9. The Laplace transform of the signal sin 5t is _____________
a) $$\frac{5}{s^2+5}$$
b) $$\frac{s}{s^2+5}$$
c) $$\frac{5}{s^2+25}$$
d) $$\frac{s}{s^2+25}$$

Explanation: We know that, sin 5t = $$\frac{(e^{j5t}-e^{-j5t})}{2j}$$
So, X(s) = $$\int_0^∞ \frac{(e^{j5t}-e^{-j5t})}{2j} e^{-st} \,dt$$
= $$\frac{5}{s^2+25}$$.

10. The Laplace transform of the signal u(t) – u(t-2) is ______________
a) $$\frac{e^{-2s}-1}{s}$$
b) $$\frac{-e^{-2s}+1}{s}$$
c) $$\frac{2}{s}$$
d) $$\frac{-2}{s}$$

Explanation: X(s) = $$\int_0^∞ x(t) e^{-st} \,dt$$
Here, the given signal is u (t).
Hence, u (t) = 1 for all t>0 and = 0 for all t<0.
Again, the given signal is u (t-2).
Hence, u (t-2) = 1 for all t>2 and = 0 for all t<2.
So, limit is from 0 to 2
∴ Laplace of signal= $$\int_0^2 e^{-st} \,dt$$
= $$\frac{-e^{-2s}+1}{s}$$.

11. The Laplace transform of the signal $$\frac{d}{dx}$$(te-t u(t)) is ______________
a) $$\frac{1}{s(s+1)^2}$$
b) $$\frac{s}{(s+1)^2}$$
c) $$\frac{e^{-s}}{s+1}$$
d) $$\frac{e^{-s}}{(s+1)^2}$$

Explanation: q (t) = t e-t u (t)
Laplace transform of q (t) is q(s) = $$\frac{1}{(s+1)^2}$$
Again, x (t) = $$\frac{d}{dx}$$ p(t)
Laplace transform of x (t) is x (s) = $$\frac{s}{(s+1)^2}$$.

12. The Laplace transform of the signal t u(t) * cos 2πt u(t) is _____________
a) $$\frac{1}{s(s^2+4π^2)}$$
b) $$\frac{2π}{s^2 (s^2+4π^2)}$$
c) $$\frac{1}{s^2 (s^2+4π^2)}$$
d) $$\frac{s^3}{(s^2+4π^2)}$$

Explanation: p (t) = t u (t)
Laplace transform of p (t), p (s) = $$\frac{1}{s^2}$$
Again, q (t) = cos 2πt u (t)
Laplace transform of q (t), q (s) = $$\frac{s}{(s^2+4π^2)}$$
Again, x (t) = p (t) * q (t)
Laplace transform of x (t), x (s) = $$\frac{1}{s(s^2+4π^2)}$$.

13. The Laplace transform of the signal t3 u(t) is _____________
a) $$\frac{3}{s^4}$$
b) $$\frac{-3}{s^4}$$
c) $$\frac{6}{s^4}$$
d) $$\frac{-6}{s^4}$$

Explanation: p (t) = t u (t)
Laplace transform of p (t), p (s) = $$\frac{1}{s^2}$$
Again, q (t) = – t p (t)
Laplace transform of q (t), q (s) = $$\frac{d}{dx}$$ P(s) = $$\frac{-2}{s^3}$$
Again, x (t) = -t q (t)
Laplace transform of x (t), x (s) = $$\frac{6}{s^4}$$.

14. The Laplace transform of the signal u(t-1) * e-2t u(t-1) is ___________
a) $$\frac{e^{-2(s+1)}}{2s+1}$$
b) $$\frac{e^{-2(s+1)}}{s+1}$$
c) $$\frac{e^{-(s+2)}}{s+2}$$
d) $$\frac{e^{-2(s+1)}}{s+2}$$

Explanation: p (t) = u (t)
Laplace transform of p (t), p(s) = $$\frac{1}{s}$$
Again, q (t) = u (t-1)
Laplace transform of q (t), q(s) = $$\frac{e^{-s}}{s^2}$$
Again, r (t) = e-2t u (t)
Laplace transform of r (t), r(s) = $$\frac{1}{s+2}$$
Again, v (t) = e-2t u (t-1)
Laplace transform of v (t), v(s) = $$\frac{e^{-(s+2)}}{s^2}$$
Again, x (t) = q (t) * v (t)
Laplace transform of x (t), x(s) = $$\frac{e^{-2(s+1)}}{s+2}$$.

15. The Laplace transform of the signal $$\int_0^t e^{-3t}$$ cos⁡2t dt is _____________
a) $$\frac{-(s+3)}{s[(s+3)^2+4]}$$
b) $$\frac{(s+3)}{s[(s+3)^2+4]}$$
c) $$\frac{s(s+3)}{(s+3)^2+4}$$
d) $$\frac{-s(s+3)}{(s+3)^2+4}$$

Explanation: Laplace transform of e-at cost u (t) = $$\frac{s+a}{(s+a)^2+1}$$
Let p (t) = e-3t cos2t u (t)
Laplace transform of p (t), p(s) = $$\frac{s+3}{(s+3)^2+4}$$
∴ Laplace transform of $$\int_{-∞}^t \,p(t)dt = \frac{1}{s} \int_{-∞}^0 \,p(t)dt + \frac{p(s)}{s}$$
Or, X(s) = $$\frac{(s+3)}{s[(s+3)^2+4]}$$.

Sanfoundry Global Education & Learning Series – Signals & Systems.

To practice all areas of Signals & Systems, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]