# Signals & Systems Questions and Answers – Common Laplace Transforms – 2

This set of Signals & Systems Puzzles focuses on “Common Laplace Transforms – 2”.

1. The Laplace transform of the function e4t + 5 is ___________
a) $$\frac{1}{s+4} + \frac{5}{s}$$
b) $$\frac{1}{s-4} + \frac{5}{s}$$
c) $$\frac{1}{s-4} – \frac{5}{s}$$
d) $$\frac{1}{s+4} – \frac{5}{s}$$

Explanation: L {e4t + 5} = $$\frac{1}{s-4} + \frac{5}{s}$$.

2. The Laplace transform of the function cos(2t) + 7sin(2t) is ____________
a) $$\frac{s-14}{s^2-4}$$
b) $$\frac{s+14}{s^2-4}$$
c) $$\frac{s-14}{s^2+4}$$
d) $$\frac{s+14}{s^2+4}$$

Explanation: L {cos (2t) + 7 sin (2t)} = $$\frac{s}{s^2+4} + \frac{7 * 2}{s^2+4}$$
= $$\frac{s+14}{s^2+4}$$.

3. Given F(s) = $$\frac{3s+5}{s^2+7}$$. The value of L-1{F(s)} is _______________
a) $$3 \,cos (\sqrt{7} t) + \frac{5}{\sqrt{7}} \,sin (\sqrt{7} \,t)$$
b) $$3 \,cos (\sqrt{7} t)- \frac{5}{\sqrt{7}} sin (\sqrt{7} \,t)$$
c) $$3 \,cos (\sqrt{7} t)$$
d) $$\frac{5}{\sqrt{7}} \,sin (\sqrt{7} t)$$

Explanation: Let, f (t) = L-1 {F(s)}
= L-1 $$\Big\{\frac{3s}{s^2+7} + \frac{5}{s^2+7}\Big\}$$
= L-1 $$\Big\{3 \frac{s}{s^2+\sqrt{7}^2} + \frac{5}{\sqrt{7}} \frac{\sqrt{7}}{s^2+\sqrt{7}^2}\Big\}$$
= $$3 \,cos (\sqrt{7} t) + \frac{5}{\sqrt{7}} \,sin (\sqrt{7} \,t)$$.

4. The Laplace transform of the function 10 + 5t + t2 – 4t3 is ___________
a) $$\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} – \frac{24}{s^4}$$
b) $$\frac{10}{s} – \frac{5}{s^2} + \frac{2}{s^3} – \frac{24}{s^4}$$
c) $$\frac{10}{s} – \frac{5}{s^2} – \frac{2}{s^3} – \frac{24}{s^4}$$
d) $$\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} + \frac{24}{s^4}$$

Explanation: L {10 + 5t + t2 – 4t3} = $$\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} – \frac{4 3!}{s^4}$$
= $$\frac{10}{s} + \frac{5}{s^2} + \frac{2}{s^3} – \frac{24}{s^4}$$.

5. The Laplace transform of the function (t2 + 4t + 2)e3t is ___________
a) $$\frac{2}{(s-3)} + \frac{4}{(s-3)^2} + \frac{2}{s-3}$$
b) $$\frac{2}{(s-3)^3} + \frac{4}{(s-3)^2} + \frac{2}{s-3}$$
c) $$\frac{2}{(s-3)^3} – \frac{4}{(s-3)^2} + \frac{2}{s-3}$$
d) $$\frac{2}{(s-3)^3} – \frac{4}{(s-3)^2} – \frac{2}{s-3}$$

Explanation: L {(t2 + 4t + 2)e3t} = L {t2 e3t + 4te3t + 2e3t}
= $$\frac{2}{(s-3)^3} + \frac{4}{(s-3)^2} + \frac{2}{s-3}$$.
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6. Given f (t) = [cos (3t)]2. The value of L {f(t)} is _______________
a) $$\frac{1}{2}(\frac{1}{s} + \frac{s}{s^2+36})$$
b) $$\frac{1}{4}(\frac{1}{s} + \frac{s}{s^2+36})$$
c) $$\frac{1}{5}(\frac{1}{s} + \frac{s}{s^2+36})$$
d) $$\frac{1}{8}(\frac{1}{s} + \frac{s}{s^2+36})$$

Explanation: We know that by trigonometric identity, cos2(3t) = $$\frac{1}{2}$$(1 + cos⁡(6t))
Also, we know that, L {cos (at)} = $$\frac{s}{s^2+a^2}$$
And L {1} = $$\frac{1}{s}$$
So, L {cos2 (3t)} = L $${\frac{1}{2}(1+cos⁡(6t))} = \frac{1}{2}(\frac{1}{s} + \frac{s}{s^2+36})$$.

7. F(t) = 0, 0≤t<6;
F(t) = 3, t≥6;
The Laplace transform of F (t) is __________
a) $$\frac{3e^{-6s}}{s^2}$$
b) $$\frac{6e^{-6s}}{s}$$
c) $$\frac{3e^{-6s}}{s}$$
d) $$\frac{6e^{-6s}}{s^2}$$

Explanation: F (t) = 3u (t)
L {3u (t)} = e-6s L {3}
= $$\frac{3e^{-6s}}{s}$$.

8. G(T) = 3, 0≤t<5;
G(T) = 10, 5≤t<8;
G(T) = 0; t≥8;
The Laplace transform of G (t) is ___________
a) $$\frac{3}{s} – \frac{7e^{-5s}}{s} + \frac{10e^{-8s}}{s}$$
b) $$\frac{3}{s} + \frac{7e^{-5s}}{s} + \frac{10e^{-8s}}{s}$$
c) $$\frac{3}{s} + \frac{7e^{-5s}}{s} – \frac{10e^{-8s}}{s}$$
d) $$\frac{3}{s} – \frac{7e^{-5s}}{s} – \frac{10e^{-8s}}{s}$$

Explanation: G (t) = 3 + (10 − 3) u (t) + (0 − 10) u (t) = 3 + 7u (t) − 10u (t)
L {3 + 7u (t) − 10u (t)} = $$\frac{3}{s} + \frac{7e^{-5s}}{s} – \frac{10e^{-8s}}{s}$$.

9. H(t) = 0, 0≤t<3;
H(t) = 6sin (t-3), t≥3;
The Laplace transform of H (t) is ___________
a) $$\frac{3e^{-3s}}{s^2-1}$$
b) $$\frac{3e^{-3s}}{s^2+1}$$
c) $$\frac{6e^{-3s}}{s^2+1}$$
d) $$\frac{6e^{-3s}}{s^2-1}$$

Explanation: H (t) = 6u (t) sin (t − 3)
L {6u (t) sin (t − 3)} = 6e−3s L {sin (t)}
= $$\frac{6e^{-3s}}{s^2+1}$$.

10. J(t) = 4, 0≤t<2;
J(t) = 4 + 5(t-2) et-2, t≥2;
The Laplace transform of J (t) is ____________
a) $$\frac{4}{s} + \frac{5e^{-2s}}{(s+1)^2}$$
b) $$\frac{4}{s} – \frac{5e^{-2s}}{(s+1)^2}$$
c) $$\frac{4}{s} – \frac{5e^{-2s}}{(s-1)^2}$$
d) $$\frac{4}{s} + \frac{5e^{-2s}}{(s-1)^2}$$

Explanation: J (t) = 4 + 5u (t) (t-2) et-2
L {4 + 5u (t) (t-2) et-2} = $$\frac{4}{s}$$ + 5 L {u (t) (t-2) et-2}
= $$\frac{4}{s}$$ + 5e-2s L {t et}
= $$\frac{4}{s} + \frac{5e^{-2s}}{(s-1)^2}$$.

11. U(t) = 0, 0≤t<7;
U(t) = (t-7)3, t≥7;
The Laplace transform of U (t) is ___________
a) $$\frac{6e^{-7s}}{s^4}$$
b) $$\frac{3e^{-7s}}{s^4}$$
c) $$\frac{6e^{-7s}}{s^3}$$
d) $$\frac{3e^{-7s}}{s^3}$$

Explanation: U (t) = u (t) (t-7)3
L {u (t) (t-7)3} = e-7s L {t3}
= $$\frac{3!e^{-7s}}{s^4} = \frac{6e^{-7s}}{s^4}$$.

12. V(t) = 5, 0≤t<1;
V(t) = t, t≥1;
The Laplace transform of V (t) is ___________
a) $$\frac{5}{s} + \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}$$
b) $$\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}$$
c) $$\frac{5}{s} – \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}$$
d) $$\frac{5}{s} – \frac{e^{-s}}{s^2} + \frac{4e^{-s}}{s}$$

Explanation: V (t) = 5 + u (t) (t-5)
L {5 + u (t) (t-5)} = $$\frac{5}{s}$$ + L {u (t) (t-5)}
= $$\frac{5}{s}$$ + e-s L {t-4}
= $$\frac{5}{s} + e^{-s} (\frac{1}{s^2} – \frac{4}{s})$$
= $$\frac{5}{s} + \frac{e^{-s}}{s^2} – \frac{4e^{-s}}{s}$$.

13. W(t) = 2, 0≤t<4;
W(t) = t2, t≥4;
The Laplace transform of W (t) is ___________
a) $$\frac{2}{s} – e-{4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})$$
b) $$\frac{2}{s} + e-{4s} (\frac{2}{s^3} – \frac{8}{s^2} – \frac{14}{s})$$
c) $$\frac{2}{s} – e-{4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})$$
d) $$\frac{2}{s} + e-{4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})$$

Explanation: W (t) = 2 + u (t) (t2-2)
L {2 + u (t) (t2-2)} = $$\frac{2}{s}$$ + L {u (t) (t2-2)}
= $$\frac{2}{s}$$ + e-4s L {(t+4)2 -2}
= $$\frac{2}{s}$$ + e-4s L {t2 + 8t + 14}
= $$\frac{2}{s} + e-{4s} (\frac{2}{s^3} + \frac{8}{s^2} + \frac{14}{s})$$.

14. The Laplace transform of the function sin (4t) cos (2t) is ____________
a) $$\frac{2}{s^2+16}$$
b) $$\frac{2}{s^2-16}$$
c) $$\frac{s^2+16}{2}$$
d) $$\frac{s^2-16}{2}$$

Explanation: L ($$\frac{1}{2}$$ sin (4t))
= $$\frac{1}{2}$$ L (sin (4t))
= $$\frac{1}{2} \frac{4}{s^2+16}$$
= $$\frac{2}{s^2+16}$$.

15. The Laplace transform of f(t) = sin(2t) cos(2t) is ____________
a) $$\frac{1}{2}{\frac{s}{s^2+16}}$$
b) $$\frac{1}{2}{\frac{4}{s^2+16}}$$
c) $$\frac{4}{s^2+16}$$
d) $$\frac{1}{2}$$

Explanation: Using trigonometric identity,
We get, sin (2t) cos (2t) = $$\frac{1}{2}$$ sin⁡(4t)
∴ L{ sin (2t) cos (2t)} = L{$$\frac{1}{2}$$ sin⁡(4t)}
We know that, L {sin at} = $$\frac{a}{s^2+a^2}$$
∴L{$$\frac{1}{2}$$ sin⁡(4t)} = $$\frac{1}{2}{\frac{4}{s^2+16}}$$.

Sanfoundry Global Education & Learning Series – Signals & Systems.

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