250+ TOP MCQs on Common Laplace Transforms – 2 and Answers

Signals & Systems Puzzles on “Common Laplace Transforms – 2”.

1. The Laplace transform of the function e^4t + 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})

Answer: B
Clarification: L {e^4t + 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})

Answer: D
Clarification: 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))

Answer: A
Clarification: 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})

Answer: A
Clarification: 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)e^3t 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})

Answer: B
Clarification: L {(t² + 4t + 2)e^3t} = L {t² e^3t + 4te^3t + 2e^3t}
= (frac{2}{(s-3)^3} + frac{4}{(s-3)^2} + frac{2}{s-3}).

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}))

Answer: A
Clarification: 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})

Answer: C
Clarification: 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})

Answer: C
Clarification: 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})

Answer: C
Clarification: 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) e^t-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})

Answer: D
Clarification: 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) e^t-2}
= (frac{4}{s}) + 5e^-2s L {t e^t}
= (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})

Answer: A
Clarification: 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})

Answer: B
Clarification: 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}))

Answer: D
Clarification: 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})

Answer: A
Clarification: 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})

Answer: B
Clarification: 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}}).