Engineering Mathematics Multiple Choice Questions on “Laplace Transform by Properties – 3”.
1. Time domain function of (frac{s}{a^2+s^2}) is given by?
a) Cos(at)
b) Sin(at)
c) Cos(at)Sin(at)
d) Sin(t)
Answer: a
Explanation: L[Cos(at)] = (frac{s}{a^2+s^2})
L-1 ([frac{s}{a^2+s^2}]) = Cos(at).
2. Inverse Laplace transform of (frac{1}{(s+1)(s-1)(s+2)}) is?
a) –1⁄2 et + 1⁄6 e-t + 1⁄3 e2t
b) –1⁄2 e-t + 1⁄6 et + 1⁄3 e-2t
c) 1⁄2 e-t – 1⁄6 et – 1⁄3 e-2
d) –1⁄2 e-t + 1⁄6 e-t + 1⁄3 e-2
Answer: b
Explanation:
Given, (F(s)=frac{1}{(s+1)(s-1)(s+2)}=frac{-1}{2(s+1)} +frac{1}{6(s-1)}+frac{1}{3(s+2)})
Hence, inverse laplace transform is (f(t)=-frac{1}{2} e^{-t}+frac{1}{6} e^t+frac{1}{3} e^{-2})
3. Inverse laplace transform of (frac{1}{(s-1)^2 (s+5)}) is?
a) 1⁄6 e – t – 1⁄36 et + 1⁄36 e-5t
b) 1⁄6 ett – 1⁄36 et + 1⁄36 e-5t
c) 1⁄6 e-tt2 – 1⁄36 e-t + 1⁄36 e5t
d) 1⁄6 e-t t-1⁄36 e-t + 1⁄36 e5t
Answer: a
Explanation:
Given, (F(s)=frac{1}{(s-1)^2 (s+5)}=frac{1}{(s-1)} left [frac{1}{(s-1)(s+5)}right ])
=(frac{1}{(s-1)} left [frac{1}{6(s-1)}-frac{1}{6(s+5)}right ])
=(frac{1}{6} left [frac{1}{(s-1)^2}-frac{1}{(s-1)(s+5)}right ])
=(frac{1}{6} left [
frac{1}{(s-1)^2} – frac{1}{6} left [frac{1}{(s-1)} – frac{1}{(s+5)}right ]right ])
=(frac{1}{6(s-1)^2}-frac{1}{36(s-1)}+frac{1}{36(s+5)})
Inverse Laplace transform is (f(t)=frac{1}{6} e^t t-frac{1}{36} e^t+frac{1}{36} e^{-5t})
4. Find the inverse laplace transform of (frac{1}{(s^2+1)(s – 1)(s + 5)}).
a) 1⁄12 et – 1⁄13 Cos(-t) – 1⁄12 Sin(-t) – 1⁄156 e-5t
b) 1⁄12 e-t – 1⁄13 Cos(t) – 1⁄12 Sin(t) – 1⁄156 e5t
c) 1⁄12 et – 1⁄13 Cos(t) – 1⁄12 Sin(t) – 1⁄156 e-5t
d) 1⁄12 et + 1⁄13 Cos(t) + 1⁄12 Sin(t) + 1⁄156 e-5t
Answer: c
Explanation:
Given , F(s)=(frac{1}{(s^2+1)(s-1)(s+5)})
F(s)=(frac{1}{6(s^2+1)}left [frac{1}{s-1}-frac{1}{s+5}right ]=frac{1}{6(s^2+1)(s-1)}-frac{1}{6(s^2+1)(s+5)})
=(frac{1}{6} left [frac{1}{2*(s – 1)}-frac{1}{2} frac{s+1}{(s^2+ 1)}right ]-frac{1}{6}left [frac{1}{26*(s + 5)}-frac{1}{26} frac{s-5}{(s^2+1)}right ])
=(frac{1}{12(s – 1)}-frac{1}{26} frac{2s+3}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12(s – 1)}-frac{1}{13} frac{s}{(s^2+ 1)}-frac{1}{12} frac{1}{(s^2+ 1)}-frac{1}{156(s + 5)})
=(frac{1}{12} e^t-frac{1}{13} Cos(t)-frac{1}{12} Sin(t)-frac{1}{156}e^{-5t})
5. Find the inverse laplace transform of (frac{s}{(s^2+ 4)^2}).
a) 1⁄4 sin(2t)
b) t2⁄4 sin(2t)
c) t⁄4 sin(2t)
d) t⁄4 sin(2t2)
Answer: c
Explanation:
Given, (Y(s)=frac{s}{(s^2+ 4)^2})
Inverse Laplace transform of (frac{1}{s^2+4})=sin(2t)
Now, (frac{d}{ds} (frac{1}{s^2+4}))=-tsin(2t)
Inverse lapalce of (frac{-2s}{(s^2+4)^2}=-frac{t}{2} sin(2t))
Inverse lapalce of (frac{s}{(s^2+4)^2}=frac{t}{4} sin(2t))
6. Final value theorem states that _________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} sX(s))
Answer: d
Explanation: Final value theorem states that
x(∞)=(lim_{xrightarrow 0} sX(s))
7. Initial value theorem states that ___________
a) x(0)=(lim_{xrightarrow ∞} sX(s))
b) x(∞)=(lim_{xrightarrow ∞} sX(s))
c) x(0)=(lim_{xrightarrow 0} sX(s))
d) x(∞)=(lim_{xrightarrow 0} sX(s))
Answer: a
Explanation: Initial value theorem states that
x(0)=(lim_{xrightarrow ∞} sX(s))
8. Find the value of x(∞) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 12⁄20
d) 2
Answer: c
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by final value theorem,
(x(∞)=lim_{xrightarrow 0} sX(s)=frac{12}{20})
9. Find the value of x(0) if (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20}).
a) 5
b) 4
c) 12
d) 2
Answer: d
Explanation:
Given, (X(s)=frac{2s^2+5s+12/s}{s^3+4s^2+14s+20})
Hence, (sX(s)=frac{2s^3+5s^2+12}{s^3+4s^2+14s+20})
Hence, by initial value theorem,
(x(0)=lim_{xrightarrow infty} sX(s)=2)
10. Find the inverse lapace of (frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]}).
a) 1⁄3 et [Cos(t) – Cos(2t)].
b) 1⁄3 e-t [Cos(t) + Cos(2t)].
c) 1⁄3 et [Cos(t) + Cos(2t)].
d) 1⁄3 e-t [Cos(t) – Cos(2t)].
Answer: d
Explanation:
Given, (Y(s)=frac{(s+1)}{[(s+1)^2+4][(s+1)^2+1]})
=(frac{s+1}{3(s^2+ 2*s + 2)}-frac{s+1}{3(s^2+ 2*s + 5)})
=(frac{s+1}{3[(s+1)^2+1]}-frac{s+1}{3[(s+1^2+4)]})
=(frac{1}{3} [e^{-t} Cos(t)]-frac{1}{3}[e^{-t} Cos(2t)])
=(frac{1}{3} e^{-t} [Cos(t)-Cos(2t)])
11. Find the inverse laplace transform of (Y(s)=frac{2s}{1-s^2}e^{-s}).
a) -e-t + 1 + et – 1
b) -e-t + 1 – et + 1
c) -e-t + 1 + et + 1
d) -e-t + 1 – et – 1
Answer: d
Explanation: Given,
Y(s)=(frac{2s}{1-s^2}e^{-s})
Let,G(s)=(frac{2s}{1-s^2}=-frac{1}{s – 1}-frac{1}{s + 1})
hence,g(t)=(-e^{-t} – e^t)
Since,Y(s)=(e^{-s} G(s)=>y(t)=g(t-1))
hence,y(t)=(-e^{-t+1}-e^{t-1})
12. Find the inverse laplace transform of (frac{1}{s(s-1)(s^2+1)}).
a) 1⁄2 e-t + 1⁄2 Sin(-t) – 1⁄2 Cos(-t)
b) 1⁄2 et + 1⁄2 Sin(t) – 1⁄2 Cos(t)
c) 1⁄2 et + 1⁄2 Sin(t) + 1⁄2 Cos(t)
d) 1⁄2 et – 1⁄2 Sin(t) – 1⁄2 Cos(t)
Answer: b
Explanation: We know that,
Given, Y(s)=(frac{1}{s(s-1)(s^2+1)})
Let, G(s)=(frac{1}{(s-1)(s^2+1)}=frac{1}{2(s^2-1)}-frac{s+1}{2(s^2+1)}=frac{1}{2*(s-1)}-frac{s}{2(s^2+1)}-frac{1}{2(s^2+1)})
Now, g(t)=(frac{1}{2}e^t-frac{1}{2}cos(t)-frac{1}{2}cos(t))
Now, Y(s)=(frac{1}{2}G(s)=>y(t)=int_0^t g(t)dt=frac{1}{2}e^t+frac{1}{2}sin(t)-frac{1}{2}cos(t))