This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Laplace Transform by Properties – 1”.
1. Laplace of function f(t) is given by?
a) F(s)=\(\int_{-\infty}^\infty f(t)e^{-st} \,dt\)
b) F(t)=\(\int_{-\infty}^\infty f(t)e^{-t} \,dt\)
c) f(s)=\(\int_{-\infty}^\infty f(t)e^{-st} \,dt\)
d) f(t)=\(\int_{-\infty}^\infty f(t)e^{-t} \,dt\)
View Answer
Explanation: Laplace of function f(t) is given by
F(s)=\(\int_{-\infty}^\infty f(t)e^{-st} \,dt\).
2. Laplace transform any function changes it domain to s-domain.
a) True
b) False
View Answer
Explanation: Laplace of function f(t) is given by F(s)=\(\int_{-\infty}^\infty f(t)e^{-st} \), hence it changes domain of function from one domain to s-domain.
3. Laplace transform if sin(at)u(t) is?
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2
View Answer
Explanation: We know that,
F(s)=\(\int_{-\infty}^\infty sin(at)u(t) e^{-st} dt=\int_0^∞ sin(at)e^{-st} dt\)
=\(\left [\frac{e^{-st}}{a^2+s^2}[-ssin(at)-acos(at)]\right ]_∞^0\)
=\(\frac{a}{a^2+s^2}\)
4. Laplace transform if cos(at)u(t) is?
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2
View Answer
Explanation: We know that,
F(s)=\(\int_{-\infty}^\infty cos(at)u(t) e^{-st} dt=\int_0^∞ cos(at)e^{-st} dt\)
=\(\left [\frac{e^{-st}}{a^2+s^2}[-scos(at)-asin(at)]\right ]_∞^0\)
=\(\frac{a}{a^2+s^2}\)
5. Find the laplace transform of et Sin(t).
a) \(\frac{a}{a^2+(s+1)^2}\)
b) \(\frac{a}{a^2+(s-1)^2}\)
c) \(\frac{s+1}{a^2+(s+1)^2}\)
d) \(\frac{s+1}{a^2+(s+1)^2}\)
View Answer
Explanation:
F(s)=\(\int_{-\infty}^\infty e^t sin(at)u(t) e^{-st} dt=∫_0^∞ sin(at)e^{-(s-1)t} dt\)
=\(\left [\frac e^{-st}{a^2+(s-1)^2} [-(s-1)sin(at)-acos(at) ]\right ]_0^∞\)
=\(\frac{a}{a^2+(s-1)^2}\)
6. Laplace transform of t2 sin(2t).
a) \(\left [\frac{12s^2-16}{(s^2+4)^4}\right ]\)
b) \(\left [\frac{3s^2-4}{(s^2+4)^3}\right ]\)
c) \(\left [\frac{12s^2-16}{(s^2+4)^6}\right ]\)
d) \(\left [\frac{12s^2-16}{(s^2+4)^3}\right ]\)
View Answer
Explanation: We know that,
\(L(t^n f(t))=(-1)^n \frac{d^n F(s)}{ds^n}\),
Here, f(t)=sin(2t)=>F(s)=\(\frac{2}{s^2+4}\),
Hence, \(L(t^2 sin(2t))=\frac{d^2}{ds^2} (\frac{2}{s^2+4})=\frac{d}{ds} \frac{(s^2+4).0-2(2s)}{(s^2+4)^2}\)
=\(-4\left [\frac{(s^2+4)^2-2s(s^2+4)2s}{(s^2+4)^4} \right ]=\left [\frac{12s^2-16}{(s^2+4)^3}\right ]\)
7. Find the laplace transform of t5⁄2.
a) \(\frac{15}{8} \frac{√π}{s^{5/2}}\)
b) \(\frac{15}{8} \frac{√π}{s^{7/2}}\)
c) \(\frac{9}{4} \frac{√π}{s^{7/2}}\)
d) \(\frac{15}{4} \frac{√π}{s^{7/2}}\)
View Answer
Explanation:
\(g(t)=t^{5/2}=\frac{5}{2} \int_0^t t^{\frac{3}{2}} dt=\frac{15}{4} \int_0^t \int_0^t √t dt dt\)
let f(t)=√t, hence, F(s)=\(\frac{\sqrt{π}}{2s^{\frac{3}{2}}}\)
hence, G(s)=\(\frac{15}{4} \,\frac{1}{s^2} \,F(s)=\frac{15}{8} \frac{√π}{s^{7/2}}\)
8. Value of \(\int_{-\infty}^\infty e^t \,Sin(t)Cos(t)dt\) = ?
a) 0.5
b) 0.75
c) 0.2
d) 0.71
View Answer
Explanation: L(Sin(2t)) = \(\int_{-\infty}^\infty e^{-st} \,Sin(2t)dt\) = 2/(s2 + 4)
Putting s=-1
\(\int_{-\infty}^\infty e^t \,Sin(2t)dt\) = 0.4
hence,
\(\int_{-\infty}^\infty e^{-st} \,Sin(t)Cos(t)dt\) = 0.2.
9. Value of \(\int_{-\infty}^\infty e^t \,Sin(t) \,dt\) = ?
a) 0.50
b) 0.25
c) 0.17
d) 0.12
View Answer
Explanation: L(Sin(2t)) = \(\int_{-\infty}^\infty e^{-st} \,Sin(t)dt\) = 1/(s2 + 1)
Putting s = -1
\(\int_{-\infty}^\infty e^t \,Sin(t)dt\) = 0.5.
10. Value of \(\int_{-\infty}^\infty e^t \,log(1+t)dt\) = ?
a) Sum of infinite integers
b) Sum of infinite factorials
c) Sum of squares of Integers
d) Sum of square of factorials
View Answer
Explanation:
\(\int_{-\infty}^\infty e^t (t-t^2/2+t^3/3-….)dt\)
\(\int_{-\infty}^\infty te^t dt=0.5 \int_{-\infty}^\infty te^t dt\)
Now,
\(\int_{-\infty}^\infty te^t dt– 1/2 \int_{-\infty}^\infty t^2 e^t dt + (1/3) \int_{-\infty}^\infty t^3 e^t dt-………\)
Now, \(\int_{-\infty}^\infty t^n e^t dt=n!/(-1)^{n+1}\)
Hence,
\(\int_{-\infty}^\infty t^n e^t dt = 1 – (1/2)(2!/(-1)^3) + (1/3)(3!/)-…….\)
\(\int_{-\infty}^\infty t^n e^t dt\) = 0! + 1! + 2! + 3! +…. = Sum of infinite factorials.
11. Find the laplace transform of y(t)=et.t.Sin(t)Cos(t).
a) \(\frac{4(s-1)}{[(s-1)^2+4]^2}\)
b) \(\frac{2(s+1)}{[(s+1)^2+4]^2}\)
c) \(\frac{4(s+1)}{[(s+1)^2+4]^2}\)
d) \(\frac{2(s-1)}{[(s-1)^2+4]^2}\)
View Answer
Explanation:
y(t)=\(\frac{1}{2} t.e^t Sin(2t)\)
Laplace transform of Sin(2t)=\(\frac{2}{s^2+4}\)
Laplace transform of tSin(2t)=\(-\frac{d}{dt} \frac{2}{s^2+4}=\frac{2(2s)}{(s^2+4)^2}=\frac{4s}{(s^2+4)^2}\)
Laplace transform of te^t Sin(2t)=\(\frac{4(s-1)}{[(s-1)^2+4]^2}\)
Laplace transform of 1/2 tet Sin(2t)=\(\frac{2(s-1)}{[(s-1)^2+4]^2}\)
12. Find the value of \(\int_0^{\infty} tsin(t)cos(t)\).
a) s ⁄ s2+22
b) a ⁄ a2+s4
c) 1
d) 0
View Answer
Explanation:
y(t)=\(\frac{1}{2} t Sin(2t)u(t)\)
Laplace transform of Sin(2t)=\(\frac{2}{s^2+4}\)
Laplace transform of tSin(2t)=\(-\frac{d}{dt} \frac{2}{s^2+4}=\frac{2(2s)}{(s^2+4)^2}=\frac{4s}{(s^2+4)^2}\)
Laplace transform of \(\frac{1}{2}tsin(2t)=\int_{-0}^{\infty} e^{-st} tsin(t)cos(t)dt=\frac{2s}{[s^2+4]^2}\)
Putting, s = 0, \(\int_0^{\infty} tsin(t)cos(t)dt=0\)
13. Find the laplace transform of y(t)=e|t-1| u(t).
a) \(\frac{2s}{1-s^2} e^s\)
b) \(\frac{2s}{1+s^2} e^{-s}\)
c) \(\frac{2s}{1+s^2} e^s\)
d) \(\frac{2s}{1-s^2} e^{-s}\)
View Answer
Explanation:
y(t)=\(e^{|t-1|}\)
Laplace transform of e|t| =\(\int_{-\infty}^\infty e^{|t|} e^{-st} dt\)
=\(\int_0^∞ e^t e^{-st} dt-\int_{-∞}^0 e^{-t} e^{-st} dt\)
=\(\int_0^∞ e^{-(s-1)t} dt-∫_{-∞}^0 e^{(-s-1)t} dt\)
Now,\(\int_0^∞ e^{-(s-1)t} dt=\left [-\frac{1}{s-1} [e^{-(s-1)t}]\right ]_∞^0\)
=\(\left [-\frac{1}{s-1} [e^{-(s-1)t} ]\right ]_∞^0=\frac{-1}{s-1}\)
Now, \(∫_{-∞}^0 e^{(-s-1)t} dt=\left [\frac{1}{-(s+1)} [e^{(-s-1)t}]\right ]_0^{-∞}\)
=\(\left [-\frac{1}{s+1} [e^{(-s-1)t} ]\right ]_0^{-∞}=-\frac{1}{(s+1)}\)
Laplace transform of |t| e|t| =\(\int_{-\infty}^\infty e^{|t|} e^{-st} dt=-\left [\frac{1}{s-1}+\frac{1}{s+1}\right ]=-\left [\frac{2s}{s^2-1}\right ]\)
Laplace transform of |t| e|t| = \(\int_{-\infty}^\infty e^{|t-1|} e^{-st} dt=\frac{2s}{1-s^2} e^{-s}\)
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