# Engineering Mathematics Questions and Answers – Laplace Transform by Properties – 1

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)=$$\int_{-\infty}^\infty f(t)e^{-t} \,dt$$

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

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

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

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) $$\frac{s+1}{a^2+(s+1)^2}$$

Explanation:
F(s)=$$\int_{-\infty}^\infty e^t sin⁡(at)u(t) e^{-st} dt=∫_0^∞ sin⁡(at)e^{-(s-1)t} dt$$
=$$\left [\frac e^{-st}{a^2+(s-1)^2} [-(s-1)sin(at)-acos⁡(at) ]\right ]_0^∞$$
=$$\frac{a}{a^2+(s-1)^2}$$
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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) $$\left [\frac{12s^2-16}{(s^2+4)^3}\right ]$$

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 t52.
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) $$\frac{15}{4} \frac{√π}{s^{7/2}}$$

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

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

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

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) $$\frac{2(s-1)}{[(s-1)^2+4]^2}$$

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

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) $$\frac{2s}{1-s^2} e^{-s}$$

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| = $$\int_{-\infty}^\infty e^{|t-1|} e^{-st} dt=\frac{2s}{1-s^2} e^{-s}$$

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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