Engineering Mathematics Multiple Choice Questions 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)
Answer: a
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
Answer: a
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
Answer: b
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
Answer: a
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})
Answer: b
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 ])
Answer: d
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})
=(-4left [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}})
Answer: b
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
Answer: c
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
Answer: a
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
Answer: b
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})
Answer: d
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
Answer: d
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})
Answer: d
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})