250+ TOP MCQs on Convolution and Answers

Ordinary Differential Equations Multiple Choice Questions on “Convolution”.

1. Find the (L^{-1} (frac{1}{s(s^2+4)})).
a) (frac{1-sin⁡(t)}{4})
b) (frac{1-cos⁡(t)}{4})
c) (frac{1-sin⁡(2t)}{4})
d) (frac{1-cos⁡(2t)}{4})
View Answer

Answer: d
Explanation: In the given question
Let p₁(s) = (frac{1}{s^2+4}) and p₂(s) = (frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{s^2+4}) = frac{sin⁡(2t)}{2})
(f_2 (t) = L^{-1} (frac{1}{s}) = 1)
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t}f_1(u) f_2(t-u) dt)
(L^{-1} (frac{1}{s(s^2+4)}) = int_{0}^{t}frac{1}{2} sin⁡(2u) du)
=(frac{1-cos⁡(2t)}{4})
Thus, the answer is (frac{1-cos⁡(2t)}{4}).

2. Find the (L^{-1} (frac{1}{s(s+4)^frac{1}{2}})), give the answer in terms of error function.
a) (frac{1}{2} erf⁡(2t))
b) (frac{1}{2} erf⁡(sqrt{t}))
c) (frac{1}{2} erf⁡(2sqrt{t}))
d) (frac{1}{2} erf⁡(4sqrt{t}))
View Answer

Answer: c
Explanation: In the given question,
Let (p_1(s) = frac{1}{(s+4)^{frac{1}{2}}}) and (p_2(s) = frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{(s+4)^{frac{1}{2}}}) = frac{e^{-4t}×sqrt{t}}{sqrtpi})
(f_2 (t) = L^{-1} frac{1}{s} = 1)
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t}f_1(u) f_2(t-u) dt)
(L^{-1} (frac{1}{s(s+4)^{frac{1}{2}}}) = int_{0}^{t}frac{sqrt{u}}{sqrtpi} e^{-4u} ,du)
(=frac{1}{sqrtpi} int_{0}^{t}sqrt{u} e^{-4u} du)
We know that error function is given by –
(erf⁡(x)=frac{2}{sqrtpi} int_{0}^{x}e^{-z^2} dz)
Applying, 4u=z² And setting the limits of u in terms of z, we get
(L^{-1} (frac{1}{s(s+4)^{frac{1}{2}}}) = frac{1}{2} erf⁡(2sqrt{t}))
Thus, the answer is (frac{1}{2} erf⁡(2sqrt{t})).

3. Find the (L^{-1} frac{s}{(s^2+4)^2}).
a) (frac{1}{4} tcos(2t))
b) (frac{1}{4} tsin(t))
c) (frac{1}{4} tsin(2t))
d) (frac{1}{2} tsin(2t))
View Answer

Answer: c
Explanation: In the given question,
Let (p_1(s) = frac{1}{s^2+4}) and (p_2(s) = frac{1}{s})
(f_1 (t) = L^{-1} (frac{1}{s^2+4}) = frac{sin⁡(2t)}{2})
(f_2 (t) = L^{-1} (frac{s}{s^2+4}) = cos⁡(2t))
By Convolution Theorem,
(L^{-1} (p_1(s)×p_2(s)) = int_{0}^{t} f_1(u) f_2(t-u) dt)
(L^{-1} left (frac{s}{(s^2+4^2)^2}right )=int_{0}^{t} sin⁡(2u)×frac{1}{2}×cos⁡(2(t-u))du)
=(frac{1}{4}[tsin(2t)-frac{cos⁡(2t)}{4}+frac{cos⁡(2t)}{4}])
=(frac{1}{4} tsin(2t))
Thus, the correct answer is (frac{1}{4} tsin(2t))
.

Global Education & Learning Series – Ordinary Differential Equations.

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