ExamVeda
Login
Home

The initial value theorem does not hold good for which of...

This question belongs to Engineering Maths Transform Theory
Transform Theory
?

The initial value theorem does not hold good for which of the following functions?

Answer & Solution
Correct Answer: Option B
Let's discuss the solution. Join the discussion
Examveda
Question posted by Examveda
Community

Join the Discussion

No comments yet Be the first to discuss this question.
Keep Learning

Related Questions on Transform Theory

The Laplace transform F(s) of the exponential function. f(t) = eat when t ≥ 0, where a is a constant and (s - a) > 0, is
A. $$\frac{1}{{{\text{s}} + {\text{a}}}}$$ B. $$\frac{1}{{{\text{s}} - {\text{a}}}}$$ C. $$\frac{1}{{{\text{a}} - {\text{s}}}}$$ D. $$\infty $$
Evaluate $$\int\limits_0^\infty {\frac{{\sin {\text{t}}}}{{\text{t}}}{\text{dt}}} $$
A. $$\pi $$ B. $$\frac{\pi }{2}$$ C. $$\frac{\pi }{4}$$ D. $$\frac{\pi }{8}$$
If the Laplace transform of $${{\text{e}}^{\omega {\text{t}}}}$$  is $$\frac{1}{{{\text{s}} - \omega }},$$  the Laplace transform of tcosh t is
A. $$\frac{{1 + {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$ B. $$\frac{{{\text{st}}}}{{\left( {{{\text{s}}^2} - 1} \right)}}$$ C. $$\frac{{1 - {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$ D. $$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$
The function f(t) satisfies the differential equation $$\frac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{t}}^2}}} + {\text{f}} = 0$$   and the auxiliary conditions, f(0) = 0, $$\frac{{{\text{df}}}}{{{\text{dt}}}}\left( 0 \right) = 4.$$  The Laplace transform of f(t) is given by
A. $$\frac{2}{{{\text{s}} + 1}}$$ B. $$\frac{4}{{{\text{s}} + 1}}$$ C. $$\frac{4}{{{{\text{s}}^2} + 1}}$$ D. $$\frac{2}{{{{\text{s}}^2} + 1}}$$
More Related Questions on Transform Theory