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ESE Electronics 2010 Paper 2: Official Paper

Option 4 : 0

ST 1: General Awareness

6945

15 Questions
15 Marks
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__Initial value theorem__:

The initial value of a function with a given Laplace transform can be obtained using the Initial value theorem as:

\(c\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} c\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sC\left( s \right)\)

It is applicable only when the number of poles of C(s) is more than the number of zeros of C(s).

__Application__:

\(F(s) = \frac{5}{{s({s^2} + 3s + 2)}}\)

Since the number of poles is greater than the number of zeros, we can apply the initial value theorem as:

\(f\left( 0 \right) = \mathop {\lim }\limits_{t \to 0} f\left( t \right) = \mathop {\lim }\limits_{s \to \infty } sF\left( s \right)\)

\(\mathop {\lim }\limits_{t \to 0} f\left( t \right) = \mathop {\lim }\limits_{s \to \infty } s. \frac{5}{s(s^2+3s+2)}\)

\(=\mathop {\lim }\limits_{s \to \infty } \frac{5}{s^2+3s+2}\)

\(=\mathop {\lim }\limits_{s \to \infty }\frac{5}{s^2(1+\frac{3}{s} +\frac{2}{s^2})}\)

f(0) = 0

__Final value theorem__:

The final value theorem states that the final value of a system can be calculated by

\(f\left( \infty \right) = \mathop {\lim }\limits_{s \to 0} sF\left( s \right)\)

F(s) is the Laplace transform of the function.

For the final value theorem to be applicable, the system should be stable in steady-state and for that real part of poles should lie on the left side of s plane.