not be the final value of v(t). For that matter v(t) may not have a final value at all. Let v(t)
be sin
ω
t. Then V(s)
ω
/(s
2
+
ω
2
) and sV(s) s
ω
/(s
2
+
ω
2
). Application of final value
theorem says that v(∞) 0. But there is no unique final value for a sinusoidal waveform.
This conflict occurs because of wrong application of the theorem. The sV(s) function in this
case has poles on j
ω
-axis and hence the final value theorem is not applicable.
15.8 SOLUTION OF DIFFERENTIAL EQUATIONS BY USING
LAPLACE TRANSFORMS
One of the important applications of Laplace transform is in solving linear constant-
coefficient ordinary differential equations with initial ...
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