Digital Signal Processing Reference
In-Depth Information
(ii) Applying the initial-value theorem, Eq. (6.27), to X 2 ( s ), we obtain
s ( s + 5)
s 3 + 5 s 2 + 17 s + 13
+
x 2 (0
) =
t 0 + x 2 ( t ) = lim
lim
s →∞ sX 2 ( s ) = lim
s →∞
2
6 s
= lim
s →∞
= 0 .
Applying the final-value theorem, Eq. (6.28), to X 2 ( s ) yields
s ( s + 5)
s 3 + 5 s 2 + 17 s + 13
x 2 ( ) = lim
t →∞ x 2 ( t ) = lim
s 0 sX 2 ( s ) = lim
= 0 .
s 0
The initial and final values of x ( t ) can be verified from the following inverse
Laplace transform of X 1 ( s ) derived in Example 6.7(ii):
x 1 ( t ) = (0 . 4e t
0 . 4e 2 t cos(3 t ) + 0 . 2e 2 t sin(3 t )) u ( t ) .
(iii) Applying the initial-value theorem, Eq. (6.27), to X 3 ( s ), we obtain
5 s
s 2 + 25
5
2 s
x 3 (0 + ) = lim
t 0 + x 3 ( t ) = lim
s →∞ sX 3 ( s ) = lim
= lim
s →∞
= 0 .
s →∞
Applying the final-value theorem, Eq. (6.28), to X 3 ( s ) yields
5 s
s 2 + 25
x 3 ( ) = lim
t →∞ x 3 ( t ) = lim
s 0 sX 3 ( s ) = lim
= 0 .
s 0
To confirm the initial and final values obtained in (iii), we determine these values
directly from the inverse transform of X 3 ( s ) = 5 / ( s 2 + 25). From Table 6.1, the
inverse Laplace transform of X 3 ( s ) is given by x 3 ( t ) = sin(5 t ) u ( t ). Substituting
t = 0 + , the initial value x 3 (0 + ) = 0, which verifies the value determined from
the initial-value theorem. Applying the limit t →∞ to x 3 ( t ), the final value
of x 3 ( t ) cannot be determined due to the oscillatory behavior of the sinusoidal
wave. As a result, the final-value theorem provides an erroneous answer. The
discrepancy between the result obtained from the final-value theorem and the
actual value x 3 ( ) occurs because the point s = 0 is not included in the ROC
of sX 3 ( s ) R 3 :Re { s > 0. As such, the expression for the Laplace transform
sX 3 ( s ) is not valid for s = 0. In such cases, the final-value theorem cannot be
used to determine the value of the function as t →∞ . Similarly, the point
s =∞ must be present within the ROC of sX 3 ( s ) to apply the initial-value
theorem.
6.5 Solution of diff erential equations
An important application of the Laplace transform is to solve linear, constant-
coefficient differential equations. In Section 3.1, we used a time-domain
approach to obtain the zero-input, zero-state, and overall solution of differ-
ential equations. In this section, we discuss an alternative approach based on
the Laplace transform. We illustrate the steps involved in the Laplace-transform-
based approach through Examples 6.16 and 6.17.
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