Monday Math 109

The Laplace Transform
Part 10: Initial and Final Value Theorems

Last time, we found the formula for the Laplace transform of the derivative of a function; if we have f(t), t>0, with Laplace transform F(s), then
.

Now, let us consider what happens to Laplace transforms as s→∞. By the definition of the Laplace transform, . Noting that as for all t>0, we see that in the limit as s→∞, the integrand goes to zero throughout the entire region of integration, except the point t=0; so long as f(t) does not have a delta function spike at t=0, the integral has to go to zero in the limit. Thus,
.
So, let us take the limit as s→∞ of our derivative formula. So long as f(t) has no step discontinuities at t=0, f’(t) will not have a delta function spike at t=0, and the above applies to its transform, so we see that
.
This is called the initial value theorem (since it involves the initial value of f).
To find the version for when there is a discontinuity at t=0, we have to remove the point at zero from the integral (replace lower limit 0 with 0+, and so we get
.

Next, let us take the limit as s→0 of the Laplace transform of f’(t). When s=0, , and so
,
and so, if we take the limit as s→0 of our derivative formula,
,
this last being the final value theorem. This tells us that the behavior of f(t) as t→∞ is reflected in the location of the poles of F(s); specifically, we see four cases:
I. If all poles are on the left side () of the s complex plane, then F(0) exists and is finite, and so ; our function f(t) decays to zero exponentially (or faster) at large t.
II. If there are poles are on the right side () of the s complex plane, such as for the Laplace transforms of the hyperbolic sine and cosine, then f(t) contains terms growing exponentially, and f(∞) does not exist.
III. If there are complex conjugate poles on the imaginary axis () of the s complex plane, such as for the Laplace transforms of the sine and cosine, then f(t) contains sinusoidal terms, and f(∞) is undefined.
IV. If there is a simple pole at the origin of the s complex plane (and none in the right side), such as for the Laplace transform of the constant function, then exists and is nonzero.

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3 Responses to “Monday Math 109”

  1. Monday Math 110 « Twisted One 151's Weblog Says:

    […] to our earlier results for the transforms of positive integer powers of t. From our previous discussion of limits, we found that for any Laplace transform F(s); thus the integral makes a more reasonable […]

  2. Monday Math 117 « Twisted One 151's Weblog Says:

    […] , where C is a constant of integration. To find the value of this constant, we turn to the initial value theorem: . Here, we see , so we find that , and we have . Recalling that , we see that our solution is […]

  3. Monday Math 117 « Twisted One 151's Weblog Says:

    […] separable: , where C is a constant, arising from our integration. To find its value, consider the initial value theorem: . We see , and so . Thus, we see that the transform of the Bessel function of the first kind of […]

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