Process Control: A First Course with MATLAB

Two theorems are now presented that can be used to find the values of the time-domain function at two extremes, t = 0 and t = ?, without having to do the inverse transform. In control, we use the final-value theorem quite often. The initial-value theorem is less useful. As we have seen from our first example in Section 2.1, the problems that we solve are defined to have exclusively zero initial conditions.
Initial-Value Theorem:
| (2.23) | |
Final-Value Theorem:
| (2.24) | |
The final-value theorem is valid provided that a final-value exists. The proofs of these theorems are straightforward. We will do the one for the final-value theorem. The proof of the initial-value theorem is in the Review Problems.
Consider the definition of the Laplace transform of a derivative. If we take the limit as s approaches zero, we find
If the infinite integral exists, [9] we can interchange the limit and the integration on the LHS to give
Now if we equate the RHSs of the previous two steps, we have
We arrive at the final-value theorem after we cancel the f(0) terms on both sides.
Consider the Laplace transform F( s) = {[6( s - 2)( s + 2)] /[ s( s + 1) ( s + 3)( s + 4)]}. What is f( t = ?)?
Consider the Laplace transform F