Properties of the Laplace Transform (example problem #4)

Given the following Laplace Transform:

$$F(s)=\frac{s^2-2s+1}{4(s-2)(s^2+2s+4)}$$

...determine the initial and final values of f(t).

1) Determine the initial value:

Recall the definition of the Initial Value Theorem:

$$f(0)=\underset{s\to \mathrm{\infty }}{lim}[sF(s)]$$ ...and apply it to the Laplace transform in our example problem.

$$\begin{array}{rl}f(0)& =\underset{s\to \mathrm{\infty }}{lim}[s(\frac{s^2-2s+1}{4(s-2)(s^2+2s+4)})]\\ & =\underset{s\to \mathrm{\infty }}{lim}[(\frac{s^3-2s^2+s}{4(s^3-8)})]\\ & =\underset{s\to \mathrm{\infty }}{lim}[(\frac{\frac{s^3}{s^3}-\frac{2s^2}{s^3}+\frac{s}{s^3}}{\frac{4s^3}{s^3}-\frac{32}{s^3}})]\\ & =\underset{s\to \mathrm{\infty }}{lim}[(\frac{1-\frac{2}{s}+\frac{1}{s^2}}{4-\frac{32}{s^3}})]\\ & =\frac{1-0+0}{4-0}\end{array}$$

$$f(0)=\frac{1}{4}$$

1) Determine the final value:

Recall the definition of the Final Value Theorem:

$$f(\mathrm{\infty })=\underset{s\to 0}{lim}[sF(s)]$$

Before we proceed, recall that the Final Value Theorem can only be used when the poles of F(s) are all in the left-half of the complex plane. (In other words, the poles must all have real components that are negative.) We find the poles by setting the denominator of the Laplace Transform function equal to zero: $$4(s-2)(s^2+2s+4)=0$$

$$s-2=0$$ $$s=2$$

$$s^2+2s+4=0$$ Continue by completing the square: $$s^2+2s+(\frac{2}{2}{)}^2-(\frac{2}{2}{)}^2+4=0$$ $$(s^2+2s+1)-1+4=0$$ $$(s+1{)}^2+3=0$$ $$(s+1{)}^2=-3$$ $$s+1=\pm j\sqrt{3}$$ $$s=-1+j\sqrt{3},-1-j\sqrt{3}$$

The poles of our function are as follows: $$2,-1+j\sqrt{3},-1-j\sqrt{3}$$ Since all of the poles are not in the left-half of the complex plane (s=2), the Final Value Theorem is in valid and no final values exist (the function oscillates indefinitely).

Continue on to Properties of the Laplace Transform example problem #5...