# Calculate the inverse Laplace transform by convolution. (5.6-26)

Synopsis:

I cannot duplicate the answer in my text although I do get somewhat close. This tells me that my method is correct but I am making another kind of error -- perhaps in my integration? The following documents in good detail the steps taken to solve for this so that the root of the error can easily be found. Your input is very graciously welcomed.

Given:

Find the inverse Laplace transform $h(t)$ by convolution for...

$$H(s)=\frac{s^2}{(s^2+1)^2}$$

As a reference the convolution is defined as...

$$h(t)=(f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau$$

My Solution:

Begin by breaking up the product $H(s)$ into terms to determine their corresponding functions...

$$F(s)=\frac{s}{s^2+1}\Rightarrow f(t)=cos(t)$$

$$G(s)=\frac{s}{s^2+1}\Rightarrow g(t)=cos(t)$$

Now insert functions into the formula $h(t)$ and simplify by switching $t$ and $\tau$ in the cosine...

$$h(t)=cos(t)*cos(t)$$

$$=\int_0^t cos(\tau)\cdot cos(t-\tau)d\tau$$

$$=\int_0^t cos(\tau)\cdot cos(\tau-t)d\tau$$

We further simplify by applying the following trigonometric product formula identity given here as a reference...

$$cos(\alpha)\cdot cos(\beta)=\frac{1}{2}[cos(\alpha-\beta)+cos(\alpha+\beta)]$$

This yields...

$$=\frac{1}{2}\int_0^t[cos(t)+cos(2\tau-t)]d\tau$$

$$=\frac{1}{2}\left[\tau\cdot cos(t)+\frac{sin(2\tau-t)}{2}\right]_0^t$$

$$=\frac{1}{2}\left[t\cdot cos(t)+\frac{sin(t)}{2}-0-\frac{sin(-t)}{2}\right]$$

$$=\frac{1}{2}\left[t\cdot cos(t)+\frac{sin(t)}{2}-0+\frac{sin(t)}{2}\right]$$

$$\frac{1}{2}\left[t\cdot cos(t)+sin(t)\right]$$

$$\frac{1}{2}\left[t\cdot cos(t)+sin(t)\right]$$

Question:

See bottom.