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.


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


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...

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

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

Now insert functions into the formula $h(t)$ and simplify by pulling out $\frac{1}{\omega}$ and switching $t$ and $\tau$ in the cosine...

$$h(t)=\frac{1}{\omega}\cdot sin(\omega t)*cos(\omega t)$$

$$=\int_0^t \frac{sin(\omega\tau)}{\omega}\cdot cos(\omega(t-\tau))d\tau$$

$$=\frac{1}{\omega}\int_0^t sin(\omega\tau)\cdot cos(\omega(\tau-t))d\tau$$

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

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

This yields...

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

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

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

$$h(t)=\frac{t\cdot sin(\omega t)}{2\omega}$$

Answer in Text:

$$h(t)=\frac{t\cdot sin(\omega t)}{2\omega}$$


See bottom.


I apologize but I actually found my error while typing this post up. That sometimes happens to me. But thank you so much for looking and I really do appreciate all the help I was given by many others in the past.


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