linear time-constant causal system

Question

I have a linear time-constant causal system with the transfer function:

And I have the insignal

How do I get the output signal?

I thought of Laplace transform the insignal and then get Y and inverse-laplace to get y(t). But I don't know how to laplace transform cos(t) without a Heaviside(t) factor.

Answer 1

The transfer function of a dynamical linear system is defined as:

$$G(s) = \frac{Y(s)}{U(s)},$$

where $U(s)$ is the Laplace transform of the input signal. The output signal can therefore be obtained just as follows:

$$Y(s) = U(s) G(s) = \mathcal{L}_s [\cos{t}] \frac{1}{(s+1)^2},$$ and hence:

\begin{align} \color{blue}{y(t)} = & \mathcal{L}^{-1} [Y(s)] \\ = & \mathcal{L}^{-1}\left[ \frac{s}{1+s^2} \frac{1}{(s+1)^2} \right] \\ = &\mathcal{L}^{-1} \left[ -\frac{1}{2 (1+s)^2}+\frac{1}{2 \left(1+s^2\right)} \right] \\ = & \color{blue}{\left(-\frac{e^{-t} t}{2} + \frac{1}{2} \sin{t} \right) H(t)},\end{align}

where $H(t)$ is the Heaviside function.

Note that the Laplace transform of $\sin{t}$ and $\cos{t}$ are tabulated and can be easily found in any textbook.

Hope this helps.

Cheers!