Linear Systems 101 - Part 4 - Preliminaires about the System's Response to Specific Inputs

In the previous post we calculated the time response of the voltage across the capacitor in an RC circuit as a function of the initial condition and an input.  Right now, we will start deriving this response for some common inputs, just for fun! Let us first recall the system equation we are studying\begin{equation}\dot v_c  = -\frac{1}{RC}v_c+\frac{1}{RC}v_i, \;\; v_c(0) = v_{c0}\end{equation}

and the resulting voltage,

\begin{equation}v_c(t) = \textcolor{blue}{e^{-t/RC}v_{c0}}+\textcolor{red}{\frac{1}{RC}\int_{0}^te^{-(t-\tau)/RC}v_i(\tau)d\tau} \end{equation}

In blue we have the natural response of the system, which is independent of the input applied and is indeed the homogeneous solution of the original differential equation. In red, we have the forced response of the system appearing due to the input signal, which comes from the particular solution of the original differential equation. 

We have briefly discussed some properties yielded by the natural response of the system in the previous posts. Arguably the most important one is stability. Note that, in this example, since R and C are positive constants, we get that\begin{equation}\lim_{t\to \infty}e^{-t/RC}v_{c0} = \lim_{t\to\infty}\frac{v_{c0}}{e^{t/RC}} = 0\end{equation}

If the recall the generic solution of the unforced scalar first order linear system,\begin{equation}x(t) = e^{-at}x_0, t \geq 0,\end{equation}

we note that\begin{equation}\lambda= -a = -\frac{1}{RC}\end{equation}

is the pole of our system. Here, since it is negative, we know for sure that the capacitor will not blow up!

Besides, note that\begin{equation}v_c(\infty) \triangleq \lim_{t\to \infty} v_c(t) =\frac{1}{RC}\int_{0}^{\infty}e^{-(t-\tau)/RC}v_i(\tau)d\tau  \end{equation}

Cool! What this equation tells us is that if we look at the steady state response of the system, the effects of the natural response will eventually vanish if the system is stable, and we will get only the effects of the forced response. Here I ask excuses from the best mathematicians for the abuse of the notation of the previous equation!

Can we be even more generic and write this response to a general system instead of an RC circuit? Sure, by defining, \begin{equation}\dot x  = -ax+bu, \;\; x(0) = x_{0},\end{equation} where x is the state and u is the input, by inspecting our expression of the voltage across the cap, we get,\begin{equation}x(t) = e^{-at}x_0+\int_0^t e^{-a(t-\tau)}bu(\tau)d\tau,\;\; t \geq 0,\end{equation}

where we note that 

1. If we set b = 0, we get the unforced response of the scalar linear system we studied in the previous posts.
2. If we set a=b = 1/RC, we get the capacitor voltage, as expected.

Right, so, let us calculate some time responses for some common inputs. With the argument that the natural response will eventually vanish, we will consider that the system is initially at rest (the initial condition is zero) and calculate in the next posts the impulse, step, and sinusoidal responses.