Linear Systems 101 - Part 3 - Forced response

 Alright, alright. So, in the previous episode, we have shown the procedure of obtaining the solution of our system and introduced the concept of Characteristic Equation. Now, let us get bold and apply an input to our system: it could be noise, an exogenous (external) input; or a known signal applied with some purpose. Recalling our RC circuit, we can write that\begin{equation}RC \dot v_c + v_c = v_i,\end{equation} where vc is the voltage across the capacitor and vi is our input. Once again, we assume that R and C are positive constants. The question is, can we find vc in function of the input vi? The well-known method of undetermined coefficients for solving forced differential equations would be,

  1. Find the solution of the homogeneous equation
  2. Propose the particular solution, which has a similar form of a known vi
  3. Evaluate the constants and provide the complete solution
As it turns out, we have already found the homogeneous solution by solving 
\begin{equation}RC \dot v_c + v_c = 0\end{equation}
which yields
\begin{equation} v_c = e^{-t/RC}v_{c0}, t \geq 0\end{equation}

Ok, but Step 2 would require the form, the pattern, the actual function plugged in vi, be it a step function, and impulse, sinusoidal signals, or whatever. Here, we want to know the general solution, regardless of what we plug in vi. For that, we will use the method of variation of parameters:
  1. Find the solution of the homogeneous equation
  2. Propose a particular solution of the form\begin{equation}v_p = \sum_{i = 1}^nc_i(t) v_{h}^{(i)},\end{equation} where n is the order of the differential equation and \begin{equation}v_h^{(i)}, \;\;i = 1,\ldots, n,\end{equation} are the homogeneous solutions without the constants, that is, the basis of the homogeneous solution
  3. Find {ci(t)} and evaluate the particular solution and 
  4. Calculate the final solution \begin{equation}v_c = v_h + v_p\end{equation} 

 Wow, this sounds very complicated! Ok, easy, let us do that in our case! The first step is already done.  In Step 2, since the order of our differential equation is n = 1, we have that\begin{equation} v_p = \sum_{i = 1}^1c_i(t) v_{h}^{(i)} = c_1(t)v_h^{(1)} = c_1(t)e^{-t/RC}.\end{equation} Note here that we took the homogeneous solution and plugged in the particular equation without its constant. Nice! Ok, let us evaluate the particular solution in Step 3: \begin{equation}v_i = RC \dot v_p + v_p= RC [\dot c_1e^{-t/RC} - c_1e^{-t/RC}/RC] + c_1(t)e^{-t/RC},\end{equation} which yields\begin{equation}RC\dot c_1e^{-t/RC} = v_i  \leftrightarrow  \dot c_1 = e^{t/RC}v_i/RC.  \end{equation}

Finally (!), by integrating from 0 to t, we get \begin{equation}c_1(t) = \frac{1}{RC}\int_{0}^t e^{\tau/RC}v_i(\tau)d\tau + C,\end{equation} where, according to the best matematicians, we could set freely the constant C to zero. However, our job is not already done! We have to find the particular solution in order to find the (in)famous (and feared and confusing and dreaded, the list of adjectives given by students goes on and on and on) convolution integral,\begin{equation}v_p = c_1(t) e^{-t/RC}=\frac{1}{RC}\int_{0}^t e^{(\tau-t)/RC}v_i(\tau)d\tau\end{equation}

Wow, that took some time. It remains to evaluate Step 4 in order to get the complete solution:\begin{equation}v_c = v_h +v_p = e^{-t/RC}v_{c0} + \frac{1}{RC}\int_{0}^t e^{-(t-\tau)/RC}v_i(\tau)d\tau, t \geq 0\end{equation}.

Ok, cool. So what? Let us have a look at some examples in the next post.

Wait, just one more minute! Why basis of the homogeneous solution? Because with \begin{equation}v_h^{(1)} = e^{-t/RC},\end{equation} we can find any homogeneous solution by simply multiplying this element by a constant (in this case, the initial condition). The concept comes from basis on linear algebra.

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