Linear Systems 101 - Part 6 - Step response

 Continuing our great journey through the lands of Linear Systems, what's next? Oh, yes, we are going to study the step response of the system. Why step response? What does that mean in physical terms? Well, let us first define the step signal as follows,\begin{equation}step(t) = \left\{\begin{array}{r}1,t \geq 0,\\0, t < 0.\end{array}\right.\end{equation}.

Well, time goes by, hours pass, seasons change, and definitions vary. The step function can be defined differently, but we will adopt the Heaviside step function definition here considering that step(0) = 1. By analyzing this definition, we can say that this signal model is a constant "something" applied to the system: something as a voltage, a force, a torque, a current, etc. In our case, it would be "constant voltage" and the equipment that we could use here is the DC power supply. At exactly t = 0, we would turn it on! Boom! One final remark, this is also known as the unity step, since its amplitude is equal to 1. It is easy to generalize this since we only need to multiply it by the constant value we would like to use.

Thus, let us recall the time response of the capacitor voltage with zero charge at the beginning of our analysis,\begin{equation}v_c(t) = \frac{1}{RC}\int_{0}^te^{-(t-\tau)/RC}v_i(\tau)d\tau\end{equation}so that\begin{equation}v_c(t) = \frac{e^{-t/RC}}{RC}\int_{0}^te^{\tau/RC}d\tau = -e^{-t/RC}\cdot e^{-\tau/RC}|_0^t  = 1 - e^{-t/RC}, t \geq0 \end{equation}

Right, so what does that tell us? For starters, we have that\begin{equation}v_c(0) = 0\text{ V }\text{ and } v_c(\infty) = 1 \text{ V }.\end{equation} Thus, we have that the capacitor voltage begins at 0 V, as expected, starts increasing, and goes to 1 V. We note that it is almost like the behavior of the step function we applied to the system with a different, but fundamental, detail: it does not instantly goes from 0 V to 1 V. This is due to the dynamics of the system! Note that the exponential term, and consequently the pole at -1/RC (or equivalently the time constant), determines how fast it reaches the final value of 1V. Cool! But how? Like this:

vc(t) [V]	0.63	0.87	0.95	0.98	0.99
t [s]	τ	2τ	3τ	4τ	5τ

That is, by taking t as multiples of τ, we have a sense of how fast the voltage increases, regardless of the numerical value of τ. For t = τ, we know that the voltage is 63% of its final value, and for t = 4τ, 98%. Let us plug some typical values to see what happens. Set R = 1kΩ and C = 100 nF so that the pole is given by -1/RC = -10,000.  In terms of time constant, we get τ = 100μs, which is quite fast. Our table becomes,

vc(t) [V]	0.63	0.87	0.95	0.98	0.99
t [μs]	100	200	300	400	500

It means that in less than half of a 1/1000 of a second, the capacitor would be practically charged! Faster than the blink of an eye? An adequate equipment which could be used for visualizing this signal is the oscilloscope in single mode or by applying a square wave by means of a function generator to the circuit, with a low frequency allowing the cap to charge and discharge completely in each cycle.

 Note also that the time constant τ depends on the system parameters. For electric circuits such as the RC circuit, τ is usually quite small so the signal transitions are quite fast, as we see in practice. However, other systems could be quite slower. It all depends here on the value of τ, or equivalently, the very nature of the system. As an example, this is the step response of a signal with τ = 1.

Finally, it remains to discuss what happens when we apply the step signal to a general first-order scalar system. Thus, \begin{equation}x(t) = \int_{0}^te^{-a(t-\tau)}bd\tau = \frac{b}{a}[1-e^{-at}], \;\; t\geq 0 \end{equation}

in which we note that\begin{equation}x(\infty) = a/b\end{equation}

That is, the final value is, a priori, different than the one applied to the system's input. In the RC circuit case, it was equal to 1 because a = b = 1/RC. Cool ain't it?