First-order systems in transient response

First order systems have input-output differential equations of the form \[\begin{aligned} \label{eq:io_ode_first_order} \tau \dfrac{d y} {d t} + y = b_1 \dfrac{d u} {d t} + b_0 u \end{aligned}\] with $\tau\in\mathbb{R}$ called the time constant of the system. Systems with a single energy storage element—such as those with electrical or thermal capacitance—can be modeled as first-order.

The characteristic equation yields a single root $\lambda = -1/\tau$, so the homogeneous solution $y_h$, for constant $\kappa\in\mathbb{R}$, is \[\begin{aligned} y_h(t) = \kappa\,e^{-t/\tau}.\end{aligned}\]

Free response

The free response $y_{\textnormal{fr}}(t)$ of a system is its response to initial conditions and no forcing ($f(t) = 0$). This is useful for two reasons:

perturbations of the system from equilibrium result in free response and
from superposition, the free response can be added to a forced response to find the specific response: $y(t) = y_\text{fr}(t) + y_\text{fo}(t)$. This allows us to use tables of solutions like to construct solutions for systems with nonzero initial conditions with forcing.

The free response is found by applying initial conditions to the homogeneous solution. With initial condition $y(0)$, the free response is \[\begin{aligned} \label{eq:free1} y_\text{fr}(t) = y(0)\, e^{-t/\tau}, \end{aligned}\] which begins at $y(0)$ and decays exponentially to zero.

Step response

In what follows, we develop forced response $y_{\textnormal{fo}}(t)$ solutions, which are the specific solution responses of systems to given inputs and zero initial conditions: all initial conditions set to zero.

If we consider the common situation that $b_1 = 0$ and $u(t) = K u_s(t)$ for some $K\in\mathbb{R}$, the solution to is \[\begin{aligned} y_\text{fo}(t) &= K b_0 (1 - e^{-t/\tau}).\end{aligned}\] The non-steady term is simply a constant scaling of a decaying exponential.

Free and forced responses and their sum for a first order system with input u(t) = K u_s(t), initial condition y(0), and b_1=0. — Figure 6.1: Free and forced responses and their sum for a first order system with input u(t) = Ku_s(t), initial condition y(0), and b₁ = 0.

A plot of the step response is shown in figure 6.1. As with the free response, within $5\tau$ the transient response is less than $1\%$ of the difference between $y(0)$ and steady-state.

Impulse and ramp responses

The response to all three singularity inputs are included in . These can be combined with the free response of using superposition.

Example 6.1

Consider a parallel RC-circuit with input current I_S(t) = 2u_s(t) A, initial capacitor voltage v_C(0) = 3 V, resistance R = 1000 Ω, and capacitance C = 1 mF. Proceeding with the usual analysis would produce the io differential equation $$\begin{aligned} C \frac{d v_C} {d t} + v_C/R = I_S. \end{aligned}$$ Use to find v_C(t).

First, we recognize that the input is u = I_S and the output is y = v_C.
Rewrite the differential equation in standard form: $$\begin{aligned} R C \frac{d v_C} {d t} + v_C = R I_S. \end{aligned}$$ From inspection: $$\begin{aligned} \tau = R C \quad \text{and} \quad f(t) = R I_S(t) = 2 R u_s(t). \end{aligned}$$
We choose to use superposition and . From , the free response is $$\begin{aligned} y_{\textnormal{fr}}(t) = v_C(0) e^{-t/\tau} = 3 e^{-t/\tau}. \end{aligned}$$ From , the characteristic response for f(t) = u_s(t) is $$\begin{aligned} y_{\textnormal{ch}}(t) = 1 - e^{-t/\tau}. \end{aligned}$$ Since our f(t) = 2Ru_s(t), the forced response is $$\begin{aligned} y_{\textnormal{fo}}(t) = 2 R y_{\textnormal{ch}}(t) = 2 R (1 - e^{-t/\tau}). \end{aligned}$$ Finally, the specific solution for y(t) = v_C(t) is $$\begin{aligned} y(t) &= y_{\textnormal{fr}}(t) + y_{\textnormal{fo}}(t) \\ &= 3 e^{-t/\tau} + 2 R (1 - e^{-t/\tau}) \\ &= 2 R + (3 - 2 R) e^{-t/\tau}. \end{aligned}$$

Online Resources for Section 6.2

No online resources.