next up previous
Next: Complete Response II - Up: First Order Systems Previous: Homogeneous Solution

Complete Response I - Constant Input

\mbox{\bf Complete Solution}=\mbox{\bf Homogeneous Solution}+\mbox{\bf Particular Solution}

The complete response of a linear system due to the external input as well as the initial condition can be found as the sum of the homogeneous and particular solutions of the DE, which is no longer homogeneous, due to the non-zero right-hand side for the external input $v(t)\ne 0$ applied to the circuit:

\begin{displaymath}v_R(t)+v_C(t)=\tau\frac{d}{dt} v_c(t)+v_c(t)=v(t) \end{displaymath}

We want to solve the DE for $v_C(t)$ as the circuit's complete response to the input $v(t)\ne 0$ as well as the initial condition $v_C(0)=V_0\ne 0$, after a switch is closed at time moment $t=0$.


Consider the RC circuit shown above where the switch is closed at $t=0$. We want to find the voltage $v_C(t)$ across $C$ (with assumed $+$ on left) and voltate $v_R(t)$ across $R$ (with assumed $+$ on top) at function of time for $t>0$.

First we consider a constant (DC) input $v(t)=V_s$ applied to the circuit at $t=0$, i.e., a step input

\begin{displaymath}v(t)=V_s u(t) =\left\{ \begin{array}{ll}V_s & t\ge 0 0 & t<0\end{array}\right. \end{displaymath}

The solution of this inhomogeneous DE is composed of two parts, The complete solution is

\begin{displaymath}v_C(t)=v'_C(t)+v''_C(t)=V_s+A e^{-t/\tau} \end{displaymath}

The constant $A$ can be determined from the initial condition $v_C(0)=V_0$:

v_C(0)=V_0=v_C(t)\big\vert _{t=0}=V_s+Ae^{0}=V_s+A,

and the complete solution is

\begin{displaymath}v_C(t)=V_s+A e^{-t/\tau}=V_s+(V_0-V_s) e^{-t/\tau} \end{displaymath}

In particular, ehen the initial condition $v_C(0)=V_0=0$, the solution is

v_C(t)=V_s+(-V_s) e^{-t/\tau}=V_s(1-e^{-t/\tau})

The current through $C$ is

i(t)=C\frac{d}{dt} v_C(t)=-\frac{C}{\tau}(V_0-V_s)e^{-t/\tau}

The voltage across $R$ is

v_R(t)=R\; i(t)=(V_s-V_0) e^{-t/\tau}

We can further verify that

\begin{displaymath}v_C(t)+v_R(t)=V_s+(V_0-V_s) e^{-t/\tau}+(V_s-V_0) e^{-t/\tau} =V_s \end{displaymath}

The plots below show $v_C(t)$ (red) and $v_R(t)$ (green) under different initial conditions (purple) and inputs (blue).

Note that after the switch is closed at $t=0$,

In general, neither the voltage across a capacitor nor the current through an inductor can be changed instantaneously as it takes time for them to build up:

v_C(t)=\frac{1}{C}\int i(t) dt,\;\;\;\;\;\;\;i_C(t)=\frac{1}{L}\int v(t) dtt

A Shortcut Method:

Observing the complete solution

v_C(t)=V_s+(V_0-V_s) e^{-t/\tau}

we see that

We can therefore generalize the complete solution obtained above to all first-order systems, i.e., their responses to a step input, a constant input that is turned on at moment $t=0$, always take the same form:

\begin{displaymath}f(t)=f(\infty)+[f(0)-f(\infty)] e^{-t/\tau} \end{displaymath}

in terms of three essential components of the system's response:
  1. $f(0)$: the initial value;
  2. $f(\infty)$: the steady state response;
  3. $\tau$: the time constant of the system.
Specifically here is how to find the three components:

In particular, note that

Example 1:

In the circuit below, $V_0=10V$, $R_1=R_2=2K\Omega$, $R_3=5K\Omega$, $C=0.5\mu F$, the circuit was in steady state at $t=0$. Find $v_C(t)$ after the switch is closed at $t=0$.



Example 1a:

In the same circuit above, find the voltages $v_1(t)$ and $v_2(t)$ across and currents $i_1(t)$ and $i_2(t)$ through $R_1$ and $R_2$, respectively.


Note that when $t=0_+$, $v_2(t)=V_c=-5V$, lower than ground voltage $V_b$!


next up previous
Next: Complete Response II - Up: First Order Systems Previous: Homogeneous Solution
Ruye Wang 2014-10-29