Complete Response I - Constant Input

The complete response of a non-homogeneous linear system due to both the external input and the initial condition can be found as the sum of the homogeneous and particular solutions of the non-homogeneous DE with a non-zero right-hand side for the external input:

Complete Solution$\displaystyle =$Homogeneous Solution$\displaystyle +$Particular Solution (106)

Specifically in the RC circuit, we have

$\displaystyle \tau\frac{d}{dt} v_c(t)+v_c(t)=v_s(t)=V_s$ (107)

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 $t=0$.


Given an RC circuit shown above where the switch is closed at $t=0$, we want to find the voltage $v_C(t)$ across $C$ and voltage $v_R(t)$ across $R$ as a 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

$\displaystyle v(t)=V_s u(t) =\left\{ \begin{array}{ll}V_s & t\ge 0\\ 0 & t<0\end{array}\right.$ (108)

The solution of this inhomogeneous DE is composed of two parts, The complete solution is sum of the homogeneous and particular solutions:

$\displaystyle v_C(t)=v'_C(t)+v''_C(t)=V_s+A e^{-t/\tau}$ (110)

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

$\displaystyle v_C(0)=V_0=v_C(t)\big\vert _{t=0}=V_s+Ae^{0}=V_s+A,
\;\;\;\;\;$i.e.,$\displaystyle \;\;\;\;A=V_0-V_s$ (111)

and the complete solution is

$\displaystyle v_C(t)=V_s+A e^{-t/\tau}=V_s+(V_0-V_s) e^{-t/\tau}$ (112)

The same result can also be obtained using the Laplace transform method.

In particular, if $v_C(0)=V_0=0$, the solution is

$\displaystyle v_C(t)=V_s+(-V_s) e^{-t/\tau}=V_s(1-e^{-t/\tau})$ (113)

The current through $C$ is

$\displaystyle i(t)=C\frac{d}{dt}\,v_C(t)=-\frac{C}{\tau}(V_0-V_s)e^{-t/\tau}
=\frac{V_s-V_0}{R}e^{-t/\tau}$ (114)

The voltage across $R$ is

$\displaystyle v_R(t)=R\; i(t)=(V_s-V_0) e^{-t/\tau}$ (115)

We can further verify that

$\displaystyle v_C(t)+v_R(t)=V_s+(V_0-V_s) e^{-t/\tau}+(V_s-V_0) e^{-t/\tau} =V_s$ (116)

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

Note that in the first case, 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:

$\displaystyle v_C(t)=\frac{1}{C}\int i(t) dt,\;\;\;\;\;\;\;i_C(t)=\frac{1}{L}\int v(t) dtt$ (117)

Therefore the capacitor can behave like a temporary voltage source, and, similarly, an inductor can behave like a temporary current source.

Example 0 (homework): When an RC circuit with zero initial voltage $v_C(0)=0$ is charged by a DC voltage $V_s=1$. Find energy $W_R$ is consumed by $R$ and energy $W_C$ is stored in $C$.

A Shortcut Method:

Observing the complete solution $v_C(t)=V_s+(V_0-V_s) e^{-t/\tau}$ obtained above, 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:

$\displaystyle f(t)=f(\infty)+[f(0)-f(\infty)] e^{-t/\tau}$ (118)

in terms of three essential components of the system's response:

In particular, note that

Example 1: $V_{in}=2\,V$, $R_1=R_2=2\,k\Omega$, $C=10^{-6}\,F=1\,\mu F$. Find $v_C(t)$.


Example 2:

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


Consider node voltage method. Applying KCL to node $c$ we get

$\displaystyle \frac{V_0-v_c(t)}{R_1}=C\dot{v}_c(t)+\frac{v_c(t)}{R_2}$ (124)


$\displaystyle \frac{R_1R_2}{R_1+R_2}\,C\;\dot{v}_c(t)+v_c(t)
=\tau\,\dot{v}_c(t)+v_c(t)=V_0\frac{R_2}{R_1++R_2}$ (125)


$\displaystyle \tau=C\frac{R_1R_2}{R_1+R_2}=C(R_1\vert\vert R_2)
=0.5\times 10^{-6}\times 10^3=5\times 10^{-4}$ (126)

To find the initial value $v_C(0)$, we assume the circuit is in steady state before $t=0$, i.e., $v_C(0_-)=-5V$. Also, as the voltage across $C$ does not change instantaneously (unless $R=0$ therefore $\tau=RC=0$), we have $v_C(0_+)=v_C(0_-)=-5\,V$. At $t=0_+$, $V_d$ drops from $10\,V$ to $0\,V$, $V_c$ also drops from $5\,V$ to $-5\,V$.

The homogeneous solution is $v_h(t)=Ae^{-t/\tau}$ and the particular (steady state) solution is $v_p(t)=V_0/2$. The complete solution is

$\displaystyle v_c(t)=v_h(t)+v_p(t)=Ae^{-t/\tau}+\frac{V_0}{2}$ (127)

The coefficient $A$ can be found by equating $v_c(t)$ evaluated at $t=0$ and the initial condition $v_c(0)=-V_0/2$:

$\displaystyle v_c(t)\bigg\vert _{t=0}=V_s+Ae^{-t/\tau}\bigg\vert _{t=0}=\frac{V_0}{2}+A
=v_c(0)=-\frac{V_0}{2}$ (128)

i.e., $A=-V_0$. Now the solution is

$\displaystyle v_c(t)=Ae^{-t/\tau}+\frac{V_0}{2}
=\frac{V_0}{2}-V_0 e^{-t/\tau}=5-10\,e^{-2000\,t}$ (129)

Example 3:

Resolve the circuit above using the short-cut method:

Example 4:

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 $V_b$!