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Next: Complete Response II - Up: First Order Systems Previous: Particular Solution

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:

\begin{displaymath}
\mbox{Complete Solution}=\mbox{Homogeneous Solution}+\mbox{Particular Solution}
\end{displaymath}

Specifically in the RC circuit, we have

\begin{displaymath}
\tau\frac{d}{dt} v_c(t)+v_c(t)=v_s(t)=V_s
\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 $t=0$.

RCcircuit.png

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

\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 $v_C(0)=V_0$:

\begin{displaymath}
v_C(0)=V_0=v_C(t)\big\vert _{t=0}=V_s+Ae^{0}=V_s+A,
\;\;\;\;\;\mbox{i.e.,}\;\;\;\;A=V_0-V_s
\end{displaymath}

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

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

The current through $C$ is

\begin{displaymath}
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}
\end{displaymath}

The voltage across $R$ is

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

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 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:

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

A Shortcut Method:

Observing the complete solution

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

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:

In particular, note that

Example 0: $V_{in}=2 V$, $R_1=R_2=2 k\Omega$, $C=10^{-6} F=1 \mu F$. Find $v_C(t)$.

1stOrderRCa.png

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 $t=0$.

BridgeCapacitor.png

completeresponse.gif

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.

BridgeCapacitor.png

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

transientDCexample.gif


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Next: Complete Response II - Up: First Order Systems Previous: Particular Solution
Ruye Wang 2016-10-31