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### Complete Response I - Constant Input

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 applied to the circuit:

We want to solve the DE for as the circuit's complete response to the input as well as the initial condition , after a switch is closed at time moment .

Consider the RC circuit shown above where the switch is closed at . We want to find the voltage across (with assumed on left) and voltate across (with assumed on top) at function of time for .

First we consider a constant (DC) input applied to the circuit at , i.e., a step input

The solution of this inhomogeneous DE is composed of two parts,
• particular solution (forced response) due to the DC input ;
• homogeneous solution (natural response) : due to the initial condition .
The complete solution is

The constant can be determined from the initial condition :

and the complete solution is

In particular, ehen the initial condition , the solution is

The current through is

The voltage across is

We can further verify that

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

• , , i.e.,

• , , i.e.,

Note that after the switch is closed at ,

• In the first case, takes a negative value if for , although the voltage source is positive (both measured with respective to the bottom wire treated as the ground). This is because right after the switch is closed, the voltage on the left side of drops from to , causing the voltage on its right side to also drop from to , lower than the ground level of .
• Voltages on both sides of go through a discontinuous transition to drop (case 1) or jump (case 2) by , however, the voltage remains the same, as the voltage across does not change instantaneously. drops from for to at , while (left side of ) drops from to .

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:

A Shortcut Method:

Observing the complete solution

we see that
• When , is the initial condition
• When , is the steady state response.

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 , always take the same form:

in terms of three essential components of the system's response:
1. : the initial value;
2. : the steady state response;
3. : the time constant of the system.
Specifically here is how to find the three components:
• Find : as discussed in previous section for steady state response.
• Find : Denote the value of immediately before and after the moment by and , respectively. If , then use for ;
• Find : When there is only one resistor in the circuit, the time constant is or . When there are multiple resistors, the time constant can be found by:
• Remove or so that the rest of the circuit () is a one port network.
• Find the equivalent resistance of the network by turning off all energy sources (short-circuit for voltage source, open-circuit for current source).
• Find time constant or .

In particular, note that

• when , the initial condition;
• when , the steady state response;
• when , the difference between the initial and the steady state values of the response decays exponentially. This term is the transient response of the system.

Example 1:

In the circuit below, , , , , the circuit was in steady state at . Find after the switch is closed at .

• Find initial value . As the circuit has been in steady state before , and voltage across a capacitor cannot not change instantaneously, we have :

• Find steady state value :

• Find equivalent resistance :

• Find time constant

• Find the complete response

• Find current through :

• Find voltages and across and , respectively:

• Find currents and through and , respectively:

• Verify current :

Example 1a:

In the same circuit above, find the voltages and across and currents and through and , respectively.

• Find
• Find and . Before , the circuit is in steady state, i.e., . However, after the switch closes at , the voltage at node d drops from 10V to 0V, and the voltage at node c drops from 5V to -5V with respect to node b as ground (voltage across a capacitor cannot change instantaneously), i.e., and ;
• Find (same as before);
• Find and :

We see that .
• Find and through and :

• Find current through capacitor :

Note that when , , lower than ground voltage !

Next: Complete Response II - Up: First Order Systems Previous: Homogeneous Solution
Ruye Wang 2014-11-02