Transient Analysis - First Order
Circuits
Switches, Transient Response,
Steady-State Response, and
Differential Equations
Kevin D. Donohue, University of Kentucky
1
Transient Response
Ø
DC analysis of a circuit only provides a description of
voltages and currents in steady-state behavior.
Ø
When the applied voltage or current changes at some
time, say t0, a transient response is produced that dies
out over a period of time leaving a new steady-state
behavior.
Ø
The circuit’s differential equation must be used to
determine complete voltage and current responses.
Kevin D. Donohue, University of Kentucky
2
Example
Describe v0 for all t. Identify transient and steady-state
responses.
VDC
t=0
R
C
t=0
+
v0
-
Show:


 − t  
VDC 1 − exp
  volts
v0 (t ) = 
RC



 0 volts

for t ≥ 0
for t < 0
For steady-state response, let t → ∞, for transient response subtract
out steady-state response.
Kevin D. Donohue, University of Kentucky
3
Instantaneous Voltage and Current
Changes in Capacitors and Inductors:
Ø
Ø
What would be the required
current, ic , in this circuit for the
voltage on the capacitor to
change instantaneously?
What would be the required
voltage, vL , in this circuit for the
current in the inductor to change
instantaneously?
ic
VDC
t=0
C
+
vC
-
L
+
vL
-
iL
IDC
t=0
Conclusion: If the source cannot produce infinite instantaneous
power, then neither the capacitor voltage, nor the inductor current can
change instantaneously.
Kevin D. Donohue, University of Kentucky
4
Switch Notation and Initial Conditions:
In order to denote the time right before t=0 (limit from the left as
t→0), and the time right after t=0 (limit from the right as t→0),
the following notation will be used:
t=0
Let t=0+ be the moment after the switch
is closed and t=0- be the moment before
the switch is closed.
For circuits with practical sources,
the voltage across a capacitor cannot
change instantaneously,
and the current in an inductor
cannot change instantaneously
v c ( 0 − ) = v c (0 + )
i L (0 − ) = i L (0 + )
Kevin D. Donohue, University of Kentucky
5
Complete Solution by the Differential
Equation Approach
5 major steps in finding the complete solution:
Ø Determine initial conditions on capacitor voltages
and/or inductor currents.
Ø Find the differential equation for either capacitor
voltage or inductor current (mesh/loop/nodal ….
analysis).
Ø Determine the natural solution (complementary
solution).
Ø Determine the forced solution (particular solution).
Ø Apply initial conditions to the complete solution to
determine the unknown coefficients in the natural
solution.
Kevin D. Donohue, University of Kentucky
6
Example
Find the complete solution for iL for vs = 10 V
t=0
+
25 Ω
0.25 H
vs
vL
-
Show for t ≥ 0:
iL = 0.4(1 − exp(−100t ))
Kevin D. Donohue, University of Kentucky
7
Example
Find the complete solution for vc when is = 1 mA
t=0
is
t=0
100 Ω
1 mF
R
Show for t ≥ 0:
+
vc
-
vc = 0.1exp( −10t )
Kevin D. Donohue, University of Kentucky
8
Step-by-Step Method
The solution of circuits containing energy storage elements can be divided into a
steady-state and transient component. In addition, when only one energy storage
element is present, the Thévenin resistance can be obtained with respect to the
terminal of the energy storage element and used to compute the time constant for the
transient component.
Ø
Ø
Ø
 t
x
(
t
)
=
K
+
K
exp
− 
Assume solution is of the form
1
2
 τ
Assume steady-state before the switch is thrown and let either vc (0− ) = vc (0+ ) or
iL (0 − ) = iL ( 0+ ) , and find initial condition for quantity of interest x( 0 + )
Let K1 = steady-state solution after switch is thrown, K 2 = x(0 + ) − K1
and τ = CRth , or τ = L Rth
Kevin D. Donohue, University of Kentucky
9
Plotting with Matlab
A solution for t> 0 is of the form
 t
x (t ) = K1 + K 2 exp − 
 τ
Can be plotted in matlab with the following statements
% Define constants for the plot (i.e. k1 = -25, k2 = 20, tau = .5)
>> k1 = -25;
>> k2 = 20;
>> tau = 0.5;
% Define a time axis about 10 time greater than the time constant tau
>> t = linspace(0, tau*10, 100) % vector of 100 point evenly spaced from 0 to tau*10
>> x = k1+k1*exp(-t/tau);
% Evaluate function at all points in t
>> plot(t,x);
>> xlabel(‘Seconds’)
>> ylabel(‘x’)
Kevin D. Donohue, University of Kentucky
10
Resulting plot:
Kevin D. Donohue, University of Kentucky
11