Second-Order Circuits
8.1 Examples of 2nd order RCL circuit
8.2 The source-free series RLC circuit
8.3 The source-free parallel RLC circuit
8.4 Step response of a series RLC circuit
8.5 Step response of a parallel RLC
1
8.1 Examples of Second
Order RLC circuits (1)
What is a 2nd order circuit?
A second-order circuit is characterized by a second-order
differential equation. It consists of resistors and the equivalent of
two energy storage elements.
RLC Series RLC Parallel RL T-config RC Pi-config
2
8.2 Source-Free Series
RLC Circuits (1)
• The solution of the source-free series RLC
circuit is called as the natural response of the
circuit.
• The circuit is excited by the energy initially
stored in the capacitor and inductor.
The 2nd d 2 i R di i
order of 2
0
expression
dt L dt LC
How to derive and how to solve?
3
8.2 Source-Free Series
RLC Circuits (3)
There are three possible solutions for the following
2nd order differential equation:
d 2 i R di i
2
0
dt L dt LC
d 2i di R 1
=>
2
2 2
0 i 0 where and 0
dt dt 2L LC
General 2nd order Form
The types of solutions for i(t) depend
on the relative values of a and w.
4
8.2 Source-Free Series
RLC Circuits (4)
There are three possible solutions for the following
2nd order differential equation:
d 2i di
2
2 2
0 i 0
dt dt
1. If a > wo, over-damped case
i (t ) A1e s1t A2 e s2t where s1, 2 0
2 2
2. If a = wo, critical damped case
i (t ) ( A2 A1t )e t where s1, 2
3. If a < wo, under-damped case
i (t ) e t ( B1 cos d t B2 sin d t ) where d 02 2
5
8.2 Source-Free Series
RLC Circuits (5)
Example 1
If R = 10 Ω, L = 5 H, and C = 2
mF in 8.8, find α, ω0, s1 and s2.
What type of natural response
will the circuit have?
• Please refer to lecture or textbook for more detail elaboration.
Answer: underdamped
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8.3 Source-Free Parallel
RLC Circuits (1)
0
1
Let i (0) I 0 v(t )dt
L
v(0) = V0
Apply KCL to the top node:
t
v 1 dv
vdt C 0
R L dt
Taking the derivative with
respect to t and dividing by C
The 2nd d 2 v 1 dv 1
order of 2
v 0
dt RC dt LC
expression
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8.3 Source-Free Parallel
RLC Circuits (2)
There are three possible solutions for the following
2nd order differential equation:
d 2v dv 1 1
2
2 02 v 0 where and 0
dt dt 2 RC LC
1. If a > wo, over-damped case
v(t ) A1 e s1t A2 e s2t where s1, 2 2 0
2
2. If a = wo, critical damped case
v(t ) ( A2 A1t ) e t where s1, 2
3. If a < wo, under-damped case
v (t ) e t ( B1 cos d t B2 sin d t ) where d 02 2
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8.4 Step-Response Series
RLC Circuits (1)
• The step response is
obtained by the sudden
application of a dc source.
The 2nd d 2 v R dv v vs
order of 2
expression dt L dt LC LC
The above equation has the same form as the equation for source-free series
RLC circuit.
• The same coefficients (important in determining the frequency
parameters).
• Different circuit variable in the equation.
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8.4 Step-Response Series
RLC Circuits (2)
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
v (t ) vt (t ) v ss (t )
The transient response vt is the same as that for source-free case
vt (t ) A1e s1t A2 e s2t (over-damped)
vt (t ) ( A1 A2t )e t (critically damped)
vt (t ) e t ( A1 cos d t A2 sin d t ) (under-damped)
The steady-state response is the final value of v(t).
vss(t) = v(∞)
The values of A1 and A2 are obtained from the initial conditions:
v(0) and dv(0)/dt. 20
8.4 Step-Response Series
RLC Circuits (3)
Example 4
Having been in position for a long time, the switch in the
circuit below is moved to position b at t = 0. Find v(t) and vR(t)
for t > 0.
• Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = {10 + [(–2cos3.464t – 1.1547sin3.464t)e–2t]} V
vR(t)= [2.31sin3.464t]e–2t V
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8.5 Step-Response Parallel
RLC Circuits (1)
• The step response is
obtained by the sudden
application of a dc source.
The 2nd d 2i 1 di i Is
order of 2
dt RC dt LC LC
expression
It has the same form as the equation for source-free parallel RLC circuit.
• The same coefficients (important in determining the frequency
parameters).
• Different circuit variable in the equation.
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8.5 Step-Response Parallel
RLC Circuits (2)
The solution of the equation should have two components:
the transient response vt(t) & the steady-state response vss(t):
i (t ) it (t ) iss (t )
The transient response it is the same as that for source-free case
it (t ) A1e s1t A2 e s2t (over-damped)
it (t ) ( A1 A2t )e t (critical damped)
it (t ) e t ( A1 cos d t A2 sin d t ) (under-damped)
The steady-state response is the final value of i(t).
iss(t) = i(∞) = Is
The values of A1 and A2 are obtained from the initial conditions:
i(0) and di(0)/dt. 27
8.5 Step-Response Parallel
RLC Circuits (3)
Example 5
Find i(t) and v(t) for t > 0 in the circuit shown in circuit shown
below:
• Please refer to lecture or textbook for more detail elaboration.
Answer: v(t) = Ldi/dt = 5x20sint = 100sint V
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