9–1
ECE2205: Circuits and Systems I
Second-Order R LC Circuit Analysis
• We have examined first-order circuits (RC and RL) in detail in the
time domain.
• We found that we can also analyze higher-order circuits in the time
domain, but the math is tedious.
• We then introduced the Laplace transform and showed how circuits
can be represented in the Laplace domain.
• This chapter of notes focuses on the analysis of second-order RLC
circuits using Laplace techniques.
• It also shows how time-domain response can be qualitatively
understood directly from the Laplace domain, without always needing
to perform an inverse Laplace transform.
The Series RLC Circuit
• We first look at the series RLC circuit,
drawn to the right.
vr (t)
vc (t)
vin (t)
• In an application, we might want to find
vr (t), v L (t), vc (t), and/or i(t).
i(t)
v L (t)
i L (0)/s
• Assuming general initial
conditions, we redraw the
circuit to the right, in the
Laplace domain.
R
Vin (s)
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
Vr (s)
a
sL
VL (s)
b vc (0)/s
Vc (s)
1
sC
9–2
ECE2205, Second-Order RLC Circuit Analysis
• Apply KCL at a:
E a (s) − Vin(s) i L (0) E a (s) − E b (s)
+
+
= 0.
R
s
Ls
• Apply KCL at b:
E b (s) − E a (s) (E b (s) − vc (0)/s) − 0
i L (0)
=
+
.
s
Ls
1/(sC)
• Re-writing in terms of Vc (s) = E b (s), VL (s) = E a (s) − E b (s), and noting
E a (s) = Vc (s) + VL (s),
¶
µ
1
Vin(s) i L (0)
1
1
VL (s) =
Vc (s) +
+
−
R
R Ls
R
s
1
i L (0)
VL (s) = Cvc (0) +
.
Ls
s
• This may be written in matrix-vector form:
 µ
¶ 

µ
¶
V
(s)
i
(0)
1
1
1
L
in
"
#
−
+

 Vc (s)


R
s
R Ls 

.
 R
¶
µ
=



i
(0)
1
L
VL (s)
Cv
(0)
+
sC
−
c
s
Ls
sC Vc (s) −
• Inverting the matrix (thank you, Matlab)

R + Ls
R
"
#
 LCs 2 + RCs + 1 LCs 2 + RCs + 1
Vc (s)
=

RLCs 2
−Ls
VL (s)
LCs 2 + RCs + 1 LCs 2 + RCs + 1


Vin(s) i L (0)
−

R
s 


i L (0)  .
Cvc (0) +
s
• Note that we can now solve for Vc (s), VL (s), and/or
Vr (s) = Vin(s) − Vc (s) − VL (s) given Vin(s), vc (0), and i L (0).
• All terms have a common denominator based on the transfer function
1
H (s) =
,
LCs 2 + RCs + 1
so we need to spend some time examining this term.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–3
ECE2205, Second-Order RLC Circuit Analysis
Second-Order System Response
• We can solve for the denominator roots using the quadratic equation:
s
p
µ ¶2
1
R
−RC ± (RC)2 − 4(LC)
R
−
=−
±
.
s1,2 =
2LC
2L
2L
LC
• There are three possible scenarios for s1,2 based on the radical being
+, −, or zero.
CASE I :
Unique real roots.
• When (RC)2 − 4LC > 0, then the circuit has two real roots s1 and s2
and s1 6= s2.
• The transfer function has poles s1 and s2 so the natural solution to the
circuit has terms of the form £
¤
yn (t) = k1es1t + k2es2t u(t).
• The circuit is said to have an overdamped response.
CASE II :
Repeated real roots.
• When (RC)2 − 4LC = 0, the circuit has two identical roots
R
s1 = s2 = − .
2L
• The natural solution has theh form
i
R
R
yn (t) = k1e− 2L t + k2te− 2L t u(t).
• The circuit is said to have a critically damped response.
CASE III :
Complex-conjugate roots.
• When (RC)2 − 4LC < 0 the radical produces an imaginary term, so
the two roots are complex conjugates of each other
s1 = σ + jω
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–4
ECE2205, Second-Order RLC Circuit Analysis
p
s2 = σ − jω,
R
4LC − (RC)2
where σ = −
and ω =
.
2L
2LC
• The natural solution is of the form
£ (σ + j ω)t
¤
(σ − j ω)t
yn (t) = k1e
+ k2e
u(t),
where k1 = |k1|e j ψ and k2 = |k1|e− j ψ . That is, k2 = k1∗, and ψ = 6 k1.
• This circuit is said to have an underdamped response.
• Note that the given form for yn (t) is not as simple as we might like
because of the “ j” terms.
• We use Euler’s equation to simplify:
£
¤
j ψ σ t j ωt
− j ψ σ t − j ωt
yn (t) = |k1|e e e + |k1|e
e e
u(t)
¡
¢
= |k1|eσ t e j (ωt+ψ) + e− j (ωt+ψ) u(t)
= 2|k1|eσ t cos(ωt + ψ)u(t).
SUMMARY:
The values of R, L, and C determine whether the natural
response is overdamped, critically damped, or underdamped.
¡ s1 t
¢
s2 t
• s1, s2 real and s1 6= s2: yn (t) = k1e + k2e u(t).
¡
¢
• s1 = s2: yn (t) = k1es1t + k2tes1t u(t).
• s1 = s2∗ = σ + jω: yn (t) = 2|k1|eσ t cos(ωt + 6 k1)u(t).
• Note that a pole-zero plot in the s-plane quickly shows qualitative
behavior. We explore this in more detail now.
Time Response vs. Pole Locations: 1st-Order Pole (stable)
• Poles qualitatively determine the behavior of the system. Zeros
quantify this relationship.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–5
ECE2205, Second-Order RLC Circuit Analysis
• For systems having only real poles, each pole contributes an output
waveform of the same general kind, all of which are added together.
1
s
←→ h(t) = e−σ t u(t).
s+σ
• If σ > 0, pole is at <e(s) < 0, stable i.e., impulse response decays,
and any bounded input produces bounded output.
EXAMPLE :
H (s) =
• If σ < 0, pole is at <e(s) > 0, unstable.
• σ is “time constant” factor: τ = 1/σ .
step([0 1],[1 1]);
impulse([0 1],[1 1]);
1
1
0.8
0.8
0.4
0.6
←−
1
e
0.4
0.2
0
0
K (1 − e−t/τ )
System response. K = DC gain
y(t) × K
e
h(t)
0.6
−σ t
Response to initial condition
−→ 0.
0.2
1
t =τ
2
3
4
5
0
0
Time (sec × τ )
1
t =τ
2
3
4
5
Time (sec × τ )
Time Response vs. Pole Locations: 2nd-Order Pole (stable)
• For systems having complex-conjugate poles, each pair contributes
an output waveform of the same general kind, all of which are added
together at the output.
• To understand, we write the second-order system in “standard form”
ωn2
b0
=K 2
H (s) = 2
s + a1 s + a2
s + 2ζ ωn s + ωn2
ζ = damping ratio.
(standard form).
ωn = natural frequency or undamped frequency.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–6
ECE2205, Second-Order RLC Circuit Analysis
where,
−1
θ = sin (ζ )
h(t) = p
=m(s)
ωn
1 − ζ2
e−σ t (sin(ωd t)) u(t),
σ = ζ ωn ,
p
ωd = ωn 1 − ζ 2 = damped frequency.
45
◦
=m(s)
30
◦
=m(s)
17.5
◦
=m(s)
ωn
σ
ωd
<e(s)
<e(s)
<e(s)
ζ = 0.707
Impulse Response
<e(s)
ζ = 0.5
ζ = 0.3
1
h(t)
0.5
e−σ t
0
Envelope of sinusoid decays as e−σ t
−e−σ t
−0.5
−1
0
5
10
15
20
25
30
Time (sec)
Impulse Responses of 2nd-Order Systems
1
Step Responses of 2nd-Order Systems
2
ζ =0
ζ =0
0.2
0.5
0.2
1.5
0.4
0.4
y(t)
y(t)
0.6
0
ζ =1
0.8
−0.5
0.6
1
0.8
0.5
1.0
−1
0
2
4
6
ωn t
8
10
12
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
0
0
2
4
6
ωn t
8
10
12
9–7
ECE2205, Second-Order RLC Circuit Analysis
• Low damping, ζ ≈ 0, oscillatory; High damping, ζ ≈ 1, no oscillations.
=m(s)
=m(s)
<e(s)
Impulse responses vs. pole locations
•0<ζ <1
•ζ =1
<e(s)
Step responses vs. pole locations
underdamped.
critically damped, ζ > 1
overdamped.
Time Response vs. Pole Locations: Higher Order Systems
• We have looked at first-order and second-order systems without
zeros, and with unity gain.
Non-unity gain
• If we multiply by K , the dc gain is K . tr , ts , M p , t p are not affected.
Add a zero to a second-order system
2
(s + 1)(s + 2)
2
2
=
−
s+1 s +2
H1(s) =
• Same dc gain (at s = 0).
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
2(s + 1.1)
1.1(s
µ + 1)(s + 2) ¶
2
0.1
0.9
=
+
1.1 s + 1 s + 2
0.18
1.64
=
+
s+1 s+2
H2(s) =
9–8
ECE2205, Second-Order RLC Circuit Analysis
• Coefficient of (s + 1) pole greatly reduced.
• General conclusion: a zero “near” a pole tends to cancel the effect of
that pole.
• How about transient response?
H (s) =
– Zero at s = −ασ.
(s/αζ ωn ) + 1
.
(s/ωn )2 + 2ζ s/ωn + 1
– Poles at <e(s) = −σ.
• Large α, zero far from poles ➠ no effect.
• α ≈ 1, large effect.
• Notice that the overshoot goes up as α → 0.
2nd-order system with zero
Overshoot versus normalized zero loc.
2
Percent overshoot
Step Response
2
1.5
1
α=1
2
4
100
0.5
0
0
2
4
ωn t
6
8
10
1.5
ζ = 0.3
0.5
0.7
1
0.5
0
0
2
4
α
• A little more analysis; set ωn = 1
s/αζ + 1
H (s) = 2
s + 2ζ s + 1
1
1
s
= 2
+
s + 2ζ s + 1 αζ s 2 + 2ζ s + 1
= Ho (s) + Hd (s).
• Ho (s) is the original response, without the zero.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
6
8
10
9–9
ECE2205, Second-Order RLC Circuit Analysis
• Hd (s) is the added term due to the zero. Notice that
1
s Ho (s).
Hd (s) =
αζ
The time response is a scaled version of the derivative of the time
response of Ho (s).
• If any of the zeros in RHP, system is nonminimum phase.
2nd-order min-phase step resp.
2nd-order nonmin-phase step resp.
2
1.5
H (s)
1.5
1
Ho (s)
H (s)
y(t)
y(t)
0.5
1
Ho (s)
0
0.5
−0.5
Hd (s)
0
−1
Hd (s)
−0.5
0
2
4
6
8
Time (sec)
10
−1.5
0
2
4
6
Time (sec)
8
10
Add a pole to a second order system
1
.
(s/αζ ωn + 1)[(s/ωn )2 + 2ζ s/ωn + 1]
• Original poles at <e(s) = −σ = −ζ ωn . New pole at s = −αζ ωn .
H (s) =
• Major effect is an increase in rise time.
Norm. rise time vs. norm. pole loc.
9
1.2
8
7
1
ζ = 1.0
0.7
0.5
6
0.8
0.6
0.4
0.2
0
0
1
2
3
4
ωn t
100
5
2
α=1
5
ωn tr
Step Response
2nd-order system with pole
1.4
5
4
3
2
6
7
8
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
1
0
2
4
α
6
8
10
9–10
ECE2205, Second-Order RLC Circuit Analysis
Summary of Higher-Order Approximations
• Extra zero in LHP will increase overshoot if the zero is within a factor
of ≈ 4 from the real part of complex poles.
• Extra zero in RHP depresses overshoot, and may cause step
response to start in “wrong direction.” Delay.
• Extra pole in LHP increases rise-time if extra pole is within a factor of
≈ 4 from the real part of complex poles.
=m(s)
Insignificant
Dominant
Unstable
region
<e(s)
• M ATLAB’s ‘step’ and ‘impulse’ commands can plot higher order
system responses.
Example (1) of Series RLC Circuit—Find vc (t) (overdamped)
• Let vc (0) = 15 V, i L (0) = 0 A, R = 8.5 kÄ, L = 1 H, C = 0.25 µF,
vin(t) = Au(t).
• From our circuit solution, we know
1
R/s
Vc (s) =
V
(s)
−
i L (0)
in
LCs 2 + RCs + 1
LCs 2 + RCs + 1
RL
RC + LCs
+
i
(0)
+
vc (0)
L
LCs 2 + RCs + 1
LCs 2 + RCs + 1
RC + LCs
A
+
vc (0)
=
s(LCs 2 + RCs + 1) LCs 2 + RCs + 1
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
ECE2205, Second-Order RLC Circuit Analysis
=
A
s(0.25 × 10−6s 2 + 2.125 × 10−3s + 1)
15(2.125 × 10−3 + 0.25 × 10−6s)
+
0.25 × 10−6s 2 + 2.125 × 10−3s + 1
= V1(s) + V2(s).
• The first term has a pole at s = 0 and poles at s1 = −500 and
s2 = −8000. (overdamped)
• The second term has the overdamped pair of nonzero real poles.
• We can take the inverse Laplace transform to find vc (t) using
partial-fraction expansion
r1
r2
r3
r4
r5
Vc (s) = +
+
+
+
s
s + 500 s + 8000 s + 500 s + 8000
r1 = sV1(s)|s=0 = A
¯
A/0.25 × 10−6 ¯¯
16A
r2 = (s + 500)V1(s)|s=−500 =
=−
s(s + 8000) ¯s=−500
15
¯
A/0.25 × 10−6 ¯¯
A
r3 = (s + 8000)V1(s)|s=−8000 =
=
s(s + 500) ¯s=−8000 15
¯
15(8500 + s) ¯¯
r4 = (s + 500)V2(s)|s=−500 =
= 16
s + 8000 ¯s=−500
¯
15(8500 + s) ¯¯
r5 = (s + 8000)V2(s)|s=−8000 =
= −1.
¯
s + 500
s=−8000
• Overall,
A 16(1 − A/15) 1 − A/15
+
−
s
s + 500
s + 8000
£
¤
vc (t) = A + 16(1 − A/15)e−500t − (1 − A/15)e−8000t u(t).
Vc (s) =
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–11
9–12
ECE2205, Second-Order RLC Circuit Analysis
SeriesRLC circuit: Example 1 with A=2
16
14
12
10
vc(t)
• Check answer at t = 0
vc (0) = A + 16(1 − A/15) − (1 − A/15)
= A − 16A/15 + A/15 + 16 − 1
= 15.
8
6
4
• Result is plotted to the right for
A = 2.
2
0
0
0.005
0.01
0.015
Time (sec)
0.02
0.025
Example (2) of Series RLC Circuit—Find vc (t) (critically damped)
• Let vc (0) = 15 V, i L (0) = 0 A, R = 4 kÄ, L = 1 H, C = 0.25 µF,
vin(t) = 0.
• From our circuit solution, with i L (0) and vin(s) substituted,
RC + LCs
Vc (s) =
vc (0)
LCs 2 + RCs + 1
1 × 10−3 + 0.25 × 10−6s
vc (0)
=
0.25 × 10−6s 2 + 1 × 10−3s + 1
4000 + s
vc (0).
= 2
s + 4000s + 4000000
• The poles are repeated in this case s1 = s2 = −2000.
• Partial-fraction expansion is
r1
r2
Vc (s) =
+
(s + 2000)2 s + 2000
¯
2
r1 = (s + 2000) Vc (s)¯s=−2000 = (s + 4000)(15)|s=−2000 = 30000
¯
¯
·
d(s + 4000)(15) ¯¯
d(s + 2000)2 Vc (s) ¯¯
=
= 15.
r2 =
¯
¯
ds
ds
s=−2000
s=−2000
• Therefore,
£
¤
vc (t) = 15e−2000t + 30000te−2000t u(t).
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
9–13
ECE2205, Second-Order RLC Circuit Analysis
Series RLC circuit: Example 2
16
14
12
10
c
v (t)
• Again, we verify that vc (0) = 15, as
expected.
8
6
• Result is plotted to the right.
4
2
0
0
0.005
0.01
0.015
Time (sec)
0.02
0.025
0.02
0.025
Example (3) of Series RLC Circuit—Find vc (t) (underdamped)
• Let vc (0) = 15 V, i L (0) = 0 A, R = 1 kÄ, L = 1 H, C = 0.25 µF,
vin(t) = 0.
• From our circuit solution, with i L (0) and vin(s) substituted,
RC + LCs
vc (0)
Vc (s) =
LCs 2 + RCs + 1
0.25 × 10−3 + 0.25 × 10−6s
vc (0)
=
0.25 × 10−6 s 2 + 0.25 × 10−3 s + 1
15s + 15000
= 2
.
s + 1000s + 1000000
√
• The poles are s1,2 = −500 ± 500 3 j. So
r1
r1∗
Vc (s) =
√ +
√ ,
s + 500 + 500 3 j s + 500 − 500 3 j
√ − j π/6
and we find r1 = 7.5 − 4.33 j = (5 3)e
.
Series RLC circuit: Example 3
16
14
• Again, we verify that vc (0) = 15.
12
10
c
v (t)
• Therefore,
√ −500t
√
vc (t) = 10 3e
cos(500 3t−π/6).
8
6
4
2
0
−2
0
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
0.005
0.01
0.015
Time (sec)
9–14
ECE2205, Second-Order RLC Circuit Analysis
Parallel RLC Circuit
• Another important second-order
circuit is the parallel RLC network
shown to the right.
• Assuming general initial
conditions, we redraw the
circuit to the right, in the
Laplace domain.
ir (t)
i in (t)
R
i c (t)
vc (0)/s
a
Iin(s)
i L (t)
sL
i L (0)/s
1
sC
• Applying KCL at a:
E a (s) − 0 E a (s) − 0 i L (0) (E a (s) − vc (0)/s) − 0
Iin(s) =
+
+
+
R
Ls
s
1/(sC)
¸
·
1
i L (0)
1
+
+ sC +
− Cvc (0)
= E a (s)
R Ls
s
·
¸
i L (0)
RLs
E a (s) = Cvc (0) −
+ Iin(s)
.
s
RLCs 2 + Ls + R
• Given values of R, L, C, i L (0), and i in(t), we can solve this equation.
• Note that the denominator is of a different form than the series RLC
circuit, but we can still get underdamped, critically damped, and
overdamped performance.
The Next Step
• Having seen first- and second-order systems in the time- and
Laplace-domains, we move on to understanding more general
systems via the “transfer function”.
c 2006, Dr. Gregory L. Plett
Lecture notes prepared by & Copyright °
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(mostly blank)