ENGE 211 Introduction to Electric
Circuits
Lecture 21:
First-order Circuit Analysis
Second-order Circuit Analysis
1
Analyzing First-Order RC Circuits
2
Practice Problem 7.14
• Find vo for t>0 if v(0) = 4 V, Rf = 50 kΩ, R1
= 10 kΩ, and C = 10 μF.
KCL at the node 1
C
Node 1
Node 2
dv
v
dv

0
 2v  0
dt
Rf
dt
C.H. : s  2  0
I.C. : v  0   4
v  t   vH  t   4e 2t V, t  0
vo  t   v  t   4e 2t V, t  0
3
Example 7.15
• Determine v(t) and vo(t) with v(0)=0 V
KCL at the node N
v1
 106
dv
v

0
dt
50k
20k
dv

 20v  50v1  0
dt
20
3
10  20
2V
where v1 
Dfferential Equation
dv
 20v  100
dt
v particular  5 s  20  0  s  20
 vH  t   Be 20t
v  t   Be 20t  5 V, t  0 with v  0   0 V
node N
 v  t   5e 20t  5 V, t  0
 vo  t   v1  t   v  t   7  5e 20t V, t  0
4
Practice Problem 7.15
• Find v(t) and vo(t).
KCL at the node 1
0  4  103
dv
v
 106

0
10k
dt
100k
dv

 10v  0.4 where v  0   0 V
dt
Dfferential Equation
dv
 10v  0.4
dt
Node 1
v particular  0.04 s  10  0  s  10
 vH  t   Be 10t
v  t   0.04  Be 10t V, t  0 with v  0   0 V
Node 2
v  t   40  1  e 10t  mV, t  0
v0  t   40 e 10t  1  mV, t  0
v  v0  0
5
Example 7.16
• Find the step response vo(t) for t>0. Let vi=2u(t) V,
R1=20 kΩ, Rf=50 kΩ, R2=R3=10 kΩ, C=2 μF.
Let’s use Thevenin Theorem!
6
Let vi=2u(t) V, R1=20 kΩ, Rf=50 kΩ, R2=R3=10
kΩ, C=2 μF.
Example 7.16
Vab  
VTh 
R3
R2  R3
VTh  
Rf
R1
vi
Vab  
R3
Rf
R2  R3 R1
vi
RTh   R3  ||  R2  Ro 
  R3  ||  R2 
R2R3

 5 k
R2  R3
10 50
2u  t   2.5u  t 
10  10 20
7
Example 7.16
•
Find the step response vo(t) for
t>0. Let vi=2u(t) V, R1=20 kΩ,
Rf=50 kΩ, R2=R3=10 kΩ, C=2 μF.
dvo
dt
 100vo  250
vo  t   2.5 e 100t  1  V, t  0
Behaves like an ideal voltage source
8
Quick Review
• The effect of a load resistor on the output voltage
0
0
V3
12V
C2
100µF
5
R2
20kΩ
3
R1
4
1
2
10kΩ
V1
4V
U1
6
3
7
0
1
5
741
2
R3
100Ω
+
-
0
4
V2
12V
0
V
U2
DC 10MOhm
0
C1
100µF
0
9
Parameter (R3) Sweeping
• Change the value of R3 from 5Ω to 1kΩ
Loading Happened!
10
Second-Order Circuits
11
Second-order RLC Circuits
• What is a 2nd order circuit?
– A second-order circuit is characterized by a
second-order differential equation. It consists of
resistors and two energy storage elements.
RLC Series
RLC Parallel
RL T-config
RC Pi-config
12
Analyzing Second-Order Circuits
Find initial conditions (IC) under DC
steady state if IC’s are not given
Identify state variables in the given
circuit & pick a state variable
Derive a differential equation for the
picked state variable
Find the solution for the differential
equation (homogeneous & particular
solution) using characteristic equation
method along with IC
13
Finding Initial and Final Values
• The capacitor voltage vC(t) and the inductor current iL(t)
– Determine the amount of energy stored in capacitor and inductor,
respectively.
– Called ‘state variable’
– Always continuous
choose ‘state variable’ for a variable in a
differential equation:
• Initial conditions
• Second-order diff. initial conditions
iL(0+), diL(0+)/dt, OR vC(0+), dvC(0+)/dt
• Two key points in determining the initial conditions
– First: careful in the polarity of voltage vC(t) and the direction of
the current iL(t)
• Sticking with ‘passive sign convention’
– Second: the capacitor voltage vC(t) is always continuous and the
inductor current iL(t) is always continuous:
• vC(0+)= vC(0-)
• iL(0+)= iL(0-)
14
Finding Initial and Final Values
• Example 8.1: The switch has been closed
for a long time. It is open at t=0. Find:
– i(0+), di(0+)/dt,
– v(0+), dv(0+)/dt,
– i(∞), v(∞)
15
Finding Initial and Final Values
• Example 8.1: The switch has been closed for a long time. It
is open at t=0. Find: i(0+), di(0+)/dt, v(0+), dv(0+)/dt, i(∞),
v(∞)
t<0
12
i  0  
 2 A  i  0 
42
2
v  0  
 12  =4 V=v  0 
42
dv  0 
dt
iC  0
12  4i  0   vL  0   v  0   0
C

di  0 
KVL in figure (b):

iC  0 
dt

t>0

2
 20 V/s
0.1
  i 0   2 A

vL  0 
L

0
 0 A/s
0.25
vL  0   12  8  4  0
t∞
i   0
v     12 V
𝑑𝑑𝑑𝑑 𝑡𝑡
𝑑𝑑𝑑𝑑 𝑡𝑡
Note that
,
, 𝑣𝑣𝐿𝐿 are
𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑
not continuous in general
16
Finding Initial and Final Values
• Example 8.2: Find: iL(0+), vC(0+), vR(0+), diL(0+)/dt,
dvC(0+)/dt, dvR(0+)/dt, iL(∞), vC(∞), vR(∞)
17
Finding Initial and Final Values
• Example 8.2: Find: iL(0+), vC(0+), vR(0+), diL(0+)/dt,
dvC(0+)/dt, dvR(0+)/dt, iL(∞), vC(∞), vR(∞)
t<0
iL  0   0 A  iL  0 
vC  0  =  20 V=vC  0 
vR  0

 =0 V
t>0
KVL in the right mesh
in figure (b):
vL  0   vC  0   20  0
diL  0 
dt

vL  0 
L
 0 A/s
vL  0   0
KCL at node a
KVL in the middle
mesh in figure (b):
vR  0   vO  0   vC  0   20  0
3
vR  0 
2

vO  0 
4
vR  0   vO  0 
vR  0   vO  0   4 V
18
Finding Initial and Final Values
• Example 8.2: Find: iL(0+), vC(0+), vR(0+), diL(0+)/dt,
dvC(0+)/dt, dvR(0+)/dt, iL(∞), vC(∞), vR(∞)
t
∞
2
3  1 A
24
4
vR    
 3A   2=4 V
24
iL    
vC     20 V
dvC  0 
dt

iC  0 
C

1
 2 V/s
0.5
3
iC  0   4/4  1 A
vO  0 
4
 iC  0   iL  0 
KCL at node b in
figure (b):
KVL in the middle
mesh in figure (b):
KCL at node a in
figure (b):
02
vR  vO  vC  20  0
vR
v
 O
2
4
dvR  0 
dt

Differentiate
& set t=0+
Differentiate
& set t=0+
dvO 0
 

dt
dvR  0 
dvR  0 
dt
dt

2
 V/s
3
dvC  0 
dt

dvO  0 
dt
dvC  0 
dt
0
 2 V/s
19
Second-Order Circuits
- LC Circuit
20
Analysis of LC Network
+ 𝑣𝑣𝐿𝐿 −
KVL around the loop
vI  vL  v  0
since vL  L
di
dt
dv
since i  C
dt
vI  L
di
v  0
dt
d  dv 
vI  L C   v  0
dt  dt 
𝑑𝑑 2 𝑣𝑣
𝐿𝐿𝐿𝐿 2 + 𝑣𝑣 = 𝑣𝑣𝐼𝐼
𝑑𝑑𝑡𝑡
Second-order
Differential Equation
21
Solving Second-Order Differential
Equation (LCCDE)
Recall, the method of homogeneous and
particular solutions:
1. Find the particular solution.
2. Find the homogeneous solution.
1)
2)
3)
4)
Assume solution of the form: 𝑣𝑣𝐻𝐻 = 𝐵𝐵𝑒𝑒 𝑠𝑠𝑠𝑠
Find the characteristic EQ.
Find the roots of the CH. EQ.
General solution: 𝑣𝑣𝐻𝐻 = 𝐵𝐵1 𝑒𝑒 𝑠𝑠1 𝑡𝑡 + 𝐵𝐵2 𝑒𝑒 𝑠𝑠2 𝑡𝑡
3. The total solution is the sum of the
particular and homogeneous
solutions. Use initial conditions to
solve for the remaining constants.
v  t   vP  t   vH  t 
22
Analysis of LC Network
This slide contains material from MIT EECS 6.002
23
Analysis of LC Network
This slide contains material from MIT EECS 6.002
24
Analysis of LC Network
Find unknowns from initial
conditions:
This slide contains material from MIT EECS 6.002
25
Analysis of LC Network
The output looks sinusoidal!
This slide contains material from MIT EECS 6.002
26
Analysis of LC Network – Plotting the
Output
This slide contains material from MIT EECS 6.002
27
Source-free Second-Order Circuits
- Series RLC Circuit
28
Source-Free Series RLC
+VL+VR-
From KVL: VR  VL  VC  0
+
VC
-
i C
dVC
dt
VL  L
Initial Conditions
1 0
v  0    idt  V0
C 
i 0  I0
di
dt
i determines the amount of
energy stored in the inductor
VC determines the amount of
energy stored in the capacitor
State Variable
continuous
29
Source-Free Series RLC
VR  Ri
+VL-
VL  L
di
dt
From KVL:VR  VL  VC  0
+VR-
Ri  L
di
 VC  0
dt
i C
+
VC
-
2
LC
d VC
dt
2
 RC
dVc
dt
dVC
dt
 VC  0
Characteristic Equation
Initial Conditions
1 0
v  0    idt  V0
C 
i 0  I0
s2 
1
R
s
0
L
LC
R
s 

2L
 R 2
   1
 2L 
LC
30
Source-Free Series RLC
There are three possible solutions for the following
2nd order differential equation:
d 2v
dt 2
d 2v
dt 2
 2SE

dv
 02v  0
dt
R dv
v

0
L dt LC
where
SE 
R
2L
and 0 
1
LC
General 2nd order Form
The types of solutions for v(t) depend
on the relative values of αSE and ω0.
31
Source-Free Series RLC
There are three possible solutions for the
following 2nd order differential equation:
2
d v
dt
2
 2SE
dv
 02v  0
dt
CH. EQ.
𝑠𝑠 2 + 2𝛼𝛼𝑆𝑆𝑆𝑆 𝑠𝑠 + 𝜔𝜔02 = 0
s  SE 
2
SE
 w 02
1. If αSE > ωo, over-damped case
v(t )  A1es1t  A2es2t
2
2
where s1,2  SE  SE  0
2. If αSE = ωo, critical damped case
v(t )  (A2  A1t )e t
where
s1,2  SE
3. If αSE < ωo, under-damped case
2
v(t )  e SE t (B1 cos d t  B2 sin d t ) where d  02  SE
32