INC 111 Basic Circuit Analysis
Week 8
RL circuits
Non-periodic Signal
There are infinite number of non-periodic signal.
This course will cover only the most basic one, a step.
A step is a result from on/off switches, which is
common in our daily life.
On switch
9V
0V
1V
I
I
1Ω
1Ω
2V
I = 1A
I = 2A
Voltage source change from 1V to 2V immediately
Does the current change immediately too?
Voltage
AC voltage
2V
1V
time
Current
2A
1A
time
1V
I
I
1Ω
1Ω
I = 1A
L
2V
I = 2A
Voltage source change from 1V to 2V immediately
Does the current change immediately too?
L
Voltage
AC voltage
2V
1V
time
Current
2A
1A
Forced Response
time
Transient Response + Forced Response
Pendulum Example
I am holding a ball with a rope attached, what is the movement of the ball if
I move my hand to another point?
Movements
1. Oscillation
2. Forced position change
• Transient Response or Natural Response
(e.g. oscillation, position change temporarily)
Fade over time
Resist changes
• Forced Response
(e.g. position change permanently)
Follows input
Independent of time passed
Forced response
Natural response
at different time
Mechanical systems are similar to electrical system
Transient Response
• RL Circuit
First-order differential equation
• RC Circuit
• RLC Circuit
Second-order differential equation
RL Circuit
i(t)
+
R
u(t)
AC
L
-
KVL
di (t )
Ri (t )  L
 u (t )
dt
First-order Differential equation
Objective: Want to solve for i(t) (in term of function of t)
consider
di (t )
Ri (t )  L
 u (t )
dt
Assume that i(t) = g(t) make this equation true.
However, g(t) alone may be incomplete. The complete answer is
i(t) = f(t) + g(t)
where f(t) is the answer of the equation
di (t )
Ri (t )  L
0
dt
Voltage source go to zero
Proof: Answer has two parts
f(t) is the answer of
this equation
therefore
g(t) is the answer of
this equation
therefore
di (t )
Ri (t )  L
0
dt
df (t )
Rf (t )  L
0
dt
------------------(1)
di (t )
Ri (t )  L
 u (t )
dt
dg (t )
Rg (t )  L
 u (t )
dt
------------------(2)
If f(t)+g(t) is also
the answer of
this equation
therefore
di (t )
Ri (t )  L
 u (t )
dt
d  f (t )  g (t ) 
R f (t )  g (t )   L
 u (t )
dt
must be true
df (t )
dg (t )
Rf (t )  Rg (t )  L
L
 u (t )
dt
dt
df (t )
dg (t )
Rf (t )  L
 Rg (t )  L
 u (t )
dt
dt
from (1)
=0
dg (t )
Rg (t )  L
 u (t )
dt
which is true
from (2)
i(t) consists of two parts
i(t )  f (t )  g (t )
Transient Response
Forced Response
Therefore, we will study source-free RL circuit first
Source-free RL Circuit
i(t)
R
+
+
L
-
Inductor L has energy stored so that
the initial current is I0
Compare this with a pendulum
with some height (potential energy) left.
height
i(t)
R
+
+
L
-
di (t )
Ri (t )  L
0
dt
di (t ) R
 i (t )  0
dt
L
There are 2 ways to solve first-order differential equations
Method 1: Assume solution i(t )  Ae st
where A and s is the parameters that we want to solve for
Substitute i(t )  Ae st
in the equation
di (t ) R
 i (t )  0
dt
L
R st
Ase  Ae  0
L
R
( s  ) Ae st  0
L
st
The term that can be 0 is (s+R/L) , therefore
The answer is in format
i(t )  Ae
R
 t
L
s
R
L
i(0)  I 0
Initial condition
from
i(t )  Ae
R
 t
L
Substitute t=0, i(t=0)=0
I 0  Ae0
I0  A
We got
i (t )  I 0 e
R
 t
L
Method 2: Direct integration
di (t ) R
 i (t )  0
dt
L
di (t )
R
  i (t )
dt
L
di (t )
R
  dt
i (t )
L
i (t )

I0
t
di (t )
R
   dt
i (t ) 0 L
t
i (t )
ln i (t ) I
0
R
 t
L 0
R
ln i (t )  ln I 0   (t  0)
L
i (t )  I 0 e
R
 t
L
i(t)
I0
i (t )  I 0 e
R
 t
L
Natural Response
of RL circuit
Approach zero
t
Natural Response only
Time Constant
Ratio L/R is called “time constant”, symbol τ
L

R
Unit: second
Time constant is defined as the amount of time used
for changing from the maximum value (100%) to 36.8%.
i(t )  I 0e
R
 t
L
 I 0e

t

e 1  0.368
i(t)
2A
Forced Response
i(t ) 1  e
R
 t
L
Natural Response
1A
Approach 1A
t
Forced response = 1A comes from voltage source 1V
Natural Response + Forced Response
Switch
Open at t =0
Close at t =0
t<0
t=0
t=0
t=0
t=0
t=0
3-way switch
Switch
Open at t =0
Close at t =0
t>0
t=0
t=0
t=0
t=0
t=0
3-way switch
t=0
t=0
R
1V
1V
R
v(t)
v(t)
1V
1V
0V
0V
t
Step function (unit)
t
t=0
R=1Ω
1V
L
2V
Will divide the analysis into two parts: t<0 and t>0
When t<0, the current is stable at 2A. The inductor acts like
a conductor, which has some energy stored.
When t>0, the current start changing. The inductor discharges energy.
Using KVL, we can write an equation of current with constant
power supply = 1V with initial condition (current) = 2A
For t>0
di (t )
Ri (t )  L
V
dt
di (t )
L
 V  Ri (t )
dt
Ldi (t )
 dt
V  Ri (t )
Ldi (t )
 V  Ri (t )   dt
L
 ln( V  Ri (t ))  t  c1
R
L
 ln( V  Ri (t ))  t  c1
R
R
R
ln( V  Ri (t ))   t  c1
L
L
V  Ri (t )  e
V  Ri (t )  e
R
 t
L
R
 t
L
Ri (t )  V  c2 e
V c2
i (t )   e
R R
e
R
 c1
L
 c2
R
 t
L
R
 t
L
R
V c2  L t
i(t )   e
R R
We can find c2 from initial condition
i(0) = 2 A
Substitute t = 0, i(0) = 2
V c2
2   1
R R
c2  V  2 R
V V  2R  L t
i (t )  
e
R
R
R
Therefore, we have
Forced Response
Substitute V=1, R=1
i(t ) 1  e
R
 t
L
Natural Response
RL Circuit Conclusion
• Force Response of a step input is a step
R
 t
L
• Natural Response is in the form k1e
where k1 is a constant, whose value depends on
the initial condition.
How to Solve Problems?
• Divide in to several periods (3 periods as shown below)
• Period 1, 3 have constant V, I -> Use DC circuit analysis
• Period 2 is transient.
Response
Period 1
Period 2
Period 3
time
Calculate Transient (period 2)
• Start by finding the current of the inductor L first
• Assume the response that we want to find is in form of
k1  k2e

t

• Find the time constant τ (may use Thevenin’s)
• Solve for k1, k2 using initial conditions and
status at the stable point
• From the current of L, find other values that the problem ask
Example
t=0
i(t)
1V
R=1Ω
L=1H
2V
The switch is at this position for a long time
before t=0 , Find i(t)
i(t )  k1  k2e

t

Time constant τ = 1 sec
i(t )  k1  k2e
At t=0, i(0) = 2 A
2  k1  k2
At t = ∞, i(∞) = 1 A
1  k1  0

t

Therefore, k1 = 1, k2 = 1
The answer is
i(t )  1  e
t
i (t )
2A
1A
i (t )  2
i(t )  1  e t
Example
+
L
-
R3
R1
R2
i(0)=5
i2(t)
R4
The current L is in form of
Time constant = R/L, find Req
Time constant
L

Req
L has an initial current of 5A at t=0
Find i2(t)
iL (t )  k1  k2e

t

R1 R2
Req  R3 
 R4
R1  R2
(Thevenin’s)
Find k1, k2 using i(0) = 5, i(∞) = 0
At t=0, i(0) = 5 A
5  k1  k2
At t = ∞, i(∞) = 0 A
0  k1  0
iL (t )  k1  k2e
iL (t )  5e
Therefore, k1=0, k2 = 5


t

t

i2(t) comes from current divider of the inductor current

t
R1
i2 (t )  5e 
R1  R2

Graph?
i2 (t )

t
R1
i2 (t )  5e 
R1  R2

Example
t=0
L stores no energy at t=0
Find v1(t)
1H
2Ω
1V
2Ω
1Ω
+
v1(t)
-
Find iL(t) first
Req  ( 2 || 2)  1  2
L
1

  0 .5
Req 2
Find k1, k2 using i(0) = 0, i(∞) = 0.25
iL (t )  k1  k2e

t

0  k1  k2
At t=0, i(0) = 0 A
At t = ∞, i(∞) = 0.25 A
0.25  k1  0
Therefore, k1=0.25, k2 = -0.25
iL (t )  0.25  0.25e 2t
v1(t) = iL(t) R
v1 (t )  0.25  0.25e 2t
Graph?
v1 (t )
v1 (t )  0
v1 (t )  0.25  0.25e 2t