Lecture 12
First-order Circuits (2)
Hung-yi Lee
Outline
• Non-constant Sources for First-Order Circuits
(Chapter 5.3, 9.1)
voc t  , isc t 
i sc
Outline
• Examples 5.12 and 5.11
• Solved by Differential Equation
• Solved by Superposition and State
Example 5.12
• RL circuit
• R=4Ω, L=0.1H
• Find i(t), t>0
• i(t)=0, if t<0
 
i 0  0
di t 
L
 Ri t   vt 
dt
vt   400 sin 280t (t>0)
di t 
0.1
 4i t   vt 
dt
Example 5.12 – Differential
Equation
di t 
0.1
 4i t   vt 
vt   400 sin 280t
i 0  0
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
iF(t): Special solution (Forced Response)
Natural Response:
diN t 
0.1
 4iN t   0
dt
iN t   Ae t  Ae 40t
t
t
0.1Ae  4 Ae  0
  40
A determined by initial condition
Ae
40 t
0
A0
i t   0 NOT iN t   0
Example 5.12 – Differential
Equation
di t 
0.1
 4i t   vt 
vt   400 sin 280t
i 0  0
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
Forced Response:
iF(t): Special solution (Forced Response)
diF t 
0.1
 4iF t   400 sin 280t
dt
k0
iF(t)=
v(t)= k1t
k 2 e at
k3 cos t  k 4 sin t
K0
Table 5.3 (P222)
K1t  K 0
K 2 e at
K 3 cos t  K 4 sin t
Example 5.12 – Differential
Equation
di t 
0.1
 4i t   vt 
vt   400 sin 280t
i 0  0
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
Forced Response:
iF(t): Special solution (Forced Response)
diF t 
0.1
 4iF t   400 sin 280t
dt
iF t   K1 cos 280t  K 2 sin 280t
0.1 280K1 sin 280t  280K 2 cos280t 
 4K1 cos 280t  K 2 sin 280t 
 400 sin 280t
28 K 2  4 K1  0
 28 K1  4 K 2  400
K1  14, K 2  2
Example 5.12 – Differential
Equation
di t 
0.1
 4i t   vt 
vt   400 sin 280t
i 0  0
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
iF(t): Special solution (Forced Response)
iN t   Ae 40t
iF t   -14 cos 280t  2 sin 280t
i t   Ae 40t - 14 cos 280t  2 sin 280t
i t   0
A - 14  0
A  14
i t   Vx
A - 14  Vx
A  14  Vx
Example 5.11 – Differential
Equation
iN t   Ae 40t
di t 
0.1
 4i t   vt 
i 0  0
vt   10e 40t
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
Change v(t)
 
iF(t): Special solution (Forced Response)
Natural Response:
40 t


iN t  Ae
diN t 
0.1
 4iN t   0
dt
Independent to the sources
This circuit always has this term.
Example 5.11 – Differential
Equation
iN t   Ae 40t
di t 
0.1
 4i t   vt 
i 0  0
vt   10e 40t
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
Forced Response:
iF(t): Special solution (Forced Response)
diF t 
0.1
 4iF t   10e  40t
dt
iF t   K 2 e 40t ?
iF t   K 2te 40t
If the form for iF(t) contains any term proportional to a
component of the natural response, then that term
must be multiplied by t.
P224 - 225
Example 5.11 – Differential
Equation
iN t   Ae 40t
di t 
0.1
 4i t   vt 
i 0  0
vt   10e 40t
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
Forced Response:
iF(t): Special solution (Forced Response)
diF t 
0.1
 4iF t   10e  40t
dt
iF t   K 2te 40t

iF t   100te 40t

0.1 K 2 e 40t  40 K 2te 40t  4 K 2te 40t  10e 40t
0.1K 2  10
K 2  100
Example 5.11 – Differential
Equation
iN t   Ae 40t
di t 
0.1
 4i t   vt 
i 0  0
vt   10e 40t
dt
i t   iN t   iF t  iN(t): General solution (Natural Response)
 
iF(t): Special solution (Forced Response)
iN t   Ae 40t
iF t   100te 40t
i t   Ae 40t  100te 40t
i 0   0
A0
Differential Equation - Summary
• List differential equation and find initial condition (from the
property of inductors and capacitors)
• 1. Find general solution (natural response)
• Exponential form: Ae-λt
• Find λ
• 2. Find special solution (forced response)
• Form: Consult Table 5.3 (P222)
• If a term in special solution is proportional to
general solution, multiplying the term by t
• Find the unknown constant
• 3. Add the general and special solution together, and
then find A in the general solution by initial condition
Example 5.12 – Superposition +
State
• RL circuits
• R=4Ω, L=0.1H
• i(t)=0, if t<0
i t   istate t   iinput t 
vt   400 sin 280t
Find istate(t)
Consider the circuit from t=0
 

i 0  0 State is zero
i
4
No state term
Only input term
i t   100 sin 280t
Example 5.12 – Superposition +
State
• Review: pulse response
i t 
R
A
iL t 
A
iL t 
i t 
 D  t
A e 1e



t

A
e D

If D is small
e  1  x (If x is small)
x
Example 5.12 – Superposition +
State
i t   100 sin 280t
100
……
0
100
The sin wav is composed of infinite tiny pulse!
Find the response of each tiny pulse and sum
them together.
t
Example 5.12 – Superposition +
State
Response of the pulse between
i t   100 sin 280t
t0  t
t0
time point t0-Δt and t0
A  100 sin 280t0
t 0 - t
i t  
t0
Δt is small
A
i t   e

t0

t t0

t0
100sin280t 0

e

t t 0

t
t
Example 5.12 – Superposition +
State
i t 
t0

100sin280t 0

e

t t0

t
t0 t1
t0
t1
t2
t2
The response of sin wave is the summation of all the
pulse responses.
Example 5.12 – Superposition +
State
Current
Source
(Input)
Current on
Inductor
(Response)
t2
t0 t1
a
Let’s focus on the
response of sin wave
at time point a
We do not have to
care the pulse
after time point a.
… ……
… … a
The function is
zero at point a
Example 5.12 – Superposition +
t t

100sin280t 0
State
t
i t  
e  t
0
0
Current on
Inductor
(Response)

… …
…
t0
t2
t1
…
a
…
100sin280t 2 
e

Value at a:
100sin280t 0

e

a t 0

D
100sin280t1

e

a t1

t3
D
a t 2

D
Example 5.12 – Superposition +
t t

100sin280t 0 
State
t
i t  
e t
0
0
Current on
Inductor
(Response)

… …
…
t0
t2
t1
…
…
a
t3
Value at a:
t a
i a   
t 0
100 sin 280t

e

t a
a t

t 

t 0
100 sin 280t

e

a t

dt
Example 5.12 – Superposition +
State
i a  
t a

100 sin 280t

t 0
RC
  L
 R
1

40
RL circuit
 R=4Ω, L=0.1H
e

a t

dt
t a
i a    100 sin 280t  40e  40a t dt
t 0
t a
 4000  sin 280t  e  40a t dt
t 0
t a
 4000e  40 a  sin 280t  e 40t dt
t 0
Example 5.12 – Superposition +
e
 sin bte dt  a  b a sin bt  b cos bt 
State
at
at
2
2
(P806)
t a
i a   4000e  40 a  sin 280t  e 40t dt
t 0
 4000e  40 a
e 40t
a


40
sin
280
t

280
cos
280
t
|
0
2
2
40  280
4000
40 sin 280a  280 cos 280a 
 2
2
40  280
1
 40 a
We can always


 4000e

280
40 2  280 2
replace “a” with
 14e
40 a
- 14 cos 280a  2 sin 280a
“t”.
Example 5.12 – Superposition +
State
i t   14e 40t - 14 cos 280t  2 sin 280t
iF t 
iN t 
Example 5.12 – Superposition +
State
• From Differential Equation
• If we have initial condition i(0)=Vx
i t   14  Vx e 40t - 14 cos 280t  2 sin 280t
• From Superposition State
• Superposition (no state)
iinput t   14e 40t - 14 cos 280t  2 sin 280t
• State
istate t   Vx e 40t
i t   istate t   iinput t 
Example 5.11 – Superposition +
State
RL circuit
 R=4Ω, L=0.1H
 Find i(t)
• i(t)=0, if t<0
i
vt   10e 40t
4
i t   2.5e 40t
Example 5.11 – Superposition +
State Current Source
Current on
Inductor
(Response)
(Input)
i t   2.5e 40t
a
Response of the pulse between
time point t0-Δt and t0
t0
A
i t   e

Its contribution at point a:
a
0

t t0

2.5e
i a  

t0
2.5e
t 

-40a
e

a t 0

t
-40t
e

t t0

t
Example 5.11 – Superposition +
State
For the response of the pulse
between time point t0-Δt and t0
Its contribution at point a is:
t a
i a   
2.5e 40 a
t 0
i a  
t a

 2.5e
 100e
 40 a
t a
a t

t 

2.5e 40 a
t 0
 40 t
t 0
e

-40a

2.5e
t0
i a  
e

40e
t a
 40  a t 
t a
a t

dt
 40 t  40  a t 

100
e
dt
dt
 e
40 a

100
e
a
1
dt

t 0

e

a t 0

t 0
t
1

40
Homework
• 5.56
• 5.60
• 5.64
Thank you!
Answer
• 5.56: iF(t) =-10te^(-20t) - 3e^(-20t)
• 5.60: vF(t) = 2 – 50te^(-25t)
• 5.64: i(t) = 0.05e^(25t) + 0.02 – 0.07e^(-25t)