Transcript Method 2 for Zero State

Lecture 14
Second-order Circuits (2)
Hung-yi Lee
Second-Order Circuits
Solving by differential equation
Second-order Circuits
• Steps for solving by differential equation
• 1. List the differential equation (Chapter 9.3)
• 2. Find natural response (Chapter 9.3)
• There are some unknown variables in the natural
response.
• 3. Find forced response (Chapter 9.4)
• 4. Find initial conditions (Chapter 9.4)
• 5. Complete response = natural response + forced
response (Chapter 9.4)
• Find the unknown variables in the natural response
by the initial conditions
Review Step 1:
List Differential Equations
v C 
R
L
v C 
1
LC
vC 
1
LC
vs
i L 
1
RC
i L 
1
LC
iL 
1
LC
is
Review Step 2:
Natural Response
2
2
2



y N t   2  y N t    0 y N t   0  1 ,  2        0
Real
 
2
2
0
λ1, λ2 is
1   2
Overdamped
1   2
2
2
  0
Critical damped
 0
Underdamped
 0
Undamped
Complex
 
2
2
0
Fix ω0, decrease α
2
2
1 ,  2        0
The position of the two roots λ1 and λ2.
α=0
Undamped
v C 
Example 9.11
i L 
R
v C 
L
R
L
1
vC 
LC
i L 
1
LC
1
LC
iL 
L  0 . 1H, R  5  , C 
vs
1
L
v s
1
640
 30 V
v s (t )  
0 V
i L t   ?
Natural response iN(t):
i N 
R
L
i N 
iN  A e
 t
1
LC
2
 t
t0
t  0
i N  50 i N  6400 i N  0
iN  0
 Ae
t0
 50  Ae
 t
 6400 Ae
 t
 0
  50   6400  0
2
1 ,  2 
 50 
50  4  6400
2
2
  25  76 j
F
v C 
Example 9.11
i L 
R
v C 
L
R
L
1
vC 
LC
i L 
1
LC
1
LC
iL 
L  0 . 1H, R  5  , C 
vs
1
L
v s
1
F
640
 30 V
v s (t )  
0 V
i L t   ?
Natural response iN(t):
 1 ,  2   25  76 j
Forced response iF(t)=0
Complete response iL(t):
i N t   e
t
a cos  d t  b sin  d t 
t0
t  0
a cos 76 t  b sin 76 t 
i L t   i N t   e
1 ,  2     j  d
y N t   e
 25 t
t0
 25 t
a cos 76 t  b sin 76 t 
Two unknown variables,
so two initial conditions.
Example 9.11
L  0 . 1H, R  5  , C 
1
F
640
 30 V
v s (t )  
0 V
i L t   ?
Initial condition
t  0
short
vC
30V

0
0
0
iL 0
open
iC
vL




 0
  30
 0
   30
t0
t0
t  0
  0
v 0   30
C v  0   0
L i  0    30
iL 0


C

C

L
Example 9.11
i L t   e
 25 t
  0
a cos 76 t  b sin 76 t 
iL 0

    30
L i L 0
i L t   be
i L 0   a  0

i L t   b  25 e
 25 t
sin 76 t  76 e
i L 0   b  76   300
i L t    3 . 95  2 e
 25 t
sin 76 t
    300

 25 t
 25 t
i L 0
sin 76 t
cos 76 t

b   3 . 95

Example 9.11
i L t    3 . 95  2 e
Textbook:
i L ( t )  3 . 95 e
 25 t
 25 t
sin 76 t
cos( 76 t  90 )

v C 
Example 9.12
R
1
v C 
L
LC
1
vC 
LC
vs
v C  10 R v C  6400 v C  6400  30
L  0 . 1H, R  ? , C 
1
F
640
0 V t  0
v s (t )  
t0
 30 V
v C t   ?
Natural response vN(t):
v N  10 R v N  6400 v N  0
 Ae
2
 t
1 ,  2 
 t
 10 R  Ae
 10 R 
vN  Ae
 6400 Ae
10 R 2
 t
t  0
t
 0
  10 R   6400  0
2
 4  6400
2
Forced response vF(t):
v F  10 R v F  6400 v F  6400  30
v F  Y ss  30
Example 9.12
1
L  0 . 1H, R  ? , C 
F
640
0 V t  0
v s (t )  
t0
 30 V
v C t   ?
Initial condition:
short
vC
0V
  0
0   0
iL 0
open


  0
iC 0

t  0
  0
v 0   0
iL 0


C
  0
C v C 0

Example 9.12
Initial condition:
  0
  0
vC 0

C v C 0

v C t   v N t   v F t 
v N t   A 1 e
1 t
 A 2e
2t
1 ,  2 
 10 R 
10 R 2
 4  6400
2
v F t   30
Different R gives different response
R  0
R  16
R
R  0?
0  R  16
R  16
Initial condition:
Example 9.12
  0
  0
vC 0

C v C 0

v C t   v N t   v F t 
v N t   A 1 e
1 t
 A 2e
2t
1 ,  2 
 10 R 
1 ,  2 
 10 R 
R  34 
10  34 2
 4  6400
  20 ,  320
2
v C ( t )  30  A1 e
v C ( t )   20 A1 e
 4  6400
2
v F t   30
Overdamped: R  16 
10 R 2
 20 t
 20 t
 A2 e
 320 t
 320 A 2 e
 320 t
A1   32 , A 2  2
A1  A 2   30
20 A1  320 A 2  0
Initial condition:
Example 9.12
  0
  0
vC 0

C v C 0

v C t   v N t   v F t 
v N t   A 1 e
1 t
 A 2e
2t
1 ,  2 
 10 R 
10 R 2
 4  6400
2
v F t   30
Critical damped: R  16 
 1 ,  2   80
v N ( t )  A1 e
v C ( t )  30  A1 e
 80 t
 A 2 te
 80 t
 30  30 e
 80 t
 A 2 te
 80 t
v C ( t )   30    80 e
 80 t
 A 2 te
 80 t
v C ( 0 )  30  A1  0

 A2 e
v C ( 0 )   30    80   A 2  0
 80 t
 80 t
   80 te
 80 t
A1   30

A 2   2400
Initial condition:
Example 9.12
  0
  0
vC 0

C v C 0

v C t   v N t   v F t 
v N t   A 1 e
1 t
 A 2e
2t
1 ,  2 
10 R 2
 10 R 
2
v F t   30
Underdamped:
v N t   e
 25 t
v C t   30  e
R  16 
a cos 76 t  b sin
 25 t
R  5
 1 ,  2   25  76 j
v C 0   30  a  0
76 t 
a cos 76 t  b sin
76 t 
a   30
v C 0    25 a  76 b  0
a cos 76 t  b sin 76 t 
 25 t
a  76    sin 76 t   b  76  cos 76 t 
e
v c t    25 e
 25 t
 4  6400
b   9 . 87
Example 9.12
v C t   30  e
v C t   30 
 25 t
a cos 76 t  b sin
a b e
2
2
 25 t
31.58
76 t 

a
cos 76 t 

2
2
a

b

cos x
v C ( t )  30  31 . 58 e
a   30
 25 t
cos( 76 t  x )
b   9 . 87

sin 76 t 
2
2
a b

b
sin x
x   161 . 8

Example 9.12
v C ( t )  30  32 e
v C ( t )  30  30 e
v C ( t )  30  31 . 58 e
 25 t
 20 t
 80 t
 2e
 320 t
 2400 te
cos( 76 t  161 . 8 )
 80 t
Non-constant Input
• Find vc(t) for t > 0 when vC(0) = 1 and iL(0) = 0
vs
1
iL
ic 
1
12
v c
12
v s  v L  ic  i L

vc

12 v c  7 v c  v c  6 v s
v L  0 . 5 i L  i c  v c
v s  i c  v c   i c  2  i c  v c dt
i L  2  i c  v c dt
1
 1

 1
vs  
v c  v c  
v c  2 
vc 
 12
 12
 12

v
 c 
Non-constant Input
12 v c  7 v c  v c  6 v s
v s  5 cos t
Natural response vN(t):
12 v N  7 v N  v N  0
v N t   A1 e
Forced response vF(t):
3t
 A2 e
1 ,  2   3 ,  4
4 t
12 v F  7 v F  v F  6 v s  6   5 sin t 
v F  K 1 sin t  K 2 cos t
12  K 1 sin t  K 2 cos t   7  K 1 cos t  K 2 sin t     K 1 sin t  K 2 cos t 
  30 sin t
 K 1  7 K 2  12 K 1   30
 K 2  7 K 1  12 K 2  0
11 K 1  7 K 2   30
7 K 1  11 K 2  0
K1  
33
17
K2 
21
17
Non-constant Input
Initial Condition:
vs
1
iL
12

vc

v C 0   1
i L 0   0
iC 0   2
iC 0   C v C 0   2
v C 0   24
Non-constant Input
v C t   v N t   v F t 
v N t   A1 e
3t
 A2 e
4 t
v F  K 1 sin t  K 2 cos t
v C t   A1 e
3t
 A2 e
4t

33
sin t 
17
3t
4t
v C t    3 A1 e  4 A 2 e

A1  A 2 
21
21
v C 0   1
K1  
cos t
17
33
cos t 
17
A1  25
17
33
17
sin t
17
1
 3 A1  4 A 2 
21
 24
A2  
429
17
33
17
v C 0   24
K2 
21
17
Second-Order Circuits
Zero Input + Zero State
& Superposition
Review: Zero Input + Zero State
y(t): voltage of capacitor or current of inductor
y(t) = general solution + special solution
=
=
= natural response + forced response
Set sources to be zero
y(t) = state response (zero input)
+ input response (zero state)
Set state vc, iL to be zero
= state response (zero input)
+ input 1 response + input 2 response ……
If input = input 1 + input 2 + ……
v C 
Example 1
1
L
H, R 
8
v s ( t )  15
State response (Zero input):
R
 
v state
L
v state 
1
LC
v state  0
  10   16  0
2
   2, 8

L
5
v C 
1
LC
, C 
4
1
vC 
1
LC
F
2
0  t  30
    3V
vC 0
R
(pulse)
  0A
iL 0

Find vC(t) for t>0
  10 v state  16 v state  0
v state
v state t   A1 e
2 t
 A2 e
8t
vs
v C 
Example 1
L
1
H, R 
8
v state t   A1 e
2 t
 A2 e
8t

  0
 2 A1   8 A 2  0
 0
C v state

vC 
LC
, C 
1
F
2
(pulse)
  0A
iL 0

v state t    4 e
A1  A 2   3

1
Find vC(t) for t>0
    3V
v state 0
5
0  t  30
    3V
State response (Zero input):
L
v C 
4
v s ( t )  15
vC 0
R
2 t
e
8t
A1   4 , A 2  1
1
LC
vs
Example 1
1
L
H, R 
8
  0
vC 0

  0
v C 0
15


(pulse)
  0A
iL 0

(set state to be zero)
…
=
v s t 
F
Find vC(t) for t>0
15
30
1
2
0  t  30
    3V
Input response (Zero State):
, C 
4
v s ( t )  15
vC 0
5
0
30
v s1 ( t )
15
0
…
30
v s 2 (t )
v C 
Example 1
15
…
0
0
v s 2 ( t ) 30
LC
2 t
2 t
vC 
  0
vC 0
 15  20 e
15
L
v C 
1
1
LC
vs
 1  10 v input
 1  16 v input 1  16  15
v input
v s 1 ( t ) 30
v input 1 t   15  A1 e
R
 A2 e
 5e
…
8t
8t

  0
v C 0

15  A1  A 2  0
A1   20
 2 A1  8 A 2  0
A2  5
t   0
0  t  30
v input
2
v input
 2  t  30 
 8  t  30 


t

15

20
e

5
e
2
t  30
Example 1
Input response (Zero State):
15
0  t  30
15
…
=
v s t 
0
30
 15  20 e
2t
-
30
0
v s1 ( t )
 5e
8t
 15  20 e
…
30
v s 2 (t )
v input 1 t 
v input t   v input 1 t 
15
v input
2t
 5e
8t
2
t   0
Example 1
Input response (Zero State):
15
t  30
15
…
=
v s t 
0
30


5 e
8t
2 t
e
e
 2  t  30 
 8  t  30 

v input 1 t  

15  20 e
 5e
…
30
v s 2 (t )
t 
2
0
v s1 ( t )
v input t   v input 1 t   v input
  20 e
30
15
8t
2t
v input
2
t  
15  20 e
 5e
 2  t  30 
 8  t  30 
v input t  
Example 1
15  20 e
2t
v input t 
 5e
8t

5 e
Input response
(Zero State)
0  t  30
t  30
v state t    4 e
8t
2t
e
e
 2  t  30 
 8  t  30 
t  30
0  t  30
State response
(Zero input)

  20 e
2 t
e
8t
v C t   v state t   v input t   15  24 e
v C t   v state t   v input t   20 e
2 t
 2  t  30 
 6e
5
8t
 8  t  30 


Example 1 – Differential Equation
1
L
H, R 
8
5
4
v s ( t )  15

F
2
0  t  30
    3V
vC 0
, C 
1
(pulse)
  0A
iL 0

Find vC(t) for t>0
15
v s t 
t0

  0
    3V
v C 0
t  30

v C 30  15 V

i L 30
  15 V
v C 30
vC 0



30
Assume 30s is
large enough

t  30


 0


 0


v C 30

Example 1 – Differential Equation
t  0
v s t   15
    3V
v  0   0
vC 0


C
t  30
v s t   0
t
   15 V
v  30   0
v C 30


C
v C  10 v C  16 v C
v C  10 v C  16 v C
 16  15
0
0  t  30 :
v C t   15  A1 e
2 t
 A2 e
8t
A1   24
A2  6
Example 1 – Differential Equation
t  0
v s t   15

v C 30
 2  30
v C 30    2 A1 e

C
C
v C 30   A1 e
t


v C t   A1 e
v s t   0
   15 V
v  30   0
    3V
v  0   0
vC 0
t  30
v C  10 v C  16 v C
v C  10 v C  16 v C
 16  15
0
2 t
 A2 e
 A2 e
 2  30
v C t   20 e
8t
 8  30
 8 A2 e
 15
 8  30
0
 2  t  30 
 5e
 8  t  30 
A1  20 e , A 2   5 e
60
240
v C 
Example 2
L
1
H, R 
8
5
1
LC
, C 
1
vC 
F
2
t0
    3V

L
v C 
4
v s (t )  t
vC 0
R
  0A
iL 0

Find vC(t) for t>0
State response (Zero input):
Changing the input will not
change the state (zero input)
response.
v state t    4 e
2 t
e
8t
1
CL
vs
v C 
Example 2
L
1
H, R 
8
5
1
LC
, C 
  0A
iL 0

Input response (Zero state):
  0

  0

v input
0

Input
(set state to be zero)
What is the response?
F
2
Find vC(t) for t>0
v input 0
1
vC 
t0
    3V

L
v C 
4
v s (t )  t
vC 0
R
4 methods
1
CL
vs
v C 
Example 2 –
Method 1 for Zero State
R
L
v C 
  0
  10 v input
  16 v input  16 t
v input
v input 0

1
vC 
LC
1
CL
vs
  0

v input
0

Natural Response:
v N  10 v N  16 v N  0
v N t   A1 e
2 t
 A2 e
8t
Forced Response:
v F  10 v F  16 v F  16 t
v F t   K 1 t  K 2
v F t   t 
5
8
10 K 1  16  K 1t  K 2   16 t
v input t   v N t   v F t   t 
5
8
 A1 e
2t
K 1  1, K 2  
 A2 e
8t
5
8
Example 2 –
Method 1 for Zero State
v input t   t 
5
8
 A1 e
 t   1  2 A1 e
v input
v input 0   
2 t
5
8
2t
 A2 e
 8 A2 e
  0
8t
v input 0
  0

v input
0
8t
A1 
 A1  A 2  0
 0   1  2 A1  8 A 2  0
v input
v input t   t 

5
8

2
3
2
3
A2  
e
2t

1
24
e
8t
1
24

Example 2 –
Method 2 for Zero State
Find the response of
each small pulse
Then sum them together
Example 2 –
Method 2 for Zero State
1
1
=
0
v t   1 
4
e
2t
4
3

3
e
1
e
v t   1 
8t
2t
4
3
e
2t
1  e  
2t 
2 t
1
3
e
8t
1
e
8t
3
8t 
8
3
4
e
 2 t   t 
3

1
3
1  e 
e
8 t
2t
e
…
t
0
3
v t   
v t  
-
…
t
1
8t
 t
e
 8 t   t 
Example 2 –
Method 2 for Zero State
1
t
v t1 t  
8
3
e
 2  t  t1 
e
 8  t  t1 
 t
v 0 t  
8
3
e
2t
e
8t
 t
Consider a point a
value of the pulse at t1 is
v t1  a 
a
t1
t1   t

8
3
e
 2  a  t1 
e
 8  a  t1 
 t
Example 2 –
Method 2 for Zero State
Consider a point a
value of the pulse at t1 is
v input t  
ta

t0

8
3
8
3

t e
 2 a t 
e
 te
t0
8a t 
3
e
 2  a  t1 
2t
dt 
8
 te dt 
3
ta
e
8 a
 te
t0
e
 8  a  t1 
 t
dt
at
ta
2a
e
v t1  a  
8
8t
dt
e
at
a
2
 at
 1
Example 2 –
Method 2 for Zero State
v input  a  
8
ta
e
2a
3
 te dt 
2t
t0
8
 te dt 
at
e
at
a
2
 at
 1
ta
e
8 a
3
 te dt
8t
t0
2t
8t




e
8
e
2a
8 a
 2 t  1   e  8 t  1 
 e 
3
 4
 3
 64

ta
8
t0
2a
8a




e
1
8
e
1
2a
8 a
 2 a  1     1   e  8 a  1     1 
 e 
3
4
64
 4
 3
 64

8
8 1
1 2a  8  1
1 8 a 
8 a  1   e 
  2 a  1  e   
3 4
4
64
 3  64

a
5
8

2
3
e
2a
-
1
24
e
8 a
We can always replace
“a” with “t”.
Example 2 –
Method 3 for Zero State
t
…
0
4  2t 1 8t 

v t    1  e  e   t
3
3


…
The value at time point a
…
…
…
…
v input  a 
ta
4  2 a t 

  1  e
3
t0 
t
1 8a t  
 e
dt  t
3

a
Example 2 –
Method 3 for Zero State
v input  a  
ta

1

1dt 

t0
ta

1dt 
a
4
4
e
2a
3
a
3
4
e

1  e  
2a
e
8a t 
dt
 2 a t 
ta
dt 
3
ta
e
2a
 e dt 
2t
t0
1
2
1
24

t0
3
t0
2
1
3
ta
t0

e
 2 a t 
3
t0
ta

4
e
2a

1 
1
1
1
8a t 
ta
e
8 a
3
e
dt
3
 e dt
8t
t0
8 a
3
1  e   a 
8 a
e
1
8
5
8
e

8a
2
3
1
e

2a
-
1
24
e
8 a
Example 2 –
Method 4 for Zero State
Source Input:
x slope ( t )  t
Input (Zero State) Response
y slope ( t )  ?
Source Input:
1
…
0
x step ( t )  u ( t )
Input (Zero State) Response
y step t   1 
4
3
e
2t

1
3
e
8t
Example 2 –
Method 4 for Zero State
y slope ( t )  ?
x slope ( t )  t
y step t   1 
x step ( t )  u ( t )
4
e
2t

3
1
e
8t
3
 t   10 y slope
 t   16 y slope t   16 x slope t 
y slope
 t   10 y step
 t   16 y step t   16 x step t 
y step
 t   10 y slope
 t   16 y slope
 t   16 x slope t   16 x step t 
y slope
 t   y step t   1 
y slope
y slope t   t 
2
3
e
2t

4
3
1
24
e
2t

1
e
8t
3
e
8t
A
A 
5
8
Example 2
Method 2:
v input  a  
ta

t0
Method 3:
v input  a  

t e
3
ta

t0
Method 4:
8
1
4
3
 2 a t 
e
e
 2 a t 

8a t 
1
e
dt
8a t 
dt
3
 t   1 
v input
4
3
e
2t

1
3
e
8t
Example 2
v C t   v state t   v input t 
v s t   t
v C t   t 
5
8

10
3
e
2t

23
e
8t
24
L
1
H, R 
8
v s (t )  t
5
4
t0
, C 
1
2
F
R
v C 
v C 
1
, C 
1
vC 
Example 2 –
L
LC
Checked by Differential Equation
L
1
H, R 
8
4
v s (t )  t
v N t   A1 e
v F t   t 
v C t   t 
A1  
2 t
 A2 e
8t
8
10
3
iL 0
   3
L i  0   3


 A1 e
, A2 
 A2 e
23
24
2
  0A

vC 0
2t
F

Find vC(t) for t>0
5
8
5
CL
t0
    3V
vC 0
5
1
8t
L

5
8
  0
C v  0   0
iL 0


C
 A1  A 2   3
1  2 A1  8 A 2  0
vs
Example 3
L
1
H, R 
8
 0

v s t    t
 30 V

    3V
vC 0

5
, C 
4
1
2
t0
0  t  30
t  30
  0A
iL 0

Find vC(t) for t>0
How to solve it?
I will show how to solve the problem by
Method 1: Differential equation
Method 2: Integrating Step Responses
Method 3: Differentiate the sources
F
Example 3
– Method 1: Differential Equation
 0

v s t    t
 30 V

    3V
vC 0
0  t  30
v C t   t 

v C 30

5

10
8
e
2t

3
23
e
v C t   1 
3
  0A
iL 0

24
C
8
e
t  30
8t

2t
0  t  30
Find vC(t) for t>0
  v 30   30  5  10 e
20

t0

23
3
e
8t
 60

3
23
e
 240
24

v C 30

  v  30   1  20 e

C
3
 60

23
3
e
 240
Example 3
– Method 1: Differential Equation

v C 30

  30 
5

8
10
e
 60
23

3
e

 240
v C 30
24

  1  20 e
 60

23
3
3
v C t    2 A1 e
 8 A2 e
e
 240
t  30
v C t   30  A1 e
2 t
v C 30   30  A1 e
 A2 e
 60
A1  
v C t   30 
2
8t
 A2 e
e
60

3
10
3
3
e
2t
v C 30    2 A1 e
 240
10

2 t
,A2 
23
24
e
8t

1
e
240

24
2
3
 60
8t
 8 A2 e
23
24
e
 2  t  30 

1
24
e
 8  t  30 
 240
Example 3
– Method 2: Integrating Step Responses
t
…
…
0
4  2t 1 8t 

v t    1  e  e   t
3
3


The value at time point a
t
If a > 30
v input  a 
t  30
…
…
…
…
4  2 a t  1 8a t  

  1  e
 e
dt
3
3

t0 
t
a
a
If a < 30
Integrating from 0 to a
If a > 30
Integrating from 0 to 30
Example 3
– Method 2: Integrating Step Responses
v input  a  
t  30
4
 1 3 e
 2 a t 



1dt 
t  30


4
t  30
e
3
t0
1dt 
 30 
4
e
4
v input t   30 
e
3
dt 
 e dt 
2t
t0
1
2t
e

t  30
e
3
2a
2
e
 2 a t 
t  30
2a
3
2
v state t    4 e
dt
1
t0
3
t0
e
8a t 
2 t
3
t0
t  30
1
60
1
24

1 
e
8t

t  30
e
8 a
3
3
2
3
dt
t0
1
1
8a t 
 e dt
8t
t0
e
8 a
1
8
e
e
 2  t  30 
240

1
1
24

e
 8  t  30 
e
8t
Example 3
– Method 3: Differentiate the sources
1
30
 0

v s t    t
 30 V

t0
0  t  30
t  30
Response: v input t   ?
0


v s t    1
0

t0
0  t  30
t  30
Response: v pulse t 
 t   v pulse t 
v input
Example 3
– Method 3: Differentiate the sources
 t   v pulse t 
v input
1
t0
0  t  30
t  30
Response: v pulse t 
4
v pulse t   1 
2
v input t   t 
30
0

v s t    1
0

t  30
v input 0  
2
e
2t
3
v input t   t 

3
1
e
8t
3
1

3

e
2t
8t
e
 A
24
1
A
 A0
8
24
2
e
2t
3
v input 30   30 
1

e
8t

24
2
3
e
 60

5
5
8
1
24
e
 240

5
8
Example 3
2
v input 30   30 
e
 60
1

3
e
 240

24
5
8
– Method 3: Differentiate the sources
 t   v pulse t 
v input
1
t0
0  t  30
t  30
Response: v pulse t 
4
v pulse t   
v input t  
30
0

v s t    1
0

0  t  30
2
e
3
2t
3
e
2t
1  e  
60
1  e  
1
60
e
1
e
8t
3
8t
24
v input t   30 

2
3
e
 2  t  30 

e
240
A   30
2t
3
1
24
240
1  e   A 
v input 30   ......  ......
2
1  e 

1
24
e
 8  t  30 
e
8t
Announcement
• 11/12 (三) 第二次小考
• Ch5. Dynamic Circuit (5.3)
• Ch9. Transient response (9.1, 9.3, 9.4)
• 助教時間：週一到週五 PM6:30~7:30
Homework
• 9.58
• 9.60
Homework
• Find v(t) for t>0 in the following circuit. Assume
that v(0+)=4V and i(0+)=2A.
Homework
• Determine v(t) for t>0 in the following circuit
Thank You!
Homework
• 9.58
v C t    12  25 e
7 t

cos 24 t  16 . 3
• 9.60
i L t   6  8 e
5t
 20 te
5t


Homework
• Find v(t) for t>0 in the following circuit. Assume
that v(0+)=4V and i(0+)=2A.
Homework
• Determine v(t) for t>0 in the following circuit
v C t    6 . 75 e
4 t
 0 . 75 e
 36 t
 16
Appendix
Higher order?
• All higher order circuits (3rd, 4th, etc) have the same
types of responses as seen in 1st-order and 2ndorder circuits
Acknowledgement
• 感謝不願具名的同學
• 指出投影片中 Equation 的錯誤
• 感謝 吳東運(b02)
• 指出投影片中 Equation 的錯誤
• 感謝 林裕洲 (b02)
• 發現課程網站的投影片無法開啟