Second-order Circuits (1)

Report
Lecture 13
Second-order Circuits (1)
Hung-yi Lee
Second-order Circuits
• A second order-circuit contains two independent
energy-storage elements (capacitors and inductors).
Capacitor + inductor
2 Capacitors
2 inductors
Second-order Circuits
• Steps for solving by differential equation (Chapter 9.3, 9.4)
• 1. List the differential equation (Chapter 9.3)
• 2. Find natural response (Chapter 9.3)
• There is 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
Solving by differential equation
Step 1: List Differential Equation
Systematic Analysis
iC  CvC
i
vL  LiL
1
vC   iC dt
C
1
iL   vL dt
L
Mesh Analysis
vs  vL  vR  vC
1
vs  Li  Ri   idt
C
1
vs  Li  Ri  i
C
1
R
1
vs  i  i 
i
L
L
LC
Systematic Analysis
iC  CvC
i
Mesh Analysis
1
R
1
vs  i  i 
i
L
L
LC
Find iL:
iL  i
vL  LiL
Find vC:
1
vC   iC dt
C
1
iL   vL dt
L
i  Cvc
1
R
1
vs  CvC CvC 
CvC
L
L
LC
1
R
1
vs  vC  vC 
vC
CL
L
LC
Systematic Analysis
v
iC  CvC
vL  LiL
1
vC   iC dt
C
1
iL   vL dt
L
Node Analysis
is  iL  iR  iC
1
v
is   vdt   Cv
L
R
1
1
is  v  v  Cv
L R
1
1
1
is  v 
v 
v
C
RC
LC
Systematic Analysis
v
iC  CvC
vL  LiL
1
vC   iC dt
C
1
iL   vL dt
L
Node Analysis
Find iL: v  LiL
1
1
1
is  v 
v 
v
C
RC
LC
1
1
1
is  LiL
LiL 
LiL
C
RC
LC
1
1
1
is  iL 
iL 
iL
LC
RC
LC
Find vC: vC=v
v1
v2
Example 9.6
Find i2
v1: is  i1  i
1
v1  v2
is   v1dt 
L1
R
1
1
1
is  v1  v1  v2
L1
R
R
v2: i  ix  i2
v1  v2 v2 1
   v2 dt
R
Rx L2
R  Rx
R
v1 
v2  v2
Rx
L2
v1
v2
Example 9.6
Find i2
1
1
1
is  v1  v1  v2
L1
R
R
R  Rx
R
v1 
v2  v2
Rx
L2
Equations for v1 and v2
1
Target: i2   v2 dt
L
Find v2 from the left
equations
Then we can find i2
v1
v2
Example 9.6
Find i2
R  Rx
R
1
1
1
v1 
v2   v2 dt
is  v1  v1  v2
Rx
L2
L1
R
R

1  R  Rx
R
R  Rx
R
v2   v2 dt 
v1 
v2  v2 is  
L1  Rx
L2

Rx
L2
1  R  Rx
R  1
Find v2
 
v2  v2   v2
R  Rx
L2  R
v1
v2
Example 9.6
Find i2

1  R  Rx
R
is  
v2   v2 dt 
L1  Rx
L2

1  R  Rx
R  1
 
v2  v2   v2
R  Rx
L2  R
 R  Rx 1 
1
R
is  v2  
 v2 
v2
Rx
L1 L2
 L1 Rx L2 
1
i2   v2 dt
L
v2  Li2
Replace v2
with i2
Example 9.7
• Please refer to the appendix
Summary
– List Differential Equations
R
1
1
vC  vC 
vC 
vs
L
LC
LC
1
1
1
iL 
iL 
iL 
is
RC
LC
LC
Solving by differential equation
Step 2: Find Natural Response
Natural Response
• The differential equation of the second-order
circuits:
2



y t   2y t   0 y t   f t 
y(t): current or voltage of an
element
α = damping coefficient
ω0 = resonant frequency
Natural Response
• The differential equation of the second-order
circuits:
2



y t   2y t   0 y t   f t 
y t   y N t   y F t 
yN t   2yN t    y N t   0
2
0
2



yF t   2yF t   0 yF t   f t 
Focus on yN(t) in this lecture
Natural Response
2



y N t   2y N t   0 y N t   0
yN(t) looks like:
t
t
y N t   Ae
  2    0
t
 Ae  2Ae   Ae  0
2

2
0
 2 
2 
2
t
2
2
0
Characteristic equation
 4
2
1     2  02
2
0
    2  02
2      
y N t   A1e 1t  A 2e 2t
2
2
0
Natural Response
1      
2
Real
 2  02
λ1, λ2 is
2
0
2      
2
1  2
2
0
Overdamped
1  2 Critical damped
2
2
  0
 0
Underdamped
 0
Undamped
Complex
 2  02
Solving by differential equation
Step 2: Find Natural Response
Overdamped Response
Overdamped Response
1      
2
2      
2
0
2
λ1, λ2 are both real numbers
 2  02
yN(t) looks like
1  2
y N t   Ae
y N t   A1e  A 2e
1t
2 t
t
2
0
Overdamped Response
y N t   A1e  A 2e
1t
A 2e
A1e
2 t
1t
A1  0
A2  0
1  0
2  0
2 t
Solving by differential equation
Step 2: Find Natural Response
Underdamped Response
Underdamped
1      
2
 
2
2      
2
0
2
j  1
2
0
1     1   
2
0
1    j   
2
0
1    jd
2
0
2
2
2     1   
2
0
2    j   
2
0
2
2
2    jd
d  02   2
Euler's formula:
e jx  cos x  j sin x
Underdamped
2    jd
1    jd
y N t   A1e  A 2e
1t
 A1e
  jd t
 t
j d t
e
y N t   e
 t
A e
1
2 t
 A 2e
  jd t
 A 2e
 j d t

A1  A 2 cos d t  j A1  A 2 sin d t 
yN(t) should be real.
A1  A 2
Euler's formula:
Underdamped
1    jd
e jx  cos x  j sin x
2    jd
y N t   e t A1  A 2  cos d t  j A1  A 2 sin d t 
jB
yN(t) should be real.
A1  A2
(no real part)
*
1
1
1
1
A1  a  j b A 2  a  jb
2
2
2
2
A1  A 2   jb
A1  A 2  a
Underdamped
1    jd
2    jd
y N t   e t A1  A 2  cos d t  j A1  A 2 sin d t 
A1  A2
*
y N t   e
 t
A1  A 2   jb
a cos d t  b sin d t 
Memorize this!
A1  A 2  a
a and b will be
determined by
initial conditions
Underdamped
y N t   e t a cos d t  b sin d t 
 a

b
y N t   e
a b 
cos d t 
sin d t 
2
2
2
2
a b
 a b

b

t  a
y N t   Le  cos d t  sin d t  L  a 2  b 2
L
L

y N t   Le t cos  cos d t  sin  sin d t 
t
2
y N t   Le
2
 t
cosd t   
L and θ will be
determined by
initial conditions
Underdamped
y N t   e t a cos d t  b sin d t 
y N t   Le
L
Lcos
Le t
 t
cosd t   
Solving by differential equation
Step 2: Find Natural Response
Undamped Response
Undamped
1    jd
2    jd
Undamped is a special case of underdamped.
 0
y N t   e
 t
y N t   Le
1  jd
2   jd
a cos d t  b sin d t 
y N t   a cos d t  b sin d t
 t
cosd t   
y N t   L cosd t   
Solving by differential equation
Step 2: Find Natural Response
Critical Damped Response
Critical Damped
 
2
2
0
 
2
2
0
 
2
2
0
y N t   A1e  A 2e
1t
Overdamped
2 t
Underdamped
y N t   e t a cos d t  b sin d t 
Critical damped
y N t   ?
1 , 2      
2
y N t   Ae ?
 t
Not complete
2
0
1  2  
y N t   A1e
 t
 A 2 te
 t
Critical Damped (Problem 9.44)
 t
 t


y N t  A1e  A 2 te
ht   A 2 te
 t
 t

h t   A2 1  t e
A2
e
 t


h t   A2 t - 2e
Solving by differential equation
Step 2: Find Natural Response
Summary
Summary
2



y t   2y t   0 y t   f t 
Fix ω0, decrease α (α is positive):
Overdamped
Critical
damped
1
R
1



vs  vC  vC 
vC
CL
L
LC
Decrease α, smaller R
Underdamped
Undamped
1
1
1
is  iL 
iL 
iL
LC
RC
LC
Decrease α, increase R
Fix ω0, decrease α (α is positive)
1 , 2     2  02
The position of the two roots λ1 and λ2.
α=0
Undamped
Homework
• 9.30
• 9.33
• 9.36
• 9.38
Thank You!
Answer
• 9.30: v1’’ + 3 v1’ + 10 v1 = 0
• 9.33: yN=a e^(-0.5t) + b te^(-0.5t)
• 9.36: yN=a e^(4t) + b e(-6t)
• 9.38: yN=2Ae^(3t) cos (6t+θ) or yN=2e^(3t) (acos6t
+ bsin6t)
• In 33, 36 and 38, we are not able to know the
values of the unknown variables.
Appendix:
Example 9.7
Example 9.7
Mesh current: i1 and ic
iL  ic  i1
vout  Kv x  KRiL  KRic  i1 
L  dic di1 
 dic di1 
Ri1  L    Ric  i1  i1      ic  i1 
R  dt dt 
 dt dt 
1
 dic di1 
KR ic  i1    ic dt  L    Ric  i1 
C
 dt dt 
2
2

di
d
i
di
d
i1 
 c
c
1
ic  RC K  1    LC  2  2 
dt 
 dt dt 
 dt
Example 9.7
(1):
(2):
(2) – (1):
L  dic di1 
i1      ic  i1 
R  dt dt 
2
2

di
d
i
di
d
i1 
 c
c
1
ic  RC K  1    LC  2  2 
dt 
 dt dt 
 dt
2
2

d ic d i1 
 dic di1 
ic  i1  RC K  1    LC  2  2 
dt 
 dt dt 
 dt
L  dic di1 
     ic  i1 
R  dt dt 
Example 9.7
 d 2ic d 2i1   L
 dic di1 
LC  2  2     RC K  1    2ic  i1   0
dt   R
 dt dt 
 dt
 d 2ic d 2i1   1 R
 dic di1  2
 2  2   
ic  i1   0
 K  1   
dt   RC L
 dt dt  LC
 dt
d 2vout
2
dt
 1 R
 dvout 2

 K  1

vout  0
 RC L
 dt LC
Appendix:
Figures from
Other Textbooks
Undamped
Acknowledgement
• 感謝 陳尚甫(b02)
• 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

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