Second-order Circuits (1)
Download
Report
Transcript Second-order Circuits (1)
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
vs Li Ri i
C
1
R
1
vs i i
i
L
L
LC
Systematic Analysis
iC CvC
i
Mesh Analysis
1
R
1
vs 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
vs 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 2y 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 2y t 0 y t f t
y t y N t y F t
yN t 2yN t y N t 0
2
0
2
yF t 2yF t 0 yF t f t
Focus on yN(t) in this lecture
Natural Response
2
y N t 2y N t 0 y N t 0
yN(t) looks like:
t
t
y N t Ae
2 0
t
Ae 2Ae 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 jd
2
0
2
2
2 1
2
0
2 j
2
0
2
2
2 jd
d 02 2
Euler's formula:
e jx cos x j sin x
Underdamped
2 jd
1 jd
y N t A1e A 2e
1t
A1e
jd t
t
j d t
e
y N t e
t
A e
1
2 t
A 2e
jd 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 jd
e jx cos x j sin x
2 jd
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 jd
2 jd
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
cosd 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
cosd t
Solving by differential equation
Step 2: Find Natural Response
Undamped Response
Undamped
1 jd
2 jd
Undamped is a special case of underdamped.
0
y N t e
t
y N t Le
1 jd
2 jd
a cos d t b sin d t
y N t a cos d t b sin d t
t
cosd t
y N t L cosd 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
ht A 2 te
t
t
h t A2 1 t e
A2
e
t
h t A2 t - 2e
Solving by differential equation
Step 2: Find Natural Response
Summary
Summary
2
y t 2y 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 KRic i1
L dic di1
dic di1
Ri1 L Ric i1 i1 ic i1
R dt dt
dt dt
1
dic di1
KR ic i1 ic dt L Ric 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 2ic 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