lecture 13 second-order circuits (1) hung-yi lee
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Lecture 13Second-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 inductors2 Capacitors
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
Mesh Analysis
i
CRLs vvvv
CC vCi
LL iv L
dtiC
v CC
1
dtvL
i LL
1
idtC
RiiLvs1
iC
iRiLvs1
iLC
iL
Riv
L s
11
Systematic Analysis
i
iLC
iL
Riv
L s
11
ii L
CCCs vLC
vL
Rvv
L C
1CC
1
CCCs vLC
vL
Rvv
L
1
C
1
CC vCi
LL iv L
dtiC
v CC
1
dtvL
i LL
1
Mesh Analysis
Find iL:
Find vC: cvCi
Systematic Analysisv
Node Analysis
CRLs iiii
vCR
vvdt
Lis
1
vCvvL
is R
11
vLC
vviC s
1
RC
11
CC vCi
LL iv L
dtiC
v CC
1
dtvL
i LL
1
Systematic Analysisv
vLC
vviC s
1
RC
11
Find vC:
LLLs iLC
iiiC
L1
LRC
1L
1
L
1
RC
11i
LCiii
LC LLs
Node Analysis
Systematic Analysis
CC vCi
LL iv L
dtiC
v CC
1
dtvL
i LL
1
vC=v
Find iL: Liv L
Example 9.6
Find i2
v1 v2
v1:
v2:
iiis 1
2iii x
R
vvdtv
Lis
211
1
1 211
1 R
1
R
11vvv
Lis
dtvLR
v
R
vv
x2
2
221 12
221 vL
Rv
R
RRv
x
x
Example 9.6
Find i2
v1 v2
2111 R
1
R
11vvv
Lis
22
21 vL
Rv
R
RRv
x
x
Target:
Equations for v1 and v2
dtvL
i 22
1
Find v2 from the left equations
Then we can find i2
Example 9.6
Find i2
v1 v2
222
2
22
21
R
1
R
1
1
vvL
Rv
R
RR
dtvL
Rv
R
RR
Li
x
x
x
xs
dtvL
Rv
R
RRv
x
x
22
21211
1 R
1
R
11vvv
Lis
22
21 vL
Rv
R
RRv
x
x
Find v2
Example 9.6
Find i2
v1 v2
222
2
22
21
R
1
R
1
1
vvL
Rv
R
RR
dtvL
Rv
R
RR
Li
x
x
x
xs
221
221
2
11v
LL
Rv
LRL
RRv
Ri
x
x
xs
dtvL
i 22
1
22 Liv
Replace 2v
2iwith
Example 9.7
• Please refer to the appendix
Summary – List Differential Equations
sCCC vL
vLC
vL
Rv
C
11 sLL i
LCi
LCii
11
RC
1L
Solving by differential equation
Step 2: Find Natural Response
Natural Response
• The differential equation of the second-order circuits:
y(t): current or voltage of an element
α = damping coefficientω0 = resonant frequency
tftytyty 202
Natural Response
• The differential equation of the second-order circuits:
tftytyty 202
tytyty FN
Focus on yN(t) in this lecture
02 N20NN tytyty
tftytyty F20FF 2
Natural Response
yN(t) looks like:
02 N20NN tytyty
tety AN
02 20
2 ttt AeAeAe 02 20
2 Characteristic equation
2
422 20
2
2
02
20
21 2
02
2
tt eety 2121N AA
Natural Response20
21 2
02
2
λ1, λ2 is
21
21
Overdamped
Critical damped
Complex0
0
Underdamped
Undamped
20
2 20
2
20
2
Real
Solving by differential equation
Step 2: Find Natural Response
Overdamped Response
Overdamped Response
20
2
21 λ1, λ2 are both real numbers
yN(t) looks like tety AN
tt eety 2121N AA
20
21 2
02
2
Overdamped Response
01 A 02 A
01 02
tt eety 2121N AA
te 11A
te 22A
Solving by differential equation
Step 2: Find Natural Response
Underdamped Response
Underdamped
20
2 22
01 1 2202 1
1j
2201 j 22
02 j
220 d
dj 1 dj 2
20
21 2
02
2
Underdamped
tt eety 2121N AA
tjtj dd ee 21 AA
tjtjt dd eee 21 AA
xjxe jx sincos Euler's formula:
tjtety ddt sinAAcosAA 2121N
yN(t) should be real.
dj 1 dj 2
21 AA
Underdamped
*21A A
bja2
1
2
1A1 jba
2
1
2
1A2
dj 1 dj 2
xjxe jx sincos Euler's formula:
tjtety ddt sinAAcosAA 2121N
yN(t) should be real.jB (no real part)
jb 21 AA a 21 AA
Underdamped
tbtaety ddt
N sincos a and b will be determined by initial conditions
dj 1 dj 2
tjtety ddt sinAAcosAA 2121N
jb 21 AA a 21 AA*21A A
Memorize this!
Underdamped
t
ba
bt
ba
abaety dd
tN sincos
2222
22
t
bt
aety dd
tN sin
Lcos
LL
ttety ddt
N sinsincoscosL
L and θ will be determined by initial conditions
tbtaety ddt
N sincos
22L ba
tety dt
N cosL
Underdamped
tety dt
N cosL
te LL
Lcos
tbtaety ddt
N sincos
Solving by differential equation
Step 2: Find Natural Response
Undamped Response
Undamped
dj 1 dj 2
0 dj 1 dj 2
tty dN cosL
Undamped is a special case of underdamped.
tety dt
N cosL
tbtaety ddt
N sincos
tbtaty ddN sincos
Solving by differential equation
Step 2: Find Natural Response
Critical Damped Response
Critical Damped
20
2 Underdamped
20
2 Overdamped
20
2 Critical damped
2120
221,
?A tN ety
Not complete tt
N eety tAA 21
tt eety 2121N AA
tbtaety ddt
N sincos
?tyN
Critical Damped (Problem 9.44)
ttN eeAty tA21 teth tA2
teAth t12
teAth 2-t2
e
A
2
Solving by differential equation
Step 2: Find Natural Response
Summary
Summary
CCCs vLC
vL
Rvv
L
1
C
1 L
1
RC
11i
LCiii
LC LLs
tftytyty 202
Fix ω0, decrease α (α is positive):
Overdamped Critical damped
Underdamped Undamped
Decrease α, smaller R Decrease α, increase R
α=0Undamped
Fix ω0, decrease α (α is positive)The position of the two roots λ1 and λ2.
20
221,
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
1iii cL Mesh current: i1 and ic
1iiKRKRiKvv cLxout
111R iiR
dt
di
dt
diLi c
c
111
1KR iiR
dt
di
dt
diLdti
Cii c
ccc
21
2
2
211
dt
id
dt
idLC
dt
di
dt
diKRCi cc
c
111 R
Lii
dt
di
dt
dii c
c
Example 9.7
21
2
2
211
dt
id
dt
idLC
dt
di
dt
diKRCi cc
c
111 R
Lii
dt
di
dt
dii c
c
11
21
2
2
21
1 1
iidt
di
dt
di
R
L
dt
id
dt
idLC
dt
di
dt
diKRCii
cc
ccc
(1):
(2):
(2) – (1):
Example 9.7
02
11
11
21
2
2
2
ii
LCdt
di
dt
diK
L
R
RCdt
id
dt
idc
cc
02
11
2
2
out
outout vLCdt
dvK
L
R
RCdt
vd
021 11
21
2
2
2
ii
dt
di
dt
diKRC
R
L
dt
id
dt
idLC c
cc
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