lecture 14 second-order circuits (2) hung-yi lee
TRANSCRIPT
Lecture 14Second-order Circuits (2)
Hung-yi Lee
Second-Order CircuitsSolving 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
sCCC vL
vLC
vL
Rv
C
11 sLL i
LCi
LCii
11
RC
1L
Review Step 2: Natural Response
λ1, λ2 is
21
21
Overdamped
Critical damped
Complex0
0
Underdamped
Undamped
20
2 20
2
20
2
Real
02 N20NN tytyty 2
02
21,
α=0Undamped
Fix ω0, decrease αThe position of the two roots λ1 and λ2.
20
221,
Example 9.11
0 V0
0 V30)(
F640
1 ,5 H,1.0
t
ttv
CRL
s
Natural response iN(t):
01
NNN iLC
iL
Ri
0? ttiL
0640050 NNN iii
tei AN 06400502 ttt AeAeAe
06400502
2
640045050,
2
21
j7625
sCCC vL
vLC
vL
Rv
C
11
sLLL vL
iLC
iL
Ri
11
Example 9.11
0 V0
0 V30)(
F640
1 ,5 H,1.0
t
ttv
CRL
s
Natural response iN(t): 0? ttiL
j7625, 21
dj 21,
tbtaety ddt
N sincos
tbtaeti tN 76sin76cos25
Forced response iF(t)=0 tbtaetiti t 76sin76cos25
NL Complete response iL(t):
Two unknown variables, so two initial conditions.
sCCC vL
vLC
vL
Rv
C
11
sLLL vL
iLC
iL
Ri
11
Example 9.11
Initial condition
30V
short
open
00 Li
300 Cv
00 Li
300 Cv
00 Ci 00
CvC
300 Lv 300
LiL
0 V0
0 V30)(
F640
1 ,5 H,1.0
t
ttv
CRL
s
0? ttiL
0t
Example 9.11
tbtaeti t 76sin76cos25L 00
Li
300 LiL 3000
Li
aiL 0 0 tbeti t 76sin25L
tetebti tt 76cos7676sin25 2525L
760L bi 300 95.3b
teti tL 76sin295.3 25
Example 9.11
teti tL 76sin295.3 25
)9076cos(95.3)( 25 teti tLTextbook:
Example 9.12
0 V30
0 V0)(
F640
1 ,? H,1.0
t
ttv
CRL
s
Natural response vN(t): 0?C ttv
tev AN
06400102 ttt AeAeRAe 06400102 R
2
640041010,
2
21
RR
sCCC vL
vLC
vL
Rv
C
11
306400640010 CCC vvRv
0640010 NNN vvRv
Forced response vF(t):
ssv YF 306400640010 FFF vvRv 30
Example 9.12
Initial condition:
0V
short
open
00 Li
00 Cv
00 Li
00 Cv
00 Ci 00
CvC
0 V30
0 V0)(
F640
1 ,? H,1.0
t
ttv
CRL
s
0?C ttv
Example 9.12
tvtvtv F NC
tt eetv 2121N AA
2
640041010,
2
21
RR
30F tv
Different R gives different response
R16R160 R
0R 16R
?0R
00 Cv 00
CvC
Initial condition:
Example 9.12
16R 34ROverdamped:
2
64004341010,
2
21
R 320,20
ttC eAeAtv 320
220
130)(
ttC eAeAtv 320
220
1 32020)(
3021 AA
032020 21 AA
2,32 21 AA
tvtvtv F NC
tt eetv 2121N AA
2
640041010,
2
21
RR
30F tv
00 Cv 00
CvC
Initial condition:
Example 9.12
16RCritical damped:
tvtvtv F NC
tt eetv 2121N AA
2
640041010,
2
21
RR
30F tv
80, 21 )( 802
801N
tt teAeAtv
30)( 802
801
ttC teAeAtv
808030)( 80802
80 tttC teeAetv
030)0( 1 AvC 301 A
3030 802
80 tt teAe
08030)0( 2 AvC 24002 A
00 Cv 00
CvC
Initial condition:
Example 9.12
tvtvtv F NC
tt eetv 2121N AA
2
640041010,
2
21
RR
30F tv
j7625, 21 16RUnderdamped: 5R
tbtaetv t 76sin76cos25N
tbtaetv t 76sin76cos30 25C
0300 avC30a
tbtae
tbtaetvt
tc
76cos7676sin76
76sin76cos2525
25
076250 bavC
87.9b
00 Cv 00
CvC
Initial condition:
Example 9.12
tbtaetv t 76sin76cos30 25C
t
ba
bt
ba
aebatv t
C 76sin76cos302222
2522
30a 87.9b
31.58cos x sin x
)76cos(58.3130)( 25 xtetv tC 8.161x
Example 9.12
)8.16176cos(58.3130)( 25 tetv tC
24003030)( 8080C
tt teetv
ttC eetv 32020 23230)(
Non-constant Input
• Find vc(t) for t > 0 when vC(0) = 1 and iL(0) = 0
Li12
1cv
sv
cc vi 12
1
LL iv 5.0
LcLs iivv
cc vi dtviiviv cccccs 2
cccccs vvvvvv
12
12
12
1
12
1
sccc vvvv 6712
dtvii ccL 2
Non-constant Input
4,3, 21 0712 NNN vvv
tKtKvF cossin 21
t
tKtKtKtKtKtK
sin30
cossinsincos7cossin12 212121
30127 121 KKK 30711 21 KK
0127 212 KKK 0117 21 KK 17
331 K
17
212 K
sccc vvvv 6712
Natural response vN(t):
Forced response vF(t):
tt eAeAtv 42
31N
tvvvv sFFF sin566712
tvs cos5
Non-constant Input
Li12
1cv
sv
10 Cv 00 Li
20 Ci 200 CC vCi
240 Cv
Initial Condition:
Non-constant Input
tvtvtv FN C
tKtKvF cossin 21 17
331 K
17
212 K
tt eAeAtv 42
31N
tteAeAtv tt cos17
21sin
17
3342
31C
10 Cv 240 Cv
117
2121 AA
tteAeAtv tt sin17
21cos
17
3343 4
23
1C
2417
3343 21 AA
251 A
17
4292 A
Second-Order CircuitsZero Input + Zero State
& Superposition
Review: Zero Input + Zero State
y(t) = general solution + special solution
= natural response + forced response
y(t) = state response (zero input) + input response (zero state)
= =
y(t): voltage of capacitor or current of inductor
If input = input 1 + input 2 + ……
= state response (zero input) + input 1 response + input 2 response ……
Set sources to be zeroSet state vc, iL to be zero
Example 1
30015)(
F2
1 ,
4
5 H,
8
1
ttv
CRL
s
Vv 30C AiL 00
State response (Zero input):Find vC(t) for t>0
01
statestatestate vLC
vL
Rv 01610 statestatestate vvv
016102 8,2
ttstate eAeAtv 8
22
1
sCCC vL
vLC
vL
Rv
C
11
(pulse)
Example 1
ttstate eAeAtv 8
22
1
Vv 30state
00state vC
321 AA
082 21 AA1,4 21 AA
ttstate eetv 824
State response (Zero input):
30015)(
F2
1 ,
4
5 H,
8
1
ttv
CRL
s
Vv 30C AiL 00
Find vC(t) for t>0
sCCC vL
vLC
vL
Rv
C
11
(pulse)
Example 1
00C v 00C v
=
0
-15
)(1 tvs
…30
15
tsv
15
0
…
)(2 tvs3030
Input response (Zero State):
30015)(
F2
1 ,
4
5 H,
8
1
ttv
CRL
s
Vv 30C AiL 00
Find vC(t) for t>0
(pulse)
(set state to be zero)
Example 1 sCCC vL
vLC
vL
Rv
C
11
0
15
)(1 tvs
…30
00C v 00C v
15161610 111 inputinputinput vvv
ttinput eAeAtv 8
22
11 15 015 21 AA
082 21 AA 52 A
201 Att ee 82 52015
15
0
…
)(2 tvs30
3083022 52015 tt
input eetv
30002 ttvinput
30t
Example 1
=
0
-15
)(1 tvs
…30
15
tsv
15
0
…
)(2 tvs3030
Input response (Zero State):
02 tvinput
tt
input
ee
tv82
1
52015
tt
inputinput
ee
tvtv82
1
52015
300 t
Example 1
=
0
-15
)(1 tvs
…30
15
tsv
15
0
…
)(2 tvs3030
Input response (Zero State): 30t
tvtvtv inputinputinput 21
3088
3022
5
20
tt
tt
ee
ee
308
302
2
5
2015
t
t
input
e
e
tv
t
t
input
e
e
tv
8
2
1
5
2015
30t300 t
Example 1
ttstate eetv 824
tt
input
ee
tv82 52015
Input response (Zero State)
State response (Zero input)
3088
3022
5
20
tt
tt
input
ee
ee
tv
300 t
30t
tt eetvtvtv 82inputstateC 62415
308302inputstateC 520 ttetvtvtv
Example 1 – Differential Equation
15 tsv
30
30015)(
F2
1 ,
4
5 H,
8
1
ttv
CRL
s
Vv 30C AiL 00
Find vC(t) for t>0
(pulse)
Vv 30C 00 Cv
0t
30t Vv 1530C
Assume 30s is large enough
030 Li
30t Vv 1530C 030
Cv
Example 1 – Differential Equation
Vv 30C
00 Cv
Vv 1530C
030 Cv
t0t 30t
15vs t 0vs t
1516
1610 CCC
vvv
0
1610 CCC
vvv
tt eAeAtv 82
21C 15
6
24
2
1
A
A:300 t
Example 1 – Differential Equation
Vv 30C
00 Cv
Vv 1530C
030 Cv
t0t 30t
15vs t 0vs t
0
1610 CCC
vvv
tt eAeAtv 82
21C
2402
601 5,20 eAeA
1530 3082
3021C eAeAv
08230 3082
3021C eAeAv
308302C 520 tt eetv
1516
1610 CCC
vvv
Example 2
0)(
F2
1 ,
4
5 H,
8
1
tttv
CRL
s
Vv 30C AiL 00
State response (Zero input):
Find vC(t) for t>0
sCCC vL
vLC
vL
Rv
C
11
ttstate eetv 824 Changing the input will not
change the state (zero input) response.
Example 2
0)(
F2
1 ,
4
5 H,
8
1
tttv
CRL
s
Vv 30C AiL 00
Input response (Zero state):
Find vC(t) for t>0
sCCC vL
vLC
vL
Rv
C
11
00input v 00input v
(set state to be zero)Input
What is the response? 4 methods
Example 2 – Method 1 for Zero State
sCCC vL
vLC
vL
Rv
C
11
tvvv 161610 inputinputinput
Natural Response:
01610 NNN vvv
21 KtKtvF tKtKK 161610 211
ttN eAeAtv 8
22
1
Forced Response:
tvvv 161610 FFF
8
5,1 21 KK
8
5ttvF
00input v 00input v
ttFNinput eAeAttvtvtv 8
22
18
5
Example 2 – Method 1 for Zero State
ttinput eAeAttv 8
22
18
5 00input v 00input v
08
50 21 AAvinput
ttinput eAeAtv 8
22
1 821
08210 21 AAvinput24
13
2
2
1
A
A
ttinput eettv 82
24
1
3
2
8
5
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
=
0
-1
…1 1
0
…t t
tt eetv 82
3
1
3
41 tttt eetv 82
3
1
3
41
tttt eeeetv 8822 13
11
3
4
tetetv tt 83
12
3
4 82 tee tt 82
3
8
Example 2 – Method 2 for Zero State
teetv tt 820 3
8
1
t
1t
teetv ttttt 11
1
82
3
8
Consider a point avalue of the pulse at t1 is
att 1
tee
av
tata
t
11
1
82
3
8
Example 2 – Method 2 for Zero State
at
t
tatainput dteettv
0
82
3
8
at
t
taat
t
ta dtteedttee0
88
0
22
3
8
3
8 1
2 at
a
edtte
atat
Consider a point avalue of the pulse at t1 is teeav tata
t 11
1
82
3
8
Example 2 – Method 2 for Zero State
1
64
118
643
81
4
112
43
8 88
22 a
eea
ee
aa
aa
aa eaea 82
64
118
64
1
3
8
4
112
4
1
3
8
aa eea 82
24
1-
3
2
8
5
at
t
taat
t
tainput dtteedtteeav
0
88
0
22
3
8
3
8
12
ata
edtte
atat
18
643
812
43
8 88
22 t
eet
ee
ta
ta
0t
at
We can always replace “a” with “t”.
Example 2 – Method 3 for Zero State
att
0
t …
teetv tt
82
3
1
3
41
dtee
avat
t
tata
input
0
82
3
1
3
41
…………
…
The value at time point a
Example 2 – Method 3 for Zero State
at
t
tatainput dteeav
0
82
3
1
3
41
at
t
at
t
taat
t
ta dtedtedt0 0
8
0
2
3
1
3
41
18
1
3
11
2
1
3
4 8822 aaaa eeeea
aa eea 82 124
11
3
2 aa eea 82
24
1-
3
2
8
5
at
t
at
t
taat
t
ta dteedteedt0 0
88
0
22
3
1
3
41
Example 2 – Method 4 for Zero State
0
1 …
ttxslope )(
?)(slope ty
Source Input:
Input (Zero State) Response
)()( tutxstep Source Input:
Input (Zero State) Response
ttstep eety 82
3
1
3
41
Example 2 – Method 4 for Zero State
txtytyty slope161610 slopeslopeslope
txtytyty step161610 stepstepstep
ttxslope )( ?)(slope ty
)()( tutxstep ttstep eety 82
3
1
3
41
txtxtytyty stepslope 16161610 slopeslopeslope
tyty stepslope tt ee 82
3
1
3
41
A24
1
3
2 82slope tt eetty
8
5A
Example 2
at
t
tatainput dteeav
0
82
3
1
3
41
at
t
tatainput dteetav
0
82
3
8Method 2:
Method 3:
Method 4: tvinput tt ee 82
3
1
3
41
Example 2
ttC eettv 82
24
23
3
10
8
5
ttvs
0)(
F2
1 ,
4
5 H,
8
1
tttv
CRL
s
tvtvtv inputstateC
Example 2 – Checked by Differential Equation
0)(
F2
1 ,
4
5 H,
8
1
tttv
CRL
s
Vv 30C AiL 00
Find vC(t) for t>0
ttC eAeAttv 8
22
18
5
ttN eAeAtv 8
22
1
sCCC vL
vLC
vL
Rv
C
11
8
5ttvF 30C v 00
Li
00C vC 30L iL
38
521 AA
0821 21 AA24
23,
3
1021 AA
Example 3 F2
1 ,
4
5 H,
8
1 CRL
Vv 30C AiL 00
Find vC(t) for t>0
30
300
0
30
0
s
t
t
t
V
ttv
How to solve it?I will show how to solve the problem by
Method 1: Differential equationMethod 2: Integrating Step ResponsesMethod 3: Differentiate the sources
Example 3 – Method 1: Differential Equation
Vv 30C AiL 00
Find vC(t) for t>0
30
300
0
30
0
s
t
t
t
V
ttv
ttC eettv 82
24
23
3
10
8
5
300 t
24060
24
23
3
10
8
5303030 eevv CC
ttC eetv 82
3
23
3
201 24060
3
23
3
2013030 eevv CC
Example 3 – Method 1: Differential Equation
tt eAeAtv 82
21C 30
24060
24
23
3
10
8
53030 eevC 24060
3
23
3
20130 eevC
tt eAeAtv 82
21C 82
2402
601C 3030 eAeAv
24
23
24
1A,
3
10
3
2A 240
260
1 ee
2402
601C 8230 eAeAv
30830282C 24
1
3
2
24
23
3
1030 tttt eeeetv
30t
Example 3 – Method 2: Integrating Step Responses
att
0
t …
teetv tt
82
3
1
3
41
dtee
avt
t
tata
input
30
0
82
3
1
3
41
…………
…
The value at time point a
If a > 30 a
If a < 30
If a > 30Integrating from 0 to a
Integrating from 0 to 30
Example 3 – Method 2: Integrating Step Responses
30
0
82
3
1
3
41
t
t
tatainput dteeav
30
0
30
0
830
0
2
3
1
3
41
t
t
t
t
tat
t
ta dtedtedt
18
1
3
11
2
1
3
430 2408602 eeee aa
ttstate eetv 824
30
0
30
0
8830
0
22
3
1
3
41
t
t
t
t
tat
t
ta dteedteedt
30830282input 24
1
3
2
24
1
3
230 tttt eeeetv
Example 3 – Method 3: Differentiate the sources
30
300
0
30
0
s
t
t
t
V
ttv
30
300
0
0
1
0
s
t
t
t
tv
1
30
tvtv pulseinput
?tvinputResponse: tvpulseResponse:
Example 3 – Method 3: Differentiate the sources
30
300
0
0
1
0
s
t
t
t
tv
1
30
tvpulseResponse:
30t tt
pulse eetv 82
3
1
3
41
tvtv pulseinput
Aeettv ttinput 82
24
1
3
2
024
1
3
20 Avinput 8
5A
8
5
24
1
3
2 82 ttinput eettv
8
5
24
1
3
23030 24060 eevinput
Example 3 – Method 3: Differentiate the sources
30
300
0
0
1
0
s
t
t
t
tv
1
30
tvpulseResponse:
300 t tvtv pulseinput 2408602 1
3
11
3
4eeeetv tt
pulse
8
5
24
1
3
23030 24060 eevinput
Aeeeetv ttinput 2408602 1
24
11
3
2
............30 inputv 30A
308302
82input
24
1
3
224
1
3
230
tt
tt
ee
eetv
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
• 9.60
3.1624cos2512 7C tetv t
tt teeti 55L 2086
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
1675.075.6 364C tt eetv
Appendix
Higher order?
• All higher order circuits (3rd, 4th, etc) have the same types of responses as seen in 1st-order and 2nd-order 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