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冷凍空調自動控制- Laplace Transform
李達生
Focusing here…
概論自動控制理論發展自控系統設計實例Laplace Transform
冷凍空調自動控制控制系統範例控制元件作動原理控制系統除錯
自動控制理論系統穩定度分析系統性能分析PID Controller
自動控制實務節能系統控制訊號擷取系統訊號雜訊處理快速溫控系統
Time Domain v.s. Frequency Domain
Time
Frequency
Amplitude
From Time Domain to Complex Plane
Complex Plane
Time
F(t)
Exp(t)
j
Time
F(t)
Exp(t) x Exp(it)
Plus
The Definition of Laplace Transform
若函數滿足下列條件
0,)( Rdtetf t Function of
Exponential Order
則可定義函數的 Laplace Transform 為
0)()()( dtetftfLsF st
Laplace Transform Example
s
eTransformatu
Transformtas
)(
1)(
22
22
1
2
cos
sin
1
!
1
11
)()(
as
sTransformat
as
aTransformat
asTransforme
t
nTansformt
sTransformt
sTransform
sFtF
at
nn
22
22
)cosh(
)sinh(
as
sTransformat
as
aTransformat
Laplace Transform Example
試證明 L[1] = 1/S
ss
e
s
est
dtedteL
ss
stst
se 1
)1(]1[
0
0
0 0
Laplace Transform Example
試證明 L[t] = 1/S2
2
0
20
0
1)1()(
)()()()()()(
][
ss
e
s
etdtte
dxxfxgxgxfdxxgxfConsider
tdtetL
st
t
stst
st
How to prove L[tn] = n!/Sn+1 ?????
Laplace Transform Example
試證明 L[eat] = 1/(s-a)
asas
e
edteeeL
tas
tasstatat
1
][
0
)(
0
)(
0
Laplace Transform of Derivation
試證明 L[f′(t)] = sF(s)-f(0)
)0()(
)0()(
)()()(
)()]([
0
00
0
fssF
fdttfes
dttfestfe
dttfetfL
st
stst
st
Laplace Transform of Second Derivation
試證明 L[f″(t)] = s2F(s)-sf(0)-f′(0)
)0()0()(
)0()}0()]([{
)0()]([)]([
2 fsfsFs
fftfsLs
ftfsLtfL
Laplace Transform of Integration
試證明
t
s
sFdttfL
0
)(])([
t
t
s
sFdttfL
tgLs
tgL
gtgsLtgL
ThendttftgIf
0
0
)(])([
)]([1
)]([
)0()]([)]([
)()(
Laplace Transform of sin
試證明 L[sinat]=a / (s2+a2)
22
22
22
2
][sin
][sin][sin
)0cos()0sin(][sin]sin[
)0()0()()]([
as
aatL
aatLsatLa
aaasatLsataL
fsfsFstfL
可得
How to prove L[cosat] = s / s2+a2 ?????
Time Domain Shift – If L[f(t)]=F(s), then L[eatf(t)] = F(s-a)
0
)(
0
)()(
)()]([
asFdttfe
dttfeetfeL
tas
atstat
Laplace Plane Shift
Time Domain Shift
若
at
atatftg
0
)()( 則 )()]([ sFetgL as
)(
)(
)()(
)(
)(0
)()()()]([
0
0
)(
0
0 00
sFe
duufee
atdatfe
dtatfe
dsatfedte
dttgedttgedttgetgL
as
suas
asats
a
st
a
a
stst
a ststst
Initial Value Problem
試證起始值定理 )(lim)(lim0
ssFtfst
)0()(lim
0)('lim
)0()()]([
0
fssFSo
tfeand
fssFtfLSince
s
st
s
Tend to Infinite
試證終值定理 )(lim)(lim0
ssFtfst
)0(
)(lim)(lim
)(lim)()(lim
)0()()]([
0
0 00f
ssFtfSo
tfdttfdttfeand
fssFtfLSince
st
t
st
s
Determine the Steady State Error by the infinite Time Value
TiKp = 0.33
Toe
b
)120)(110(
50
ss
Controller System
Feedback
12
1
s
P
P
i
o
Ksss
sK
sT
sT
50)120)(110)(12(
)12(50
)(
)(
System Transfer Function 可表示為
Determine the Steady State Error by the infinite Time Value
外界溫度變化輸入設為
P
Po
i
Laplace
i
Ksss
sK
s
TsT
s
TsTTtT
50)120)(110)(12(
)12(50)(
)()(
可得
系統穩態誤差可以終值定理求得
)(lim)(lim0
ssEteest
SS
Determine the Steady State Error by the infinite Time Value
系統穩態誤差可以終值定理求得
33.0sin0571.0
50111
501
50)120)(110)(12(
)12(50lim
)()(lim
0
0
P
P
P
P
P
s
ois
SS
KceT
K
KT
Ksss
sK
s
T
s
Ts
sTsTse
系統穩態誤差以百分比表示為 %71.50571.0
T
T
T
e
i
SS
Inverse Laplace Transform
逆轉換可由正轉換導出
atInverseas
s
atInverseas
a
eInverseas
tInverses
n
tInverses
Inverses
tfsF
at
nn
cos
sin
1
!
1
11
)()(
22
22
1
2
)(
)(1
cosh
sinh
22
22
atuInverses
etInverse
atInverseas
s
atInverseas
a
as
Convolution
若 L-1[G(s)] = g(t), L-1[F(s)]=f(t) 則
tgfduutgufsGsFL
0
1 )()()]()([
How to prove?????
How to get time response
控制系統 Transfer Function 可由下式表示
)84)(8(
)4(8
)(
)()(
2
sss
s
sR
sCsT
當輸入函數為單位步階訊號時 , 輸出訊號為何 ?
225
25
310
12
1
2
2
2)2(
)2(
8
)84)(8(
)4(8)(
)84)(8(
)4(81
)(
s
s
ss
ssss
ssC
sss
s
s
sC
How to get time response
各個 Item 均有其對應時間項
225
25
310
12
1
2
2
2)2(
)2(
8
)84)(8(
)4(8)(
)84)(8(
)4(81
)(
s
s
ss
ssss
ssC
sss
s
s
sC 2
1
Inverse L
aplace
te 810
1
tte t 2sin
5
12cos
5
32
C(t)
