12 第二章 數學基礎

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第二章 數學基礎

2-1 拉普拉斯轉換

拉普拉斯轉換(Laplace Transformation)主要功用為將常微 分方程式變換為等效代數方程式。

2×3 0.3010

0.4771

0.7781 6

對數轉換

加法

反對數轉換 直接乘法

微分方程式 轉換方程式

轉換解 解

拉氏轉換

代數運算

反拉氏轉換 特定積分

Σ fn(t)

f(t)=f1(t)+ f2(t)+…

F(s)=B(s)/A(s)

F(s)=F1(s)+ F2(s)+…

L

因式分解 特定積分

L-1

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2-1-1 複變數的概念

複變數及複變函數在拉氏轉換中之應用,是為古典控 制系統理論之基礎,故先介紹複變數概念如下:

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2-2 拉普拉斯轉換之定義

f(t) = 時間 t 之函數,但 t<0 時,f(t) = 0

s = 複變數,稱為拉氏運算子,即 s = σ+jω。

F(s) = f(t)的拉氏轉換,則 F(s) = L[f(t)] =

0

stdt e ) t (

f

例 1:單位階級函數(unit step function) us(t) f(t) = 0, t﹤0

= us(t)= 1 t≧0 F(s) = L[f(t)] =

0

stdt e

1 = 1/s

1

單位階級函數

f(t)

t

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例 2:正弦函數 f(t) = 0, t﹤0 = sinωt , t≧0

F(s) = L[f(t)] =

0

stdt e t sin

∵ ejwt = cosωt + j sinωt e-jwt = cosωt - j sinωt

∴ sinωt = (ejwt - e-jwt )/2j

∴ F(s) = (1/2j)

0

( e

jwt

e

jwt

) e

st

dt

= (1/2j)

0

( e

(sjw)t

e

(sjw)t

) dt

= (1/2j) [ 1/(s-jω) - 1/(s+jω)]

= ω/(s2 + ω2)

cosωt 之拉氏轉換可以同法求得

正弦函數

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2-3 拉氏轉換重要定理:

1.L [

A

f(t)] = A×L[f(t)] = A F(s)

2.L [f1(t)+f2(t)] = L[f1(t)] +L [f2(t)] = F1(s)F2(s) 3.時間平移定理

令 t-α = τ

F(s) = ∫f(τ) e-sτ

= ∫f(t-α) e-s(t-τ) d (t-α) = ∫f(t-α) e-s(t-τ) d t = ef(t-α) e-st d t = eL[f(t-α)]

∴ L[f(t-α)] = e-sα F(s)

例 1 脈波函數(Pulse function) f(t) = 0, t ﹤0, t ﹥t0

= A = const. , 0≦t≦t0

f(t) = AU(t) – AU(t- t0)

∴L[f(t)] = L[AU(t)] - L[AU(t- t0)]

= A/s - A e-st /s = A[1- e-st ]/s

f(t)

t

A

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例 2 脈衝函數(Impulse function) f(t) = 0, t ﹤0, t ﹥t0

= lim (A/ t0 ), 0≦t≦t0

t0→0

f(t)面積為 A,當 t0→0,(A/ t0 ) →∞

L[f(t)] = lim A[1- e-st ]/(s t0) t0→0

A(d/dt0)[ 1- e-st ] =

(d/dt0)[ s t0] = As/s = A

假如 A=1,f(t) = δ(t) ← 單位脈衝函數 F(s) = 1

脈衝函數

A/t

0

t0

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4. f(t)與 e-αt的乘積

L[e-αt f(t)] = ∫e-αt f(t) e-st d t = ∫f(t) e-(s+α)t d t

= F(s+α)

例: L[sinωt] = ω/(s22) = F(s)

L[ e-αt sinωt] = F(s+α) = ω/[(s+α)22]

5.時間尺度之改變

若 t 改為 t/α,f(t)→f(t/α) L[f(t/α)] = ∫f(t/α) e-st d t 令 t/α = t1,αs=s1

L[f(t/α)] = ∫f(t1) e-s1t1 d(α t1) =α∫f(t1) e-s1t1 d t1

=α F(s1) = α F(αs) 例: f(t) = e-t,f(t/5) = e-0.2 t

L[f(t)] = L[e-t] = F(s) = 1/(s+1) L[f(t/5)] = 5 F(5s) = 5/(5s+1) L[e-0.2 t] = 1/(s+0.2) = 5/(5s+1)

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6.微分定理

L[df(t)/dt] = L[f’(t)] = sF(s) – f(0)

L[d2f(t)/dt2] = L[f’’(t)] = s2F(s) –s f(0)–f’(0) 其中

f’(t)及 f’’(t)分別為 f(t)的第一次導數及第二次導數,

f(0)及 f’(0)分別為 f(t)及 f’(t)的初始條件。

例:g(t) = Acosωt,F(s) = L[sinωt]= ω/(s22) 試求 L[g(t)]。

L[g(t)] = L[Acosωt] = A L[cosωt] = A L[d(sinωt/ω)/dt]

= (A/ω) L[d(sinωt)/dt] = (A/ω)[sF(s)-f(0)]

= (A/ω)[sω/(s22) - 0] = As/(s22) 7.終值定理

lim f(t) = lim sF(s)

t→∞ s→0

pf: lim ∫[df(t)/dt] e-st d t = lim [sF(s)-f(0)]

s→0 s→0

∵ lim e-st = 1,∫[df(t)/dt] d t = f(t)∣ = f(∞) – f(0)

s→0

f(∞) – f(0) = lim sF(s) - f(0)

s→0

f(∞) = lim f(t) = lim sF(s)

t→∞ s→0

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8.初值定理

f(0) = lim f(t) = lim sF(s)

t→0 s→∞

pf :

L[df(t)/dt] = sF(s) – f(0)

∵lim ∫[df(t)/dt] e-st d t = 0 s→∞

lim L[df(t)/dt] = 0 = lim[sF(s)-f(0)] = limsF(s) – f(0)

s→∞ s→∞ s→∞

∴ f(0)= lim sF(s)

s→∞

9.積分定理

L[∫f(t)dt] = F(s)/s + f –1(0)/s

相關拉氏轉換常用公式及基本定理歸納附於本章末之表 2-6 及表 2-7,可作為相關運算之參考。

2-4 反拉氏轉換 L-1[F(s)] = f(t)

以部份分數展開法求反拉氏轉換 F(s) = F1(s) + F2(s) +…….+ Fn(s)

L-1[F(s)] = L-1[F1(s)] + L-1[F2(s)] +……+ L-1[Fn(s)]

= f1(t) + f2(t) +……+ fn(t)

其中若 F(s) = B(s)/A(s),則須知 A(s)之根,即

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K(s+z1) (s+z2) ……(s+zn) F(s) = B(s)/A(s) =

(s+p1) (s+p2) ……(s+pn) 依據 A(s)之根的不同有三種情況

1. F(s)僅包含不重複單極之部份分式展開法

F(s) = B(s)/A(s) = a1 /(s+p1) + a2 /(s+p2) +……+ an/(s+pn)

a

k

= [

F(s)( s+pk)

]

s=-pk

L-1[ak /(s+pk)] =

a

k

e

-Pkt

∴f(t) =

a

1

e

-P1t

+ a

2

e

-P2t

+……+ a

n

e

-Pnt

例 1 求下列函數之反拉氏轉換 s+3

F(s) =

(s+1)(s+2) sol:

F(s) = (s+3)/(s+1)(s+2) = a1 /(s+1) + a2 /(s+2) a1 = [(s+3)/(s+1)(s+2)‧(s+1)]s=-1 = 2

a2 = [(s+3)/(s+1)(s+2)‧(s+2)]s=-2 = -1

∴ f(t) = L-1[F(s)] = L-1[2/(s+1)] + L-1[-1/(s+2)]

= 2e-t – e-2t

另可由比較係數法求得

s+3 = a1 (s+2) + a2(s+1) = (a1 + a2 )s +2 a1 + a2

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a1 + a2 = 1 a1 = 2 2 a1 + a2=3 a2 = -1

2. F(S)包含共軛複極之部份分數展開法

F(s) = B(s)/A(s) = (α1s+α2)/[(s+p1) (s+p2)] +a3 /(s+p3) +…

……+ an/(s+pn)

將上式兩邊乘上(s+p1) (s+p2),令 s = -p1

[

F(s)( s+p1)( s+p2)

]

s=-P1

={

1s+α2) +[a3 /(s+p3)] ( s+p1)( s+p2)+……

+[ an/(s+pn)] ( s+p1)( s+p2)} s=-P1

1s+α2) s=-P1

= [

F(s)( s+p1)( s+p2)

]

s=-P1

由實部與虛部比較可得α1,α2

s+1

例2 求 F(s) = 的反拉氏轉換。

s(s2+s+1) sol: F(s) = (α1s+α2)/ (s2+s+1) + a/s

s2+s+1 = (s+0.5+0.866j)(s+0.5-0.866j) 以上式乘 F(s)兩邊,可得

(s+1)/s = a(s2+s+1)/s +α1s+α2

令 s = -0.5 – 0.866j

(0.5-0.866j)/(-0.5-0.866j) =α1(-0.5-0.866j) +α2 化簡為:

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(0.5-0.866j) =α1(0.25+0.866j-0.75) +α2(-0.5-0.866j)

∴-0.5α1 – 0.5α2 = 0.5 → α12 = -1 0. 866α1 – 0.866α2 = -0.866 → α1 –α2 = -1

∴ α1 = -1,α2 = 0

a = [s(s+1)/s(s2+s+1)]s=0 = 1

∴ F(s) = -s/ (s2+s+1) + 1/s = 1/s – (s+0.5)/[(s+0.5)2+0.8662] + 0.5/[(s+0.5)2+0.8662]

∴ f(t) = 1 – e-0.5tcos(0.866t) + 0.578 e-0.5tsin(0.866t) 其中 0.578 = 0.5/0.866

3. F(s)包含重複極點之部份分數展開法

F(s) = B(s)/A(s) = B(s)/[ (s+p1)r (s+pr+1)……(s+pn)]

= br /(s+p1)r + br -1/(s+p1)r-1 + ……+ b1/(s+p1) + ar+1/(s+pr+1) + ar+2/(s+pr+2) +……+ an/(s+pn)

br = [F(s)(s+p1)r]s=-P1

br-1 = {(d/ds)[F(s)(s+p1)r]s=-P1 .

. .

br-j =(1/j!) {(dj/dsj)[F(s)(s+p1)r]s=-P1

. . .

b1 =[1/(r-1)!] {(dr-1/dsr-1)[F(s)(s+p1)r]s=-P1

ar+1 ,ar+2 ……an 由下式求得

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ak = [F(s)(s+pk)]s=-Pk

s2+2 s+3

例3 求 F(s) = 的反拉氏轉換。

(s+1)3

sol: F(s) = b3 /(s+1)3 + b2/(s+1)2 + b1/(s+1)

b3 = [F(s)(s+1)3]s=-1 = (s2+2 s+3)s=-1 = 2 b2 = {(d/ds)[F(s)(s+1)3]s=-1 = (2 s+2)s=-1 = 0 b1 =(1/(3-1)!) {(d2/ds2)[F(s)(s+1)3]s=-1

=(1/2) [(d2/ds2)(s2+2s+3)]s=-1 = (1/2)‧2 = 1

∴ f(t) = L-1[F(s)] = L-1[2/(s+1)3] + L-1[1/(s+1)]

= (t2+1) e-t (t≧0)

2-5 拉氏轉換法求解線性常微分方程式

以拉氏轉換法求解線性常微分方程式的步驟說明於下:

1. 利用拉氏轉換基本定理,將微分方程式轉換為代數方程 式。

2. 以代數四則運算整理代數方程式,並將整理後得到的總 函數作部份分數展開。

3. 利用拉氏轉換公式將部份分式作反拉式轉換,使總函數 轉換成原微分方程式的時域解。

例1 求解 mx’’ + kx = f(t),f(t)為外力函數。

Sol:

L[mx’’] = m[s2X(s)-sx(0)-x’(0)]

L

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(ms2+k)X(s) – msx(0) – mx’(0) = F(s)

∴X(s) = F(s)/ (ms2+k) +[msx(0) + mx’(0)]/ (ms2+k)

∴x(t) = L-1[F(s)/ (ms2+k)] + L-1 {[msx(0) + mx’(0)]/ (ms2+k)}

假如 F(s) = 1/s (單階函數),且初始條件 x(0) =x’(0)=0 x(t) = L-1[1/s (ms2+k)] = (1/k){1-cos[(k/m)0.5t]}

例2 求下列微分方程式之拉氏轉換 x’’+3x’+6x = 0,x(0)=0,x’(0)=3 並求 x(t)。

sol: L(x’’+3x’+6x) = 0

s2X(s) – sx(0) – x’(0) + 3sX(s) – 3x(0) + 6X(s) = 0 X(s) = 3/( s2 + 3s + 6)

=(2√3/√5){(√15/2)/[(s+1.5) 2+(√15/2)2]}

∴x(t) = (2√3/√5) e-1.5tsin[(√15/2)t]

例3 右圖之機械系統,設受一單擊外力,

求所產生之振盪運動,假設此系統初時靜止 sol: 此系統為一脈衝輸入所驅動,故

mx’’+ kx = δ(t) L(mx’’+ kx) = L[δ(t)] = 1 由例 1 知,X(s) = 1/ (ms2+k)

∴x(t) = (1/√mk)sin[(√k/m)t] ← 簡諧運動 m

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2.6

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2.7

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