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
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：
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
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
stdt
= (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
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 dτ
= ∫∞f(t-α) e-s(t-τ) d (t-α) = ∫∞f(t-α) e-s(t-τ) d t = esα ∫∞f(t-α) e-st d t = esα L[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
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
0t0
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／(s2+ω2) = F(s)
L[ e-αt sinωt] = F(s+α) = ω／[(s+α)2+ω2]
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)
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/(s2+ω2) 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/(s2+ω2) - 0] = As/(s2+ω2) 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
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
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=-pkL-1[ak /(s+pk)] =
a
ke
-Pkt∴f(t) =
a
1e
-P1t+ a
2e
-P2t+……+ a
ne
-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
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={
(α1s+α2) +[a3 /(s+p3)] ( s+p1)( s+p2)+……+[ an/(s+pn)] ( s+p1)( s+p2)} s=-P1
∴
(α1s+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：
(0.5-0.866j) =α1(0.25+0.866j-0.75) +α2(-0.5-0.866j)
∴-0.5α1 – 0.5α2 = 0.5 → α1 +α2 = -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
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
(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
2.6
2.7
