Transcript 上課資料
416
Chapter 7 The Laplace Transform
作用：把微分變成乘法
Chapter 4 曾經提過
k
d y t
k
dt
可寫成 D k y t
k
Laplace transform可以將
s Y s s
k
k 1
y 0 s
k 2
d y t
k
dt
y 0
變成
sy
(k 2)
0 y ( k 1) 0
417
Section 7-1 Definition of the Laplace Transform
7-1-1 Definitions
Laplace Transform of f(t)
L f ( t )
e
st
f ( t ) dt
0
經常以大寫來代表 transform 的結果
F s L f ( t )
e
0
st
f ( t ) dt
418
Laplace Transform is one of the integral transform
transform:
把一個 function 變成另外一個 function
integral transform: 可以表示成積分式的 transform
F s
b
a
K s , t f ( t ) dt
kernel
對 Laplace transform 而言
K s, t e
st
,
a = 0, b
註：Chap. 14 將教到的 Fourier transform, 也是一種 integral
transform
419
7-1-2 Linear Property
e
0
st
f ( t ) g ( t ) dt
e
st
f ( t ) dt
0
e
st
g ( t ) dt
0
L f ( t ) g ( t ) L f ( t ) L g ( t )
事實上，所有的 integral transform 都有linear property
420
7-1-3 The Laplace Transforms of Some Basic Functions
f(t)
F(s)
1
s
n!
1
tn
exp(at)
sin(kt)
cos(kt)
sinh(kt)
cosh(kt)
s
n 1
1
sa
k
2
2
s k
s
2
2
s k
k
2
2
s k
s
2
2
s k
(彼此密切相關）
421
L 1
Example 1
L 1
e
st
dt e
0
(1)
e
(text page 275)
s
s
st
s
e
0
s
s
( e
s 0
)1
s
s
比較正式的寫法是 lim e
(2) 這裡假設 s > 0, 所以
b
e
s
s
0
s b
s
L t
Example 2
L t
te
st
b
b
a
a
u t v t dt u t v t
dt
0
te
st
s
e
s
e
s
s
12
s
2
0
s
e
0
e
st
dt
s
s 0
0e
s
s 0
s
422
(text page 275)
2
st
e 2
s
0
b
a
u t v t dt
Example 3
L e
3t
s 1 3
423
(text page 275)
Pole (分母為0 的地方) 在複數平面左半邊
stable
Pole 在複數平面右半邊
unstable
Im(s)
s = -3
Re(s)
Example 4 L sin(2 t )
424
(text page 276)
除了課本的解法之外，
另一個解法
i 2t
i 2t
sin(2 t ) 1 e e
2i
i2t
i2t
L sin(2 t ) 1 L e 1 L e
1 1 1 1
2i
2i
2i s i 2 2i s i 2
s i 2 ( s i 2)
1
1 2i 4 2 2
2 i ( s i 2)( s i 2 ) 2 i s 4 s 4
Example 5
(text page 277)
L 1 5 t L 1 5 L t 1 52
s s
L 4e
3t
10 sin 2 t 4 L e
3t
10 L s in 2 t s 4 3 s 20 4
2
425
7-1-4 When Does the Laplace Transforms Exist?
Constraint 1 for the existence of the Laplace transform :
For a function f(t), there should exist constants c, M > 0, and T > 0
such that
f t Me
ct
for all t > T
In this condition, f(t) is said to be of exponential order c
Fig. 7.1.2
426
Example： f(t) = t, e−t, 2cost 皆為 exponential order 1
Fig. 7.1.3
(a)
(b)
(c)
補充：其實，對一個function 而言， exponential order c 不只一個
例子： f(t) = tn 為 exponential order c, c > 0
n
There exists an M such that t c t M
e
if c > 0
Example： f(t) =
exp(t2)
時，並不存在一個 c 使得
f t Me
ct
for all t > T
Fig. 7.1.4
只要有一個 c 使得 f t M e c t for all t > T
我們稱 f(t) 為 of exponential order
否則，我們稱 f(t) 為 not of exponential order
427
428
Constraint 2 for the existence of the Laplace transform :
f(t) should be piecewise continuous on [0, )
在任何 t [a, b] 的區間內 (0 a b < )
f(t) 為 discontinuous 的點的個數為有限的
稱作是「piecewise continuous」
Fig. 7.1.1
注意： 1/t 不為 piecewise continuous
429
Constraints 1 and 2 are “sufficient conditions”
若滿足
Laplace transform 存在
若不滿足
Laplace transform 未必不存在
反例： f(t) =
t−1/2
不為 piecewise continuous
但是 Laplace transform 存在 F s s 1 / 2
補充說明： f(t) = t−1/2 不為 piecewise continuous 是因為
f(0)
所以 f(t) 在 t = 0 附近有無限多個不連續點
事實上，只要 f(t1) , |t1| is not infinite，
f(t) 必定不為 piecewise continuous
430
431
Theorem 7.1.3
If f(t) is piecewise continuous on [0, ) and of exponential order, then
lim F s 0
s
432
7-1-5 Section 7-1 需要注意的地方
(1) Laplace transform of some basic functions 要背起來
(2) 記公式時，一些地方要小心 sin, sinh, 1/tn
sin kt
沒有平方
k
2
2
s k
有平方
(3) 熟悉(a) 包含 exponential function 的積分
以及 (b)
b
a
u t v t dt u t v t a
b
b
a
u t v t dt
的積分技巧
(4) 要迅速判斷一個式子當 t
(5) 小心正負號
時是否為 0
433
附錄七：充分條件和必要條件的比較
If A is satisfied, then B is also satisfied :
A is the sufficient conditions of B (充分條件)
A
B
If B is satisfied, then A is bound to be satisfied :
A is the necessary conditions of B (必要條件)
B
A
B is satisfied if and only if A is be satisfied :
A is the necessary and sufficient conditions of B
(充分且必要的條件)
Section 7-2 Inverse Transforms and
Transforms of Derivatives
本節有兩大部分：
(1) inverse Laplace transform 的計算 (7-2-1 ~ 7-2-3)
(2) 將微分變成 Laplace transform 當中的乘法 (7-2-4 ~ 7-2-6)
434
435
7-2-1 Inverse 方法一： One-to-One Relation
When (1) f1(t) and f2(t) are piecewise continuous on [0, ), and
(2) f1(t) and f2(t) are of exponential order, then
if f1(t) f2(t)
then F1(s) F2(s)
換句話說，在這種情形下，Laplace transform 是 one-to-one
的運算。
If the Laplace transform of f1(t) is F1(s),
then the inverse Laplace transform of F1(s) must be f1(t).
436
Table of Inverse Laplace Transforms
F(s)
L−1{F(s)}
1
s
n!
1
s
n 1
1
sa
k
2
2
s k
s
2
2
s k
k
2
2
s k
s
2
2
s k
tn
exp(at)
sin(kt)
cos(kt)
sinh(kt)
cosh(kt)
437
s
Example 1 (text page 282) (a) L 1 1
5
L
1
1 1 L 1 4 ! 1 t 4
5
5
4!
4!
s
s
s
Example 2 (text page 283) L 1 22 s 6
L
1
4
1
1
2 s 6 L 1 2 s 6
s
2
2
L
3
L
2
2
2
2
2
s 4
s 4 s 4
s 4
s 4
2 cos( 2 t ) 3 sin(2 t )
438
7-2-2 Inverse 方法 (二） Decomposition of Fractions
Example 3 (text page 283)
L
1
s 6s 9
( s 1)( s 2)( s 4)
2
s 6s 9
A B C
( s 1)( s 2)( s 4) s 1 s 2 s 4
2
L
1
t
2t
4 t
s 6s 9
Ae Be Ce
( s 1)( s 2)( s 4)
2
問題：A, B, C 該如何算出？
2
A ( s 2)( s 4) B ( s 1)( s 4) C ( s 1)( s 2)
s 6s 9
( s 1)( s 2)( s 4)
( s 1)( s 2)( s 4)
s 6 s 9 ( A B C ) s (2 A 3 B 3 C ) s 8 A 4 B 2 C
2
2
太麻煩
7-2-3 計算分數分解係數的快速法
s 6s 9
A B C
( s 1)( s 2)( s 4) s 1 s 2 s 4
2
兩邊各乘上 (s − 1)
s 6 s 9 A ( s 1) B ( s 1) C
( s 2)( s 4)
s2
s4
2
16
把 s = 1 代入 5 A
這二個步驟可以合併
左式乘上 (s − 2)後，把 s = 2 代入
s 6s 9
B
( s 1)( s 2)( s 4)
2
s2
25
6
左式乘上 (s + 4)後，把 s = −4 代入
s 6s 9
C
( s 1)( s 2)( s 4 )
2
s 4
1
30
439
通則：要將一個 fraction 分解
K s
( s a1 )( s a 2 )
440
(Cover up method)
AN
s aN
A1
A2
Q s
s
a
s
a
(s aN )
1
2
a1, a2, …., aN 互異
(1) 用多項式的除法算出 Q(s)
K s
( s a1 )( s a 2 )
(s aN )
商
Q s
餘式
K1 s
( s a1 )( s a 2 )
(s aN )
使得 order of K1(s) < N
(2) 算出 An
An
K1 s
( s a1 )( s a 2 )
( s a n 1 )( s a n )( s a n 1 )
( s a N ) sa
n
例子： s 8 s 31 s 36 s 20
4
3
2
( s 1)( s 2 )( s 3)
s 8 s 31 s 36 s 20
4
3
2
3
1
s 2 s 3s 2
3
2
( s 1)( s 2)( s 3)
2
s 1
s 2 s 3s 2
3
A2
s 2 s 3s 2
2
( s 1)( s 2 )( s 3)
A1
2
A3 A4 ( s 3)
A1
A2
Q (s)
2
s 1 s 2
( s 3)
2
( s 1)( s 2 )( s 3)
8
2
2
4
2
( s 1)( s 2)( s 3)
24
2
s2
3
2
s 2 s 3 s 2 ( s 3) 2 A1 ( s 3) 2 A2 A A ( s 3)
3
4
( s 1)( s 2 )
s 1
s2
s 2 s 3s 2
3
A3
2
( s 1)( s 2)
s3
56
2
28
Q(s) = 1
441
442
3
2
s 2 s 3 s 2 ( s 3) 2 A1 ( s 3) 2 A2 A A ( s 3)
3
4
( s 1)( s 2 )
s 1
s2
d s 2 s 3s 2
3
A4
2
( s 1)( s 2)
ds
(3 s 4 s 3)( s 1)( s 2) ( s 2 s 3 s 2)(2 s 3)
2
3
( s 1) ( s 2)
2
s3
2
2
21
s 8 s 31 s 36 s 20
4
3
2
( s 1)( s 2)( s 3)
2
1
2 24 28 21( s 3)
2
s 1 s 2
( s 3)
s3
443
小技巧：其實，如果只剩下一個未知數，我們可以將 s 用某個數
代入， 快速的將未知數解出
例如，前面的例子，將 s = 0 代入原式
A2 A3 3 A4
A1
2
9
18
2
9 A2
A4 ( 1 9 A1
A3 ) / 3 21
2
444
2
A 2 s A3
A1
s
2s 3
例子：
2
2
( s 1)( s 2 s 2 ) s 1 s 2 s 2
s 2s 3
A1
2
( s 1)( s 2 s 2)
2
s 1
6
5
s 2s 3
s 2s 3
6/5 1
1 2 s 3
2
2
( s 1)( s 2 s 2) s 1 5 ( s 1)( s 2 s 2) 5 s 2 s 2
2
2
s 2s 3
6/5 1 2 s 3
2
( s 1)( s 2 s 2) s 1 5 s 2 s 2
2
445
7-2-4 Transforms of Derivatives
L f ( t )
e
st
f ( t ) dt e
st
f (t )
0
0
s e
st
f ( t ) dt
0
0 f (0 ) sL f ( t ) sL f ( t ) f ( 0 )
b
a
u t v t dt u t v t a
b
b
a
u t v t dt
L f ( t ) sL f ( t ) f (0) s sL f ( t ) f (0) f ( 0 )
2
s L f ( t ) sf (0) f (0)
3
2
L f ( t ) sL f ( t ) f (0) s L f ( t ) s f (0) sf (0) f ( 0)
446
Theorem 7.2.2 Derivative Property of the Laplace Transform
L f
(n)
t s n F s s n 1 f 0 s n 2 f 0
sf
(n2)
0 f ( n 1) 0
447
7-2-5 Solving the Constant Coefficient Linear DE by Laplace
Transforms
an y
(n)
x a n 1 y ( n 1) x
a1 y ( x ) a 0 y g x
Laplace transform
a n s Y s s
n
a n 1 s
n 1
y 0 s
Y s s
n 1
n2
n2
y0
y 0 s
n3
y0
a1 sY s y 0 a 0Y s G s
sy
(n2)
0 y ( n 1) 0
sy
( n3)
0 y ( n 2 ) 0
a n s Y s s
n
a n 1 s
n 1
y 0 s
Y s s
n 1
n2
n2
y0
y 0 s
n 1
sy
y0
(n2)
0 y
sy
( n3)
( n 1)
0
0 y ( n 2 ) 0
a1 sY s y 0 a 0Y s G s
P s Y s Q s G s
P s a n s a n 1 s
n
Q s a n s
n 1
a n 1
n 1
y 0 s
s
a1 s a 0
(auxiliary)
n2
y0
sy
n2
y 0
sy
(n2)
( n3)
0 y ( n 1) 0
0 y ( n 2 ) 0
a2
a1
sy 0
y 0
y 0
448
Y s
Q s
P s
G s
P s
449
Y s W sQ s W sG s
G(s): Laplace transform of the input
Q(s): caused by initial conditions
Y(s): Laplace transform of the response
W(s): transform function
L−1[W(s)Q(s)]: zero-input response or state response
L−1[W(s)G(s)]: zero-state response or input response
W s
1
P s
450
Example 4 (text page 285)
y ( t ) 3 y ( t ) 13 sin 2 t
y (0) 6
(Step 1) Laplace Transform
sY ( s ) y 0 3Y ( s ) 13
( s 3)Y ( s ) 6
2
2
s 4
26
s 4
2
(Step 2) Decompose
Y (s)
Y (s)
Y (s)
6
26
s 3 ( s 3)( s 2 4 )
8 2 s 6
2
s3
s 4
8 2 s 3 2
2
2
s3
s 4
s 4
(Step 3) Inverse Laplace Transform
y t 8e
3t
2 cos 2 t 3 sin 2 t
26
2
( s 3)( s 4)
s 3
26 2
13
26
2
2
( s 3)( s 4 ) s 3
2 s 2 1 8 22 s 6
( s 3)( s 4 )
s 4
2
451
Example 5 (text page 286)
y ( t ) 3 y ( t ) 2 y ( t ) e
(Step 1)
Laplace
(Step 2)
Decompose
4 t
y (0) 1
y (0 ) 5
快速法
( s 3 s 2)Y s s 2
1
s4
1
Y s 2 s 2
2
s 3 s 2 ( s 3 s 2 )( s 4 )
2
s2
1
( s 1)( s 2 ) ( s 1)( s 2 )( s 4 )
3 4 1 1 1 1 1 1
s 1 s 2 5 s 1 6 s 2 30 s 4
16 1 25 1 1 1
5 s 1 6 s 2 30 s 4
(Step 3)
Inverse
t
2t
4 t
y t 16 e 25 e 1 e
5
6
30
452
7-2-6 快速法
k
d sk
k
dt
(A) 求 P(s)
y ( t ) 3 y ( t ) 2 y ( t ) P s s 3 s 2
2
很像 Sec. 4-3 的…
(B) 求 Q(s)
Q s a n s
n 1
a n 1
y 0 s
s
n2
y0
sy
n2
y 0
sy
(n2)
( n3)
0 y ( n 1) 0
0 y ( n 2 ) 0
a2
a1
sy 0
y 0
y 0
s
an
n 1
y 0
s
n2
y 0
1
0
( n3)
y
0
0
(n2)
y
0
y 0
y 0
y
y 0
a n 1
s
a2
(n2)
y
( n 1)
y 0
a1
相加
例如，page 451的例子
1
3
s
1
s
1
1
5
1
5
3
1
s
2
Q(s)
453
454
7-2-7 Section 7.2 需要注意的地方
(1) 熟悉分數分解
(2) 可以簡化運算的方法，能學則學
鼓勵各位同學多發揮創意，多多研究能簡化計算的快速法
數學上…….並沒有標準解法的存在
(3) Derivative 公式 initial conditions 的順序別弄反
L f
(n)
t s n F s s n 1 f 0 s n 2 f 0
sf
(n2)
0 f ( n 1) 0
Section 7-3 Operational Properties I
介紹兩個可以簡化 Laplace transform 計算的重要性質
First Translation Theorem (translation for s)
L e
at
f t F s a
Second Translation Theorem (translation for t)
L f t a u t a e
as
F s
u(t): step function
(注意兩者之間的異同)
455
7-3-1 First Translation Theorem (Translation for s)
L e
at
f t F s a
Proof:
L e
at
f t
e
0
st
e
at
f ( t ) dt
e
0
( sa )t
f ( t ) dt F s a
456
7-3-1-1 Inverse of “Translation for s”
When f(t) is piecewise continuous and of exponential order
L
1
F s a e a t f t
(一對一)
註： Sections 7-3 和 7-4 其他的定理亦如此
457
458
7-3-1-2 Examples
Example 1 (text page 290)
(a) L e 5 t t 3 L t 3
3!4
s s5
s
s s5
6
4
( s 5)
(b) L e 2 t cos 4 t L cos 4 t
s s ( 2 )
s
s 16
2
s s 2
s2
2
( s 2) 1 6
459
Example 2 (text page 291)
2 ( s 3) 11
1
1
1
1
2
L
11
L
(a) L 1 2 s 52 L 1
2
2
s
3
(
s
3)
(
s
3)
(
s
3)
2e
(b)
L
1
3t
11 t e
3t
s / 2 5 / 3 L 1 s / 2 5 / 3 L 1 ( s 2) / 2 2 / 3
2
2
2
s 4s 6
(
s
2)
2
(
s
2)
2
1
1
2
2
s
2
1
L
L
2
2
2
3
( s 2) 2
( s 2) 2
2 t
1 e cos
2
2t
2 e 2 t sin
3
2t
460
Example 3 (text page 292)
y 6 y 9 y t e
2
1
2
6
17
2
2
17
12
2s
L t e
2
3t
L t
2
s s 3
5
23
s
s s 3
( s 6 s 9 )Y s 2 s 5
2
y (0) 2,
3t
2
3
( s 3)
2
3
( s 3)
y (0) 17
461
( s 6 s 9 )Y s 2 s 5
2
2
3
( s 3)
2 ( s 3) 1 1
2
2
Y s 2 s 52
5
2
5
( s 3)
( s 3)
( s 3)
( s 3)
4!
2
11
1
s 3 ( s 3) 2 1 2 ( s 3) 5
y t 2e
3t
11te
3t
4 3t
1 t e
12
462
7-3-2 Step Function
u(t): unit step function
u(t) = 1 for t > 0
u(t) = 0 for t < 0
t-axis
t=0
u(t−a)
u(t−a) = 1 for t > a
u(t−a) = 0 for t < a
t=a
The unit step function acts as a switch (開關).
t-axis
463
Any piecewise continuous function can be expressed as the
unit step function for t 0
Example 5 (text page 294)
20 t
f t
0
for 0 t < 5
for t > 5
f t 20t u t 20t u t 5
Fig. 7.3.5
464
In general,
h1 ( t )
f t
h2 ( t )
for 0 t < a
for t > a
f t h1 t u t ( h 2 t h1 t ) u t a
7-3-3 Second Translation Theorem (Translation for t)
L f t a u t a e
或
L g t u t a e
as
F s
a>0
L g t a
a>0
as
f(t-a)u(t-a)
465
466
Proof:
L f t a u t a
e
e
s ( t1 a )
0
as
e
0
e
0
f t1 dt1
s t1
f t1 d t1 e
st
f t a u t a dt
令 t1 = t − a
as
F s
e
a
st
f t a dt
1
Example 7(a) (text page 295) L
L
1
4t
1
e ,
s4
L
Example 8 (text page 295)
1
s 1 4 e
2 s
1 e 2 s e 4 ( t 2 )u t 2
s4
L co s t u ( t )
L cos( t ) L cos( t )
L cos t u ( t )
s
s 1
2
e
s
s
s 1
2
467
468
Example 9 (text page 296)
y y f t
y 0 5
for 0 t <
0
f t
3 cos t
f t 3 cos t u t
for t
L cos( t ) L cos( t )
L 3 cos t u ( t )
( s 1)Y s 5
Y s
3s
s 1
2
3s e s
2
s 1
5 3 1 1 s e s
s 1 2 s 1 s 2 1 s 2 1
e
s
s 1
2
s
s
Y s 5 3 1 21 2 s e
s 1 2 s 1 s 1 s 1
e
t
sin ( t ) co s( t )
y t 5e
t
( t )
3 e
sin( t ) cos( t ) u ( t )
2
5e
t
( t )
3 e
cos( t ) sin( t ) u ( t )
2
469
7-3-4 本節需要注意的地方
(1) 套用 “translation for t” 的公式時，
先將 input 變成 g( t + a ) 再作 Laplace transform
(例如 Example 7)
(2) Second translation theorem (translation for t) 當 a > 0 時才適用
(3) 套用公式時，注意「順序」
470
Section 7-4 Operational Properties II
471
7-4-1 Derivatives of Transforms
L t f ( t ) ( 1) d n F s
ds
n
n
n
比較：
L f
(n)
t s n F s s n 1 f 0 s n 2 f 0
微分
Laplace
乘 sn
乘 tn
Laplace
微分
sf
(n2)
0 f ( n 1) 0
472
Proof of the Theorem of Derivatives of Transforms:
F s
e
st
d F s d
ds
ds
n
f ( t ) dt
0
e
st
f ( t ) dt
0
n
d F s d
n
n
ds
ds
0
e
0
st
f ( t ) dt
d [ e st ] f ( t ) dt e st tf ( t ) dt L tf ( t )
0
ds
0
n
d [ e st ] f ( t ) dt
n
ds
L ( t ) f ( t ) ( 1) L t f ( t )
n
n
n
e
0
st
( t ) f (t ) d t
n
Example 1 (text page 302)
L sin kt
練習：為何
k
s k
2
2
L t sin kt
k
2 ks
L t sin kt d 2
ds s k 2 (s 2 k 2 )2
L t co s kt
s k
2
2
(s k )
2
2
2
473
474
7-4-2 Convolution (旋積)
Definition of convolution:
f t g t
f ( ) g ( t ) d
(標準定義)
* 代表 convolution
f t g t
t
0
f ( ) g ( t ) d
(課本關於 convolution 的定義)
When f(t) = 0 for t < 0 and g(t) = 0 for t < 0 ,
上方的式子可以簡化為下方的式子
Convolution 的物理意義 (重要)
當 y t
t
475
f ( ) g ( t ) d
0
Input f() 對 output y(t) 的影響為 g(t−)
g(t−) 只和 t 與 之間的差有關
Input f() 對 output y(t) 的影響，決定於 t 與 之間的差
例如： f() 是在 這個時間點上太陽照射到某個地方的熱量
g(t-) 可想像成是經過了t- 的時間之後，還未幅射回外太
熱量比例
y(t) 可想像成是溫度
476
7-4-3 Convolution Theorem
L f t g t L f t L g t F s G s
Multiplication
Convolution
Proof:
F sG s
e
s
f ( ) d
0
0
e
s ( )
e
s
g ( )d
0
f ( ) g ( ) d d
0
note (A) 令 t = +
見後頁說明
note (B )
見後頁說明
e
st
f ( ) g ( t ) dtd
0
e
0
st
t
[ f ( ) g ( t )d ] dt L f g
0
477
note (A)
定理：
dxdy
1
C dw dv
note (B)
積分範圍的改變：
w
x
C det
v
x
t
det
Fig. 7.4.1
w
y
v
y
t
1
det
0
1
1
1
478
Example 3 (text page 303)
L
t
0
e sin( t ) d L e sin t
t
1
1
s 1 s2 1
Example 4 (text page 304)
1
1 sin kt sin kt 1
L 2
2
2
2 2
k
(s k ) k
1
t
0
sin k sin k ( t ) d
479
7-4-4 Integration
L
t
0
f d
L f ( t ) 1
F (s)
s
(想成 “負一次微分”)
480
Example:
1
L
2
s ( s 1)
1
t
sin d cos t 1
0
Example:
L1
L1
di t
dt
di t
dt
Ri t
Ri t
Q t
C
Q 0
C
E t
1
C
L1 sI s L1i 0 R I s
t
i d
0
Q 0 1
C
E t
I s
1
L E t
s C s
481
7-4-5 Transform of a Periodic Function
Theorem 7.4.3 When f(t + T) = f(t)
then
Proof:
1
L f ( t )
sT
1 e
令 f1(t) = f (t)
f1(t) = 0
T
e
0
st
f t dt
when 0 t < T
otherwise
f (t) = f1(t) + f1(t − T) + f1(t − 2T) + f1(t − 3T) + …………….
= f1(t) + f1(t −T)u(t − T) + f1(t −2T) u(t −2T ) + f1(t −3T) u(t −3T )
+ …………….
L{f (t)} = L{f1(t)} + L{f1(t −T)u(t − T)} +L{ f1(t − 2T) u(t − 2T)}
482
+L{ f1(t − 3T) u(t − 3T)} + …………….
L f1 ( t )
T
e
0
st
f 1 t dt
L f1 ( t T ) u ( t T ) e
L f1 ( t )
sT
L f1 ( t 2T ) u ( t 2T ) e
L f1 ( t 3T ) u ( t 3T ) e
:
:
2 sT
L f1 ( t )
3 sT
L f1 ( t )
1
L f ( t )
sT
1 e
T
e
0
st
f t dt
483
Example 7 (text page 307)
Square Wave (方波) 的例子
1
E t
0
for 0 t < 1
for 1 t < 2
E t 2 E t
Fig. 7.4.4
484
1
st
1
L E ( t )
1
e
dt
2 s 0
1 e
1 e
1
2 s
s
1 e
s
2
1
0e
st
dt
1 e
1
s
s
s
(1 e )(1 e )
s
1
s
s (1 e )
485
Example 8 (text page 308)
L1 d i R i E ( t )
dt
i (0) 0
E(t) 為 page 483 之方波
sL1 I s L1i 0 R I s
I s
1
s
s (1 e )
1 / L1
( s R / L1 ) s (1 e
s
)
486
1/ L
I s
1 s 1/ R 1/ R
1 s
(s R / L)s 1 e
s
s R / L 1 e
s
2 s
3 s
1/ R 1/ R
1 e e
e
s
sR/L
長除法
1 1 x1 x 2 x 3
1 x
s
e
1
1
e
e
R s R/L s R/L s R/L s R/L
s
2 s
I s 1 1 e e
R s
s
s
e
3 s
s
L
1
1 1
s
L
1
2 s
s u t k
e
ks
k = 0, 1, 2, 3, …..
3 s
487
ks
ks
e
1
e
s R/L
s R/L
先使用 L
1
再算出 L 1
R
t
1
e L
s R/L
ks
的公式
R
(tk )
e
e L
u t k
s R/L
ks
註：雖然也可以用
e
e
s R/L
但是較麻煩且容易出錯
k (sR / L )
s
s s R / L
來算
7-4-6 Section 7.4 要注意的地方
(1) 注意代公式的順序 (例：Page 487 例子)
(2) 熟悉 convolution
(3) 變成積分時，別忘了加上 initial value
t
如 Q t Q 0 1 i d
0
C
C
C
(4) 一定要記熟幾個重要的 properties (7 大性質)
488
Section 7.5 The Dirac Delta Function
7-5-1 Unit Impulse
a t t0
0
1
2 a
for t < t0−a or t > t0+a
for t0−a t t0+a
稱作 unit impulse
高= 1/2a
面積 = 1
t0−a
t0+a
t-axis
489
490
7-5-2 Dirac Delta Function
t t 0 lim a t t 0
a 0
a t t0
0
Fig. 7-5-2
for t = t0
for t t0
7-5-3 Properties of the Dirac Delta Function
(1) Integration
(2) Sifting
Proof:
q
p
q
p
t t 0 dt 1
f t t t 0 dt f t 0
f t t t 0 dt lim
a 0
f t 0 lim
a 0
q
p
q
p
when t0 [p, q]
f t a t t 0 dt
a t t 0 dt f t 0
當 a 很小的時候，f(t) f(t0)
for t0−a t t0+a
491
(3) Laplace transform of (t − t0)
L t t 0 e
Proof:
e
0
st
t0 s
when t0 > 0
( t t 0 ) dt
(4) Relation with the unit step function
t
t 0 d u t t 0
d u t t t
0
0
dt
492
493
7-5-4 Example
Example 1(a) (text page 313)
y y 4 t 2
s Y s s Y s 4e
2
2 s
2 s
Y s 2 s 4 e2
s 1
s 1
L
y t cos t 4 sin( t 2 ) u ( t 2 )
cos t 4 sin t u ( t 2 )
cos t
y t
cos t 4 sin t
y (0) 0
y (0) 1
0 t < 2
t 2
1
1
sin t
2
s 1
494
7-5-5 幾個名詞
P s Y s Q s G s
Y s W sQ s W sG s
where
W s
1
P s
(1) w(t) = L−1{W(s)} 稱作 weight function 或 impulse response
Note: When Q(s) = 0 (no initial condition) and G(s) = 1 (g(t) = (t)),
Y(s) = W(s), y(t) = w(t).
(2) 許多文獻把 Dirac delta function (t − t0) 亦稱作 delta function ，
impulse function，或 unit impulse function
495
7-5-6 本節要注意的地方
(1) Dirac delta function 不滿足 Theorem 7.1.3
(2) 幾個定理記熟，本節即可應付自如
Section 7-6 Systems of Linear Differential
Equations
Chapter 7 的應用題
比較：類似的問題，也曾經在 Section 4-8 出現過
7-6-1 雙彈簧的例子
496
497
7-6-2 電路學的例子
L
di1 t
dt
q3
C
i2 R 2 E t
i2 R
i1 i2 i3
(由第2, 3 個式子)
i1 i2 d q 3 i2 d R C i2
dt
dt
di1 t
i2 R 2 E t
Fig. 3-3-4
L
Fig. 7-6-2
R C d i2 i2 i1 0
dt
dt
L
di1 t
dt
498
i2 R 2 E t
R C d i2 i2 i1 0
dt
Example 2 (text page 317)
E(t) = 60 V, L = 1 h, R = 50 , C = 10−4 f, i1(t) = i2(t) = 0
di1 t
dt
50 i2 60
0.005 d i2 i2 i1 0
dt
sI 1 s 50 I 2 s 60
s
……….. (式1)
I 1 s (0.005 s 1) I 2 s 0 …...
(式1) × 1 + (式2) × s
2
(0.005 s s 50) I 2 s 60
s
( s 200 s 10000) I 2 s 12000
s
2
(式2)
120
6/5
I 2 s 12000 2 6 / 5
2
s
s 100
s ( s 100 )
( s 100 )
複習：分子如何算出？
100 t
100 t
i2 t 6 120 te
6e
5
5
將 I2 s 6 / 5
s
120
6/5
2
s 100
(s 100)
代入式 (1)
sI 1 s 6 0 6 0 6 0 0 0 2 6 0
s
s
s 100
(s 100)
I1 s
6000
60
2
s(s 100)
s(s 100)
499
b c ( s 100)
6000
60
a
I1 s
2
2
s
(
s
100)
s
s ( s 100)
( s 100)
a
6000
60
2
( s 100)
( s 100)
60( s 100)
b 6000
s
s
s0
500
6
5
60
s 100
60( s 100)
c d ( 6000
)
6000
6000
2
2
ds
s
s
s
s
s 100
60
6/5
I1 s 6 / 5
2
s
s 100
(s 100)
100 t
100 t
i1 t 6 60 te
6e
5
5
s 100
6
5
501
7-6-3 Double Pendulum (雙單擺) 的例子
( m1 m 2 ) l1 1 m 2 l1l 2 2 ( m1 m 2 ) l1 g 1 0
2
m 2 l 2 2 m 2 l1l 2 1 m 2 l 2 g 2 0
2
( m1 m 2 ) l 1 m 2 l1l 2 2 ( m1 m 2 ) l1 g 1 0
2
1
m 2 l 2 2 m 2 l1l 2 1 m 2 l 2 g 2 0
2
Example 3 (text page 318)
m1 = 3, m2 = 1, l1 = 12 = 16, 1(0) = 1, 2(0) = 1,
'1(0) = 0, '2(0) = 0
1024 1 256 2 64 g 1 0
16 1 4 2 g 1 0
256 1 256 2 16 g 2 0
16 1 16 2 g 2 0
Laplace
16 s 1 s 4 s 2 s g 1 s 16 s 4 s 12 s
2
2
16 s 1 s 16 s 2 s g 2 s 16 s 16 s 0
2
2
502
(16 s g ) 1 s 4 s 2 s 12 s
………….. (式1)
16 s 1 s (16 s g ) 2 s 0
………….. (式2)
2
2
2
2
503
(式1) × (16s2 + g) (式2) × 4s2
[(16 s g ) 64 s ] 1 s 12 s (16 s g )
2
2
4
2
[192 s 32 s g g ] 1 s 12 s (16 s g )
4
1 s
2
2
2
12 s (16 s g )
2
192 s 32 s g g
4
2
192 s 12 gs
3
2
(24 s g )(8 s g )
2
2
as b cs d
2
2
24 s g 8s g
(8 a 24 c ) s (8 b 24 d ) s ( a c ) gs bg dg 192 s 12 gs
3
8 a 24 c 192
a c 12
2
a 6
c 6
3
8 b 24 b 0
bd 0
b 0
d 0
6s
6s
s
1 s
6 2
6 2 s
2
2
2 4 s g 8 s g 24 s ( g / 24 ) 8 s ( g / 8)
g 3
g
1
1 t cos
t cos
t
4
4
2 6
2 2
將
1 s
12 s (16 s g )
2
192 s 32 s g g
4
2
2
代入 (式2)
192 s
as b cs d
2 s 162 s
1 s
2
2
2
2
16 s g
(24 s g )(8 s g ) 24 s g 8 s g
2
3
直接用之前的式子
8 a 24 c 192
ac0
a 12
c 12
8 b 24 b 0
bd 0
b 0
d 0
504
505
2 s
12 s
s
122 s 1 2 2
12 2 s
2
2 4 s g 8 s g 2 4 s ( g / 2 4 ) 8 s ( g / 8)
g 3
g
1
2 t cos
t cos
t
2
2
2 6
2 2
分式分解快速驗算技巧
將 s = 0, s = 1, 或其他的值代入，看等號是否成立
506
7-6-4 本節需要注意的地方
(1) 正負號勿寫錯
(2) 要熟悉聯立方程式的變數消去法
(3) 多學習，甚至多「研發」簡化計算的技巧
507
Review of Chapter 7
(1) Laplace transform 定義
F s
e
st
f ( t ) dt
0
Inverse Laplace transform
If
f ( t ) F s
L aplace
and f(t) is piecewise continuous
then F ( s ) f t
inverse L aplace
(2) 7 大transform pairs
(看講義 page 420)
of exponential order
508
transform pairs 補充
509
f(t)
F(s)
2 ks
2
2 2
(s k )
2
2
s k
2
2 2
(s k )
2 ks
2
2 2
(s k )
2
2
s k
2
2 2
(s k )
tsin(kt)
tcos(kt)
tsinh(kt)
tcosh(kt)
u(ta)
e
as
t
1
0
a
/s
1
0
b
f (t) = f (t+2a)
e
as
e
s
bs
1
as
s (1 e )
510
(3) 7 大 properties
input
(1) Differentiation (Sec 7-2)
(n)
f
t
(2) Multiplication by t (Sec7-4)
n
t f (t )
(3) Integration (Sec 7-4)
t
0
f d
(續)
Laplace transform
s F s s
n
sf
(n2)
n 1
f 0 s
0 f ( n 1) 0
n
( 1) d n F s
ds
n
n2
F (s)
s
f 0
511
input
Laplace transform
(4) Multiplication by exp (Sec7-3)
F s a
e f t
at
(5.1) Translation (Sec 7-3)
e
f t a u t a
(5.2) Translation (Sec 7-3)
g t u t a
(6) Convolution (Sec 7-4)
t
f ( ) g ( t ) d
e
as
as
F s
L g t a
F sG s
0
(7) Periodic Input (Sec 7-4)
f(t) = f(t + T)
1
sT
1 e
T
0
e
st
f t dt
512
Properties 補充
input
Laplace transform
Scaling
aF(as)
f(t / a)
Multiple Integrations
t
n
0
0
3
2
0
0
f 1 d 1 d 2
Integration for s
f(t) / t
F (s)
s
d n 1 d n
s
n
F s1 ds1
(4) 簡化運算的方法
分式分解 (see pages 439-444)
Initial conditions (see pages 452, 453)
(5) Delta function 的四大性質
Pages 491, 492
513
514
(6) General solutions
Laplace transform 的 general solution，可以用 initial conditions
來表示。
例子： f t 4 f t 0
用Sec. 4-3 的
方法解出
f t c1e
2t
用 Laplace transform ：
2
s F s sf 0 f 0 4 F s 0
F s
sf 0 f 0
s 4
2
f t f 0 cosh 2 t
f 0 2
s
f 0 2
2 s2 4
s 4
f 0
2
sinh 2 t
c2e
2 t
515
和 Section 4-3 的解互相比較
f t
2 f 0 f 0
4
e
2t
2 f 0 f 0
4
e
2 t
f 0 c1 c 2
f 0 2 c1 2 c 2
將
f 0 c1 c 2
f t c1e
2t
c2e
f 0 2 c1 2 c 2
2 t
代入
516
Exercise for Practice
Sec. 7-1:
5, 9, 10, 12, 30, 33, 34, 38, 41, 50, 55, 56
Sec. 7-2:
11, 18, 23, 27, 28, 30, 37, 38, 41, 42, 43
Sec. 7-3:
10, 12, 16, 19, 26, 34, 35, 42, 56, 58, 62, 64, 70, 74, 83
Sec. 7-4:
2, 13, 24, 25, 26, 38, 40, 43, 52, 53, 54, 59, 61, 63, 66
Sec. 7-5:
2, 5, 6, 11, 12, 15
Sec. 7-6:
6, 10, 11, 14, 15
Review 7:
12, 24, 25, 29, 36, 39, 42, 43, 44