ページ1:

Laplace transform
1+ / 02 / 2565
INTRODUCTION
f(r)
L(f)
f(1)
L(f)
"Oliver Heaviside" was an English
mathematician and physicist
s
1
1
1/s
7
cos of
Laplace transforms are invaluable for any
engineer's mathematical toolbox as they make
solving linear ODEs and related initial value
problems, as well as systems of linear ODEs, much
easier.
S
Applications: electrical networks, springs, mixing
problems, signal processing, and other areas of
engineering and physics.
C
HE
R
2
ww
Driving force
Dashpot (damping)
I
1/32
w
8
sin of
3
2
21/53
9
cosh at
52 – 2
e
10
sinh at
2-a2
4
(n-0, 1,-)
+1
ra
T(a+1)
5
11
=
(a positive)
@+1
6
sa
12
s-a
cos et
(s-a)²+2
w
12
et sin of
(s-a)2+2
2
PROPERTIES OF LAPLACE TRANSFORM
0 (Equilibrium
1. Linear operator
L{of, (t)+bf,(t)} = al { f,(t)} + b² {f,{e}}
position)
Q
Where a, b is constant coefficients
f(t)=4t2-3cos(t)+5e
L{4t
2-3 cost+5e"} = 4L {t2}-3L {cost} +5L {e}
8
F(s) =
3s
s+1
5
s+1
DEFINITION OF THE LAPLACE TRANSFORM
Laplace transforms is a mathematical operation that
is used to "transform" a variable (such as x, or y, or z,
or t) to a parameter (s)- transform ONE variable at a
time. Mathematically, it can be expressed as:
f(t)
function
F(s)
function
-st
L{f(t)} = F(s) = =Se™ f(t)dt
0
2. Shifting property L{1,(t)}=L{e"ƒ,(t)} = F(s-a)
f(t) = et sin(at)
Where a is the shifting factor
➡L{f(t)}=L{e sin(at)}
3. Change of scale property {f(at)} = F
+0
f(t)=sin(3t)
a
F(s)=
(s-2)²+a2
Where a is scale factor for the change
L{f(t)}=L{sin(3t)} F(s)=1.
1
3
s²+9
+1
Examples
1 f(t) 1,t≥0
L₤1] =
=
=
2 f(t) t2,t≥0
=
f(t) eat,t≥0
Lɛ뺪¡ · Ïëˆê'¡ dt - je
f(t) coswt,t≥0
(-stat
.
-st
Lfcoswt = cos wit
43
t≥o
11 / 02 / 2565
INVERSE LAPLACE TRANSFORM
(570)
From here
=
(570)
(-s+a)
Laplace transform
00
-st
to here
L{f(t)} = √e¯* f(t)dt = F(s)
0
S
(579)
to here
s-a
-st
e
(-scosit + wsin wt)
(-5)²+10²
I
(570)
Inverse Laplace transform
From here
80
-st
L{f(t)} = √e* f(t)dt = F(s)
0

ページ2:

Laplace transform
Inverse Laplace transform L'FIS?
Step 4
-t
fit) =
case 3
-2+
2e2te20
Complex or Imaginary
-2t
SOLUTION OF INITIAL VALUE PROBLEM ODES
To solve initial value problems for linear differential equations
with constant coefficients
F(5) =
2 < (4)(4)(5) => complex
mm
5(5²+25+5)
S
step 1
patial
fraction
F(S) = 3
= k1
+
5(53725+5)
K25 + k>
52+23+5
①
Driving force
Dashpot (damping)
2
step 2
And k₁ K2 K3
900;
K₁ = ;
3
3
=
=
K₁ =
k1 (5²+25+5)+(k₂s + ₤3) (5)
3 =
5k1
Differential equation y(t)
L
Initial value
Algebraic equation Y(s)
Solve
y=0 (Equilibrium
position)
Differential equation y(t)
Initial value
=
3 =
0 =
(5+25+5)+(25 +375
35 +5 +3
+ K25 +135
(+2)² + (+3)5 +3
(3+k²)² + (b +ks)5
Algebraic equation Y(s)
3+k2
K₂ = -3
K1 K2 K3 to ①
step 3
F61=
F(57=
S
(5+2)
5²+25+5
3
[(5+1)+()(2)]
-
5
5+17²+2²
2
L-1
Y(S)
y(t)
Solution
(n-1)
aytay
++ay" +ay'+ay=r(t)
when a, a, a, and a, are the constant
y(0)=K₁,y'(0) =K₁‚y”(0)=K₂, y²
(n-1) (0) = K
aL{y}+{y}+_+₁₂L{y"}+a‚L{y}+a¸L{y} = L{r(t)}
The derivatives of y't)
(2)
({y}{y(t)}-y(0)-sy (0)-(0)
The first and second derivatives of y(t)
(2)
L{y}={y(t)}-sy(0) -y (0)
(3)
Lyst{y(t))-y(0)
STEPS OF THE LAPLACE TRANSFORM METHOD
t-space
Given problem
y"-y=t
y(0)=1
y'(0) =1
s-space
Subsidiary equation
(82-1)Y 8+1+1/82
step 4
Inverse Laplace transfrom LF(S)
f(t)=
Solution of given problem
y(t) = a + sinh tật
Solution of subsidiary equation
Y
L{y(t)) =Y(s)
S

ページ3:

Laplace transform
PARTIAL FRACTION METHOD
The expression of F(s) to be inversed should be in partial
fractions as:
Examples
case 1
Real and Distinct
P(s)
F(s) =
Q(s)
F(S) =
(5+1)(5+2)
step 1
partial fraction
*** where polynomial P(s) is at least one order less than
the order of polynomial Q(s)
F (S) =
2
K1
ke
(5+1)(5+2)
(5+17
(5+2)
S
F(s)=
s² + 2s² + 6s+7
s²+s+5
P(s)
bs+bs
m-1
+...+b₁s+b₁₂
steps Find ka and k₂
F(s)=
2
Q(s)
as tas
+...+as+a₁
2
m≤n
= (K1+k² ) (5+1) (S+2).
5+1 5+2
2
P(s)
A₁
A₂
A
F(s) =(s+1)+
F(s)=
+
+.... +
s²+s+5
Q(s) s-a₁ S-a,
s-a
S = -1 j
2
==
x-3x+7
0 – 16
Proper
x-5
Not proper
www.
2x²+7' x 2
x-5
- งของ 2 ส่วน
Jurino
สวยอะค่ะ ส่วน
=
ยู
ใช้ธนบ
เยาว
+10
X-52+0x+
ง
=
K₁(S+2)+ k₂CS+1)
K16-1+27
k1
=
2
5= -2
2
K2(-2+1)
k2
=
-2
Step 3
k and k₂ to ①
FC57 =
2
5+1
(-2)
5+2
ถ้ากําลังของเศษมากกว่าสาน
เราต้องน่าไปหารยาวก่อน
0 + 1 + 1 คน เข้าใช้ ไม่ได้ ตอนที่ 15
ไป
จน
step 4
F(t)
=
2ce7
f(+1
=
Inverse Laplace tranform [F(S)
(-27(02)
case 2
Real and Repeated
P(s)
L'[F(s)]=L'
=[¹
+L₁
Ą₂
A
2+...+[1
F(57
2
Q(s)
s-a₁
s-a
(5+1)(5+272
step 1
Patial fraction
Case 1. Roots of the Denominator of F (s) are Real and Distinct
F($)
2
k1
N(s)
F(s) =
(s + p₁ ) (s + p₂ ) ... (s +p₁)
step 2
(5+1715+232
S+1
+ kz
(5+212
+ k3
S+2
เปิด รา
2
Case 2. Roots of the Denominator of F (s) are Real and Repeated
F(s)=
N(s)
(s+p)
Case 3. Roots of the Denominator of F (s) are Complex or Imaginary
2
=
5=-1;
2
find ka kz k3
=K₁(-1+23
+ K₂(5+1) + K3 (5+1) (5+2)
K₁
< 2
5=-2;
2 ==
K2(-2+1)
N(s)
s₁ = -σ±jw
F(s) =
s(s² +as+c)
as+bs+c
b² 4 ac
K₂
4
-2
S=1, k₁ = 2, K2 = -2 ;
2
= 2(1+2)+ (-27(1+1) + k3 (1+1) (1+2)
2
= 18 + (-4) + 6k3
K₂ • -2
step 3 K1 K2 K3
to ①
F(5)
= 2
5+1
+(-2) +(-2)
(5+272
(5+2)
S

ページ4:

Laplace transform
y" + ay' + by = rlt) when y(0) = ko, y'(0) = k₁
step 1 setting up the subsidiary equation
y"+ y = sin (2t)
9107=29'107 =
= 1
(Y(S) 5² - 54(0) - 4' (0)) + 9(Y(STS-420)) + b (Yes)) = R(S)
(s+ as + b) Y(s) (s+a) y(0) - y' (0) = R(S)
(s²+as+b)Y(S)
= (s+α) y(0) + y' (o) + RS
Step2 solution of the subsidiary equation by algebra
LEY"}+
+4{4}
S² YES? SY(0) 4'10) + Y(S)
-
S
step 3
=
L{ sin (2+)}
2
✓
5²+4
2
Y(S)
=
(sta)go + y'co)
5²+4
5+95 +6
RCS)
5²+as+b
(5+174257
-25-1 = 2
transfer function
5274
Inverse Laplace Transfrom YES? Yet?
(5²+974(57
=
2 + 25 +1
5²+4
51541573
+
=
+
5795+6
s²+as+b
(5²+174(57
=
2
+
23(3+1+1+4)
5²+4
Examples
YCS7=
= 0
y" - y' - 2y
= 0 410)=1 9'607=0
step 1 setting up the subsidiary equation
|(5² (Y(S))-5(Y (0) - y' (07) - ((57 Y(S)-4(0)) - 2 (Y(57)
(5-5-2) YCS) + (1-5)410)-4 (0)
(5-5-2)4157
= 0
=
-(1-5)4(0) + 4'10)
4057
YEST
2+(25+1)(5+47
(8²+4)(S²+17
=
25+5²+85+6
(54+4)(5+1)
32
25+5+89+6
(5²+4)(5²+17
253+5785+6
283+$²+85+6
25++85+1
= AS+B +C+D
$2+4
5²+1
= (A≤+BX$²+1) +(CS+D) (5²+4)
= AS³+AS+ Bs²+B+ ¿s³+ ACS + DS² + 4D
= (A+)s³² + (B+D)² + (A+4d)S+B+4D
step2 solution of the subsidiary equation by algebra
17
25³ = (A+)5³
31
85 (A+adis
Y(S)
=
-(1-5)4(0) 4'10)
Atd 23
A+AC 8 (3)
52-5-2
-(1-5)(1) +
2)
$² (B+P)s
4)
B+ 4D (4)
0
B+D = 1-2)
52-5-2
Y(s)
=
S-1
(5-2715+17
A = 0
B =
2=2
D =
Step 3 Inverse Laplace Transfrom YCS) Yet?
Y(5) =
0+(-1) +
5²+4
25+ ½
Y(S) =
5-1
(5-2)(5+1)
A
+
B
5²+1
(5-2)
(5+1)
2
YC57
=
1(4) + 2(1)+3(4)
5²+1
5-1
= (S+1)A + (5-2)B
S= -1 to 2
= (-1-2)B
B
=
S= 2 to 1
j
2-1
= (2+1)A
A
=
then
Y(S7
=
Yct)
(5-2)
+
(5+1)
2t
-t
=
+
e
Ans
=
+2
=
f
- ½ sin(2t) + 2 cos(t) + sin(t)
37
Y(t)
=
- ½ sin(2t) + cos(t) +
sin(t)
2
S

ページ5:

S
2
y
Laplace transform
.(4)
y =
0
; y(07= 0, y'10>= 1, y "10700, y" 107=0
6241414-4243
= 0
54 4157-534(07-5² 407-54" (0) - 4" (107 - YEST
(54-9745-5407-517-5807-0
(54-174(5)-52 = 0
YC57
= 0
1354-17
0
4(5) = 5²
(52+1753-11
Y(S)
(52+97152-17
AS+B
+
5²+1
ds +D
5²-1
§²
=
=
05²+ 5² + 03+0
=
11
Os³ = (A+)5³
A+ C = 0― (17
(AS+B)(s²-17 + (ds +D) (5²+1)
As²- AS + Bs²=B +
253 +25+ DS40
(A +ε 753³+ (B+D75² + (-A+C75-B+D
37 OS = (-A+C)s
-A+C = 0 (3)
2
27
$²= (B+D)S
42
-B+D
-(4)
B+D=1(2)
Ann (1), (2), (3), (4) Dilo A=0
Y(57
=
05+2
5²+1
+
B = 1/2 c = 0 D = 1
Os+
S²-1
1 (1)
1
YC57
53341573
Ycti
=
=
=
111)
2151
sin(t)
+
+
+
sinht
2
S

ページ6:

Laplace transform
TABLE 6.2.1
Elementary Laplace Transforms
f(t) = L−¹ {F(s)}
F(s) = L {f(t)}
Notes
1
1. 1
s> 0
Sec. 6.1; Ex. 4
1
2. eat
S
sa
Sec. 6.1; Ex. 5
s-a
2
n!
3. ", na positive integer
s>0
Sec. 6.1; Prob. 24
1+45
2
4. tp, p>-1
5. sin(at)
6. cos(at)
7. sinh(at)
8. cosh(at)
9. eat sin(bt)
10. eat cos(bt)
г(p+1)
s>0
SP+1
Sec. 6.1; Prob. 24
a
s>0
s² + a²°
Sec. 6.1; Ex. 7
s>0
Sec. 6.1; Prob. 5
a
s>|a|
Sec. 6.1; Prob. 7
$2
a²
S
$2 a²'
s>|a|
Sec. 6.1; Prob. 6
b
(s-a)²+b2'
s-a
(s-a)²+b2'
s> a
Sec. 6.1; Prob. 10
s> a
Sec. 6.1; Prob. 11
11. tea, na positive integer
n!
(s-
a)"+1"
s> a
Sec. 6.1; Prob. 14
12. u(t)=
0 I<c
e-cs
(1 1 2 c
s>0
Sec. 6.3
S
13. u(t)f(tc)
14. et f(t)
15. f(ct)
16. (f*g)(t) = f f(1-T)g(T) dT
17. 8(tc)
18. f(") (t)
19. (-1)" f(t)
e-cs F(s)
Sec. 6.3
F(s-c)
Sec. 6.3
S
LF (²),
c> 0
Sec. 6.3; Prob. 17
F(s)G(s)
e-cs
Sec. 6.6
Sec. 6.5
s" F(s)-s" f(0) f(n-1) (0) Sec. 6.2; Cor. 6.2.2
F(n) (s)
Sec. 6.2; Prob. 21