Laplace Transforms
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Transcript of Laplace Transforms
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Laplace Transforms
1. Standard notation in dynamics and control (shorthand notation)
2. Converts mathematics to algebraic operations
3. Advantageous for block diagram analysis
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0
)( dtf (t)e=tf -stL
Laplace Transform
Example 1:Example 1:-
00
- - - -( ) ( )
00 0
-
0
( ) 0
1 1( ) -
( ) 0
st st
bt bt st b s t b s t
st
a a aa ae dt e
s s s
e e e dt e dt eb s s b
df dff e dt s (f) f( )
dt dt
L
L
L L L
Usually define f(0) = 0 (e.g., the error)
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?
00
00
0
where
2
2
2
=dt
fd
fsfsF= s
ffssF= s
ssΦ
dt
df φ
dt
dφ=
dt
fd
n
n
L
LL
![Page 4: Laplace Transforms](https://reader033.fdocuments.in/reader033/viewer/2022061612/568138af550346895da06e12/html5/thumbnails/4.jpg)
22
22222
111
2
1
2cos
ωs
s=
ωs
jωs
ωs
jωs=
jωsjωs=
ee=t
tjt-j
LL
Note that
cos sin
cos sin
1
j t
jwt
e t j t
e t j t
j
![Page 5: Laplace Transforms](https://reader033.fdocuments.in/reader033/viewer/2022061612/568138af550346895da06e12/html5/thumbnails/5.jpg)
2 2
sin2
j t j te et
j
s
L L
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Unit Step Function:
0 0
1 0
1
tS t
t
S ts
L
![Page 7: Laplace Transforms](https://reader033.fdocuments.in/reader033/viewer/2022061612/568138af550346895da06e12/html5/thumbnails/7.jpg)
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apte
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Difference of two step inputs S(t) – S(t-1)
Note that S(t-1) is the step starting at t = 1.
By Laplace transform
1( )
seF s
s s
Can be generalized to steps of different magnitudes(a1, a2).
![Page 8: Laplace Transforms](https://reader033.fdocuments.in/reader033/viewer/2022061612/568138af550346895da06e12/html5/thumbnails/8.jpg)
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0
1 1( ) (1 )
h st hsF s e dt eh hs
Let h→0, f(t) = δ(t) (Dirac delta)
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0
0 0
Impulse Function: t
11
1
1
lim
lim lim
hs
h
hs hs
h h
t ehs
e se
hs s
tL
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Laplace transforms can be used in process control for:
1. Solution of differential equations (linear)
2. Analysis of linear control systems (frequency response)
3. Prediction of transient response for different inputs
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Please see Table 3.1 in Text
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Example 3.1
Solve the ODE,
5 4 2 0 1 (3-26)dy
y ydt
First, take L of both sides of (3-26),
25 1 4sY s Y s
s
Rearrange,
5 2
(3-34)5 4
sY s
s s
Take L-1,
1 5 2
5 4
sy t
s s
L
From Table 3.1,
0.80.5 0.5 (3-37)ty t e
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Example 2:Example 2:
system at rest (s.s.)
Step 1 Take L.T. (note zero initial conditions)
00
0000
2461162
2
3
3
at t=dt
du
)=(y)=(y)=y(
udt
duy
dt
dy
dt
yd
dt
yd
3 26 11 6 ( ) 4 2s Y(s)+ s Y(s)+ sY(s) Y s = sU(s) + U(s)
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To find transient response for u(t) = unit step at t > 0
1. Take Laplace Transform (L.T.)2. Factor, use partial fraction decomposition3. Take inverse L.T.
![Page 15: Laplace Transforms](https://reader033.fdocuments.in/reader033/viewer/2022061612/568138af550346895da06e12/html5/thumbnails/15.jpg)
Rearranging,
Step 2a. Factor denominator of Y(s)
Step 2b. Use partial fraction decomposition
Multiply by s, set s = 0
input) (1
1
6116
2423
stepunits
U(s)
ssss
s+Y(s)=
))(s+)(s+)=s(s+s++s+s(s 3216116 23
321321
24 4321
s
α
s
α
s
α
s
α
))(s+)(s+s(s+
s+
3
1
321
2
321321
24
1
0
4321
0
α
s
α
s
α
s
αsα
))(s+)(s+(s+
s+
ss
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For 2, multiply by (s+1), set s=-1 (same procedurefor 3, 4)
3
531 432 , α, αα
3
13
53
3
1 32
) y(tt
eeey(t)= ttt
Step 3. Take inverse of L.T.
You can use this method on any order of ODE, limited only by factoring of denominator polynomial(characteristic equation)
Must use modified procedure for repeated roots, imaginary roots
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)3
3/5
2
3
1
1
3
1(
ssssY(s)=
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One other useful feature of the Laplace transform is that one can analyze the denominator of the transform to determine its dynamic behavior. For example, if
the denominator can be factored into (s+2)(s+1).Using the partial fraction technique
The step response of the process will have exponential terms e-2t and e-t, which indicates y(t) approaches zero. However, if
We know that the system is unstable and has a transient response involving e2t and e-t. e2t is unbounded for large time. We shall use this concept later in the analysis of feedback system stability.
23
12 ss
Y(s)=
1221
s
α
s
αY(s)=
))(s(sssY(s)=
21
1
2
12
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Properties of Laplace Transform
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Linearity
where and are constants.
af t bg t
a f t g t
aF s bG s
a b
L
L +bL
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Sifting Theorems
0 0
0 0
-
(a) Dead time
( )
(b) Multiplication by
t s t s
bt
bt
g t f t t S t t
g t e f t e F s
e
e f t F s b
L L
L
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Final Value Theorem“offset”
Example 3: step responseExample 3: step response
offset (steady state error) is a.
sY(s))=y(s 0lim
aτs
a
τs
asY(s)
s
a
τsY(s)
s
1lim
1
1
1
0
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Initial Value Theorem
by initial value theorem
by final value theorem
sY(s)y(t)=st limlim
0
))(s+)(s+s(s+
s+=For Y(s)
321
24
00 )=y(
3
1)=y(C
hap
ter
3
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Transform of Time Integration
0
t F sf d
s L
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Differentiation of F(s)
1n
n nn
d F st f t
ds L
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Integration of F(s)
s
f tF d
t
L