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Unformatted text preview: 1. Standard notation in dynamics and control
(shorthand notation) = how you communicate
with other engineers.
other engineers
2. Converts differential calculus mathematics to
algebraic operations, very useful for:
Initial Value Problems (IVPs). Chapter 3 Chapter 3 Laplace Transforms For “simple”
simple
Y(s)  maybe
a short cut ! 3. Advantageous for block diagram analysis = a
“systems” way to analyze more complex systems.
(study subsystems, then combine to large system)
subsystems then combine to large system) Other Transforms Laplace Transform Definition ⎡d2 f
L⎢ 2
⎣ dt ∞ L [f (t )]= ∫ f (t )e st dt = F ( s )
0 = s [ sF ( s )  f (0) ] − f ′(0) ∞
a
⎡ a⎤ a
L [a]= ∫ ae dt = − ⎡e −st ⎤ = 0 − ⎢ − ⎥ =
⎣ ⎦0
s
⎣ s⎦ s
0
st ∞ ∞ 0 0 L [ebt ]= ∫ ebt est dt = ∫ e(b+s)t dt = ∞
1
1
⎡ − (b +s)t ⎦ =
⎤
⎣e
0
b+s
s+b
∞ df
⎡ df ⎤
and a theorem ... L [f ′ ] = L ⎢ ⎥ = ∫ est dt = sF ( s ) − f (t = 0)
⎣ dt ⎦ 0 dt Very convenient for zero initial condition f(0) = 0
An example is when f = the “error” in a process control IVP,
example is when
the error
process control IVP
i.e at t = 0 the process is at a “nominal design” steadystate. Chapter 3 Chapte 3 Two derivations (“a” is a constant, f ’ is derivative wrt time):
(“a”
∞ ⎤
df
⎡ dg ⎤
where we assign
⎥ = L ⎢ dt ⎥ where we assign g = dt = f ′(t )
⎣⎦
⎦
= sG ( s )  g (0)
= s 2 F ( s ) − sf (0) − f ′(0) n
etc. for higher orders d f
n dt
⎡ e jωt + e + jωt ⎤
L [cos (ωt )] = L ⎢
⎥
2
⎣
⎦
1⎛ 1
1⎞
=⎜
+
⎟
2 ⎝ s + jω s − jω ⎠
1 ⎛ s − jω s + jω ⎞
= ⎜2
+
⎟
2 ⎝ s + ω 2 s2 + ω 2 ⎠
s
=2
s + ω2 ⎡ e + jωt  e − jωt ⎤
L [sin (ωt )] = L ⎢
⎥
2j
⎣
⎦
=
Note: ω
s + ω2
2 e jωt = cos ωt + j sin ωt
e − jwt = cos ωt − j sin ωt
j = −1 2. Examples 2
Numerator( s )
=
3s + 4s + 1 Quadratic Denominator( s ) b2 − 4ac = 16 − 12 = 4 > 0 , with a > 0, b < 0
so this is an overdamped system = 2 exponential decays
negative real roots s = −b ± b − 4ac
.
2a
2 3s 2 + 4s + 1 = (3s + 1)( s + 1) = 3( s + 1 )( s + 1)
3
t/3, et (real roots)
Transforms to “pieces”: e 3
3
j )( s + 0.5 −
j)
2
2
3
3
e −0.5t sin
t , e −0.5t cos
t
2
2 s 2 + s + 1 = ( s + 0.5 + 2 Chapter 3 Chapter 3 1. Y ( s ) = 2+s
b2 − 4ac = 1 − 4 = −3 < 0 , an underdamped system.
s + s +1
2 Transforms to
(oscillation, complex roots to Denom(s) = 0)
From Table 3.1, line 17
L
e − bt sin ω t ←⎯
→ 2
=
s + s +1
2 (no oscillation, neg. real roots to Denom.(s) = 0) ω
(s + b)2 + ω 2
2 ⎛ 3⎞
( s + 0.5) + ⎜
⎟
⎝2⎠ L
e − bt cos ωt ←⎯
→ 2 2 s
( s + b) 2 + ω 2 ω = omega Difference of two step inputs S(t) – S(t1)
( S(t1) is step starting at t = h = 1 ) Let h→0, f(t) = δ(t) (Dirac delta)
f(t)
delta) F (s) = ∫ h 0 1 − st
1
e dt = (1 − e − hs )
h
hs If h = 1, rectangular pulse input L(δ) = 1
L( Chapter 3 Chapter 3 By Laplace transform
F ( s) = 1 1 −s
−e
ss Can be generalized to steps of different magnitudes
be generalized to steps of different magnitudes
(a1, a2). Please see Table 3.1 pg 4243 in SEMD3 for
examples of “pairs” of functions : f(t) and F(s)
f( Laplace transforms can be used in process
control for: 2. Analysis of linear control systems
(frequency response)
(frequency response)
3. Prediction of transient response “pieces” for
different inputs, e.g. stability analysis.
ff Chapter 3 Chapter 3 1. Solution of differential equations (linear) Please see Table 3.1 pg 4243 in SEM for
examples of “pairs” of functions : f(t) and F(s)
f( Chapter 3 Chapter 3 Please see Table 3.1 pg 4243 in SEMD3 for
examples of “pairs” of functions : f(t) and F(s)
f( 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 Example 3.1
dy
5 + 4y = 2
y (0) = 1
(326)
dt
First, take Laplace of both sides of (326),
2
5 ( sY ( s ) − 1) + 4Y ( s ) =
s
Rearrange,
5s + 2
(334)
Y (s) =
s ( 5s + 4 )
Find Standard form,
⎡ 5s + 2 ⎤
⎤
s + 0.4
−1 ⎡
y ( t ) = L−1 ⎢
⎥=L ⎢
⎥
⎣ s ( 5s + 4 ) ⎦
⎣ ( s + 0) ( s + 0.8 ) ⎦
Using Table 3.1 #11 form,
y ( t ) = 0.5 + 0.5e−0.8t Y(s)= Chapter 3 Chapter 3 Solve the ODE,
the ODE (337) E (s) = sE (s) = a
τs + 1 lim s→ 0 a
= a = E( ∞ )
τs + 1 The “offset” = steady state error = a.
B. Timeshift theorem
y(
y(t)=0 t < to
Ly The step response of the process will have exponential terms
step response of the process will have exponential terms
e2t and et, which indicates y(t) approaches zero. However, if 1
1
=
s − s − 2 (s + 1 )(s − 2 )
2 lim y(t)= lim sY(s)
t →0 Useful in evaluating steadystate
in evaluating steady
error E(t) as time goes to infinity. 1
a
τs + 1 s
w h a t is th e f in a l v a lu e o f E a t th e n e w s te a d y s ta te ? If α1
α
+2
s + 2 s +1 C. Initial value theorem
Initial
theorem ( (t  θ )) = e θ s Y(s)
Y(s) Chapter 3 Chapter 3 s →0 Y(s)= We know that the system is unstable and has a transient
response involving
response involving e2t and et. e2t is unbounded for
unbounded
large time. We shall use this concept later in the analysis
of feedback system stability. Final value theorem
y(∞)= lim sY(s) the denominator can be factored into (s+2)(s+1).
Using the partial fraction technique Y(s) = Other applications for y(t) = E(t)
= the process control error
A. 1
s + 3s + 2
2 s →∞ For Y(s)= y( 0 )=0
y(∞)= 1
3 4 s+2
s(s+1 )(s+2 )(s+3 ) by initial value theorem
by final value theorem,
Since roots of denominator
of sY(s)=0 are 1, 2 and 3,
i.e. all have negative real
parts. For Y(s)=Num(s)/Denom(s) inverse transformed to yield y(t), the roots
of Denom(s)=0 determine the kinds of “pieces” in y(t).
Denom
determine the kinds of pieces in
If the roots are: 1) Individual real = the corresponding piece decays if root<0, it grows if
real th
if
it
if Chapter 3 root >0. A constant “piece” arises when only one root is at 0. 2) Repeated real = polynomial in t times exponential decay or growth
Repeated real polynomial in
exponential decay or growth
corresponding to the real part of root <0 or real part of root >0
respectively. If a repeated root is at s = 0, response is not convergent. 3) Complex (conjugate) = Re + Im j and Re – Im j yields growing
oscillation (Re>0) or decaying oscillation (Re<0). Always in “pairs”. Purely
imaginary pair gives response that does not converge due to continuous
oscillation as sine or cosine.
ill 4) Repeated complex = polynomial in t times form from 3). Repeated
imaginary pairs response is unbounded sine/cosine times polynomial
imaginary pairs, response is unbounded, a sine/cosine times a polynomial
in time, thus does not converge. ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.
 Winter '08
 Staff

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