Transcript Slide 1

Example 2:
Chapter 3
d3y
d2y
dy
du
 6 2  11  6 y  4  2u
3
dt
dt
dt
dt
y( 0 )=y( 0 )=y( 0 )=0
du
 0 at t=0
dt
system at rest (s.s.)
To find transient response for u(t) = unit step at t > 0
1. Take Laplace Transform (L.T.)
2. Factor, use partial fraction decomposition
3. Take inverse L.T.
Step 1
Take L.T. (note zero initial conditions)
s3Y(s)+ 6s2Y(s)+11sY(s)  6Y (s) = 4sU(s) + 2U(s)
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Partial Fraction Expansions
Chapter 3
Basic idea: Expand a complex expression for Y(s) into
simpler terms, each of which appears in the Laplace
Transform table. Then you can take the L-1 of both sides of
the equation to obtain y(t).
Example:
Y s 
s5
 s  1 s  4 
Perform a partial fraction expansion (PFE)
1
2
s5


 s  1 s  4  s  1 s  4
(3-41)
(3-42)
where coefficients 1 and  2 have to be determined.
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To find 1 : Multiply both sides by s + 1 and let s = -1
s5
 1 
s4
4

s 1 3
Chapter 3
To find  2 : Multiply both sides by s + 4 and let s = -4
s5
 2 
s 1
s 4
1

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A General PFE
Consider a general expression,
Y s 
N s
Ds

N s
n
  s  bi 
(3-46a)
i 1
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Here D(s) is an n-th order polynomial with the roots  s  bi 
all being real numbers which are distinct so there are no repeated
roots.
The PFE is:
Chapter 3
Y s 
N s
n
  s  bi 
n

i 1
i
(3-46b)
s  bi
i 1
Note: D(s) is called the “characteristic polynomial”.
Special Situations:
Two other types of situations commonly occur when D(s) has:
i) Complex roots: e.g., bi  3  4 j
ii) Repeated roots (e.g., b1  b2  3 )
j
1

For these situations, the PFE has a different form. See SEM
text (pp. 61-64) for details.
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Example 3.2 (continued)
Chapter 3
Recall that the ODE, y  6 y  11y  6 y  1, with zero initial
conditions resulted in the expression
Y s 

1
s s  6s  11s  6
3
2

(3-40)
The denominator can be factored as


s s3  6s 2  11s  6  s  s  1 s  2  s  3
(3-50)
Note: Normally, numerical techniques are required in order to
calculate the roots.
The PFE for (3-40) is
Y s 




1
 1  2  3  4 (3-51)
s  s  1 s  2  s  3 s s  1 s  2 s  3
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Step 2b. Use partial fraction decomposition
Chapter 3
α
4s+2
α
α
α
 1 2  3  4
s(s+1 )(s+2 )(s+3 ) s s  1 s  2 s  3
Multiply by s, set s = 0
α
4s+2
α 
 α
 α1  s  2  3  4 
(s+1 )(s+2 )(s+3 ) s 0
 s  1 s  2 s  3  s 0
2
1
 α1 
1 2  3
3
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Chapter 3
For 2, multiply by (s+1), set s=-1 (same procedure
for 3, 4)
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α2  1, α3  3, α4 
3
1
1
3
5/3
Step 3. Take inverse of L.T. (Y(s)= 


)
3s s  1 s  2 s  3
Take L-1 of both sides:
1
1 1/ 6 
1  1/ 2 
1  1/ 2 
1  1/ 6 
L Y  s    L 
L 
L 
L 



 s 
 s  1
s  2
 s  3 
From Table 3.1,
1 t
5  3t
 2t
y(t)=  e  3e  e
3
3
1
t
y(t) 
3
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Chapter 3
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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Important Properties of Laplace Transforms
1. Final Value Theorem
Chapter 3
It can be used to find the steady-state value of a closed loop
system (providing that a steady-state value exists.
Statement of FVT:
lim y  t   lim  sY  s 
t 
s0
providing that the limit exists (is finite) for all
Re  s   0, where Re (s) denotes the real part of complex
variable, s.
2. Initial Value Theorem
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Example:
Suppose,
Chapter 3
Y s 
5s  2
s  5s  4 
(3-34)
Then,
 5s  2 
y     lim y  t   lim 
 0.5

t 
s 0  5s  4 
3. Time Delay
Time delays occur due to fluid flow, time required to do an
analysis (e.g., gas chromatograph). The delayed signal can be
represented as
y  t  θ  θ  time delay
Also,
L  y  t  θ    eθsY  s 
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