Business
• Next HW due Monday with SP10
– Not Friday!!!
• Divide into lab groups (of 4)
Class 22
– Hand it to the TAs
• Sign up for tour of BYU Heating Plant
Complicated Transfer Functions
– To see process control equipment up close
– Class is too big!
– This Friday (Oct. 22)
• 1 pm
• 2 pm
• 3 pm
Road Map for 2nd Order Equations
G (s ) =
Standard Form
Step
Response
Sinusoidal
Response
Other Input
Functions
(long-time only)
(5-63)
-Use partial
fractions
Critically
damped
ζ=1
(5-50)
Underdamped
0<ζ<1
(5-51)
What About Higher Order Systems?
Overdamped
ζ>1
(5-48, 5-49)
or
30 s 2 + 6s + 7
G (s ) =
3
24 s + 20 s 2 + 10s + 2
• Transfer function can usually be represented as a
ratio of two polynomials in the Laplace variable s
along with a possible delay term:
where
and
Z ( s) −θs
G ( s) =
e
P( s)
m
Z ( s) = bm s + bm−1 s m−1 +...+ b1 s + b0
P( s) = a n s n + a n −1 s n −1 +...+ a1 s + a 0
Roots of Z(s) = “zeros”
Roots of P(s) = “poles”
A polynomial in the numerator
is called a lead element
A polynomial in the denominator
is called a lag element
Method to describe stability behavior of
system using simple analysis of transfer
function
Relationship between
OS, P, tr and ζ, τ
(pp. 119-120)
Poles and Zeros
30
24 s 3 + 20s 2 + 10s + 2
Different Forms of G(s)
G ( s) =
bm ( s − z1 )( s − z 2 )... ( s − z m )
a n ( s − p1 )( s − p 2 )... ( s − p n )
e −θ s
so z1, z2, …, zn are the zeros
and p1, p2, …, pn are the poles
Alternatively, in time constant form,
G ( s) =
bm ( − z1 )( − z 2 )... ( − z m )(τ l1 s + 1)(τ l 2 s + 1)... (τ lm s + 1)
a n ( − p1 )( − p 2 )... ( − p n )(τ 1 s + 1)(τ 2 s + 1)... (τ n s + 1)
=K
(τ s + 1)(τ
(τ s + 1)(τ
l1
l2
1
2
s + 1)... (τ lm s + 1)
s + 1)... (τ n s + 1)
e −θ s
e −θ s
n ≥ m to be physically
realizable
so -1/τl1, -1/τl2, …, -1/τln are the zeros
and -1/τ1, -1/τ2, …, -1/τn are the poles
1
So Who Cares?
What Do Zeros Tell Us?
• Grey area (positive poles)
means unstable behavior
Imaginary axis
Complex conjugate pair
of poles -- stable but
oscillatory
Stable
region
Unstable
region
Unstable pole
“ka-boom”
x
x
x
less
oscillations
Stable pole
very fast
time constant
x
more
Stable pole
long time
constant
x
• Distance from origin means
Real axis
– More oscillations (y
direction)
– Faster response due to
shorter time constant
(x direction)
• Pole on origin means
integrating process
• Poles on y axis mean pure
oscillatory behavior with no
exponential damping
• Zeros have no effect on system stability.
• Zero in right half plane: may result in an inverse response to a
step change in the input
Chapter 6
• Poles show the stability of the process
• Zeros show some dynamics (lead-lag)
• Plot poles on real vs imaginary axes with “×”
0
(initially negative)
A certain amount of mathematical analysis (see Exercises 6.4, 6.5,
and 6.6) will show that there are three types of responses involved
here:
The response of this system to a step change in input is
⎛ τ −τ
⎞
τ −τ
y ( t ) = KM ⎜1 + a 1 e−t / τ1 + a 2 e−t / τ 2 ⎟
τ 2 − τ1
⎝ τ1 − τ 2
⎠
⇒ y
Note that y ( t → ∞ ) = KM as expected; hence, the effect of
including the single zero does not change the final value nor does
it change the number or location of the response modes. But the
zero does affect how the response modes (exponential terms) are
weighted in the solution, Eq. 6-15.
Chapter 6
Chapter 6
Solution
axis
• Zero in left half plane: may result in “overshoot” during a step
response (see Fig. 6.3).
(6-14)
calculate the response to the step input of magnitude M and plot
the results qualitatively.
Real
Inverse response
For the case of a single zero in an overdamped second-order
transfer function,
K ( τ a s + 1)
( τ1s + 1)( τ2 s + 1)
o
t
Example 6.2
G(s) =
Imaginary axis
(6-15)
Case a:
τ a > τ1
Case b:
0 < τ a ≤ τ1
Case c:
τa < 0
Example Problem
G (s ) =
30
15
15
⇒
⇒
(3s + 1)(4 s 2 + 2s + 1)
24 s 3 + 20s 2 + 10s + 2 12 s 3 + 10s 2 + 5s + 1
Chapter 6
Put in pole-zero format:
15
1⎞ ⎛
1
1⎞
⎛
3⎜ s + ⎟4⎜ s 2 + s + ⎟
3⎠ ⎝
2
4⎠
⎝
Convert to sine-cosine form:
G (s ) =
Inverse Response
G (s ) =
Find poles:
See page 134 for examples of inverse response
• Increase of steam to reboiler initially causes frothing/spillage on first trays
(-1/3,0)
15
15
=
2
1⎞ ⎛
1
1⎞
⎡
⎤
⎛
3⎜ s + ⎟4⎜ s 2 + s + ⎟ 12⎛⎜ s + 1 ⎞⎟ ⎢⎛⎜ s + 1 ⎞⎟ + 3 ⎥
3⎠ ⎝
2
4⎠
⎝
3 ⎠ ⎣⎢⎝
4 ⎠ 16 ⎦⎥
⎝
⎛ 1 3 ⎞ ⎛ 1
3
⎜− ,
⎟ ⎜
⎜ 4 4 j⎟ ⎜− 4 , − 4
⎝
⎠ ⎝
⎞
j ⎟⎟
⎠
2
Now Plot
Example: Problem 6.1
– Exponential decay
(good!)
0.4
0.3
0.2
0.1
0
-0.4
-0.3
-0.2
-0.1
• One imaginary
conjugate pair
-0.1 0
0.7(s 2 + 2 s + 2 )
(s5 + 5s 4 + 2s3 − 4s 2 + 6)
G( s) =
• All left-hand half
0.5
Find zeros and poles (use Mathcad)
1.5
– Oscillatory behavior
Zeros
-0.3
-1
-1
0.5
-0.4
Poles
Zeros
0
-0.5
-6
-4
-2
0
2
-1
-1.5
Polynomial Approximations to e −θs
Pade Approximations
theta = 2
Wanted: polynomial approximations to
Why: Analysis of transfer functions
Two widely used approximations are:
Chapter 6
θ 2 s 2 θ3s 3 θ 4 s 4
= 1 − θs +
−
+
+K
2!
3!
4!
(6-34)
The approximation is obtained by truncating after only a few
terms.
2. Padé Approximations:
Many approximations are available. For example, the 1/1
approximation is,
θ
1− s
2
e − θs ≈
(6-35)
θ
1+ s
2
(6-57)
(use for numerator terms)
• An alternative first-order approximation consists of the transfer
function,
e −θ0 s =
1
eθ 0 s
≈
1
1 + θ0 s
(6-58)
5
0
-2
-1
-5 0
1
2
s
• Time delays are very bad for control because they involve a
delay of information
• Pade approximation often used when e-θs is in denominator
Skogestad’s “half rule”
• Skogestad (2002) has proposed a related approximation method
for higher-order models that contain multiple time constants.
Chapter 6
Chapter 6
• For small values of s,
exp(-theta*s)
"1/1"Pade
"2/2"Pade
10
Implications for Control:
Taylor Approximation of Higher-Order
Transfer Functions
Goal: Approximate high-order transfer function models with
lower-order models that have similar dynamic and steadystate characteristics.
15
exp(-theta*s)
e − θs :
Chapter 6
20
1. Taylor Series Expansion:
e −θ0 s ≈ 1 − θ 0 s
1
-1
• Two poles in unstable area
(kaboom)
• Any input or disturbance
action will cause growth
beyond bounds
-0.5
e
0.000
0.583
0.585
-0.585
-0.583
1
-0.2
− θs
Poles
-4.345
0.756
-1.083
-1.083
0.756
• He approximates the largest neglected time constant in the
following manner.
¾One half of its value is added to the existing time delay (if
any) and the other half is added to the smallest retained time
constant.
¾Time constants that are smaller than the “largest neglected
time constant” are approximated as time delays using (6-58).
What does
this mean?
(use for denominator terms for non-dominant time constants)
3
Example 6.4
Skogestad Revisited
Consider a transfer function:
(sounds like a movie)
G (s) =
•
Keep it
2. Find 2nd largest time constant (τ2)
•
•
Add half of τ2 to τ1
Add the other half of τ2 to the time delay
(in numerator)
Chapter 6
1. Find largest time constant (τ1)
3. All other τ’s
•
K ( −0.1s + 1)
( 5s + 1)( 3s + 1)( 0.5s + 1)
(6-59)
Derive an approximate first-order-plus-time-delay model,
Ke−θs
G% ( s ) =
τs + 1
using two methods:
(6-60)
(a) The Taylor series expansions of Eqs. 6-57 and 6-58.
Add to time delay (in numerator)
(b) Skogestad’s half rule
Example
please!
Compare the normalized responses of G(s) and the approximate
models for a unit step input.
(b) To use Skogestad’s method, we note that the largest neglected
time constant in (6-59) has a value of three.
Solution
(a) The dominant time constant (τ = 5) is retained. Applying
the approximations in (6-57) and (6-58) gives:
and
(6-61)
(numerator)
1
≈ e −3s
3s + 1
1
≈ e−0.5 s
0.5s + 1
(6-62)
(denominator terms)
Substitution into (6-59) gives the Taylor series
approximation, G%TS ( s ) :
Ke −0.1s e −3s e −0.5s Ke −3.6 s
=
G%TS ( s ) =
5s + 1
5s + 1
(6-63)
Chapter 6
Chapter 6
−0.1s + 1 ≈ e −0.1s
G (s) =
Ke −2.1s
G% Sk ( s ) =
6.5s + 1
Chapter 6
G1(s)
Figure 6.10 Comparison of the actual and approximate models for Example 6.4.
(6-64)
Example:
Parallel Processes
(6-64)
Awesome!
I prefer
FOPDT
(6-59)
• According to his “half rule”, half of this value is added to the
next largest time constant to generate a new time constant
τ = 5 + 0.5(3) = 6.5.
• The other half provides a new time delay of 0.5(3) = 1.5.
• The approximation of the RHP zero in (6-61) provides an
additional time delay of 0.1.
• Approximating the smallest time constant of 0.5 in (6-59) by
(6-58) produces an additional time delay of 0.5.
• Thus the total time delay in (6-60) is,
θ = 1.5 + 0.1 + 0.5 = 2.1
and G(s) can be approximated as:
and G(s) can be approximated as:
Ke −2.1s
G% Sk ( s ) =
6.5s + 1
K ( −0.1s + 1)
( 5s + 1)( 3s + 1)( 0.5s + 1)
R(s)
X(s)
+
+
G2(s)
Y(s)
Q(s)
G1(s) is 1st order
G2(s) is 2nd order
Skogestad’s method provides better agreement with the actual
response.
4
Parallel Process (cont.)
Y ( s)
= Goverall ( s ) = G1 ( s ) + G2 ( s )
X ( s)
Y ( s)
K1
K2
=
+
X ( s ) τ 1s + 1 τ 22 s 2 + 2ζτ 2 s + 1
=
K1 (τ 22 s 2 + 2ζτ 2 s + 1) + K 2 (τ 1s + 1)
(τ 1s + 1)(τ 22 s 2 + 2ζτ 2 s + 1)
=
K1τ 22 s 2 + (K 2 + 2ζτ 2 )s + K1 + K 2
(τ 1s + 1)(τ 22 s 2 + 2ζτ 2 s + 1)
Now put in standard form:
Homework Hint on Prob 6.7
See online hint, because I
changed the problem a little bit!
Moral:
2 systems in parallel
K ′(as 2 + bs + 1)
give lead-lag and
=
(τ 1s + 1)(τ 22 s 2 + 2ζτ 2 s + 1) complicated pole-zero form
5