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Unformatted text preview: CHE 461: Class Discussion Outline Notes
Read Chapter 11 in SEM2 textbook.
(1)
Start thinking about "teams" for design project
(2) Boeing 747 example: openloop zeros are also zeros of the closedloop
zero at +6 is in closedloop Y/Ysp, but single slow pole dominates and setpoint response
look about like a 7 sec 1storder response. Go over general formula of Gc and Gp as
numerator/denominator form to show zeros/poles of closed loop. (3) Idea that the "roots" of characteristic equation (CE) are the poles of the closed transfer
function which arise from the loop structure. Y/D still has the openloop pole(s) of Gd. (4) Stability idea > demo of 3rdorder Gp and P controller. Loop goes unstable if Kc is too
large (2 and 6 are ok, 15 is too large and output overshoots more than 100 %!).
Cause of instability = lags in series, here 3 “enough” to make effect of manipulated
variable sluggish  also plot of larger overshoot for a little additional time delay (0.05)
"Real" instability (bangbang of control valves) vs. "theoretical" (unconstrained)
instability with exponential growth to infinity! Note that hitting a constraint DOES NOT
“cause” the instability in the sense that you can often initially hit the constraint after a
setpoint change, but still be stable after manipulated variable “comes off the constraint”. (5) Three "tools" to analyze stability in Laplace domain (finite # of poles and zeros)
(1) = Algebra, either Routh array or direct substitution
(2) = Graphical, Root Locus sketches and "rules"
(3) = graphical brute force, computer generated root loci (6) C.E. = characteristic equation, polynomial in s = 0, highest order coefficient in s > 0
(greater than 0) is the "standard" form here. Then if any coefficient is < 0 (less than 0),
the equation has roots which are either on the imaginary axis or in the RHP ( > BIBO
unstable poles !). Missing coefficient counts as a nonpositive coefficient. A zero
coefficient is neither necessary nor sufficient to have purely imaginary roots (roots on the
imaginary axis).
missing coefficient but no purely imaginary root = unstable
(s1)(s+1) = s21
(s2+1)(s+1) = s3+s2+s+1 no missing coefficient but 2 purely imaginary roots = unstable (7) Routh array example for fixed Gc = 11 and Gp = 1/(s+1)3, C.E. is s3 + 3s2 + 3s + 12 = 0
1 3 0 3
(3•3)(1•12)
3
12 12
0 0
0 0 0 Looking at the first column of the Routh array, all numbers in that column must be > 0 for the
CE roots to have negative real parts (stable closedloop system). Here we see the third number
from the top is (912)/3 = 1, thus a gain of 11 results in an unstable closed loop. Chapter 11 Stability of ClosedLoop
Control Systems Fig. 11.8 “Standard 6block” diagram of a feedback control system.
(“splits” Gp TF in 3block diagram into Gv Gp and Gm and adds Km) Rule of Thumb:
Closedloop response becomes less oscillatory and “more stable” by
decreasing Kc or increasing τI . Chapter 11 General Stability Criterion
Consider the “characteristic equation,” 1 + GC GV GPGM = 0
Note that the lefthand side is merely the denominator of the
closedloop transfer function.
The roots (poles) of the characteristic equation (s  pi) determine
the type of response that occurs:
Complex roots ⇒ oscillatory response
All real roots ⇒ no oscillations
***All roots in left half of complex plane = stable system 1 Chapter 11 Roots on the
imaginary axis
are unstable. Fig. 11.25 Stability regions in the complex plane
For roots of the characteristic equation. Stability Considerations Chapter 11 • Feedback control can result in oscillatory or even unstable
closedloop responses.
• Typical behavior (for different values of controller gain, Kc).
(f diff 1 Figure 11.23 pg. 277 SEM2 text 5s+1 D Ysp 1
1
Km Ysp_t Sum1 E P
PIDalpha U 1
2s+1 Gc = Kc Gd Gv 1
5s+1
Gp 1 Yd
Yu
Sum2 Mux
Y_Ysp_gra s+1
Gm 2 Chapter 11
Chapter 11 Fig. 11.23 Effect of controller gains on closedloop response
to a unit step change in set point. Fig. 11.26 Contributions of
characteristic equation
roots to closedloop
response.
Roots of (s values such that):
1 + GcGvGpGm(s) = 0 3 Routh Stability Criterion
Characteristic equation Chapter 11 1 + GC GV GPGM = 0 rearranged into a polynomial in s = 0. an s n + an −1s n −1 + (1121) Where an >0 . According to the Routh criterion, if any of the
other coefficients a0, a1, …, an1 are negative or zero, then at
least one root of the characteristic equation lies in the RHP,
and thus the system is unstable. On the other hand, if all of
th th
th
if
the coefficients are positive, then one must construct the
Routh Array shown next to determine “yes” or “no” on the
stability of the closed loop. an s n + an −1s n −1 + Chapter 11 + a1s + a0 = 0 + a1s + a0 = 0 the charcteristic equation b1 = an−1an−2 − an an−3
an−1 For stability, all elements in the first column must be
positive. Note an was required to be positive at the start. 4 Chapter 11 The first two rows of the Routh Array are comprised of the
coefficients in the characteristic equation. The elements in the
remaining rows are calculated from coefficients by using the
formulas: b1 =
b2 = an −1an − 2 − an an −3
an −1
an −1an − 4 − an an −5
an −1 (1194)
(1195) b1an −3 − an −1b2
b1
b1an −5 − an −1b3
c2 =
b1
c1 = (1196)
(1197) (n+1 rows must be constructed; n = order of the characteristic eqn.) Application of the Routh Array:
GP = GL = 8
, GV = GM = 1, GC = KC ≥ 0 since K P = 1 < 0
( s + 2)3 Chapter 11 The Characteristic Equation is 1 + GC GV GPGM = 0
8KC
1+
=0
( s + 2)3 + 8K C = 0
( s + 2)3 s 3 + 6s 2 + 12s + 8 + 8 K C = 0
We want to know what value of Kc causes instability, i.e., at least one
root of the above equation has positive real part. Using the Routh array, 1
6
6(12) − (1)(8 + 8KC )
6
8 + 8KC 12
8 + 8KC n = 3, check 1st colm, 4 numbers 0
0 Condition for Stability with positive controller gain is then: 72 − (8 + 8 KC ) = 64 − 8KC > 0, so upper limit on KC < 8 5 Chapter 11 Figure 11.29
Flowchart for
performing a
stability analysis. 1 − 0.5θ s RHP zero
=
= firstorder Padé
1 + 0.5θ s LHP pole
page 138 SEM2 e −θ s ≈ More details than stability analysis: Chapter 11 Calculate the closedloop poles
… or sketch how they move as
you increase the magnitude of Kc. Fig. 11.27 “Root locus
diagram”  here for thirdorder
system.
G OL ( s ) = 4KC
( s + 1)( s + 2)( s + 3) 6 Additional Stability Criteria (frequency domain) Chapter 11 With time delays, there are an infinite number of closedloop poles ! 1. Direct Substitution – Chap. 11
Substitute s = jω in the characteristic equation
and solve two equations in the unknows Kc and
ω
Real part = 0, Imaginary part = 0
2. Bode Stability Criterion
•
Chap. 14  can exactly analyze systems with
time delays IF  AR and phase angle
“monotonically” decrease as frequence increases
3. Nyquist Stability Criterion
•
Chap. 14 – can exactly analyze systems with time
delays without the restriction required for the
Bode Stability Criterion 7 Routh_DS 1/2 CHE 461: Routh Array / Direct Substitution Examples
1. Routh array example for fixed P controller
For this example, Gc = Kc = 11 and Gp = 1/(s+1)3.
1 ( M21 M12 ) Then b1 = ( M11 M22 ) M21 12 0 b2 0 12 M11
For b1 we use the M matrix M
21 0 b1 The Routh array for this problem is then 3 3 Characteristic equation is "CE" = s3 + 3s2 + 3s + 12 = 0 0 0 M12 1
= M22 3 = (3 3) 12 3 ( 1 12 )
3 = 1 Note the pattern for the M matrix  it’s first column is made from the first elements of the two rows
above b1. The second column is made from the elements from the two rows above and one position
to the right of b1. Then b1 is equal to { (the product of elements "off" the diagonal)  (the product of
elements "on" the diagonal) } all divided by the lower left corner element of the matrix. This pattern
is used for each element bi in the row. For example the M matrix for b2 would have the same first 1 0 column , but zeros in the second column 3 0 and thus b2 = 0.
Looking at the first column of the Routh array, all numbers in that column must be > 0 for the CE
roots to have negative real parts (stable closedloop system). Here we see the third number from the
top is 1, thus a gain of Kc = 11 would result in an unstable closed loop. 2. Routh array example with Kc as a variable
(2) By letting the controller gain (Kc) appear as a variable in the CE, a further level of analysis allows
us to answer the question of "What is the upper limit on Kc for a stable closedloop system? For this
problem, the characteristic equation is:
CE = s3 + 3s2 + 3s + (Kc + 1) = 0
Now the Routh array with K c for the problem is 1
For b1 we use 3 3
1 Kc which yields b =
1 1 3 3 1 b1
1 [3 3] Kc 0 0 [ 1 (1
3 0 0 Kc 0 0
Kc ) ] = 8 Kc
3 Routh_DS 2/2 For small Kc > 0 we easily see that b1 > 0. The upper limit on Kc is 8, thus we require Kc < 8 for
stability but we also require Kc > 0 for performance. For example, even though Kc = 0.5 would yield
a stable closed loop, the closedloop gain would then be = 1, which means that the output would
move in the wrong direction when R is changed to a new desired value (C > R instead of R !).
At Kc = 8 the third row contains all zeros and there are two roots on the imaginary axis, since for
larger Kc we would have two sign changes in the first column. The two elements of the last nonzero
row, [k1 k2] = [3 9] here, can be used to find the location of these purely imaginary roots in the s
plane as follows. The imaginary roots will be the solutions of k1 s2 + k2 = 0 = 3 s2 + 9 for this
example. Here the imaginary roots are at ± √3 j. 3. Direct Substitution Method
In this method we assume that for some particular value of Kc, one or more roots are purely imaginary
by setting s = j ω in the characteristic equation and then solving for the required ω and Kc which
would make this occur.
Then for CE = s3 + 3s2 + 3s + (Kc + 1) = 0
with s=jω and s2 = (j ω)2 =  ω2 so .... ( ω3 j )
which means and 3 ( ω2 )
ω3 3 (j ω) 3ω = 0 and s3 = (j ω)3 =  ω3 j
1 Kc = 0 j 3 ω2 1 0 Kc = 0 so ω = 3 and K c = 8 so that the purely imaginary roots are s = ± 3 j. 4. RHP Zero Effect on Stability
Now suppose we have the same process poles, but add a RHP zero at +1. Then Gp = (s+1)/(s+1)3 and
we have ... CE = s3 + 3s2 + ( 3  Kc ) s + ( Kc + 1 ) = 0. From the coefficients of the CE we can see
that Kc > 3 would yield an unstable closed loop, but we don’t know if the closed loop would be stable
for Kc = 2.99. 1
3 Kc
0 3
1 Kc
0 The new Routh array with K c for this problem is b1
0
0 1K
0
0
c 1
For b1, use 3 3
1 Kc to get b =
1
Kc [ 3 (3 Kc ) ] [ 1 (1
3 Kc ) ] = 8 4 Kc
3 Now we see that the upper limit on Kc is only 2, thus Kc > 0 for performance and Kc < 2 for stability
when we have this RHP zero. This is often expressed as " the RHP zero "causes" the closedloop to
be "more unstable", since a lower Kc results in an unstable closed loop ". Exercise 11.1 pg. 289 SEM2 CHE 461 In this problem you translate the temperature control system for a distillation column P&I
Diagram (the P&ID) into a block diagram. The temperature controller manipulates the signal to
the control valve which affects the molar reflux flow rate, R , as the manipulated process input
variable, U. You must draw a block diagram containing transfer function models to show
relationships. You must label your block diagram showing Tsp, T, Tm, R, xF, F and P variables. The “standard  block” negative feedback control loop block diagram is shown in Fig. 11.9
reproduced below. But in that figure there is only one identified disturbance. Here we have 2
disturbances, the feed composition (mole fraction of light component in the feed), xF , and the
total molar feed rate F.
Thus we need to add a second disturbance and a second transfer function block to indicate that
the Y variable, here the temperature T at that point in the distillation column, depends on three
input variables. Exercise 11.20 pg. 296 SEM2 CHE 461 Given the following GOL(s), determine values of Kc for which the closedloop system is stable
using two approaches.
0.5K c e −3s
GOL ( s ) =
10 s + 1
For Part (a) if you look in the textbook's Index on pg 710 you will find that the Pade'
approximation of time delay is discussed on pg. 138, equation (635) in Chapter 6. In Part (b) you do not use the Pade’ approximation, but you will use the Euler identity, which
can also be located in your book by checking the Index. ...
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