Chemical Process Control a First Course with Matlab

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❏ To define a transfer function object, we use tf(), which takes the numerator and denominator polynomials as arguments. For example, we define G(s) = s s 2 –5s+4 with M2 - 4 G1 = tf([1 0], [1 -5 4]) We define G(s) = 6s 2 –12 (s 3 +s 2 –4s–4) with G2 = tf([6 0 -12], [1 1 -4 -4]) We can also use the zero-pole-gain function zpk() which takes as arguments the zeros, poles and gain factor of a transfer function. Recall our comments after zp2tf(). This gain factor is not the steady state (or dc) gain. For example, we define G(s) = 4 s(s+1)(s+2)(s+3) with G3 = zpk([],[0 -1 -2 -3], 4) % the [] means there is no zero ❏ The tf() and zpk() functions also serve to perform model conversion from one form to another. We can find the polynomial form of G3 with tf(G3) and the pole-zero form of G2 with zpk(G2) ❏ The function pole() finds the poles of a transfer function. For example, try: pole(G1) pole(G2) You can check that the results are identical to the use of roots() on the denominator of a transfer function. ❏ We may not need to use them, but it is good to know that there are functions that help us extract the polynomials, or poles and zeros back from an object. For example: [q,p]=tfdata(G1,'v') [z,p,k]=zpkdata(G3,'v') % option 'v' for row vectors ❏ The addition and multiplication operators are overloaded and we can use them to manipulate or synthesize transfer functions. This capability will come in handy when we analyze control systems. For now, let's consider one simple example. Say we are given 1 G 1 = s + 1 , and G 2 2 = s + 2 We can find G 1 + G 2 and G 1 G 2 easily with G1=tf(1,[1 1]); G2=tf(2,[1 2]); G1+G2 G1*G2 % or we can use zpk(G1+G2) % or we can use zpk(G1*G2)
M2 - 5 This example is simple enough to see that the answers returned by MATLAB are correct. ❏ With object-oriented programming, an object can hold many properties. We find the associated properties with get(G1) Among the MATLAB result entries, we may find the properties InputName, OutputName, and Notes. We can set them with 3 G1.InputName = 'Flow Rate'; G1.OutputName = 'Level'; G1.Notes = 'My first MATLAB function'; You'll see the difference if you enter from now on: G1 get(G1) MATLAB can use symbolic algebra to do Laplace transform. Since this skill is not crucial to solve control problems, we skip it here. You can find a brief tutorial on our Web Support, and you are encouraged to work through it if you want to know what symbolic algebra means. 3 We are using the typical structure syntax, but MATLAB also supports the set() function to perform the same task.
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P.C. Chau © 2001 Table of Contents
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P.C. Chau © 2001 10. Multiloop Sys
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Stephanopoulos (1984), of collectin
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1 - 2 Desired pH + - Error pH Contr
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P.C. Chau © 2001 ❖ 2. Mathematic
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2 - 3 2.2 Laplace transform Let us
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2 - 5 on complex variables, our Web
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2 - 7 2. Dead time function (Fig. 2
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2 - 9 lim s → 0 0 ∞ df(t) dt e
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2 - 11 α 3 = 6s 2 - 12 (s + 1) (s
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2 - 13 f(t) = 1 2 (1 - j) e( - 2+3j
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2 - 15 2.6 Transfer function, pole,
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2 - 17 final change in y(t) relativ
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2 - 19 Large a Small τ Small a Lar
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2 - 21 At steady state, Eq. (2-35)
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2 - 23 We now raise a second questi
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2 - 25 where Laplace transform give
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2 - 27 steady state and reformulati
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2 - 29 V dC' dt + Q in,s C' = C in,
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2 - 31 Y R = G 1+GH (2-63) The RHS
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2 - 33 ✎ Example 2.16. Derive the
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2 - 35 We now do the expansion: y(t
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P.C Chau © 2001 ❖ 3. Dynamic Res
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3 - 3 The Laplace transform of Eq.
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3 - 5 3.2 Second order differential
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3 - 7 (1) ζ > 1, overdamped. The r
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3 - 9 3.3 Processes with dead time
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3 - 11 q o c o c1 V 1 V 2 c n-1 q n
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3 - 13 approximation is all we use
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3 - 15 We do not need to carry the
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3 - 17 where (a) τ 1 is roughly th
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3 - 19 p2=conv([1 1],[3 1]); % Seco
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4-2 ✎ Example 4.1: Derive the sta
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4-4 With Example 4.1 as the hint, w
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4-6 now take the Laplace transform
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4-8 In this case where U and Y are
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4-10 ✎ Example 4.7A: Repeat Examp
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4-12 We first take the inlet glucos
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4-14 state space representation. 4.
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4-16 (also phase variable canonical
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4-18 The time domain solution vecto
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5 - 2 which is the negative of the
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5 - 4 5.1.2 Proportional-Integral (
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5 - 6 I action General qualitative
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5 - 8 point T sp (or reference R)
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5 - 10 5.2.3 Synthesis of a single-
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5 - 12 (2) We use a valve with line
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5 - 14 We now take a formal look at
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5 - 16 There are two noteworthy ite
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5 - 18 transmitted to a controller,
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5 - 20 point, we now want to reduce
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5 - 22 6. When we developed the mod
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6 - 2 Process L = 0 Step input P =
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6 - 4 ITAE = ∞ t e'(t) dt 0 (6-5)
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6 - 6 While the calculations in the
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6 - 8 For set point change: K c = a
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6 - 10 where τ c is the system tim
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6 - 12 guideline suggests that we n
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6 - 14 6.2.3 Internal model control
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6 - 16 Substitution of (E6-5) and (
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6 - 18 ❐ Review Problems 4. Repea
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Table 6.3. Summary of methods to se
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P.C. Chau © 2001 ❖ 7. Stability
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7 - 3 Once again, if all three pole
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7 - 5 The two additional constraint
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7 - 7 ✎ Example 7.2A: Apply direc
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7 - 9 function y=f(x) y = 5*x + tan
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7 - 11 parts of the controller func
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7 - 13 1+K 1 (s +3) (s + 2) (s + 1)
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7 - 15 7.5 Root Locus Design In ter
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7 - 17 out, and the response in thi
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8 - 2 y(t) = AK p τ p ω τ p 2 ω
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8 - 4 transfer function can be "bro
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8 - 6 other hand, is easier to inte
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8 - 8 is obvious that the 70.7% com
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8 - 10 figure(1), bode(G); figure(2
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8 - 12 polar(phase,mag) ✎ Example
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P.C. Chau © 2001 8 - 14 8.3 Stabil
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8 - 16 margin between 30° and 45°
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8 - 18 ✎ Example 8.12. Derive the
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8 - 20 for a PI controller. From a
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8 - 22 closed-loop system apply onl
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8 - 24 underdamped time response cu
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8 - 26 angle and the Nyquist criter
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8 - 28 taui=3; taud=0.5; gc=tf([tau
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P.C. Chau © 2001 ❖ 9. Design of
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9 - 3 With the same reasoning that
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Page 181 and 182: 9 - 11 A note of caution is necessa
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Page 185 and 186: 9 - 15 x e = (A ee - K er A 1e )x ~
Page 187 and 188: 9 - 17 x = Fx ~ + Gu + Hy Find Eq.
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Page 191 and 192: 10-3 The remaining task to derive t
Page 193 and 194: 10-5 τ = 0.05 s I τ = 0.5 s I τ
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Page 205 and 206: 10-17 ✑ 10.6.3 Relative gain arra
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Page 209 and 210: 10-21 G 11 G 12 G 21 G 22 d 11 d 12
Page 211 and 212: 10-23 K = 0.07 - 0.05 0.1 - 0.15 Wi
Page 213 and 214: 10-25 µ 2 = m 1 Find the gain matr
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Page 217 and 218: M1 - 2 It is important to know that
Page 219 and 220: M1 - 4 Of course, we can solve the
Page 221 and 222: So you say wow! But MATLAB can do m
Page 223 and 224: P.C. Chau © 2000 MATLAB Session 2
Page 225: F(s) = 6s 2 -12 6(s- 2)(s+ 2) (s 3
Page 229 and 230: M3 - 2 The functions also handle mu
Page 231 and 232: M3.2 LTI Viewer M3 - 4 Features cov
Page 233 and 234: M4 - 2 From X 2 U = 1 (s 2 +2ζω n
Page 235 and 236: M4 - 4 If we examine the values of
Page 237 and 238: poly(A) % Check the characteristic
Page 241 and 242: M5 - 3 M5.2 Control toolbox functio
Page 243 and 244: eig(ssm.a) M5 - 5 For fun, we can r
Page 245 and 246: M6 - 2 The point of the last two ca
Page 247 and 248: M6 - 4 A graphics window with pull-
Page 249 and 250: op_zero=[-0.5 -1.8]; % one zero in
Page 253 and 254: G=tf(1,[1 0.4 1]); bode(G) % Done!
Page 255: M7 - 5 There is no magic in the fun
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Chemical Process Control a First Course with Matlab

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