Chemical Engineering Process Control CHEAT SHEET Final Exam 5 Col 4pt

Closed Loop ‘Servo, Regulatory’ Regulatory’ S-S Gain Set all G(s) = 0 The Proportional Controller  Case 1: Closed Loop Response to Unit Step change in ys.p.

Note that with a P-contr the offset is ALWAYS nonzero. As Kc ↑, offset ↓ but will never be zero. Kc ↑, the closed loop response becomes faster  Case 2: Regulatory Closed Loop Unit Step Response As Kc ↑, offset ↓

The Closed-Loop Response of a First Order Process with the PI Controller 

Case 1: Servo Response

The I-action adds one zero and one pole to the closed loop transfer function. The presence of the additional pole will make the system response sluggish (pole) and the additional zero may cause an overshoot in the closed loop response. Offset SUMMARY: The P-action speeds up the closed loop response (at

high Kc values) but always results in an offset. The I- action eliminates the offset but adds a pole and a zero to the closed loop transfer function. Therefore, it slows down the response and may lead to instability at high Kc/τI. Similarly, it can be shown that the D-action adds one zero to the closed loop transfer function; and therefore, it speeds up the response but the offset remains with the D action alone.

Open-Loop Step Test Method

In this method the ‘combined’ transfer functions of the valve,  K   K  process and the measuring element element is obtained as, for EX, a first G (s ) = φ  =  AR = order plus time delay model, and the Controller Tuning Relations,  s + 1 τ 2ω 2 + 1 such as those given in Table 12.1 and 12.3 of Seborg et al. are Note that we can find Gc(s) because Gp(s) is known and we can used to find the controller settings. Advantages of the Closed-Loop Closed-Loop Tuning Method specify the desired closed loop servo transfer function. The simplest and yet realistic specification for the CL transfer function i. The process remains under control Disadvantages Disadvanta ges of the Closed-Loop Tuning Method is that the closed-loop system should behave like a first order lag i. Time consuming with a small time constant (fast closed-loop response) and a ii. The process must be brought to the verge of instability (may steady-state gain equal to 1 (to ensure zero offset, that is the results in off spec product) output reaches the set-point at steady-state).  K K  Choose the correct Control Action → reverse or direct =  AR → The ‘controller action’ is chosen after the correct action of the τ 2ω 2 τ ω  control valve has been selected. Why? Because of safety, the valve action must be chosen first. The controller action is chosen such that the overall steady state gain of the FB control system is positive. KcKvKpKm>0 . Km is often positive; depending on the sign of the process gain and the chosen ‘control valve action’, the 1 Gc = 1/G(1/tc*s) controller action is ultimately selected to make the overall gain of   K  =  K  ω n = The Internal Model Control (IMC) Method for Tuning PID the FB control system positive τω  τ  EX. For the FB control of liquid level in a tank using the outlet flow rate as the manipulated manipulated variable, the process gain is negative since as the flow out of the tank increases, the level decreases. For safety consideration a ‘fail open’ (FO) or ‘air-to-close’ valve is needed, since we want to avoid spilling over. Therefore, if the controller output which is the actuating pressure signal t o the valve increases the valve will close further since the valve is ‘airtoclose’, i.e. if P ↑ then x ↓. Therefore, the valve gain is negative , 1 Kv0. This means the controller is ‘reverse acting’.

In gener 

n

Recall that P (t ) = P +Kc (ls. ls p. . p. −l ) So, if l becomes larger that the set-point we need to increase the

flow out of the tank. Since the valve is ‘air-to-close’, we need to reduce the pressure applied to the valve, and that requires a positive Kc . In analog controllers, there is a switch to make the controller gain Kc positive (reverse acting) or negative (direct acting). In digital controllers this is achieved simply by changing the ‘sign’ of  Kc .

If any of the coefficients n−1 a to 0 a is negative when n a is positive, the system is unstable. Therefore, this first test is a ‘necessary’ BUT not ‘sufficient’ condition for stability. If this test is satisfied the second Routh test must also be satisfied to ensure the ‘sufficient’ condition for system stability. For an ‘n-th’ order C.E. polynomial, there will be n+1 rows.

For the system to be stable, all the coefficients in the 1 st column must Note that with this choice of the controller  C(s), perfect control is feasible because Y(s) = Ys.p.(s), under all conditions. This is true be even if the model of the process is not perfect. positive. The problem is that C (s) cannot always be chosen as the inverse of the plant model, G (s), due to the presence of non-inverting elements, such as a positive zero or time delay terms existing in the process model, G (s) . Therefore, C(s) is chosen as the inverse of the ‘invertible’ portion of the plant model. The ‘invertible’ ‘invertible’ portion of the process model does NOT involve delay time or  positive zeros. EX 12.3. Divide the following model t o invertible and non-invertible non-invertible parts 



Direct Substitution Method

We use the property that the closed-loop system is marginally stable if the closed-loop C.E. has a pair of pure imaginary roots. Therefore, if we replace s = j ω in the closedloop C.E., the corresponding corresponding controller proportional gain will be the maximum or  the ultimate gain (Kult = Kc,max). The advantage of this method To get rid of the offset, we retain the S-S part of G(0)+

or ultimate controller gain for the system given below that has a time delay:

This transfer function has 2 lead terms. It is physically not realizable In order to improve the controller performance performance we add an rth order filter to the controller 

log AR 

Chapters 13 and 14: Frequency Response of Linear Systems

8

The study of steady state response of linear systems to sinusoidal A ROL = Kc   input change. Theorem: If the input to a linear system is changed 90 0ω 2 + 1 9ω 2 + 1 100ω 2 + 1 by a sine wave the output is also a sine wave at steady state (as t ∞ ∞ φ OL = − tan−1 ( 3ω ) − tan −1 ( 30ω ) − tan −1 (10ω ) = −π  − j t  − st  →∞ ) with the same frequency but different amplitude and with a y (t ) e d t   y (t ) e dt   Y (s ) 8 = 0∞ G (s ) = phase difference (a lag or a lead). Both the amplitude ratio and G ( j  ) = 0∞ ω  = 0.218rad / s AOR  st  L =1= uK lt     U (s ) the phase difference depend on the frequency of the input sine u (t ) e − d t   u (t )e − j t d t   c 9 c 2 + 1 900 c 2 + 1 100 0 0 wave. The ratio of output amplitude to input amplitude is called ∴  K ult = K c ,max =23.8 Amplitude Ratio (AR). AR=B/A. The AR and then phase T y T  y difference are strong functions of  ω . The application application of  Definitions of Gain Margin (GM) and Phase Margin (PM) y ( t ) cos ( t ) d t − j y( t) sin ( t ) d t   0 frequency response is for: [1] process identification (empirically These two measures of stability, determine how far the closedG ( j  ) = 0Tu T  u obtain AR and φ and compare to AR and φ of known low order  order  loop system is from the ‘marginally-stable’ condition. condition. The larger  u (t ) cos ( t ) d t − j u(t ) sin ( t )d t   0 0 transfer functions, and therefore, approximate an unknown system the GM and PM are, the more stable and slower the closed-loop with a low order model). [2] controller design. ( ) − j2A( ) system will be. 1A G ( j  ) = GM = 1/AROLat thecriticalfrequency   K   A    K  A ( ) − j2B( ) 1B G (s ) = U (s ) = 2 Y (s ) =    2 2   2 PM = 180 o + OLwhere AR of OpenLoopis 1  s + 1  s +   s + 1   s +   For any given ω , you can calculate  A1, A2 , B1, B2  from measured −1 input and output pulses L {Y (s )} = y '(t ) 

∫ ∫

t 

y'( )t =

2

2

K A− e

+1

 B  A

=

2

) +1

(  KA

'y( → t ∞) =

 AR =

KA

+

2

) +1

(

t+ tan −

sin {

sin {

1

(−

The Root Locus Diagram

+t tan   ( − ) }

( τω ) + 1

   )    B    d    (   e    d   u    t    i   n   g   a    M

 Im      Re  

φ  = tan −1 

- 45o

 z  = a

+b

2

   )   g   e    d    (   e   s   a    h    P

-720 -1080 -1440

∠ z  = tan

−1

-3

-2

10

-1

10

10

0

10

1

10

Frequency (rad/sec)

G (s ) =

 K e −

 s a nd  K  , , an

=  z { cos( ∠ z ) +  j sin( ∠ z ) } =  z  exp{ j ( ∠ z ) }

2 0 lo log

A R→0 = 20 lo log

≈K18∴ =K

a nltoi g 18 / 2 0≈ 7.9

 j ( ∠ z )

locus diagram of the system described in EX 11.7. First Order Filter f(s) [r=1] The breakaway point  of the branches of the root locus diagram from the real axis can be found from the following equation

And the angle of the asymptotes of the branches going to infinity can be found by with k taking values of 0,1and 2 they are pi/3, pi, 5pi/3 Control of Unstable Processes and application of the RootLocus Diagram:

An unstable process has at least one positive pole in its process transfer function: EX 11.9.

≈

∠ z  = ∠ z 1 + ∠ z 2

Alternatively, Alternatively, let z be a complex number which is the ratio of two other complex numbers, z1 and z2

EX 11.8. Using the above ‘rules’, construct the approximate root

 z =

 z 1

 z  =

 z 1

G (s ) ≈

∠ z = ∠ z 1 − ∠ z 2

7.9 e −3 s

8.3 s

 z 2

The obtained AR and Phase difference are used for system identification identification and ‘controller design’ using graphical presentation of the AR and φ in the form of Bode, Nyquist and Nichols plots. Graphical presentation presentation of AR and φ in the form of graphs such as Compare the above equation with the transfer function of an ideal BODÉ and NYQUIST diagrams, will help interpret the results. PID controller to get the tuning parameters of the PID controller  NYQUIST Diagram: Polar representation of this information (AR and φ ) on the G(jω ) plane. For any given value of  ω , calculate the AR and φ . This will represent represent one point on the G(jω ) plane with a distance from the origin equal to the AR and an angle with the real axis equal to φ . Methods to construct Bodé and Nyquist Diagrams [1] Graphical − approximate plots − useful for the fundamental fundamental understanding of the concepts and

C .L .C .E . = 1+G c (s )G v ( s)G p (s )

= 1 +G OL(s ) = 0

Routh Test (Chapter 11), Direct Substitution (Chapter 11), Root Locus Diagram (Chapter 11), Stability Test in the Frequency Domain → Bode and Nyquist stability Statement of the Stability Theorem in the Frequency Domain

A system is closed-loop stable if the amplitude ratio of its open AR OL loop ( AR  OL) system is less than one at the critical frequency (frequency at which the OL phase difference, φ  OL, is −180°).

system that ensures a GM of 2 and a PM of 30o, whichever is more conservative G  p ( s ) =

50 0.016 1 Gv ( s ) = G m (s ) = G c ( s ) = K c 30 s + 1 3 s + 1 10 s + 1

 50   0.016   1  K   c  30 s+ 1   3 s+ 1   10 s+ 1    

G OL ( s ) = 

8

 AROL = K c

φ OL

= − tan

GM

y s .p .

=

As in(

+ 1 900ω 2 + 1 100ω 2 + 1 ( 3ω ) − tan−1 ( 30ω ) − tan −1 ( 10ω ) = −π  9ω 2

−1

1

=2=

 AROL

AR

φOL = -

atcriticalfrequency

OL atcriticalfrequency

ω c

= 0.218rad / s

∴ A ROL atcriticalfrequency =1 / 2

= 1/ 2 =

c

K 2

9

c

+1

8

 

90 900

2

c

+1

10 0 c 2

+1

Solution. = K  c

8

2

9 (0 (0 .2 18 18)

+1

2

90 0( 0( 0. 0. 21 21 8) 8)

+

1 1 00 00 (0 (0 .2 .2 18 182 ) + 1

∴  K c = 0.5 (1.19)(6.54)(2.39) =1.17 8

Note that this value of  K c c  is much smaller than t he maximum value of K c,max  =23.8 obtained for the system in EX 14.1.The design c ,max =23.8 based on the given PM results in another  K c c  value and we choose the one which is more conservative or smaller. PM = 30 o = 180o + Φ  OL where AR of Open Loop is 1 Φ OL where AR of Open Loop is 1=-180 o+30 o=-150 o=- (5/6) π = - 2.616 rad = − tan −1 ( 3 ) − tan −1 (30 ) − tan −1 (10

Qualitative Proof  t)  

G OL (s ) = G c ( s) G v ( s) G p (s ) G m (s ) [1] Graphical Construction of Approximate Bodé and Nyquist  B OLwhere AR of Open Loop is 1 plots Steps involved: [A] Obtain expressions for the AR and φ A R G ( j    ) = = OL OL OL = ∠ G OL ( j  )  A [B] Obtain expressions for the AR as ω→ 0 and for  φ as ω→ 0 Fund ω  by trial and error (remember that your calculator should We know that both  AR OL OL and φ  OL are functions of frequency, ω  . → low frequency asymptote (lfa), steady-state be in radian mode and NOT in the degrees mode): ° = −π , the 180 °  If we vary ω  such that the  AR OL OL=1 and φ  OL= −  [1] Obtain expressions for the AR as ω→∞ and for  φ asω→∞ ≈ 0.14 r ad ad / s measured output y m will be: → high frequency asymptote (hfa) 8 AOL R = 1 = cK   [1] Using the ‘hfa’, determine log AR → log ω to get the slope of   y m  A sin ω t  π   A sin ω t  2 2 2 9(0.14) +1 900(0.14) +1 100(0.14) +1 the ‘hfa’ [2] Obtain the intersection of ‘lfa’ with hfa to obtain ω n, At this point, we close the loop and set y s.p. . Note that the error  error  s.p.= 0  the corner frequency 8  A signal is still the same as when the loop was open and y s.p. was ( A = K c EX 13.1. Draw the ‘approximate’ Bode plots and Nyquist plot for a ). Under these conditions, the closed-loop system oscillates (1.08)(4.31)(1.72) First Order System and then compare them with the Matlab generated sinω  t ). with constant amplitude (that is, the closed-loop system is at the (1.08)(4.31)(1.72) Compare the above equation equation to the transfer function of an ideal PI 2  AR OL verge of instability). instability). If  AR   K c = = 1.005 OL is increased slightly (for EX, by G (s ) = controller to get the tuning parameters of the PI controller  8 5 s + 1 increasing the controller gain) the CL system becomes unstable  plots for 

=

CHAPTER 12 Direct Synthesis Method for Tuning the PID Controllers

have a GM between 1.7 to 4 and a PM between 30o and 45o. EX 14.2 . Design a FB controller (obtain Kc) for the following

Stability Tests for CL FB Control Systems

processidentification [2] Numerical − Obtain AR( ω ) and φ (ω ) from G(s), form a table table of AR, φ , ω , and draw AR versus ω on a log-log plot and φ versus ω on a semi-log plot. − Alternatively, use MATLAB  y m =  B sin(ω t + φ )

Closed-loop stability requires that 2 − Kc > 0 and Kc − 1 > 0, Therefore: Kc < 2 Kc > 1 1 < Kc < 2. Therefore, for the closed loop system to be stable the controller gain must be larger than 1 and smaller than 2. That is, there will be a minimum and a maximum controller gain, Kc,min and Kc,max, to ensure closed-loop system stability.

GUIDLINE. A heuristic approach to design a FB controller is to

+1

Stability Analysis and Design of FB Control Systems using Frequency Response Technique The Closed-Loop TF of a SISO System, assuming Gm=1, is Theorem: If we replace s = jω in the transfer function of a linear  ( ) ( ) ( ) G d  (s ) system, we obtain a complex number G(j ω ) whose magnitude | Y (s ) = G c s G v s G p s Y  s . p .(s ) + D (s ) 1 + G c (s )G v (s )G p ( s) 1 +G c( s )G v(s)G p( s) G(jω )| is equal to the amplitude ratio (AR) and whose argument ∠ G(jω ) is the phase difference ( φ ).

 z 2

log

 s

+1

 z  =  z e Let z be a complex number which is the product of two other  complex numbers, z 1 and z2

 z  =  z 1 ×  z 2

-90o

-1800

 b       a  

hfa slope = Magnitud lfa slope = For a first

 z =  z 1 × z 2

φ

-10

-360

)  + j Im G ( j  )  2

0

-30 0

= ∠ G ( j  )

a process whose model can be approximated by a first order plus a =  z  cos( ∠ z ) b =  z sin( ∠ z ) time delay. Use a Pade approximation approximation and a Taylor series approximation approximation for the delay term. Then, rearrange the IMC  z = a +  jb controller equation on as a PI or PID controller and obtain the =  z  cos( ∠ z ) +  j z  sin( ∠ z ) corresponding corresponding tuning of the PI or PID controller.

• The angle of asymptotes of branches approaching infinity can be found from the following equation

Tu

-20

Review of Complex Numbers: G(jω ) has a real part and an

.

in degrees

+ Im 2

10

)

imaginary part G ( j ) = Re G ( j

EX 12.5. Tune a PID controller by designing an IMC controller for 

2

20

φ  = tan −1 ( − τω )

2

 z  = a +  jb Rules for the construction of the approximate Root-Locus Diag: • There will be  p branches of the root locus diagram. • Branches start from the open loop poles and end at the open loop zeros OR at ∞. • The real axis is a part of the root locus diagram if the total number of the OLTF poles and zeros to the right of that point (on the real axis) is an odd number (1, 3, 5, ...). • The breakaway point of the braches from the real axis

+1

Bode Diagram

K 

AR = G ( j  

Re

2

c

φ

Conclusion: Conclusion: With one single experiment, one can determine AR(ω ) and φ (ω ) of an unknown process, and construct the Bodé diagrams. The Bode plots can then be compared with the Bodé diagrams of known low order transfer functions to estimate the approximate approximate transfer function of the unkown system. Ex 13.2. Using the Bodé Plots shown below, determine the TF of  the unknown process

−1

Shortcut to Determine AR( ) and ( ) of Linear Systems Theorem: If we replace s = jω in the transfer function of a linear  system, we obtain a complex number G(j ω ) whose magnitude | G(jω )| is equal to the amplitude ratio (AR) and whose argument ∠ G(jω ) is the phase difference ( φ ).

This is a powerful method for designing controllers. The Root Locus Diagram is the locus of the closed-loop characteristic equation roots (or the  poles of the closed-loop system) as Kc varies from 0→∞. Graphical Method. .

∫  ∫ 

) = Re( ) ± j  Im( )  AR =

G (j

)}

AR=0, Φ = - 9

∫  ∫ 

∫  ∫ 

the order of the numerator polynomial should be at least the same as the order of the denominator polynomial. polynomial. τ c is the filter time constant and works as the IMC ‘tuning parameter’. It is the only tuning parameter of the IMC controller.

ω = 0.532 rad/s Kc,max = Kult = 0.92

log

Nyquist Plot: Note that as ω goes to zero, AR approaches K with Φ = 0. At high frequencies AR approaches approaches the origin with an angle of approach ( Φ ) equal to – 90 o. Therefore the Nyquist Nyquist diagram which is the location of the tip of the vector whose magnitude is the AR and its angle with the real axis is Φ Implementation of the Theorem [1] Obtain the expressions for  Avoid Derivative and Proportional Kicks the AR OL OL and φ  OL. [2] Find the critical freq, ω  c , by letting φ  OL= → In the case of an analog controller (PID), avoid introducing introducing step resembles a half circle and with its mirror image it forms a complete circle: −π . [3] For a given controller gain, K c c  , substitute ω  c  in the AR OL OL changes in the set-point. Instead, use a ramp change. expression and determine if  AR OL OL is 1 → Never use the D-action if the measured signal is noisy. (closed-loop unstable). unstable). [4] If you want to find the maximum If the D-action is required, then use a filter (digital and analog). analog). controller gain ( K c,max  c,max = K  ult  ult ) find the value of  K  c  c  that gives  AR  OL OL=1 For a digital PID, use the measured signal instead of the error  (marginally stable). signal for the P and D actions EX 14.1. For the process conditions given below, find the Avoid Anti-reset Windup Reset ≡ integral action maximum or the ultimate controller gain, K ult  ult , that results in a If the error signal persists for a long time, the controller output P(t) closed-loop system which is marginally stable. eventually hits the limit → saturates. This saturation is referred to 50 0.016 1 as reset windup and should be avoided. How to avoid reset ( ) ( ) ( ) = = = G c ( s ) = K c G s G s G s  p v m windup? If we have an analog PID, we have to order an anti-reset 30 s + 1 3 s + 1 10 s + 1 windup option. If it is a digital PID, this can be achieved by Alternatively, use of a ‘pulse’ signal for process inserting 2 ‘IF’ statements. identification:  50   0.016   1  K   if (Pn.gt.Pmax) then Pn = Pmax G OL ( s ) =       if (Pn.lt.Pmin) then Pn = Pmin  30 s+ 1   3 s+ 1   10 s+ 1  c 

r is chosen such that the product of  f(s)C(s) becomes “proper” → j sinθ , we have: Using the Euler equation, e− j θ = cosθ − j sinθ

Bodé

Bodé P

Stability of Closed Loop Control Systems Routh Test

is that it can be applied to a closed-loop C.E. which has a ‘time delay’. EX 11.6. Using the direct substitution method, find the maximum

Presentation of the Stability Theorem on Bodé and Nyquist Diagrams

− tan −1 (τω )

(

− )=−

+

B

(oscillates with an increasing amplitude). amplitude). If  AR OL OL is decreased, however, the system will be closed-loop stable (oscillates with decreasing decreasing amplitude until it settles at a steady-state value).

)=

fre ε=Asin

φ

Therefore the choice of t he controller gain according to the design specification given in this problem is dictated by the PM which

asks for a smaller controller gain, K c c  = 1.005, which is virtually 1. This results in a more stable (slower) closed-loop response.

Y These processes have at least one positive pole in their OL transfer function. If PID controllers are used, include the D-action. Y Control of Non-Linear Processes If the relationship between u and y is mildly non-linear, a linear 

Chapter 15, FF and Ratio Control

A FFC controller is used if large and frequent disturbances affect the controlled variable, especially in a slow process. The FFC should not be used alone. alone. The FB and FF controllers are used together to correct for variations in the feed flowrate (F) and feed temperature (T F). The Feed Forward Control Equation

Y (s ) = G p (s )U (s ) + G d (s )D (s )

In general, The function of a FF controller is to ensure that y’(t) tracks y’ sp(t). Therefore, to derive the ‘ideal’ FF controller equation, we replace Y(s) by Ysp(s) in the above equation and rearrange the equation to obtain:

Chapter 16, Enhanced Single Loop C ontrol Strategies Cascade Control: Ex 16.1. Temp control in a jacketed CSTR.

1 2

(s )

 (s )

G ( s) =  11 G 21 ( s )

G1 2 (s )  U 1 (s )

G22 ( s)  U2 (s )

λ 11 = K 11h11

4   20 10 = 5    = =  14   14 7

U1 & Y1 is the correct pairing. U2 & Y2 is dangerous dangerous because the two loops fight each other. EX 18.5. The Bristol RGA of a process is given below, suggest the correct pairing of Us and Ys.

controller (PID) will work fine. However, in the presence of strong non-linearity non-linearity (which is very common in chemical processes), a linear controller will fail to perform well. We have three alternatives: 1. Use a linear controller (such as PID) with frequent retuning (commonly practiced). 2. Use an adaptive controller. 3. Use a non-linear controller (Research).

 0.8 Λ = 0.05 0.15

Problem with FB only: There are usually fluctuations in the steam supply pressure. The fluctuations act as disturbances disturbances affecting the Adaptive Controllers Adaptive controllers involve on-line system identification plus steam flow rate (manipulated (manipulated variable). controllerdesign. If fluctuations in steam pressure are large, a simple FB There are many types of adaptive controllers: control will not perform well . a. Gain Gain sche schedu duliling ng or or Pro Progr gram am Ada Adapt ptiv ive e The Block Diagram of a Cascade Control system Controller 

0.1

0.1

0.85

 0 

0.05 0.9

↔ ↔ ↔

Control of MIMO Processes in the Presence of Interaction

Note that if G 12(s) and G 21(s) are zero, then loops are independent and there would be no interactions between the two loops. Otherwise, Y 1 will be influenced by both U 1 and U2. The same is true for Y 2. What is the Bristol R GA (relative gain array matrix)?

λ  Λ =  11 λ 21

Feed Forward Controller Block Diagram

λ 12 



λ 22 

−

Definition of Individual Entries of  Λ :

Definition

The steady state gain is (K c Kv Kp Km = Koverall). The overall gain constant for good performance. K m and Kv Note that the output of the primary controller is the set-point of the should be maintained constant Note that the FF controller only receives measurements from may be constant for linear measuring element and valve secondary controller  disturbance(s) disturbance(s) and is unaware of the actual value of the controlled ↔ characteristics. cs. We may only change the controller gain to offset variable. changes in the process gain. Therefore, as the process gain FF and FB Block Diagram changes, the controller gain (K c) must be changed to maintain the The numerical value of  λ 11 will determine if there is any overall gain constant. Requires parameter parameter estimation block (K p = interaction between the two loops. ∆ y/∆ u). b. Model Reference Adaptive Controllers Controllers

Ysp1

In general, it can be shown that the elements of each column or  each row of  Λ matrix add up to 1 How to find the RGA Matrix? The Closed Loop Transfer Function of the combined FF and FB Control System. Let us assume that Km = 1.

The Overall Closed-Loop Transfer Function for the Cascade Controller is:

Assuming that Gm =1 and Gv =1, we will have the required FF controllerequations:

The cascade controllers are very common in industry. In a cascade controller  always use a P-controller as the secondarycontroller. 1. The ‘secondary controlled variable’ does NOT have to be maintained at its setpoint, therefore there is no need for the I-action in the secondary loop. 2. A proportional controller with a large Kc allows Gc2 to nullify the disturbance effects, very quickly. For the master controller, use a PI or a PID controller. TUNING of the slave controller is done first, then the master controller.

11

=

+

Explicit STR: Identify the process in an on-line on-line manner. Based 1 − λ 11  on the prior information about the plant we have to select the  λ  Λ =  11 process model model structure. For EX, a FOPTD. This process ID  1 − λ 11 λ 11  block must calculate K, θ , and τ for the model G(s) = Ke −θ s/ (τ s+1). This block will use a recursive parameter estimation method (such as the least squares) to estimate K, θ , and τ . The 1 0 Λ=  controller block designs designs the new controller (PID, use ZN or IMC 0 1 tuning method). Implicit STR: STR: The process model is not identified identified explicitly

Ysp

Y (s )

The Approximate Model (Transfer Function) of the process with Time Delay

= G (s )U ( s )

Y 1 (s ) = G 11(s )U 1(s ) + G 12(s ) U 2( s ) +

5 3

 K  = 

+ G n1 (s )U n ( s)



Y2

s = 0.

Advantages of a FFC

The FF controller compensates for the effect of large disturbances affecting the controlled variable. Perfect control is ‘theoretically’ y’ possible. Disadvantages of a FFC

Requires extra instrumentation to measure the major  disturbances. It only corrects for the measured disturbances, and it does not take any action for the other disturbances. Since the controlled variable is not fed back to the FFC, it may (and usually does) suffer from drifts. It takes some actions without knowing their consequence. consequence. Therefore, always use the FF controller with a FB controller. The FF controller requires a lot of engineering engineering effort to develop. Accurate transfer functions for  Gp (s) and Gd (s) are required. The Ratio Controller 

The Ratio controller is a special steady-state FF controller used to control the ratio of flowrates of two streams. EX 15.2. Control of the air to fuel f low ratio in a furnace

A ratio controller can be implemented using two different configurations.

The second configuration is:

= (G 11 + G 12G I 2 )v 1 + (G 12 + G 11G I  1) v 2 = (G 21 + G 22G I 2 )v 1+ (G 22 + G 21G I  1)v 2

desired 1 reference0.5 controllers model