Control System Synthesis and Compensation Techniques

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Explore various chapters discussing topics like direct synthesis, closed-loop transfer functions, PI controllers, time delay compensation, and Smith predictor approach in control systems. Learn about modeling feedback controllers, closed-loop performance, and response specifications.


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  1. Chapter 12

  2. Direct Synthesis G + G Y =1 C ( G includes Gm, Gv) Y G G sp C 1. Specify closed-loop response (transfer function) Chapter 12 Y Y sp d 2. Need process model, (= GPGMGV) G 3. Solve for Gc, Y Y 1 G (12-3b) sp = G d C Y Y 1 sp d

  3. Specify Closed Loop Transfer Function s Y Y e = (12 6) + 1 s sp c d Chapter 12 (first order response, no offset) ( ) = , speedof response = process time delay in G c But other variations of (12-6) can be used (e.g., replace time delay with polynomial approximation) 1 G s 1 If = 0, then (12-3b) yields G = ) (12-5) c c K s +1 K s 1 For G = , G = = + (PI) c s +1 K K s c c c

  4. Derivation of PI Controller for FOPTD Process Consider the standard first-order-plus-time-delay model, s Ke ( ) = (12-10) G s Chapter 12 + s 1 Specify closed-loop response as FOPTD (12-6), but approximate - s e - - s. 1 Substituting and rearranging gives a PI controller, ( ) 1 1/ , c c I G K s = + 1 , c K + with the following controller settings: = = K (12-11) c I

  5. Time Delay Compensation Model-based feedback controller that improves closed-loop performance when time delays are present Chapter 16 Effect of added time delay on PI controller performance for a second order process ( 1 = 3, 2 = 5) shown below

  6. Chapter 16

  7. Chapter 16 = = * * s s G G G G + G G e G e Y Y No model error: = + C G = C G ( ) C C 1 * s 1 1 G G e sp C * s G G e G G G G + Y Y = = (16 22) C + C * * 1 1 G G sp C C (sensitive to model errors > +/- 20%)

  8. Chapter 16

  9. Direct Synthesis Approach (Smith Predictor) G G Y + 1 Y = = s ( ) C G P G Q s e sp C P Assume time delay between set-point change and controlled variable (same as process time delay, ). = Chapter 16 s G e ) s ( P If P C 1 1 R C = = G then C P G 1 s Pe P R Q 1 C = G From Block Diagram, 1 + s P Pe G C Q = GC Equating... 1 ( P Q )

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