Data Based Design of 3Term Controllers. Data Based Design of 3 Term Controllers p. 1/10

Data Based Design of 3Term Controllers Data Based Design of 3 Term Controllers p. 1/10

Data Based Design of 3 Term Controllers p. 2/10 History Classical Control - single controller (PID, lead/lag) is designed based on transfer function or frequency response data

Data Based Design of 3 Term Controllers p. 2/10 History Classical Control - single controller (PID, lead/lag) is designed based on transfer function or frequency response data Modern/Post Modern Control (H, H 2, l 1 ) - optimal controller of high order based on state space model Fuzzy Neural Control- model free approach but no guarantee of stability or performance Our Recent Results - complete set of PID controllers achieving stability and performance based on transfer function model

Data Based Design of 3 Term Controllers p. 3/10 Drawbacks of Identification States have no physical significance

Data Based Design of 3 Term Controllers p. 3/10 Drawbacks of Identification States have no physical significance Parameters in the model have no physical significance

Data Based Design of 3 Term Controllers p. 3/10 Drawbacks of Identification States have no physical significance Parameters in the model have no physical significance State variables have no meaningful dimensions or units

Data Based Design of 3 Term Controllers p. 4/10 Main Idea: To synthesize the set of all stabilizing PID/First Order controllers from experimental data of the plant rather than from mathematical models. To synthesize the set of all PID/First Order controllers satisfying the given performance requirements from the experimental data of the plant.

Data Based Design of 3 Term Controllers p. 5/10 Assumption The plant is stabilizable (i.e., N(s) and D(s) are coprime). where P(s) = N(s) D(s)

Data Based Design of 3 Term Controllers p. 5/10 Assumption The plant is stabilizable (i.e., N(s) and D(s) are coprime). where P(s) = N(s) D(s) The plant has no jω poles.

Data Based Design of 3 Term Controllers p. 5/10 Assumption The plant is stabilizable (i.e., N(s) and D(s) are coprime). where P(s) = N(s) D(s) The plant has no jω poles. Only available information

Data Based Design of 3 Term Controllers p. 5/10 Assumption The plant is stabilizable (i.e., N(s) and D(s) are coprime). where P(s) = N(s) D(s) The plant has no jω poles. Only available information P(jω) for ω [0, )

Data Based Design of 3 Term Controllers p. 6/10 Preliminaries Consider a real rational function R(s) = A(s) B(s) where A(s) and B(s) are polynomials of real coefficients with degrees m and n, respectively. Assume that A(s) and B(s) have no zeros on the jω axis. Write R(jω) = R r (ω) + jr i (ω) where R r (ω) and R i (ω) are real rational functions in ω. Note that R r (ω) and R i (ω) have no real poles for ω (,+ ).

Data Based Design of 3 Term Controllers p. 7/10 Real Hurwitz Signature Lemma Define sgn[x] = 8 >< >: +1 if x > 0 0 if x = 0 1 if x < 0

Data Based Design of 3 Term Controllers p. 7/10 Real Hurwitz Signature Lemma Define If n m is even sgn[x] = 8 >< >: +1 if x > 0 0 if x = 0 1 if x < 0 σ(r) =! l 1 sgn[r r (ω + 0 )] + 2 X ( 1) j sgn[r r (ω j )] + ( 1) l sgn[r r (ω l )] j=1 ( 1) l 1 sgn[r i ( )] If n m is odd! Xl 1 σ(r) = sgn[r r (ω 0 )] + 2 ( 1) j sgn[r r (ω j )] ( 1) l 1 sgn[r i ( )] j=1

Data Based Design of 3 Term Controllers p. 8/10 Complex Hurwitz Signature Lemma Consider a complex rational function Q(s) = D(s) E(s), Q(jω) = Q r(ω) + jq i (ω) where Q r (ω) and Q i (ω) are real rational functions. Q r (ω) and Q i (ω) have no real poles for ω (,+ ).

Data Based Design of 3 Term Controllers p. 9/10 PID controller design + C(s) P(s)

Data Based Design of 3 Term Controllers p. 9/10 PID controller design + C(s) P(s) Let the PID controller be of the form C(s) = K i + K p s + K d s 2 s(1 + st), T > 0

Data Based Design of 3 Term Controllers p. 10/10 Main Results The complete set of stabilizing PID gains for a given LTI plant can be found from the frequency response data P(jω) and the knowledge of the number of RHP poles. Using the result above, the subset of the PID gains that satisfy the several given performance requirements.

Data Based Design of 3 Term Controllers p. 11/10 PID Controller Design Let us consider the plant and PID controller pair of the form: P(s) = N(s) D(s) C(s) = K i + K p s + K d s 2 s(1 + st) where deg[d(s)] = n and deg[n(s)] = m., T > 0 Then consider the rational function F(s) = s(1 + st) + `K i + K p s + K d s 2 P(s) Closed-loop stability is equivalent to the condition that zeros of F(s) lie in the LHP. This is also equivalent to the condition σ(f(s)) = n + 2 (p p + )

Data Based Design of 3 Term Controllers p. 12/10 PID Controller Design Continue... Now consider the rational function F(s) = F(s)P( s)

Data Based Design of 3 Term Controllers p. 12/10 PID Controller Design Continue... Now consider the rational function F(s) = F(s)P( s) Note that σ ` F(s) = σ(f(s)) + σ(p( s)) = n + 2 (p p + ) + (z + z ) (p + p ) = n + 2 z + + z = n m {z } 2z + + 2 relative degree of P(s)

Data Based Design of 3 Term Controllers p. 13/10 PID Controller Design Continue... Write F(jω) = jω(1 + jωt)p( jω) + (K i + jωk p ω 2 K d )P(jω)P( jω) = F r (ω) + j F i (ω) where F r (ω) = (K i K d ω 2 ) P(jω) 2 ω 2 TP r (ω) + ωp i (ω) F i (ω) = ω `K p P(jω) 2 + P r (ω) + ωtp i (ω)

Data Based Design of 3 Term Controllers p. 14/10 Lemma: Determining the required signature Relative Degree and Net Phase Change: A. In the Bode magnitude plot of the LTI system P(jω), the high frequency slope is (n m)20db/decade where n m is the relative degree of the plant P(s). B. The net change of phase of P(jω), ω [0, ), denoted 0 (φ) is: 0 (φ) = ˆ(n m) 2(p + z + ) π 2 where p + and z + are numbers of RHP poles and zeros of P(s), respectively.

Data Based Design of 3 Term Controllers p. 15/10 Proof of Lemma The statement A is obvious from the property of Bode magnitude plot.

Data Based Design of 3 Term Controllers p. 15/10 Proof of Lemma The statement A is obvious from the property of Bode magnitude plot. Let p and z are number of LHP poles and zeros of P(s). Then the net phase change of P(jω) for ω [0, ) is 0 (φ) = ˆ(z z + ) (p p + ) π 2 = ˆ(n m) 2(p + z + ) π 2.

Data Based Design of 3 Term Controllers p. 16/10 Determining z + and p + Assume that a known feedback controller C(s) stabilizes the plant P and the closed-loop response can be measured and denoted by G(jω), for ω [0, ) Then P(jω) is the computed frequency response of the unstable plant P(jω) = G(jω) C(jω)(1 G(jω))

Data Based Design of 3 Term Controllers p. 17/10 Determining z + and p + (continue...) Lemma: z + = 1 2 ˆ rp r C 2z + c σ(g) p + = 1 2 [σ(p) σ(g) r C] z + c where z + c denotes the number of RHP zeros of C(s).

Data Based Design of 3 Term Controllers p. 18/10 Proof of Lemma G(s) = P(s)C(s) 1 + P(s)C(s) since G(s) is stable, σ(g) = `z + z c `z+ + z + c (n + nc ) = r P r C 2z + c 2z+ which implies z + = 1 2 ˆ rp r C 2z + c σ(g). From σ(p) applied to P(s), we have p + = z + + 1 2 σ(p) + 1 2 r P. Then we now have p + = 1 2 [σ(p) σ(g) r C] z + c.

Data Based Design of 3 Term Controllers p. 19/10 Determining thezerosof F i (ω) = 0 Recall F i (ω) = ω(k p P(jω) 2 + P r (ω) + ωtp i (ω)) = 0

Data Based Design of 3 Term Controllers p. 19/10 Determining thezerosof F i (ω) = 0 Recall F i (ω) = ω(k p P(jω) 2 + P r (ω) + ωtp i (ω)) = 0 Then for ω 0, or K p = P r(ω) + ωtp i (ω) P(jω) 2 cos φ(ω) + ωt sin φ(ω) K p = P(jω)

Data Based Design of 3 Term Controllers p. 20/10 PID Controller Design Continue... The set of PID stabilizing controllers can be found as follows: Fix K = K p and set F i (ω, K p ) = 0.

Data Based Design of 3 Term Controllers p. 20/10 PID Controller Design Continue... The set of PID stabilizing controllers can be found as follows: Fix K = K p and set F i (ω, K p ) = 0. Let i t {+1,0, 1} and j {+1, 1}. Determine strings of integers {i 0, i 1, } satisfying: For n m even: o ni 0 2i 1 + 2i 2 + + ( 1) i 1 2i l 1 + ( 1) l i l ( 1) l 1 j = n m + 2z + + 2

Data Based Design of 3 Term Controllers p. 21/10 PID Controller Design Continue... For each string (sign sequence of appropriate number of terms) satisfying the signature formula, the conditions for stability are: sgn [R(ω t, K i, K d )] i t > 0, for t = 0, 1, 2,,

Data Based Design of 3 Term Controllers p. 22/10 Example Consider the stable plant 0 Bode Diagram 10 20 Magnitude (db) 30 40 50 60 70 800 Phase (deg) 180 360 540 10 2 10 1 10 0 10 1 10 2 Frequency (rad/sec) 6 π 2 = (n m) {z } =2 2( p + z + ) {z} =0! π 2 z+ = 2

Data Based Design of 3 Term Controllers p. 23/10 Example (continue...) The required signature for stability can now be determined and is σ( F) = (n m) + 2z + + 2 = (2) + 2(2) + 2 = 8.

Data Based Design of 3 Term Controllers p. 23/10 Example (continue...) The required signature for stability can now be determined and is σ( F) = (n m) + 2z + + 2 = (2) + 2(2) + 2 = 8. Since n m is even, we have i 0 2i 1 + 2i 2 2i 3 + 2i 4 + ( 1) l i l = 8, and it is clear that at least four terms are required to satisfy the above.

Data Based Design of 3 Term Controllers p. 24/10 Example (continue...) From the figure it is easy to see that Kp solutions and therefore we have has at most three positive frequencies as i 0 2i 1 + 2i 2 2i 3 + i 4 = 8. 40 30 20 necessary range of ω 10 0 feasible range of K p 10 20 0 1 2 3 4 5 6 7 8 9 10 ω

Data Based Design of 3 Term Controllers p. 25/10 Example (continue...) Fix K p = 1 and compute the set of ω s that satisfies cos φ(ω) + ωt sin φ(ω) P(jω) = 1. 40 30 ω 1 =0.742 ω 2 =1.865 ω 3 =7.854 20 K p 10 K p =1 0 10 20 0 1 2 3 4 5 6 7 8 9 10 ω

Data Based Design of 3 Term Controllers p. 26/10 Example (continue...) This leads to the requirement i 0 2i 1 + 2i 2 2i 3 = 7

Data Based Design of 3 Term Controllers p. 26/10 Example (continue...) This leads to the requirement i 0 2i 1 + 2i 2 2i 3 = 7 giving the feasible string F = {i 0, i 1, i 2, i 3 } = {1, 1, 1, 1}.

Data Based Design of 3 Term Controllers p. 26/10 Example (continue...) This leads to the requirement i 0 2i 1 + 2i 2 2i 3 = 7 giving the feasible string F = {i 0, i 1, i 2, i 3 } = {1, 1, 1, 1}. Thus, we have the following set of linear inequalities for stability: 0.0138K i > 0 0.1390 + 0.0364K i 0.0201K d < 0 0.2791 + 0.0229K i 0.0797K d > 0 0.1349 + 0.0003K i 0.0182K d < 0

Data Based Design of 3 Term Controllers p. 27/10 Example: Stabilizing PID Setfor K p = 1 6 4 2 0 Kdbar 2 4 6 8 0 1 2 3 4 5 6 7 Kibar

Data Based Design of 3 Term Controllers p. 28/10 Example: Entire Stabilizing PID Set 5 0 Kpbar 5 10 8 6 4 2 0 Kdbar 2 4 6 8 0 2 4 Kibar 6 8 10

Data Based Design of 3 Term Controllers p. 29/10 Performance Specifications Many performance attainment problems can be cast as stabilization of families of real and complex plants. For example, The problem of achieving a gain margin is equivalent to stabilizing the family of real plants P c (s) = {KP(s) : K [K min, K max ]}. The problem of achieving prescribed phase margin θ m is equivalent to stabilizing the family of complex plants P c (s) = n o e jθ P(s) : θ [0, θ m ].

Data Based Design of 3 Term Controllers p. 30/10 The problem of achieving an H norm specification on the sensitivity function S(s), that is, W(s)S(s) < γ is equivalent to stabilizing the family of complex plants (" # ) P c 1 (s) = 1 + 1 P(s) : θ [0,2π]. γ ejθ W(s) The problem of achieving an H norm specification on the complementary sensitivity function T(s), that is, W(s)T(s) < γ is equivalent to stabilizing the family of complex plants P c (s) = j» P(s) 1 + 1 γ ejθ W(s) ff : θ [0,2π].

Data Based Design of 3 Term Controllers p. 31/10 Assumption The only information available to the designer is: Knowledge of the frequency response magnitude and phase, equivalently, P c (jω), ω (, + ). Knowledge of the number of RHP poles, p +.

Data Based Design of 3 Term Controllers p. 32/10 Determining Performance Set The complete set of stabilizing PID gains for a given complex LTI plant can be found from the frequency response data P c (jω) and the knowledge of the number of RHP poles The set of stabilizing PID gains can be computed by the following procedure: Determine the relative degree n c m c from the high frequency slope of the Bode magnitude plot where n c and m c are degrees of numerator and denominator of P c.

Data Based Design of 3 Term Controllers p. 33/10 Set ω 0 =, ω l = + and determine all strings of integers i t {+1,0, 1} and j { 1,+1} such that l 1 X ( 1) l 1 r i r j = n c m c + 2z c + + 2 r=1 where n c and m c denote the numerator and denominator degrees of P c (s) and z + c the number of RHP zeros. For the fixed K p = K p chosen in Step 1, solve for the stabilizing (K i, K d ) from: for t = 0,1,.» K i K d ωt 2 + ω t sin φ(ω t ) ωt 2T cos φ(ω t) P c i t > 0 (jω)

Data Based Design of 3 Term Controllers p. 34/10 Performance Example Taking the same frequency domain data set P(jω) used in the previous example, we consider the problem of achieving an H norm specification on the complementary sensitivity function T(s), that is, W(s)T(s) < 1 where W(s) = s + 0.1 s + 1. By solving the complex stabilization problem, we have the stabilizing PID controller parameter region that satisfies the given H norm specification.

Data Based Design of 3 Term Controllers p. 35/10 Thecomplete set ofstabilizingpidgainsfor H specification when K p = 1 6 4 2 K d 0 2 Stabilizing region for P c (s) 4 Stabilizing region for P(s) 6 8 0 1 2 3 4 5 6 7 K i

Data Based Design of 3 Term Controllers p. 36/10 Nyquist plot of W(s)T(s) By selecting a point, we verify that the point selected satisfied the given H specification. 1.5 1 0.5 imag 0 0.5 1 1.5 1.5 1 0.5 0 0.5 1 1.5 real

Data Based Design of 3 Term Controllers p. 37/10 Entireset of stabilizingpidgainssatisfyingthe H specification By sweeping K p, we have the entire stabilizing PID gains that satisfy the given H specification as shown in Figure. 2 0 K p 2 4 6 4 2 0 K d 2 4 0 1 K i 2 3 4

Data Based Design of 3 Term Controllers p. 38/10 Example with Gain/Phase Margin Specification Consider the following nonminimum phase plant: P(s) =» s 3 4s 2 + s + 2 s 5 + 8s 4 + 32s 3 + 46s 2 + 46s + 17 e s We wish to find the entire set of controllers that simultaneously satisfies the following specifications: PID controllers must stabilize the given plant with a delay The closed-loop system must guarantee the following gain and phase margins:

Data Based Design of 3 Term Controllers p. 39/10 (A) All Stabilizing PID Controllers The feasible ranges of k p given are: [ 19.1, 8.5], [ 8.5,4.23], [4,23, ]. We can easily conclude that the only region that contains the solutions is [ 8.5,4.23]. The complete set of stabilizing PID controllers is: K p 4 2 0 2 4 6 8 Kp= 6.49 Kp= 7.16 3 Kp= 7.83 2 1 0 1 2 Kp= 5.82 Kp= 5.15 Kp= 4.48 Kp= 3.80 Kp= 2.46 Kp= 3.13 2 Kp= 1.12 Kp= 1.79 4 Kp= 3.56 Kp= 2.89 Kp= 0.21 Kp= 0.46 Kp= 2.22 Kp= 1.55 Kp= 0.88 6 3 4 0 k i k d

Data Based Design of 3 Term Controllers p. 40/10 (B)All PID Satisfying thegm and PM Requirement We obtained the final result. 3 2 1 Kp= 2.22 Kp= 1.55 Kp= 0.88 k p 0 1 2 Kp= 0.21 Kp= 0.46 Kp= 1.12 Kp= 1.79 3 Kp= 3.13 Kp= 2.46 3 4 2 1 0 k d 1 2 3 0 1 k i 2

S C k p 2-D Regions with k p = 0.88 S kp : stable set with a fixed k p S C k p : unstable set S kp (G) : set satisfying the GM requirement S kp (φ) : set satisfying the PM requiremen S kp (G) S kp (φ) set satisfying both gain and phase margin requirements 5 Kp= 0.88 4 3 2 1 3 S kp (G) S kp Kd 1 0 1 4 S kp (G) S kp (φ) 2 2 3 S kp (φ) 4 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Ki Data Based Design of 3 Term Controllers p. 41/10

First Order Controller Design Data Based Design of 3 Term Controllers p. 42/10

Data Based Design of 3 Term Controllers p. 42/10 First Order Controller Design Consider the First order controller of the form for an LTI plant. C(s) = x 1s + x 2 s + x 3

Data Based Design of 3 Term Controllers p. 42/10 First Order Controller Design Consider the First order controller of the form for an LTI plant. C(s) = x 1s + x 2 s + x 3 Assumption The plant has no jω poles or zeros. Available frequency domain data P(jω) for ω [0, ). Knowledge of p +.

Data Based Design of 3 Term Controllers p. 43/10 Root Invariant Regions Let P(jω) = P r (ω) + jp i (ω) F(s) = (s + x 3 ) + (sx 1 + x 2 )P(s)

Data Based Design of 3 Term Controllers p. 43/10 Root Invariant Regions Let P(jω) = P r (ω) + jp i (ω) F(s) = (s + x 3 ) + (sx 1 + x 2 )P(s) For closed-loop stability, it is necessary and sufficient that σ(f(s)) = n + 1 (p p + ) Let F(s) = F(s)P( s) and the stability condition is F(s) = (s + x 3 )P( s) + (sx 1 + x 2 )P(s)P( s)

Data Based Design of 3 Term Controllers p. 44/10 Root Invariant Region (continue...) In other words, F(jω, x 1, x 2, x 3 ) = x 2 P(jω) 2 + ωp i (ω) + x 3 P r (ω) {z } F r (ω,x 1,x 2,x 3 ) +jω `x 1 P(jω) 2 x 3 P i (ω) + P r (ω). {z } F i (ω,x 1,x 2,x 3 ) The curves F r (ω, ) = 0 and F i (ω, ) = 0, 0 ω < along with the F(0, ) = 0 and F(, ) = 0 partition the (x 1, x 2, x 3 ) parameter space into signature invariant regions.

Data Based Design of 3 Term Controllers p. 45/10 Procedure for First Order Controller Design 1. Determine the relative degree n m from the high frequency slope of the Bode magnitude plot.

Data Based Design of 3 Term Controllers p. 45/10 Procedure for First Order Controller Design 1. Determine the relative degree n m from the high frequency slope of the Bode magnitude plot. 2. Let 0 [φ(ω)] denote the net change of phase P(jω) for ω [0, ). Determine z + from knowledge of p + and 0 [φ(ω)] = ˆ(n m) 2(p + z + ) π 2. 3. Plot the curves below in the (x 1, x 2 ) plane for a fixed x 3. x 3 + x 2 P(0) = 0 8 < : x 1 (ω) = 1 P(jω) sin φ(ω) x ω 3 cos φ(ω), for 0 < ω < x 2 (ω) = 1 P(jω) (cos φ(ω)x 3 + ω sin φ(ω)), 1 + P( )x 2 = 0. for 0 < ω <

Data Based Design of 3 Term Controllers p. 46/10 Procedure for First Order Controller Design (continue...) 4. The curves x 1 (ω) and x 2 (ω) partition the (x 1, x 2 ) plane into disjoint signature invariant regions. The stabilizing regions correspond to those for which F(s) has a signature of n m + 2z + + 1.

Data Based Design of 3 Term Controllers p. 47/10 Example For illustration, we have collected the frequency domain (Nyquist-Bode) data of a stable plant: P(jω) = {P(jω) : ω (0,10) sampled every 0.01}. The Nyquist plot of the plant obtained is: 50 Nyquist Diagram 5 Nyquist Diagram 40 4 30 3 20 2 Imaginary Axis 10 0 10 Magnification shown in Fig. 3 Imaginary Axis 1 0 1 20 2 30 3 40 4 50 30 20 10 0 10 20 30 40 Real Axis 5 14 12 10 8 6 4 2 Real Axis

Data Based Design of 3 Term Controllers p. 48/10 Example (continue...) From the data P(jω), we have P(0) = 13.333 and P( ) = 0. Then it is easy to see that the straight line is not applicable.

Data Based Design of 3 Term Controllers p. 48/10 Example (continue...) From the data P(jω), we have P(0) = 13.333 and P( ) = 0. Then it is easy to see that the straight line is not applicable. After fixing x 3 = 0.2, the data points representing the straight line and the curve are depicted. By testing a point for each root invariant region, we have the following. 0.8 0.7 0.6 0.5 0.4 x 2 0.3 0.2 Stabilizing regions 0.1 0 0.1 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 x 1

Data Based Design of 3 Term Controllers p. 49/10 Example (continue...) We now consider the problem of determining the entire set of first order stabilizing controllers satisfying the required closed-loop performance described by the requirement on the H norm of the weighted complementary sensitivity function: W(jω)T(jω) < γ, for all ω

Data Based Design of 3 Term Controllers p. 50/10 Example (continue...) In this problem, we let γ = 1 and x 3 = 2.5. We superimpose on the top of the stabilizing region shown (left) for the real plant, the stabilizing sets for the complex plant families P c (jω, θ) for θ = 0, π 3, 2π 3, π, 4π 3, 5π, 2π are plotted (right). 3 0 0 0.5 0.5 x 2 1 1.5 x 2 1 1.5 2 2 2.5 stabilizing first order set 2.5 3 3 4 3 2 1 0 1 2 x 1 3.5 4 3 2 1 0 1 2 x 1

Data Based Design of 3 Term Controllers p. 51/10 Example (continue...) To verify, a number of points inside the performance region, construct the corresponding controllers, and Nyquist plots of W(s)T(s) have been plotted as shown. These points are shown as * in (right). We observe from the Nyquist plots and every test set satisfies the H performance requirement. 1 0.8 0.6 0.4 0.2 imag 0 0.2 0.4 0.6 0.8 1 1 0.5 0 0.5 1 real

Data Based Design of 3 Term Controllers p. 52/10 Example of FO Controller Design Consider an unstable plant with time delay. G(s) =» s + 1 s 4 + 8s 3 + 48s 2 + 46s 1 e s The design specifications are given as follows: the closed-loop system must satisfy the following gain and phase margin requirements:

(A) The Complete Set of Stabilizing FO Controllers The feasible range is found as: x 3 [ 6, ]. We chose to execute the algorithm for x 3 [ 6,70]. Then we obtain the the set of FO Controller parameters so that the each and every corresponding closed-loop system satisfies the given gain and phase margin requirements. X3 70 60 50 40 30 20 X3=70.00 X3=61.55 X3=53.11 X3=44.66 X3=36.22 X3=27.77 X3=19.33 10 X3=10.88 6000 0 3000 2000 1000 0 X3= 2.44 1000 2000 2000 4000 Data Based Design of 3 Term Controllers p. 53/10

Data Based Design of 3 Term Controllers p. 54/10 (B) A Set of FOC that Meets the Desired Overshoot and Settling Time Requirements Computing a feasible range of the generalized time constant based on the CRA, the feasible range of the time constant is obtained as: τ [2.692, 6.258]. 70 70.0 60 61.5 50 53.1 40 44.7 x 3 36.2 30 27.8 20 19.3 10 0 1000 500 10.9 2.4 1000 500 0 500 0 x 2 x 1 1500 2000

Data Based Design of 3 Term Controllers p. 55/10 Stepresponseswith FO Controllers in S Step Responses 1 0.8 y(t) 0.6 0.4 0.2 0 0 5 10 15 20 25 30 35 40 t[s]

2-D setof FO Controllers, S x 3,when x 3 = 53.1 To proceed, we pick a FO Controller from the controller set S and examine various performance of the corresponding closed-loop system. In this example, we first select x 3 = 53.1 and the corresponding 2-D set is depicted as in figure. 4000 X3=53.1 3500 3000 S x3 2500 Sx3 (G) S x3 (φ) S x3 X2 2000 S x3 (G) S x3 (φ) S x3 (T) 1500 1000 500 0 2500 2000 1500 1000 500 0 500 1000 1500 2000 2500 X1 Data Based Design of 3 Term Controllers p. 56/10

Data Based Design of 3 Term Controllers p. 57/10 Various characteristics of the CL system with the selected FOC A selection of the controller point from the set S x 3 with the fixed x 3 enables us to display the following six figures displaying various characteristics of the closed-loop system with the FO Controller chosen. X3=53.1 Step Reponse 1000 1 X2 800 y(t) 0.5 600 2 1 600 400 200 0 200 400 X1 0 0 10 20 30 40 t[s] Roots Space 100 50 Im 0 Imag 0 1 2 2 1 0 1 2 Re Gain and Phase Margin Gm=( 24.8, 11.0) [db] Pm=( 180.0, 66.4) [deg] 50 100 3 2 u(t) 1 0 50 40 30 20 10 0 10 Real Control Input 1 t[s] 0 10 20 30 40

Data Based Design of 3 Term Controllers p. 58/10 The top left figure shows the controller set S x 3 and + indicates the selected controller of the parameter values: x 1 = 83.3, x 2 = 831, x 3 = 53.1.

Data Based Design of 3 Term Controllers p. 58/10 The top left figure shows the controller set S x 3 and + indicates the selected controller of the parameter values: x 1 = 83.3, x 2 = 831, x 3 = 53.1. The top right figure shows the step response of the closed-loop system. Two figures in the middle show the Nyquist plot, and the closed-loop poles and zeros. The bottom two figures show the values of gain and phase margins, and the control input signal, respectively.

Data Based Design of 3 Term Controllers p. 59/10 Data-Robust Design (An Example) What we know: P(jω) ± 20% for all ω

Data Based Design of 3 Term Controllers p. 59/10 Data-Robust Design (An Example) What we know: P(jω) ± 20% for all ω Number of RHP zeros is z + = 2

Data Based Design of 3 Term Controllers p. 59/10 Data-Robust Design (An Example) What we know: P(jω) ± 20% for all ω Number of RHP zeros is z + = 2 Relative degree is r P = 1 due to high frequency slope (-20db/decade) Signature required is r P + 2z + + 2 = 1 + 2(2) + 2 = 7. It requires at least 3 frequencies to satisfy K p = g(ω). Notations: P(jω) = Family of response shown in Bode plot P max r (ω) = max P(jω) P(jω) P r(ω), P min r (ω) = min P(jω) P(jω) P r(ω), P max i (ω) = max P(jω) P(jω) P i(ω), P min i (ω) = min P(jω) P(jω) P i(ω), for every ω for every ω for every ω for every ω

Data-Robust Design (An Example) Continue... Data Based Design of 3 Term Controllers p. 60/10

Data Based Design of 3 Term Controllers p. 60/10 Data-Robust Design (An Example) Continue... K = P r(ω) + ωtp i (ω) P(jω) 2 := g(ω)

Data Based Design of 3 Term Controllers p. 60/10 Data-Robust Design (An Example) Continue... K = P r(ω) + ωtp i (ω) P(jω) 2 := g(ω) We overbound the g(ω) family: 8 < g max (ω) = max : g(ω) : Pr max 8 < g min (ω) = min : g(ω) : Pr max (ω), P min r (ω), Pi max P(jω) max, P(jω) min (ω), Pi max P(jω) max, P(jω) min (ω), P min r (ω), Pi min (ω), (ω), Pi min (ω), 9 = ; 9 = ;

Data Based Design of 3 Term Controllers p. 61/10 g(ω) graph: K p = 5 20 ω 2 + 15 ω 2 10 K p =5 g(ω) 5 0 ω 1 ω 3 ω 3 + 5 10 0 10 1 10 2 ω 1 + ω

Data Based Design of 3 Term Controllers p. 62/10 Interval Linear Programming y mx c > 0, m [m, m + ], c [c, c + ].

Data Based Design of 3 Term Controllers p. 62/10 Interval Linear Programming y mx c > 0, m [m, m + ], c [c, c + ]. c c+ c m m+ m

Data Based Design of 3 Term Controllers p. 62/10 Interval Linear Programming y mx c > 0, m [m, m + ], c [c, c + ]. c c+ y (m +, c + ) (m +, c ) c m m+ m x (m, c + ) (m, c )

Data Based Design of 3 Term Controllers p. 62/10 Interval Linear Programming y mx c > 0, m [m, m + ], c [c, c + ]. c c+ y (m +, c + ) (m +, c ) c m m+ m x (m, c + ) (m, c ) 8 >< >: y m x c > 0 y m x c + > 0 y m + x c > 0 y m + x c + > 0

Data Based Design of 3 Term Controllers p. 63/10 A set of linear programmings b t = ω tp i (ω t ) + ω 2 t TP r(ω t ) P(jω t ) 2

Data Based Design of 3 Term Controllers p. 63/10 A set of linear programmings b t = ω tp i (ω t ) + ω 2 t TP r(ω t ) P(jω t ) 2 8 >< >: K i (ω 1 )2 K d b 1 > 0 K i (ω 2 )2 K d b 2 < 0 K i (ω 3 )2 K d b 3 > 0 K i (ω 4 )2 K d b 4 < 0

Data Based Design of 3 Term Controllers p. 63/10 A set of linear programmings b t = ω tp i (ω t ) + ω 2 t TP r(ω t ) P(jω t ) 2 8 >< K i (ω 1 )2 K d b 1 > 0 K i (ω 2 )2 K d b 2 < 0 8 >< K i (ω 1 )2 K d b + 1 > 0 K i (ω 2 )2 K d b + 2 < 0 >: K i (ω 3 )2 K d b 3 > 0 K i (ω 4 )2 K d b 4 < 0 >: K i (ω 3 )2 K d b + 3 > 0 K i (ω 4 )2 K d b + 4 < 0 8 >< K i (ω + 1 )2 K d b 1 > 0 K i (ω + 2 )2 K d b 2 < 0 >: K i (ω + 3 )2 K d b 3 > 0 K i (ω + 4 )2 K d b 4 < 0

Data Based Design of 3 Term Controllers p. 63/10 A set of linear programmings b t = ω tp i (ω t ) + ω 2 t TP r(ω t ) P(jω t ) 2 8 >< K i (ω 1 )2 K d b 1 > 0 K i (ω 2 )2 K d b 2 < 0 8 >< K i (ω 1 )2 K d b + 1 > 0 K i (ω 2 )2 K d b + 2 < 0 >: K i (ω 3 )2 K d b 3 > 0 K i (ω 4 )2 K d b 4 < 0 >: K i (ω 3 )2 K d b + 3 > 0 K i (ω 4 )2 K d b + 4 < 0 8 >< K i (ω + 1 )2 K d b 1 > 0 K i (ω + 2 )2 K d b 2 < 0 8 >< K i (ω + 1 )2 K d b + 1 > 0 K i (ω + 2 )2 K d b + 2 < 0 >: K i (ω + 3 )2 K d b 3 > 0 K i (ω + 4 )2 K d b 4 < 0 >: K i (ω + 3 )2 K d b + 3 > 0 K i (ω + 4 )2 K d b + 4 < 0

Data Based Design of 3 Term Controllers p. 64/10 Stabilizing set for K p = 5 7 For k p =5 6 5 4 3 K d 2 1 0 1 2 5 0 5 10 15 20 25 30 K i

Digital ControlDesign Data Based Design of 3 Term Controllers p. 65/10

Data Based Design of 3 Term Controllers p. 66/10 Digital Control Design: Objectives To synthesize the set of all stabilizing PID controllers from experimental data of the discrete-time plant rather than from mathematical models. To synthesize the set of all stabilizing First Order controllers from experimental data of the discrete-time plant rather than from mathematical models.

Data Based Design of 3 Term Controllers p. 67/10 Preliminaries: Tchebyshev Decomposition Consider a real polynomial in z, P(z) = a n z n + a n 1 z n 1 + a n 2 z n 2 + + a 1 z + a 0.

Data Based Design of 3 Term Controllers p. 67/10 Preliminaries: Tchebyshev Decomposition Consider a real polynomial in z, P(z) = a n z n + a n 1 z n 1 + a n 2 z n 2 + + a 1 z + a 0. Since z k z=e jθ = coskθ + j sinkθ, we have P(e jθ ) = (a n cos nθ + + a 1 cos θ + a 0 ) + j (a n sinnθ + + a 1 sinθ)

Data Based Design of 3 Term Controllers p. 68/10 Tchebyshev Decomposition (continue...) We now have P(e jθ ) u= cos θ = R(u) + j p 1 u 2 T(u)

Data Based Design of 3 Term Controllers p. 68/10 Tchebyshev Decomposition (continue...) We now have P(e jθ ) u= cos θ = R(u) + j p 1 u 2 T(u) where R(u) = a n c n (u) + a n 1 c n 1 (u) + + a 1 c 1 (u) + a 0 T(u) = a n s n (u) + a n 1 s n 1 (u) + + a 1 s 1 (u) k c k (u) s k (u) 1 u 1 2 2u 2 1 2u 3 4u 3 + 3u 4u 2 1 4 8u 4 8u 2 + 1 8u 3 + 4u 5 16u 5 + 20u 3 5u 16u 4 12u 2 + 1

Data Based Design of 3 Term Controllers p. 69/10 Preliminaries (continue...) Consider a real rational function Q(z) = P 1(z) P 2 (z) where P 1 (z) and P 2 (z) are polynomials of real coefficients with degrees m and n, respectively. Assume that P 1 (z) and P 2 (z) have no zeros on the unit circle. Write Q(z) z= u+j 1 u 2 = R q(u) + j p 1 u 2 T q (u) where R q (u) = R(u) D(u), T q(u) = T(u) D(u)

Data Based Design of 3 Term Controllers p. 70/10 Preliminaries (continue...) Recall Q(z) z= u+j 1 u 2 = R(u) D(u) + j 1 u 2 T(u) D(u)

Data Based Design of 3 Term Controllers p. 70/10 Preliminaries (continue...) Recall Q(z) z= u+j 1 u 2 = R(u) D(u) + j 1 u 2 T(u) D(u) Since D(u) = R 2 2(u) + ( 1 u 2) T 2 2 (u) > 0 for all u [ 1, 1], the zeros of T q (u) are identical to those of T(u). Let t 1,,t k denote the real distinct zeros of T(u) of odd multiplicity ordered as 1 < t 1 < t 2 < < t k < +1.

Data Based Design of 3 Term Controllers p. 71/10 Schur Signature Lemma Let Q(z) be a real rational function with i z zeros and i p poles, respectively, inside the unit circle C and no zeros/poles on the unit circle and Q(z) z= u+ 1 u 2 = R(u) + j 1 u 2 T(u). Then the net change in phase of Q(e jθ ) as θ runs from 0 to π: π 0 Q(e jθ ) = π (i z i p ) = πσ[q(z)]

Data Based Design of 3 Term Controllers p. 72/10 Assumption The plant is stabilizable (mathematically, N(z) and D(z) are coprime). where P(z) = N(z) D(z)

Data Based Design of 3 Term Controllers p. 72/10 Assumption The plant is stabilizable (mathematically, N(z) and D(z) are coprime). where P(z) = N(z) D(z) The plant has no poles on the unit circle.

Data Based Design of 3 Term Controllers p. 72/10 Assumption The plant is stabilizable (mathematically, N(z) and D(z) are coprime). where P(z) = N(z) D(z) The plant has no poles on the unit circle. Only available information:

Data Based Design of 3 Term Controllers p. 73/10 PID Controller Design for Discrete-time Systems + C(z) P(z)

Data Based Design of 3 Term Controllers p. 73/10 PID Controller Design for Discrete-time Systems + C(z) P(z) Let the PID controller be of the form C(z) = K P + K ITz z 1 + K D(z 1) Tz

Data Based Design of 3 Term Controllers p. 73/10 PID Controller Design for Discrete-time Systems + C(z) P(z) Let the PID controller be of the form equivalently, C(z) = K P + K ITz z 1 + K D(z 1) Tz C(z) = K 2z 2 + K 1 z + K 0 z(z 1)

Data Based Design of 3 Term Controllers p. 74/10 PID Controller Design (continue...) The closed-loop characteristic polynomial is δ(z) := z(z 1)D(z) + `K 2 z 2 + K 1 z + K 0 N(z).

Data Based Design of 3 Term Controllers p. 74/10 PID Controller Design (continue...) The closed-loop characteristic polynomial is δ(z) := z(z 1)D(z) + `K 2 z 2 + K 1 z + K 0 N(z). Closed-loop stability requires σ(δ) = n + 2. Consider Π(z) = δ(z) D(z) = z(z 1) + `K 2 z 2 + K 1 z + K 0 P(z) Consequently, the stability requires that σ(π) = n + 2 i p where i p is the number of poles of the plant located inside unit circle.

Data Based Design of 3 Term Controllers p. 75/10 Stability Condition with PID Controllers Let P(z) be the plant with relative degree r. Let the PID controller be C(z) = K 2z 2 + K 1 z + K 0 z(z 1).

Data Based Design of 3 Term Controllers p. 75/10 Stability Condition with PID Controllers Let P(z) be the plant with relative degree r. Let the PID controller be C(z) = K 2z 2 + K 1 z + K 0 z(z 1) Then the closed-loop system is stable if and only if. σ(ν) = r + o z + 1 where ν(z) = z 1 P(z 1 )Π(z) Π(z) = z(z 1) + `K 2 z 2 + K 1 z + K 0 P(z) and o z is the number of non-minimum phase zeros of P(z).

Data Based Design of 3 Term Controllers p. 76/10 PID Controller Design (continue...) Now ν(z) z= u+jv = z 1 P(z 1 )Π(z) z= u+jv = ˆz 1 P(z 1 ) + `K 0 z 1 + K 1 + K 0 z P(z)P(z 1 ) z= u+jv = ( u 1 + jv) (R p (u) jvt p (u)) + [K 0 ( u jv) + K 1 + K 2 ( u + jv)] m 2 (u) where v = p 1 u 2, m(u) = P(z) z= u+jv.

Data Based Design of 3 Term Controllers p. 77/10 PID Controller Design (continue...) R ν (u, K 0, K 1, K 3 ) = (u 1)R p (u) (1 u 2 )T p (u) + K 1 m 2 (u) u(2k 0 + K 3 )m 2 (u) T ν (u, K 3 ) = R p (u) (u + 1)T p (u) + K 3 m 2 (u). Note that P(z) z= u+j 1 u 2 = R p(u) + j p 1 u 2 T p (u) R p (u) and T P (u) can be obtained directly from the experimental data P(e jθ ) u= cos θ,

Data Based Design of 3 Term Controllers p. 78/10 Algorithm: PID Controller Design For fixed K 3 = K 3, solve K 3 = R p(u) + (u + 1)T p (u) m 2 (u) = g(u) determine the roots u 1, u 2, u 0 = 1 < u 1 < u 2 < < u l < u l+1 = +1 Develop linear inequalities corresponding to stability as follows. Let I j = n o i j 0, ij 1,, ij l, ij l+1 denote a string where i j i {0,1, 1} such that i j 0 2ij 1 + 2ij 2 + ( 1)l+1 i j l+1 = r + o z + 1.

Data Based Design of 3 Term Controllers p. 79/10 Algorithm: PID Controller Design (continue...) For each string I j satisfying the above, we have the set of inequalities sgn [R ν (u t, K 0, K 1, K 3)] i j t > 0 which is a set of linear inequalities in K 0, K 1 space for fixed K 3. By constructing these inequalities for each string satisfying the above, we obtain the stabilizing set for K 3 = K 3.

Data Based Design of 3 Term Controllers p. 80/10 An Example: PID Design Available Information: Frequency domain data P(e jωt ) := j P(e jωt ), ω = 2π T ff sampled every T = 0.01.

Data Based Design of 3 Term Controllers p. 80/10 An Example: PID Design Available Information: Frequency domain data P(e jωt ) := j P(e jωt ), ω = 2π T ff sampled every T = 0.01. The plant is stable. In other words, the number of unstable poles of the plant is 0, that is, o p = 0.

Data Based Design of 3 Term Controllers p. 81/10 An Example: PID Design (continue...) The Nyquist plot of the plant P(z): 1.5 Nyquist Diagram 1 0.5 Imaginary Axis 0 0.5 1 1.5 1 0.5 0 0.5 1 1.5 Real Axis

Data Based Design of 3 Term Controllers p. 82/10 An Example: PID Design (continue...) Net phase: π 0 P(e jθ ) = π [r + (o z o p )] := 2π. o z = 2 r +o p = 2 2+0 = 0.

Data Based Design of 3 Term Controllers p. 82/10 An Example: PID Design (continue...) Net phase: π 0 P(e jθ ) = π [r + (o z o p )] := 2π. o z = 2 r +o p = 2 2+0 = 0. The stability requirement is equivalent to σ [ν(z)] = 2 + 0 + 1 = 3. Applying this to the signature lemma,! 1 2 sgn[t( 1)] sgn[r( 1)] 2sgn[R(t 1)] + 2sgn[R(t 2 )] sgn[r(1)] := 3 where t i are the zeros of g(u) in g(u) for fixed K 3.

Data Based Design of 3 Term Controllers p. 83/10 An Example: PID Design (continue...) It is easy to see that at least two zeros t i are required and also that the only feasible string of sign sequences is: sgn of T( 1) R( 1) R(t 1 ) R(t 2 ) R(1) 1 1 1 1 1 The feasible range of K 3 values is that corresponding to the requirement of two zeros in T(u). Plot the right hand side of g(u): g(u) =! 1 P c (u) 2 P r (u) (1 + u)p i (u) = K 3 where P(e jωt ) ωt=θ = P(e jθ ) u= cos θ = P r (u) + j p 1 u 2 P i (u).

Data Based Design of 3 Term Controllers p. 84/10 An Example: PID Design (continue...) 2 1 0 fesible region of K 3 values 1 K 3 2 3 4 5 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 u

Data Based Design of 3 Term Controllers p. 85/10 An Example: PID Design (continue...) At K 3 = 1.3, it is found from the graph that u = 0.4736 := t 1, u = 0.0264 := t 2.

Data Based Design of 3 Term Controllers p. 85/10 An Example: PID Design (continue...) At K 3 = 1.3, it is found from the graph that u = 0.4736 := t 1, u = 0.0264 := t 2. Then the set of linear inequalities corresponding to K 3 = 1.3 is T( 1) = 1 R( 1) = 2.3111 + 1.7778K 1 + 3.5556K 2 > 0 R( 0.4736) = 0.6939 + 0.7473K 1 + 0.7078K 2 < 0 R( 0.0264) = 0.7226 + 0.6403K 1 + 0.0338K 2 > 0 R(1) = 0.3556 + 1.7778K 1 3.5556K 2 < 0

Data Based Design of 3 Term Controllers p. 86/10 An Example: PID Design (continue...) By sweeping K 3 over ( 0.7,1.4), we have the stabilizing PID parameter regions shown: 2 1 K 3 0 1 3 2 K 2 1 0 1 1.5 1 0.5 K 1 0 0.5 1

Data Based Design of 3 Term Controllers p. 87/10 1st Order Controllers for Discrete-time Systems Consider the frequency response of the discrete-time plant P : P(z) z= u+jv = R p (u) + jvt p (u).

Data Based Design of 3 Term Controllers p. 87/10 1st Order Controllers for Discrete-time Systems Consider the frequency response of the discrete-time plant P : P(z) z= u+jv = R p (u) + jvt p (u). Note that R P (u) and T p (u) for 1 u 1 are immediately available from the given frequency response data points provided by P(e jθ ) for θ [0, π].

Data Based Design of 3 Term Controllers p. 88/10 Stability Condition with First Order Controllers Let P(z) be the plant with the number of unstable poles being o p. Let the first order controller be C(z) = x 1z + x 2 z + x 3. Then the closed-loop system is stable if and only if σ(π) = o p + 1 where Π(z) = (z + x 3 ) + (zx 1 + x 2 ) P(z) and o p is the number of non-minimum phase poles of P(z).

Data Based Design of 3 Term Controllers p. 89/10 1st Order Controllers Design Consider Π(z) = (z + x 3 ) + (zx 1 + x 2 ) P(z) and Π(z) z= u+jv = ( u + x 3 + jv) + ux 1 + x 2 + jvx 1! R p (u) + jvt p (u)! = (x 3 u) + R p (u)x 2 ur p (u) + v 2 T p (u)! x 1 +jv "! # R p (u) ut p (u) x 1 + T p (u)x 2 + 1.

Data Based Design of 3 Term Controllers p. 90/10 Stability Conditions For complex root crossing, we now have the expression of the curve in (x 1, x 2 ) space for every fixed x 3. 2 6 4 ur p (u) + v 2 T p (u)! v R p (u) ut p (u)! R p (u) vt p (u) 3 2 4 x 1(u) 7 5 x 2 (u) 3 5 = 2 4 (x 3 u) v 3 5. {z } A(u) Since det[a(u)] = `R p(u) 2 + v 2 Tp 2 (u) 2 v = v P(e jθ ),

Data Based Design of 3 Term Controllers p. 91/10 Stability Conditions (continue...) The two straight lines representing the real root crossing can be obtained by letting u = 1 and u = 1, equivalently letting θ = 0 and θ = π. (x 3 1) + P(e j0 )(x 2 x 1 ) = 0 (x 3 1) + P(e jπ )(x 2 x 1 ) = 0.

Data Based Design of 3 Term Controllers p. 92/10 An Example: First Order Controller Design Available Information: Frequency domain data P(e jωt ) := np(e jωt ), ω = π o T sampled every T = 0.01.

Data Based Design of 3 Term Controllers p. 92/10 An Example: First Order Controller Design Available Information: Frequency domain data P(e jωt ) := np(e jωt ), ω = π o T sampled every T = 0.01. The plant is stable, i.e., the number of poles outside the unit circle is 0, that is, o p = 0.

Data Based Design of 3 Term Controllers p. 93/10 An Example: First Order Controller Design (continue...) At x = 0.75, the following region is obtained. 1 0.8 0.6 0.4 Region 4 Region 3 0.2 x 2 0 0.2 Region 2 0.4 Region 1 0.6 0.8 1 1 0.8 0.6 0.4 0.2 0 0.2 0.4 0.6 0.8 1 x 1 Each separated region represents a set of controller parameters that gives a fixed number of unstable poles of the closed-loop system.

Data Based Design of 3 Term Controllers p. 94/10 An Example: First Order Controller Design (continue...) The following figure shows the Nyquist plots with selected controllers from the four specified regions. 6 A controller from Region 1 0.5 A controller from Region 2 4 Imag 2 Imag 0 0 2 5 0 5 10 Real A controller from Region 3 2 0.5 1.5 1 0.5 0 0.5 Real A controller from Region 4 5 Imag 0 2 4 Imag 0 5 6 4 2 0 2 4 Real 10 10 5 0 5 Real

Data Based Design of 3 Term Controllers p. 95/10 An Example: First Order Controller Design (continue...) The Nyquist plot with a controller from Region 1 shows that the encirclement around 1 point is 2π, i.e., N = 2. Since o p = 0, the corresponding closed-loop system will have 2 poles outside the unit circle. Similarly, corresponding closed-loops system with controllers from Region 3 and 4 will have 2 and 3 poles outside the unit circle, respectively.

Data Based Design of 3 Term Controllers p. 96/10 An Example: First Order Controller Design (continue...) By sweeping over x 3, we have the entire first order stabilizing controllers for the given plant. 1 0.5 x 3 0 0.5 1 0.5 x 2 0 0.5 0.2 0 0.2 0.4 0.6 0.8 1 x 1

Data Based Design of 3 Term Controllers p. 97/10 Data Measurement by Impulse Response Typically data is obtained from the measurement by a sinusoidal input.

Data Based Design of 3 Term Controllers p. 97/10 Data Measurement by Impulse Response Typically data is obtained from the measurement by a sinusoidal input. Alternative input - Impulse Output - Markov parameters y[k] = [m 0, m 1, m 2,, m k, ] Y (z) = m 0 + m 1 z 1 + + m k z k + By truncating the terms, we have approximation of the frequency response of the system. P(z) z= u+jv = Y n (z) z= u+jv where n is the number of the terms taken from Y (z).

Data Based Design of 3 Term Controllers p. 98/10 Example: Data Measurement by Impulse Input Markov parameter obtained y[k] = [0, 0, 1, 0, 0.25, 0, 0.0625, 0, 0.015625, 0, 0.00390625, ]

Data Based Design of 3 Term Controllers p. 98/10 Example: Data Measurement by Impulse Input Markov parameter obtained y[k] = [0, 0, 1, 0, 0.25, 0, 0.0625, 0, 0.015625, 0, 0.00390625, ] Stabilizing region with K 3 = 1.2 for n = 3,5,7,10, 20 2.5 2 n=3 K 2 1.5 n=7 1 0.5 n=5 n=10,20 0 1.5 1 0.5 0 0.5 1 K 1

Data Based Design of 3 Term Controllers p. 99/10 Data Measurement by Step Input Alternative input - Step Input Step Response parameters y s [k] = [y 0, y 1, y 2,, y k, ]

Data Based Design of 3 Term Controllers p. 99/10 Data Measurement by Step Input Alternative input - Step Input Step Response parameters y s [k] = [y 0, y 1, y 2,, y k, ] =» z 1 H(z) z= u+jv = Y s (z) z z= u+jv y 0 + y 1 z 1 + y 2 z 2 + + y k z k + `1 z 1 z= u+jv = y 0 + (y 1 y 0 )z 1 + (y 2 y 1 )z 2 + + (y k y k 1 )z k + = m 0 + m 1 z 1 + m 2 z 2 + + m k z k + where m 0, m 1,, m k, are the Markov parameters.

Data Based Design of 3 Term Controllers p. 100/10 Example: Data Measurement by Step Input The step response and the Markov parameters computed y s [k] = [0, 0, 1, 1, 1.25, 1.25, 1.3125, 1.3125, 1.328125, 1.328125, 1.33203125, m(k) = [0, 0, 1, 0, 0.25, 0, 0.0625, 0.015625, 0, 0.00390625, ]

Example: Data Measurement by Step Input The step response and the Markov parameters computed y s [k] = [0, 0, 1, 1, 1.25, 1.25, 1.3125, 1.3125, 1.328125, 1.328125, 1.33203125, m(k) = [0, 0, 1, 0, 0.25, 0, 0.0625, 0.015625, 0, 0.00390625, ] Stabilizing resion with K 3 = 1.2 for n = 5 2.5 For K 3 =1.2 2 1.5 K 2 1 0.5 0 1.5 1 0.5 0 0.5 1 K 1 Data Based Design of 3 Term Controllers p. 100/10

Data Based Design of 3 Term Controllers p. 101/10 Computer Aided Design using Labview The algorithm for the design of a PID Controller from the frequency response data of the system has programmed in LabVIEW due to its user-friendly graphical environment. The Virtual Instrument (VI) in Labview have a front panel that is displayed to the user and a block diagram, where the computations are performed. The inputs to the program are the frequency response data and the number of RHP poles of the system.

Data Based Design of 3 Term Controllers p. 102/10 CAD using LABVIEW(continue...) Left: The file containing the frequency response data from a stable system (number of right hand plane poles equals zero) is fed into the program through the file path box located at the top. Right: The front panel shows stabilizing sets of K p, K i and K d.

Data Based Design of 3 Term Controllers p. 103/10 CAD using LABVIEW(continue...) Left: T is set to a very small number.

Data Based Design of 3 Term Controllers p. 103/10 CAD using LABVIEW(continue...) Left: T is set to a very small number. g(ω) versus frequency generated from the inputs is used to compute ω values associated with a particular value of K p. Using the values of ω obtained above, linear equations are solved by the program to compute stabilizing sets of K i and K d for a fixed K p. As the selected K p is changed, the K i, K d graph changes dynamically to show the new stabilizing ranges of K i and K d for the selected K p.

Data Based Design of 3 Term Controllers p. 104/10 CAD using LABVIEW: 3D graph of stabilizing sets The stabilizing set of K p, K i and K d for the given system is shown.

Data Based Design of 3 Term Controllers p. 105/10 CAD using LABVIEW(continue...) Front Panel of VI showing performance indices for specific K p, K i and K d.

Data Based Design of 3 Term Controllers p. 106/10 CAD using LABVIEW(continue...) Front Panel of VI that satisfies multiple performance index specifications

Data Based Design of 3 Term Controllers p. 106/10 CAD using LABVIEW(continue...) Front Panel of VI that satisfies multiple performance index specifications The generated set of points satisfies multiple performance indices simultaneously. Performance indices used in the program: Gain Margin= 3db, Phase Margin= 45 o, Overshoot= 30%.