PID Controllers for Systems with Time-Delay 1
General Considerations CHARACTERISTIC EQUATIONS FOR DEALY SYSTEMS u(t) Delay T y(t) =u(t T ) u(t) _y(t) Integrator y(t) _y(t)+ay(t T )=u(t) a Delay T 2
u(t) u(t T ) Delay T _y(t) Integrator y(t) a _y(t) + ay(t) = u(t T ) u(t) Delay T _y(t) Integrator y(t) a _y(t) = ay(t T )+u(t T ) 3
u(t) Äy(t) Integrator _y(t) Integrator y(t) a 1 Delay T 1 a 0 Delay T 0 Äy(t) + a 1 _y(t T 1 ) + a 0 y(t T 0 ) = u(t) Let y(t) = x 1 (t); _y(t) = x 2 (t) Then _x 1 (t) _x 2 (t) = 0 1 x1 (t) 0 0 x1 (t T + 0 ) 0 0 x 2 (t) a 0 0 x 2 (t T 0 ) x1 + 0 0 x 1 (t T 1 ) + 0 u(t) : 0 a 1 x 2 (t T 1 ) 1 4
Stability of Delay Systems Let y(t)=e st be a proposed solution of Then we have y(t) Ä(t) + a 1 y(t _(t T 1 ) + a 0 y(t T 0 ) = 0 s 2 + a 1 e st 1 s + a 0 e st 0 e st 0 so that s must satisfy s 2 + a 1 se st 1 + a 0 e st 0 =0 Characteristic equation of the delay system. The location of its zeros determine the stability of the system. If any roots lie in the closed RHP, the system is unstable as the solution grows without bound. 5
Consider a LTI system with l distinct delays, _x(t) =A 0 x(t)+ lx Ai x(t T i )+Bu(t) i=1 The corresponding characteristic equation is Ã! lx ±(s) :=det si A 0 e st i A i = P 0 (s)+ i=1 n 1 n 1 and P 0 (s) =s n + X a i s i ; P k (s) = X (b k ) i s i i=0 i=0 mx P k (s)e L ks k=1 (Retarded Delay Systems) Äy(t)+a 1 _y(t T 1 )+a 0 y(t T 0 )=u(t) (Neutral Delay System) Äy(t T 2 ) + a 1 _y(t T 1 ) + a 0 y(t T 0 ) = u(t) 6
Roots of Characteristic ti Equations Retarded Systems: There can only be a finite number of RHP roots. The stability of retarded systems is equivalent to the absence of closed RHP roots. The fact that retarded systems have a finite number of RHP roots means that t one can count the number of roots crossing into the RHP through the stability boundary and keep track of the number of RHP roots as some parameter vary. Neutral Systems: Certain root chains can approach the imaginary axis from the LHP and thus destroy stability. If delays are multiples of a common delay, we have ±(s) = a s 2 s + k s 0 (s) + a 1 (s)e + a 2 (s)e + + a k (s)e 7
THE PADE APPROXIMATION AND ITS LIMITATIONS e sl» N r (sl) = D r (sl) where N r (sl) = D r (sl) = rx k=0 r rx k=0 (2r k)! k!(r k)! ( sl)k (2r k)! k!(r k)! (sl)k For example, the 3 rd order Pade approximation is given by N 3 (sl) D 3 (sl) = L3 s 3 +12L 2 s 2 60Ls +120 L 3 s 3 +12L 2 s 2 +60Ls +120 8
PID Stabilization ti of a Delay Systems Using a 1 st Od Order Pade Approximation (An Example) sl 1 2 Ls st Order Pade approximation e» = 2+Ls μ k sl ) k ( Ls +2) Plant G(s) = e» = Ts+1 (Ts+1) (Ls +2) With the PID controller (k p,k i,k d ), the closed-loop characteristic polynomial becomes ±(s; k p ; k i ; k d ) = s(t s + 1)(Ls + 2) + (k 2 i + k p s + k d s )(k)( Ls + 2) = (Ts 2 + s)(ls +2)+(kds 0 2 + ki)( Ls 0 +2)+kps( Ls 0 +2) where k 0 d = kk d ;k 0 i = kk i ;k 0 p = kk p : 9
Using the PID Design Algorithm, we have and 8 >< >: k i > 0 4(1 + kk p ) 2(1 + kk p)(2t + L kk p L) k i k d < L(4T + L kk p L) kl(4t + L kk p L) 1 k < k p < 1 k μ 1+ 4T L k d < T k For a fixed k p, it becomes the set of linear inequalities in terms p of k i, k d and can be solved by LP. 10
Question: Does the 1 st order Pade approximation accurately capture the actual set of stabilizing PID parameters for the original time-delay system? The stabilizing set of (k i,k d ) values for a fixed k p. Next Example 11
Example Plant G(s) = 1:6667 1 + 2:9036s e 0:2475s Plant with the 1 st order Pade approximation G m (s) = 1:6667 ( 0:1238s +1) (1 + 2:9036s) (0:1238s + 1) Compute the entire stabilizing PID parameter values. 12
k p = 8:4467 k i = 60 k d = 1:5 Time-response of the closed-loop loop system The stabilizing (k i,k d ) values at k p =8.4467 Showing unstable behavior! 13
Tried with the 2 nd, 3 rd, and 5 th order Pade approximation While the 2 nd order Pade approximation fails to capture the actual stabilizing set, the 3 rd and 5 th order Pade approximations apparently do a better job. 14
Example with large delay Plant G(s) = 1 1+s e 10s PID Controllers for Systems with Time-Delay Approximate the time-delay term using the 1 st, 2 nd, 3 rd, 5 th, 7 th, and 9 th order Pade approximations 1 st and 2 nd order approximations 3 rd, 5 th, 7 th, and 9th order approximations 15
Observations For small values of the time-delay, the approximate sets easily converge to the possible true sets. However, the convergence becomes more difficult as the value of the time-delay increases. The convergence of the approximate set to a possible true set improves with increased order of the Pade approximation. The Pade approximation is not a satisfactory tool for ensuring the stability of the resulting control design. It is not a priori i clear as to what order of the approximation will yield a stabilizing set of parameters accurately approximating the true set. 16
Pontryagin s results PID Controllers for Systems with Time-Delay THE HERMITE-BIEHLER THEOREM FOR QUASI-POLYNOMIALS Let f(s; t) be a polynomial in two variables with real or complex coefficients defined as follows: f(s; t) = MX N X ahks h t k h=0 k=0 Definition f(s; t) ) is said to have a principal term if there exists a nonzero coefficient a hk where both indices have maximal values. Without loss of generality, we will denote the principal term as a MNs M t N : This means that for each other term a hk s h t k ; for a hk =0; ; we have either M>h;N>k;orM = h; N > k; orm>h;n= k: Example f(s; t) ) =3s + t 2 does not have a principal term but the polynomial f(s; t) = s 2 + t +2s 2 t does. 17
Theorem (Pontryagin) f(s; t) If the polynomial function F (s) = f(s; e s ) large positive real parts. does not have a principal term, then the has an infinite number of zeros with arbitrarily f(s; t) If does have a principal term, the main result of Pontryagin is to show that the Hermite-Biehler Theorem extends to the class of functions F (s) =f(s; e s ): 18
Study of the zeros of functions of the form g(s,cos(s),sin(s)) Let g(s,u,v) be a polynomial with real coefficients: MX NX g(s; u; v) = s h Á (k) h (u; v) h=0 k=0 Á (k) h (u; v) is a polynomial l of degree k, k homogeneous in u and v. Assume that Á (k) 2 2 h (u; v) is not divisible by u + v : Á (k) h (1; j) 6= 0 Let Á (N) (u; v) = NX k=0 Á (k) M (u; v) the coefficient of s M 19
Consider G(s) = g(s; cos(s); sin(s)) Let (N) (s) := Á (N) (cos(s); ( sin(s)) THEOREM Let g(s,u,v) be a polynomial l with principal i term given by s M Á (N) M (u;v): If η is such that (N) ( + j!) does not take the value zero for real ω, then starting from some sufficiently large value of l; the function G(s) will have exactly 4lN + M zeros in the strip 2l¼ + Re[s] 2l¼ + : Thus for the function G(s) to have only real roots, it is necessary and sufficient that in the interval 2l¼ + Re[s] 2l¼ + ; it has exactly real roots starting ti with some sufficiently i large l 4lN + M l: 20
Consider f(s; t) = MX h=0 NX a hk s h t k = s M X (N) (t)+ k=0 M 1 X h=0 NX a hk s h t k k=0 X (N) (t) = NX amkt k k=0 Definition s Let F (s) = f(s; e s ): where f(s,t) is a polynomial l with a principal i term, and F (j!) = F r (!) + jf i (!) Let ω r1, ω r2, ω r3, denote the real roots of F r (ω), and let ω i1, ω i2, ω i3, denote the real roots of F i (ω), both arranged in ascending order of magnitude. The we say that the roots of F r (ω) and F i (ω) interlace if they satisfy the following property:! r1 <! i1 <! r2 <! i2 < 21
THEOREM (HB Theorem to quasi-polynomial) PID Controllers for Systems with Time-Delay If all the roots of F(s) lie in the open LHP, then the roots of F r (ω) and F i (ω) are real, simple, interlacing, and F 0 i (!)F r (!) F i (!)Fr(!) 0 > 0 ( ) for each ω (-, ), where F r (ω) and F i (ω) denote the first derivative with respect to ω of F r(ω) and F i(ω), respectively. Moreover, in order that all the roots of F(s) lie in the open LHP, it is sufficient that one of the following conditions be satisfied: 1. All the roots of F r (ω) and F i (ω) are real, simple, and interlacing and the inequality (*) is satisfied for at least one value of ω; 2. All the roots of F r (ω) are real and for each root, (*) is satisfied, i.e., F i (! r )F 0 r(! r ) < 0; 3. All the roots of F i (ω) are real and for each root, (*) is satisfied, i.e., F 0, i i (! i )F r (! i ) > 0: 22
THEOREM If the function X (N) (e s ) has roots in the open RHP, then the function F(s) has an unbounded set of zeros in the open RHP. If all the zeros of the function X (N) (e s ) lie in the open LHP, then the function F(s) can only have a bounded set of zeros in the open RHP. 23
Application to Control Theory Classes of Quasi-polynomials: Retarded-type (or delay-type) Quasi-polynomials: This class consists of quasi-polynomials whose asymptotic chains go deep into the open LHP. Neutral-type quasi-polynomials: This class consists of quasipolynomials that along with delay-type chains contain at least one asymptotic chain of roots in a vertical strip of the complex plane. Forestall-type quasi-polynomials: This class consists of quasipolynomials with at least one asymptotic chain that goes deep into the open RHP. 24
Definition A delay-type quasi-polynomial is said to be stable iff all its roots have negative real parts. Definition A neutral-type quasi-polynomial is said to be stable if there exists a positive number σ such that the real parts of all its roots are less than σ. 25
THEOREM Let ± (s) =e sl m d(s)+e s(l m L 1 ) n 1 (s)+e s(l m L 2 ) n 2 (s)+ + n m (s) and write ± (j!)=± r (!)+j± i (!): Under the following conditions (A1) (A2) deg[d(s)] = q and deg[n i (s)] q for i =1; 2;:::;m; 0 <L 1 <L 2 < <L m ± (s) is stable iff 1. ± r (!) and ± i (!) have only simple, real roots and these interlace, 2. ± 0 i (! o)± r (! o ) ± i (! o )± 0 r (! o) > 0; for some! o 2 ( 1; 1): 26
Example Plant G(s) = 1 ; C(s) =k p + k i = k ps + k i 2s + 1 s s With k p =1.8, k i =0.2, we have ±(s) =2s 2 +2:8s +0:2 and it is stable. Consider G(s) = 1 2s + 1 e 10s With k p =1.8 and k i =0.2, the characteristic equation of the closed-loop cosed oop system is: ±(s) =2s 2 + s +(1:8s +0:2)e 10s =0 For analyzing the stability, consider ± (s) =e 10s ±(s) =(2s 2 + s)e 10s +1:8s +0:2 27
The real and imaginary parts are given by ± r (!) = 0:2! sin(10!) 2! 2 cos(10!) ± i (!) =![1:8+cos(10!) 2! sin(10!)] : Shows interlacing. Shows instability 28
Analysis PID Controllers for Systems with Time-Delay 1. The example illustrates the case of a time-delay system that satisfies the interlacing and monotonic phase increase properties but fails to be stable. 2. The reason for this behavior lies in the nature of the roots of real and imaginary parts of the polynomial: they are not all real. THEOREM (Pontyagin) Let M and N denote the highest powers of s and e s, respectively, in δ * (s). Let η be an appropriate constant such that the coefficients of terms of highest degree in δ r (ω) and δ (ω) i do not vanish at ω=η. Then for the equations δ r r( (ω)=0 or δ i i( (ω)=0 to have only real roots, it is necessary and sufficient that in each of the intervals 2l¼ +! 2l¼ + l = l o ;l o +1;l o +2; 4lN + M l 0 : δ r (ω) or δ i (ω) have exactly real roots for a sufficiently large 29
Let ^s =10s ^± (^s) = (0:02^s 2 +0:1^s)e^s +0:18^s +0:2 The real and imaginary parts of the new quasi-polynomial is ^± r (^!) = 0:2 0:1^! sin(^!) 0:02^! 2 cos(^!) ^± i (^!) = ^![0:18 + 0:1cos(^!) 0:02^! sin(^!)] : The roots of ^± i (^!) =0 ^! o =0; ^! 1 =8:0812; ^! 2 =8:8519; ^! 3 =13:5896; ^! 4 =15:4332; ^! 5 =19:5618; ^! 6 =21:8025; Choose = ¼ 4 30
1. ± ^± i (^) (!) has only one real root in [0, 2π-π/4]; the root at the origin. 2. Since ^± i (^!) is an odd function, in the interval [-7π/4, 7π/4], ± ^ i (^!) will have only one real root. 3. ^± i (^!) has no real roots in the interval [7π/4, 9π/4]; ± ^ i (^!) has only one real root in [-2π+π/4, 2π+π/4] which does not sum up to 4N + M = 6 for l 0 = 1: 4. Let l 0 =2so the requirement on the number if real roots is 8N+M=10. ^±i (^!) hasonlyfiverealrootsin[-4π+π/4, roots in 4π+π/4]. 5. Following the same procedure for l =3; 4;::: we see that the number of real roots of ^± i (^!) in [-2 lπ+π/4, 2 π+π/4] l is always less than 4lN + M =4l +2: 6. We conclude that the roots of ± ^ i (^!) are not all real. 31
STABILITY OF SYSTEMS WITH A SINGLE DELAY Consider the characteristic equation ±(s; L) =d(s)+n(s)e Ls =0 Problem: Determine the ranges of values of L for which all the roots of the characteristic equation lie in the LHP. A systematic procedure to analyze the behavior of the roots of the characteristic ti polynomial l as L increases from 0 to. 32
Walton and Marshall ss Procedure Step 1: Examine the stability at L=0. Step 2: Examine the behavior of the roots as increasing L from 0 to an infinitesimally small and positive. The number of roots changes from being finite to infinite. For an infinitesimally small L, the new roots must come in at infinity. Otherwise, e -Ls 1 and no new roots. Determine where in complex plane these new roots arise. If deg[n(s)]<deg[d(s)], the roots s is large iff e -Ls is large (i.e., Re[s]<0) New roots occur in the open LHP If deg[n(s)]=deg[d(s)], the location of the roots is determined by the sign of W(ω( 2 ) for large ω. 33
Step 3: Examine potential crossing points on the imaginary axis (we separately consider the case s=0) Consider ½ d(j!)+n(j!)e jl! =0 d( j!)+n( j!)e jl! =0 d(j!)d( j!) n(j!)n( j!) =0 W(! 2 ) := d(j!)d( j!) n(j!)n( j!) Remark If no positive roots of W(ω 2 )=0, then no values of L for which ±(j!;l) =0 If deg[n(s)]<deg[d(s)] and W(ω 2 ) has no positive real roots, then there is no change in stability: The system will be stable for all L 0 if the system is stable at L=0. L The system will be unstable for all L 0 if the system is unstable at L=0. 34
Case when s=0 In this case, we have only one equation d(0) ( ) + n(0) ( ) =0 ) d(0) ( ) + e L0 n(0) ( ) =0; ; for all nite L The system is unstable for all values of L and for analysis this solution can be ignored. To find L, d(j!)+n(j!)e jl! =0 ) e jl! = d(j!) n(j!) := cos(l!) j sin(l!) Once we have found a value of L at which there is a root of the characteristic equation on the imaginary axis, we need to determine if the root crosses the imaginary axis and in which direction or if it merely touches the imaginary axis. 35
ds ds ds Re > 0 Re < 0 Re = 0 dl dl dl destabilizing stabilizing Necessary to consider high-order derivatives After some manipulations, we have S =sgn W 0 (! 2 ) = ½ 1; destabilizing +1; stabilizing 36
Example ±(s; L) =s +2e Ls PID Controllers for Systems with Time-Delay 1. Examine δ(s,0)=s+2, so the system is stable for L=0. 2. Since deg[d(s)]=1>deg[n(s)]=0, we skip step 2. 3. From d(s)=s, n(s)=2, we have W(ω 2 )= ω 2-4. W (ω 2 )=1>0. Since S=sgn[W (ω 2 )]=1, the root is destabilizing. The corresponding values of L are 8 >< cos(l!) =Re j! =0 2 >: sin(l!) =Im j! =1 2 L =(4k +1) ¼ ;k =0; 1; 2; 4 37
At L=π/4, L two roots of δ(s,l)=0 0 cross from left to right of the imaginary axis. At L=5π/4, two more roots cross from left to right of the imaginary axis and so on. Conclusion The region of stability is 0 L< π/4 38
FIRST ORDER SYSTEMS WITH TIME-DELAY Plant: G(s) = k 1 + T s e Ls PID Controller: C(s) =k p + k i s + k ds Stability Conditions for Delay free Systems Characteristic Polynomial without time-delay: ±(s) =(T + kk d )s 2 +(1+kk p )s + kk i Assuming k>0, we have ½ k > 1 ;k i > 0; k d > T p k k ¾ or ½ k < 1 ;k i < 0; k d < T ¾ p k k 39
Characteristic Polynomial with time-delay: ±(s) =(kk i + kk p s + kk d s 2 )e Ls +(1+Ts)s Write e Ls ±(s) =kk i + kk p s + kk d s 2 +(1+Ts)se Ls =: ± (s) Substituting s=jω, ± (j!) =± r (!)+j± i (!) where ± r (!) = kk i kk d! 2! sin(l!) T! 2 cos(l!) ± i (!) =! [kk p +cos(l!) T!sin(L!)] p We now separately treat the two cases: open-loop stable and open-loop unstable plants. 40
Open-loop Stable Plant Plant: G(s) = k 1 + T s e Ls T>0 (for stable plants) ± (j!) = ± r (!) + j± i (!) ± r (!) = kk i kk d! 2! sin(l!) T! 2 cos(l!) ± i (!) =! [kk p +cos(l!) T!sin(L!)] k p only affects δ i ω. k i and k d affect δ r ω. Parameters appear affinely in δ r ω and δ i ω. For stability, δ r ω andδ i ω must have all real roots and these roots must interlace. 41
Lemma The imaginary part of δ* jω has only simple real roots iff T 1 k <k p < 1 T k L 1 sin( 1 ) cos( 1 ) where α 1 is the solution of the equation tan( ) = T T + L in the interval (0, π. This lemma gives the ranges of k p. 42
Let z=ωl 0, L then ± r (z) = k L 2 z2 [ k d + m(z)k i + b(z)] where m(z) = L2 z 2 ; b(z) = L kz sin(z)+ TL z cos(z) Lemma For each value of kp in the range, the necessary and sufficient conditions on k i and k d for the roots of δ r (z) and δ i (z) to interface is the following infinite set of inequalities: k i > 0; k d > m 1 k i + b 1 ; k d < m 2 k i + b 2 ; k d > m 3 k i + b 3 ; k d <m 4 k i + b 4 ; where the parameters m j and b j for j=1,2,3, are given by m j := m(z j ); b j := b(z j ): 43
Theorem The range of k p values for which a given open-loop stable plant, with transfer function considered, can be stabilized using a PID controller is given by T 1 k <k p < 1 T k L 1 sin( 1 ) cos( 1 ) where α 1 is the solution of the equation tan( ) = T T + L in the interval 0,π. For k p values outside this range, there are no stabilizing i PID controllers. The complete stabilizing i region is given by: 44
For each k p 1/k,1/k, /k the cross section of the stabilizing region in the k i, k d spaceisthe trapezoid T; For k p=1/k, the cross-section of the stabilizing region in the (k i,k d ) space is the triangle Δ; For each k p 1/kk 1/k,k u, the cross section of the stabilizing region in the k i,k d space is the quadrilateral Q. m j = L2 ; zj 2 b j = L kz j! j = z j kl sin(z j )+ T L z j cos(z j ) sin(z j )+ T z j(cos(z j )+1) L where z j are the real, positive solutions of kk p +cos(z) T z sin(z) =0 L 45
Algorithm for Determining Stabilizing PID Parameters 1. Initialize k p =-1/k and step=(k u +1/k)/(N+1) where N is the desir ed number of points and k u = 1 T k L 1 sin( 1 ) cos( 1 ) 2. Set K p =k p +step; 3. If k p <k u, then go to 4. Else terminate the algorithm. 4. Find the roots z 1 and z 2 of kk p +cos(z) T z sin(z) =0: L 5. Compute the parameters m j and b j, j=1,2 associated with the z j. 6. Determine the stabilizing region in the (k i,k d ) space. 7. Go to 2. 46
Example (Location of Z-N solution in the set) 0:1 G(s) ( ) = 0:01s 01 + 1 e 0:1s Stabilizing parameter set obtained by Ziegler-Nichols step response method. Stabilizing region Stabilizing parameter set with the largest stability radius. K p =1.2 47
Example (Set using Pade Approximation vs. Set using a True Delay System) Set from the true delay system 3D stabilizing set Set from the 1 st order Pade approximation (It contains destabilizing parameters) 48
Open-loop Unstable Plant Plant: G(s) = k 1 + T s e Ls T<0 (for unstable plants) Lemma For T/L >0.5, δ i jω) has only simple real roots iff 1 T 1 sin( 1 ) cos( 1 ) <k < 1 p k L k where α 1 is the solution of the equation T tan( ) = T + L in the interval (0,π. In the special case of T/L =1, we have α 1 π/2. For T/L 0.5, the roots of δ i jω) are not all real. 49
Let z=ωl 0, ± r (z) = k L 2 z2 [ k d + m(z)k i + b(z)] where m(z) = L2 z 2 ; b(z) = L kz sin(z)+ TL z cos(z) Lemma For each value of k p in the range, the necessary and sufficient conditions on k i and k d for the roots of δ r (z) and δ i (z) to interlace are the following infinite set of inequalities: k i < 0; k d <m 1 k i + b 1 ; k d>m 2 k i + b 2 ; k d <m 3 k i + b 3 ; k d > m 4 k i + b 4 ; where the parameters m j and b j for j=1,2,3, are given by m j := m(z j ); b j := b(z j ) j j j j 50
Theorem A necessary and sufficient condition for the existence of a stabilizing PID controller for the open-loop unstable plant considered is T/L >0.5. If this condition is satisfied, then the range of k p values for which a given open-loop unstable plant, with transfer function considered, can be stabilized using a PID controller is given by 1 T k L 1 sin( 1 ) cos( 1 ) where α 1 is the solution of the equation tan( ) = T T + L <k p < 1 k in the interval (0,π). In the special case of T/L =1, we have α 1 π/2. For k p values outside this range, there are no stabilizing PID controllers. Moreover, the complete stabilizing region is given: 51
For each k p k l, 1/k, the cross section of the stabilizing region in the k i,k d space is quadrilateral Q. The stabilizing region of (k i,k d ) for k l <k p <-1/k where k l := 1 T 1 sin( 1 ) cos( 1 ) k L 52
Example Consider a process defined by dy(t) = 0:25y(t) 0:25u(t 0:8) G(s) = 1 e 0:8s dt 1 4s The stabilizing region of (k p,k i,k d ) values for the PID controllers. (-8.6876<k p <-1) 53
ARBITRARY LTI SYSTEMS WITH A SINGLE TIME-DELAY Tsypkin proposed a method to extend the Nyquist criterion to deal with time-delay systems (1946). This may lead to misleading conclusions unless care is taken. Example G(s) = 2s +1 s +2 The closed-loop system is stable with unity negative feedback. According to Tsypkin, theclosed-loop loop system should tolerate a time-delay upto 3.7851. However, when we add a 1 second delay to the nominal transfer function, the closed-loop system becomes unstable. 54
The Nyquist plot intersects the unit circle at ω 0 =1. The closed-loop system should tolerate a time-delay upto L 0 = ¼ +argg(j! 0) =3:7851:! 0 The closed-loop system is unstable with a 1 second delay. 55
Pontryagin s Theory vs. the Nyquist Criterion PID Controllers for Systems with Time-Delay Let h(z,t) be a polynomial in the two variable z and t with constant coefficients, h(z; t) = X a m n mn z t m;n The term a rs z r t s is called the principle term of the polynomial if a rs 0 and r and s each attain their maximum. Write h(z;t) = (s) r (t)z r +  (s) r 1(t)z r 1 + +  (s) 1 (t)z +  (s) 0 (t); where  j (t); j = 0; 1; 2; : : : ; r are polynomials in t with degree at most equal to s.  (s) 56
Two Theorems of Pontryagin to Clarify Nyquist Criterion Based Conditions for Systems with Time-delay Theorem If the polynomial h(z;t) = X a mn z m t n has no principal term, then the function m;n H(z) = h(z;e z ) has an unbounded number of zeros with arbitrary large positive real part. Theorem Let H(z)=h(z,e z ) where h(z,t) is a polynomial with principal term a rs z r t s. If the function χ s r e z has roots in the open RHP, then the function H(z) has an unbounded set of zeros in the open RHP. If all the zeros of the function χ s r e z lie in the open LHP, then the function H(z) has no more than a bounded d set of zeros in the open RHP. 57
Conditions which h should be satisfied when using the Nyquist criterion with the conventional Nyquist contour Theorem Suppose that we are given a unity feedback system with an open-loop transfer function N(s) ) Ls Ls G(s) = G 0 (s)e = e D(s) where N(s) and D(s) are real polynomials of degree m and n, respectively and L is a fixed delay. Then we have the following conclusions: 1. If n<m, or, n=m and b n /a n 1 where an, bn are the leading coefficients of D(s) and N(s), respectively, the Nyquist criterion is not applicable and the system is unstable according to Pontryagin s theorems. 2. If n>m, or, n=m and b n /a n <1, the Nyquist criterion is applicable and we can use it to check the stability of the closed-loop system. 58
It is appropriate p to point out that most likely Typkin assumed the plant to be strictly proper, though he did not state it explicitly in the literature. Attaching a PID controller to a proper or strictly proper plant opens up the very real possibility of ending up with an improper or a proper open-loop transfer function. This is the reason that the above investigation had to be undertaken. 59
Solution Approach PID Controllers for Systems with Time-Delay 1. Find the complete set of k s which stabilize the delay-free plant P 0 (s) and denote this set as S 0. 2. Define the set S N,whichisthesetofk s is set k s such that C(s,k)P 0 (s) is an improper transfer function or lim jc(s; k)p 0(s)j 1 s!1 Note that the elements in S N make the closed-loop system unstable after the delay is introduced. Exclude S N from S 0 and denote the new set by S 1, that is, S 1 = S 0 ns N 3. Compute the set S L : S L = k j k 62 S N and 9L 2 [0;L 0 ];! 2 R; s:t:c(j!)p 0 (j!)e jl! = 1 ª S L is the set of k s which make C(s,k)P(s) have a minimal destabilizing delay that is less than or equal to L 0. S R = S 1 ns L 4. The set is the solution 60
Theorem The set of controllers C(s,k) denoted by S R is the complete set of controllers in the unity feedback configuration that stabilize the plant P(s) with delay L from 0 up to L 0. Proportional Controllers N(s) Plant and controller: P (s) = P Ls Ls 0 (s)e = e ; C(s) = k p D(s) To implement the method, the key is to find S L. 61
The point the Nyquist curve crossing (-1,0): 10) Find L and ω satisfying i C(j!)P 0 (j!)e jl! = 1 arg[k p P 0 (j!)] L! = 2h¼ ¼; h 2 Z jk p P 0 (j!)j = 1: L(!;k p ) = arg[k pp 0 (j!)] + ¼! 1 k p (!) = jp 0 (j!)j : 62
For k p >0, L(!;k p )=L(!) = arg[p 0(j!)] + ¼! Solve L(ω) L 0 to get a set of ω: Ω Set of k p >0 corresponding to Ω : S L S L+ consists of all the positive k p s that make the system have poles on the imaginary axis for certain L L 0. For K p <0, Ω : a set of ω for L(ω) L 0 S L : a set of k p <0 corresponding to Ω The complete set S = + S L : S L S L [ S L 63
Algorithm for P Controllers 1. Compute the delay-free stabilizing k p set, S 0 2. Find S N If deg[n(s)]>deg[d(s)], S N =R. i.e., S R = If deg[n(s)]<deg[d(s)], S N = If deg[n(s)]=deg[d(s)], ½ ¾ S an N = k p j jk p j bn ; where an, b n are the leading coefficients of D(s) and N(s). 3. Compute S 1 = S 0 ns N 4. Compute S L 5. Compute S R = S 1 ns L 64
Example P (s) = s 2 +3s 2 e Ls with delay up to L s 3 +2s 2 0 =1:8 +3s +2 For the delay-free plant, the stabilizing k p range S 0 =(-0.4093,1). Since deg[n(s)]=2<deg[d(s)], S N = and S 1 =S 0 For k p >0, Ω 1.5129, 65
The corresponding S L + = [0.4473, + 66
For k p <0, Ω 0.7359, p 1.3312 2.6817, The corresponding S L 0.6025, 0.4135, 1.3691 S R = S 1 ns L = ( 0:4093; 1)n([0:4473; +1) [ [ 0:6025; 0:4135] [ ( 1; 1:3691]) = ( 0:4093; 0:4473) 67
PI Controllers PI Controller: C(s) =k p + k i s = k ps + k i s Open-loop transfer function: G(s) =C(s)P (s) =C(s)P 0 (s)e Ls = G 0 (s)e Ls μ kp s + k i N(s) Consider G 0 (s) =C(s)P 0 (s) = s D(s) = (k p s + k i ) N(s) sd(s) {z } R 0 (s) Magnitude and phase conditions arg[(k i + jk p!)r 0 (j!)] L! = ¼¼ j(k i + jk p!)r 0 (j!)j = 1 68
Rewrite the magnitude and phase conditions, L(!; k p ;k i ) = arg[(k i + jk p!)r 0 (j!)] + ¼ s! 1 k i = jr 0 (j!)j 2 k2 p! 2 : Fix k p, then M(!) = 1 jr 0 (j!)j 2 k2 p!2 k i = p M(!) Note that only those ω s with M(ω) 0 need consideration when computing S L. 69
Algorithm for PI Controllers 1. Compute S 0 2. Compute S N If deg[n(s)]>deg[d(s)], S N =R 2, i.e., S R = If deg[n(s)]<deg[d(s)], S N = ½ If deg[n(s)]=deg[d(s)], S N = (k p ;k i )jk p ;k i 2 R and jk p j 3. Compute where a n, b n are leading coefficients of D(s) and N(s). S 1 = S 0 ns N 4. For a fixed k p, find S R,kp p, R,kp Determine the sets Ω and Determine the sets Ω and S + L;k p : S L;k p : a n b n ¾ 70
5. Compute S L;kp = S + L;k p [S L;k p S R;kp = S 1;kp ns L;kp p p p 6. By sweeping over k p, the complete set of PI controllers that stabilize all plant with delay up to L 0 S R = [ k p S R;kp k p 71
PID Controllers for an Arbitrary LTI Plant with Delay where G(s) =C(s)P 0 (s)e Ls = G 0 (s)e Ls G 0 (s) =C(s)P 0 (s) = k ds 2 + k p s + k i N(s) s D(s) = (k d s 2 N(s) + k p s + k i ) sd(s) {z } R 0 (s) The magnitude and phase conditions: arg[(k i k d! 2 + jk p!)r 0 (j!)] L! = ¼ j(k i k d! 2 + jk p!)r 0 (j!)j = 1 72
Rewrite the phase and magnitude conditions, PID Controllers for Systems with Time-Delay L(!;k p ;k i ;k d ) = ¼ +arg([(k i k d! 2 )+jk p!] R 0 (j!)) s! k i k d! 2 1 = jr 0 (j!)j (k p!) 2 : 2 For fixed k p, 1 M(!) = jr0 (j!)j (k p!) 2 2 k i k d! 2 = p M(!) Similar to the PI case, we only need to consider ω s with M(ω) 0 when computing S L. 73
Algorithm for PID Controllers 1. Compute S 0 2. Compute S N If deg[n(s)]>deg[d(s)]-1, S N =R 3, i.e., S R = If deg[n(s)]<deg[d(s)]-1, S N = If deg[n(s)]=deg[d(s)]-1, ½ ¾ a n S N = (k p ; k i ; k d )jk p ; k i ; k d 2 R and jk d j b n 1 3. Compute where a n, b n 1 are leading coefficients of D(s) () and N(s). () S 1 = S 0 ns N 4. For a fixed k p, determine the set S R,kp 74
Determine the set Ω and S + Lk L,kp ½ Ð + =! j!>0andm(!) 0and p L(!) = ¼ +argf[ M(!) + jk p!] R 0 (j!)g! ½ (k i ; k d ) j (k i ; k d ) =2 S N;kp and 9! 2 Ð + S + L;k p = ( i ; d) j ( i ; d) = N;kp such that k i k d! 2 = p ¾ M(!) : ¾ L 0 Note that S L,kp + is a set of straight lines in the (k i,k d ) space. Determine the sets Ω and S L,kp Compute S L;kp = S + L;k p [ S L;kp and S R;kp = S 1;kp ns L;kp 5. By sweeping over k p, the complete set of PID controllers that stabilize all plants with delay up to L 0 : S R = [ k p S R;kp 75
Example P (s) = k Ts+1 e Ls ; L 2 [0;L 0 ] PID Controllers for Systems with Time-Delay The stabilizing PID parameters for the delay-free plant are: ½ S 0 = (k p ;k i ;k d ) j k p > 1 ;ki > 0;k d > T or kp < 1 ;ki < 0;k d < T k k k k ¾ Since deg[d(s)]-deg[n(s)]=1, ½ ¾ T T S N = (k p ;k i ;k d ) j k p ;k i ;k d 2 R and jk d j k Assuming k>0, we have 8 < (kp ;k i ;k d ) j k p > 1;k T k i > 0; >k ª k d > T for T>0 k S 1 = S 0 ns N = : ª (kp ; k i ; k d ) j k p < 1; k k i < 0; T < k k d < T for T < 0 k 76
For T>0, with different k p values, the stabilizing regions of (k i,k d )take on different but simple shapes: For -1/k<k p 1/k, S R,kp is a trapezoid. (a) For k p >1/k, S R,kp is a quadrilateral. (b) and (c) 77
Example P (s) = s 3 4s 2 + s +2 s 5 +8s 4 +32s 3 +46s 2 +46s +17 e Ls with L up to L 0 =1, that is, for all L 0, 1. Fix k p =1, compute the stabilizing k i, k d values for the delay-free plant, say S 0,kp. Stabilizing region of (k i,k d ) with k p =1 for delay-free system S 0;kp =1 78
Since deg[d(s)]-deg[n(s)]>1, ()] d [ ()] S N = and S 1 S 0. For k i k d! 2 = p M(!) > 0; the set of ω where L(ω) L 0 is Ω 0.524825, 0.742302 2.57318, L(ω) vs. ω with k i k d! 2 = p M(!) 79
The corresponding values of p M(!) ) S L,kp + : the straight lines defined by k 2 p + i k d! = M(!) for! 2 Ð PID Controllers for Systems with Time-Delay p M(!) vs:!with kp =1 80
p For k i k d! 2 = M(!) ) < 0; Ω - = [1.35894, 1.8659] 4.37326, L(!) vs:!with k i k d! 2 = p M(!) 81
Finally, we exclude S + Lk L,kp and S Lk L,kp from S 1k 1,kp to get S Rk R,kp Stabilizing region of (k i, k d )with k p =1 for plant with delay up to 1. 82