PID Controllers for S ystems Systems with with Time- Time Delay
|
|
- Jack Cannon
- 6 years ago
- Views:
Transcription
1 PID Controllers for Systems with Time-Delay 1
2 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
3 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
4 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 ) x x 1 (t T 1 ) + 0 u(t) : 0 a 1 x 2 (t T 1 ) 1 4
5 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
6 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
7 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
8 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
9 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 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
10 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
11 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
12 Example Plant G(s) = 1: :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
13 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 = Showing unstable behavior! 13
14 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
15 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
16 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
17 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
18 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
19 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
20 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
21 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
22 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
23 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
24 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
25 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
26 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
27 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
28 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
29 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
30 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
31 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
32 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
33 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
34 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
35 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
36 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
37 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 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
38 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
39 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
40 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
41 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
42 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
43 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
44 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
45 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
46 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
47 Example (Location of Z-N solution in the set) 0:1 G(s) ( ) = 0:01s 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
48 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
49 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
50 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
51 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
52 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
53 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. ( <k p <-1) 53
54 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 However, when we add a 1 second delay to the nominal transfer function, the closed-loop system becomes unstable. 54
55 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
56 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  (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
57 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
58 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
59 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
60 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 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
61 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
62 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
63 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
64 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
65 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 =( ,1). Since deg[n(s)]=2<deg[d(s)], S N = and S 1 =S 0 For k p >0, Ω , 65
66 The corresponding S L + = [0.4473, + 66
67 For k p <0, Ω , p , The corresponding S L , , 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
68 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
69 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
70 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
71 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
72 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
73 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
74 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
75 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
76 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
77 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
78 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
79 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 , , L(ω) vs. ω with k i k d! 2 = p M(!) 79
80 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
81 p For k i k d! 2 = M(!) ) < 0; Ω - = [ , ] , L(!) vs:!with k i k d! 2 = p M(!) 81
82 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
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
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control
More informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationThe Complete Set of PID Controllers with Guaranteed Gain and Phase Margins
Proceedings of the th IEEE onference on Decision and ontrol, and the European ontrol onference 5 Seville, Spain, December -5, 5 ThA7.5 The omplete Set of PID ontrollers with Guaranteed Gain and Phase Margins
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closed-loop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s) - H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closed-loop system when the gain K changes from 0 to 1+ K G ( s)
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Routh-Hurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More informationLaplace Transform Analysis of Signals and Systems
Laplace Transform Analysis of Signals and Systems Transfer Functions Transfer functions of CT systems can be found from analysis of Differential Equations Block Diagrams Circuit Diagrams 5/10/04 M. J.
More informationControl Systems. Frequency Method Nyquist Analysis.
Frequency Method Nyquist Analysis chibum@seoultech.ac.kr Outline Polar plots Nyquist plots Factors of polar plots PolarNyquist Plots Polar plot: he locus of the magnitude of ω vs. the phase of ω on polar
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationPID CONTROLLER SYNTHESIS FREE OF ANALYTICAL MODELS
PID CONTROLLER SYNTHESIS FREE OF ANALYTICAL MODELS L.H. Keel,1 S.P. Bhattacharyya, Tennessee State University, Nashville, Tennessee, USA Texas A&M University, College Station, Texas, USA Abstract: In this
More informationOn Stabilization of First-Order Plus Dead-Time Unstable Processes Using PID Controllers
On Stabilization of First-Order Plus Dead-Time Unstable Processes Using PID Controllers Chyi Hwang Department of Chemical Engineering National Chung Cheng University Chia-Yi 621, TAIWAN E-mail: chmch@ccu.edu.tw
More informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationLABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593
LABORATORY INSTRUCTION MANUAL CONTROL SYSTEM I LAB EE 593 ELECTRICAL ENGINEERING DEPARTMENT JIS COLLEGE OF ENGINEERING (AN AUTONOMOUS INSTITUTE) KALYANI, NADIA CONTROL SYSTEM I LAB. MANUAL EE 593 EXPERIMENT
More informationStability Margin Based Design of Multivariable Controllers
Stability Margin Based Design of Multivariable Controllers Iván D. Díaz-Rodríguez Sangjin Han Shankar P. Bhattacharyya Dept. of Electrical and Computer Engineering Texas A&M University College Station,
More informationSTABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to control-system design: 1 Stability, 2 Steady-state response, 3 Transient response
More informationChapter 7 : Root Locus Technique
Chapter 7 : Root Locus Technique By Electrical Engineering Department College of Engineering King Saud University 1431-143 7.1. Introduction 7.. Basics on the Root Loci 7.3. Characteristics of the Loci
More informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
More informationTopic # Feedback Control
Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability
More informationControl Systems. Root Locus & Pole Assignment. L. Lanari
Control Systems Root Locus & Pole Assignment L. Lanari Outline root-locus definition main rules for hand plotting root locus as a design tool other use of the root locus pole assignment Lanari: CS - Root
More informationIdentification Methods for Structural Systems. Prof. Dr. Eleni Chatzi System Stability - 26 March, 2014
Prof. Dr. Eleni Chatzi System Stability - 26 March, 24 Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Intro Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /5/27 Outline Closed Loop Transfer
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steady-state Steady-state errors errors Type Type k k systems systems Integral Integral
More informationChapter 6 - Solved Problems
Chapter 6 - Solved Problems Solved Problem 6.. Contributed by - James Welsh, University of Newcastle, Australia. Find suitable values for the PID parameters using the Z-N tuning strategy for the nominal
More informationECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8
Learning Objectives ECE 345 / ME 380 Introduction to Control Systems Lecture Notes 8 Dr. Oishi oishi@unm.edu November 2, 203 State the phase and gain properties of a root locus Sketch a root locus, by
More informationControl System Design
ELEC ENG 4CL4: Control System Design Notes for Lecture #22 Dr. Ian C. Bruce Room: CRL-229 Phone ext.: 26984 Email: ibruce@mail.ece.mcmaster.ca Friday, March 5, 24 More General Effects of Open Loop Poles
More informationME 375 Final Examination Thursday, May 7, 2015 SOLUTION
ME 375 Final Examination Thursday, May 7, 2015 SOLUTION POBLEM 1 (25%) negligible mass wheels negligible mass wheels v motor no slip ω r r F D O no slip e in Motor% Cart%with%motor%a,ached% The coupled
More informationChapter 6: The Laplace Transform. Chih-Wei Liu
Chapter 6: The Laplace Transform Chih-Wei Liu Outline Introduction The Laplace Transform The Unilateral Laplace Transform Properties of the Unilateral Laplace Transform Inversion of the Unilateral Laplace
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Bounded-input bounded-output (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationControl Systems I. Lecture 7: Feedback and the Root Locus method. Readings: Jacopo Tani. Institute for Dynamic Systems and Control D-MAVT ETH Zürich
Control Systems I Lecture 7: Feedback and the Root Locus method Readings: Jacopo Tani Institute for Dynamic Systems and Control D-MAVT ETH Zürich November 2, 2018 J. Tani, E. Frazzoli (ETH) Lecture 7:
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Basic Feedback Analysis & Design
AERO 422: Active Controls for Aerospace Vehicles Basic Feedback Analysis & Design Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University Routh s Stability
More information26 Feedback Example: The Inverted Pendulum
6 Feedback Example: The Inverted Pendulum Solutions to Recommended Problems S6. Ld 0(t) (a) Ldz6(t) = g0(t) a(t) + Lx(t), Ld (t) dt - ga(t) = Lx(t) Taking the Laplace transform of both sides yields szlo(s)
More informationProblem Set 4 Solutions 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6245: MULTIVARIABLE CONTROL SYSTEMS by A Megretski Problem Set 4 Solutions Problem 4T For the following transfer
More informationQUANTITATIVE L P STABILITY ANALYSIS OF A CLASS OF LINEAR TIME-VARYING FEEDBACK SYSTEMS
Int. J. Appl. Math. Comput. Sci., 2003, Vol. 13, No. 2, 179 184 QUANTITATIVE L P STABILITY ANALYSIS OF A CLASS OF LINEAR TIME-VARYING FEEDBACK SYSTEMS PINI GURFIL Department of Mechanical and Aerospace
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a linear-time-invariant
More informationUncertainty and Robustness for SISO Systems
Uncertainty and Robustness for SISO Systems ELEC 571L Robust Multivariable Control prepared by: Greg Stewart Outline Nature of uncertainty (models and signals). Physical sources of model uncertainty. Mathematical
More informationA system that is both linear and time-invariant is called linear time-invariant (LTI).
The Cooper Union Department of Electrical Engineering ECE111 Signal Processing & Systems Analysis Lecture Notes: Time, Frequency & Transform Domains February 28, 2012 Signals & Systems Signals are mapped
More informationOutline. Control systems. Lecture-4 Stability. V. Sankaranarayanan. V. Sankaranarayanan Control system
Outline Control systems Lecture-4 Stability V. Sankaranarayanan Outline Outline 1 Outline Outline 1 2 Concept of Stability Zero State Response: The zero-state response is due to the input only; all the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This
More informationCourse Summary. The course cannot be summarized in one lecture.
Course Summary Unit 1: Introduction Unit 2: Modeling in the Frequency Domain Unit 3: Time Response Unit 4: Block Diagram Reduction Unit 5: Stability Unit 6: Steady-State Error Unit 7: Root Locus Techniques
More informationHomework 7 - Solutions
Homework 7 - Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More information(Continued on next page)
(Continued on next page) 18.2 Roots of Stability Nyquist Criterion 87 e(s) 1 S(s) = =, r(s) 1 + P (s)c(s) where P (s) represents the plant transfer function, and C(s) the compensator. The closedloop characteristic
More informationFEL3210 Multivariable Feedback Control
FEL3210 Multivariable Feedback Control Lecture 5: Uncertainty and Robustness in SISO Systems [Ch.7-(8)] Elling W. Jacobsen, Automatic Control Lab, KTH Lecture 5:Uncertainty and Robustness () FEL3210 MIMO
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOT-LOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOT-LOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the root-locus
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
More information9.5 The Transfer Function
Lecture Notes on Control Systems/D. Ghose/2012 0 9.5 The Transfer Function Consider the n-th order linear, time-invariant dynamical system. dy a 0 y + a 1 dt + a d 2 y 2 dt + + a d n y 2 n dt b du 0u +
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More informationRobust Loop Shaping Controller Design for Spectral Models by Quadratic Programming
Robust Loop Shaping Controller Design for Spectral Models by Quadratic Programming Gorka Galdos, Alireza Karimi and Roland Longchamp Abstract A quadratic programming approach is proposed to tune fixed-order
More informationPower series solutions for 2nd order linear ODE s (not necessarily with constant coefficients) a n z n. n=0
Lecture 22 Power series solutions for 2nd order linear ODE s (not necessarily with constant coefficients) Recall a few facts about power series: a n z n This series in z is centered at z 0. Here z can
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationCourse Outline. Closed Loop Stability. Stability. Amme 3500 : System Dynamics & Control. Nyquist Stability. Dr. Dunant Halim
Amme 3 : System Dynamics & Control Nyquist Stability Dr. Dunant Halim Course Outline Week Date Content Assignment Notes 1 5 Mar Introduction 2 12 Mar Frequency Domain Modelling 3 19 Mar System Response
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Dynamic Response
.. AERO 422: Active Controls for Aerospace Vehicles Dynamic Response Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. . Previous Class...........
More informationNyquist Criterion For Stability of Closed Loop System
Nyquist Criterion For Stability of Closed Loop System Prof. N. Puri ECE Department, Rutgers University Nyquist Theorem Given a closed loop system: r(t) + KG(s) = K N(s) c(t) H(s) = KG(s) +KG(s) = KN(s)
More informationComputation of Stabilizing PI and PID parameters for multivariable system with time delays
Computation of Stabilizing PI and PID parameters for multivariable system with time delays Nour El Houda Mansour, Sami Hafsi, Kaouther Laabidi Laboratoire d Analyse, Conception et Commande des Systèmes
More informationChemical Process Dynamics and Control. Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University
Chemical Process Dynamics and Control Aisha Osman Mohamed Ahmed Department of Chemical Engineering Faculty of Engineering, Red Sea University 1 Chapter 4 System Stability 2 Chapter Objectives End of this
More informationClass 12 Root Locus part II
Class 12 Root Locus part II Revising (from part I): Closed loop system K The Root Locus the locus of the poles of the closed loop system, when we vary the value of K Comple plane jω ais 0 real ais Thus,
More informationLecture 4 Stabilization
Lecture 4 Stabilization This lecture follows Chapter 5 of Doyle-Francis-Tannenbaum, with proofs and Section 5.3 omitted 17013 IOC-UPC, Lecture 4, November 2nd 2005 p. 1/23 Stable plants (I) We assume that
More informationDigital Control: Part 2. ENGI 7825: Control Systems II Andrew Vardy
Digital Control: Part 2 ENGI 7825: Control Systems II Andrew Vardy Mapping the s-plane onto the z-plane We re almost ready to design a controller for a DT system, however we will have to consider where
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuous-time, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationRichiami di Controlli Automatici
Richiami di Controlli Automatici Gianmaria De Tommasi 1 1 Università degli Studi di Napoli Federico II detommas@unina.it Ottobre 2012 Corsi AnsaldoBreda G. De Tommasi (UNINA) Richiami di Controlli Automatici
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationChapter Stability Robustness Introduction Last chapter showed how the Nyquist stability criterion provides conditions for the stability robustness of
Lectures on Dynamic Systems and Control Mohammed Dahleh Munther A Dahleh George Verghese Department of Electrical Engineering and Computer Science Massachuasetts Institute of Technology c Chapter Stability
More informationControls Problems for Qualifying Exam - Spring 2014
Controls Problems for Qualifying Exam - Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationEECE 460 : Control System Design
EECE 460 : Control System Design SISO Pole Placement Guy A. Dumont UBC EECE January 2011 Guy A. Dumont (UBC EECE) EECE 460: Pole Placement January 2011 1 / 29 Contents 1 Preview 2 Polynomial Pole Placement
More informationMULTILOOP PI CONTROLLER FOR ACHIEVING SIMULTANEOUS TIME AND FREQUENCY DOMAIN SPECIFICATIONS
Journal of Engineering Science and Technology Vol. 1, No. 8 (215) 113-1115 School of Engineering, Taylor s University MULTILOOP PI CONTROLLER FOR ACHIEVING SIMULTANEOUS TIME AND FREQUENCY DOMAIN SPECIFICATIONS
More informationOutline. Introduction to Time-Delay Systems. lecture no. 2. What is changed for delay systems? Nyquist stability criterion
Outline ntroduction to Time-Delay Systems Nyquist criterion for time-delay systems e sh Roots of quasi-polynomials: general observations lecture no. Delay sweeping (direct method) Leonid Mirkin Commensurate
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31-3-17 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationLecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design.
ISS0031 Modeling and Identification Lecture 5: Linear Systems. Transfer functions. Frequency Domain Analysis. Basic Control Design. Aleksei Tepljakov, Ph.D. September 30, 2015 Linear Dynamic Systems Definition
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency Response-Design Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response- Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationProblem Set 5 Solutions 1
Massachusetts Institute of Technology Department of Electrical Engineering and Computer Science 6.245: MULTIVARIABLE CONTROL SYSTEMS by A. Megretski Problem Set 5 Solutions The problem set deals with Hankel
More informationIdentification Methods for Structural Systems
Prof. Dr. Eleni Chatzi System Stability Fundamentals Overview System Stability Assume given a dynamic system with input u(t) and output x(t). The stability property of a dynamic system can be defined from
More informationREAL TIME CONTROL FOR TIME DELAY SYSTEM
REAL TIME CONTROL FOR TIME DELAY SYSTEM Rihem Farkh Rihem.Farkh@enit.rnu.tn Kaouther Laabidi Kaouther.Laabidi@enit.rnu.tn Mekki Ksouri Mekki.Ksouri@insat.rnu.tn Research unit on Systems Analysis and Control
More informationSolutions to Skill-Assessment Exercises
Solutions to Skill-Assessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationDesign Methods for Control Systems
Design Methods for Control Systems Maarten Steinbuch TU/e Gjerrit Meinsma UT Dutch Institute of Systems and Control Winter term 2002-2003 Schedule November 25 MSt December 2 MSt Homework # 1 December 9
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationDidier HENRION henrion
GRADUATE COURSE ON POLYNOMIAL METHODS FOR ROBUST CONTROL PART II.1 ROBUST STABILITY ANALYSIS: POLYTOPIC UNCERTAINTY Didier HENRION www.laas.fr/ henrion henrion@laas.fr Cité de l espace with Ariane launcher
More informationEE 380. Linear Control Systems. Lecture 10
EE 380 Linear Control Systems Lecture 10 Professor Jeffrey Schiano Department of Electrical Engineering Lecture 10. 1 Lecture 10 Topics Stability Definitions Methods for Determining Stability Lecture 10.
More informationSTABILITY OF CLOSED-LOOP CONTOL SYSTEMS
CHBE320 LECTURE X STABILITY OF CLOSED-LOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 10-1 Road Map of the Lecture X Stability of closed-loop control
More informationAlireza Mousavi Brunel University
Alireza Mousavi Brunel University 1 » Control Process» Control Systems Design & Analysis 2 Open-Loop Control: Is normally a simple switch on and switch off process, for example a light in a room is switched
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequency-response design method Goal: wrap up lead and
More informationRobust fixed-order H Controller Design for Spectral Models by Convex Optimization
Robust fixed-order H Controller Design for Spectral Models by Convex Optimization Alireza Karimi, Gorka Galdos and Roland Longchamp Abstract A new approach for robust fixed-order H controller design by
More information5 Root Locus Analysis
5 Root Locus Analysis 5.1 Introduction A control system is designed in tenns of the perfonnance measures discussed in chapter 3. Therefore, transient response of a system plays an important role in the
More informationLecture 13: Internal Model Principle and Repetitive Control
ME 233, UC Berkeley, Spring 2014 Xu Chen Lecture 13: Internal Model Principle and Repetitive Control Big picture review of integral control in PID design example: 0 Es) C s) Ds) + + P s) Y s) where P s)
More informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science - Bangalore A Fundamental Problem in Control Systems Poles of open
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Stability Routh-Hurwitz stability criterion Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More informationCh 14: Feedback Control systems
Ch 4: Feedback Control systems Part IV A is concerned with sinle loop control The followin topics are covered in chapter 4: The concept of feedback control Block diaram development Classical feedback controllers
More informationECE 388 Automatic Control
Controllability and State Feedback Control Associate Prof. Dr. of Mechatronics Engineeering Çankaya University Compulsory Course in Electronic and Communication Engineering Credits (2/2/3) Course Webpage:
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture : Different Types of Control Overview In this Lecture, you will learn: Limits of Proportional Feedback Performance
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationSYSTEMTEORI - ÖVNING Stability of linear systems Exercise 3.1 (LTI system). Consider the following matrix:
SYSTEMTEORI - ÖVNING 3 1. Stability of linear systems Exercise 3.1 (LTI system. Consider the following matrix: ( A = 2 1 Use the Lyapunov method to determine if A is a stability matrix: a: in continuous
More informationME 304 CONTROL SYSTEMS Spring 2016 MIDTERM EXAMINATION II
ME 30 CONTROL SYSTEMS Spring 06 Course Instructors Dr. Tuna Balkan, Dr. Kıvanç Azgın, Dr. Ali Emre Turgut, Dr. Yiğit Yazıcıoğlu MIDTERM EXAMINATION II May, 06 Time Allowed: 00 minutes Closed Notes and
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
More informationAircraft Stability & Control
Aircraft Stability & Control Textbook Automatic control of Aircraft and missiles 2 nd Edition by John H Blakelock References Aircraft Dynamics and Automatic Control - McRuler & Ashkenas Aerodynamics, Aeronautics
More information