Routh Hurwitz (RH) Stability Test. M, results with an output c( t)

Transcription

1 Routh Hurwitz (RH) Stbility Test A little history: erouth.htm Geerl Sttemet A BIBO system is stble if for ll time,t, iput r( t) M, results with output c( t) P for fiite M d P. It is ecessry coditio for stbility tht ll poles of the trsfer fuctio be locted i the --LHP A system with poles o ---jw-----is defied s mrgilly stble system. Motivtio (Rtiol) Cosider the followig closed loop feedbck cotrol system. Correctio: H = /(s+) ot /(s+) Let us study the trsiet respose for three differet vlues of the forwrd gi, K.

2 Imgiry Axis The Mtlb code I used is: cler ll K =; um = []; de=[ 0]; um =[0]; de=[ 0]; um=cov(um,um); de=cov(de,de); G=K*tf(um,de); % This is the ope loop trsfer fuctio umh=[]; deh=[ ]; H=tf(umH,deH); GCL=feedbck(G,H); % This is the closed loop trsfer fuctio pole(gcl) disp('hit y key to cotiue...') puse step(gcl) % ote: My use pzmp commd to plot the poles d zeros. A- K = Pole-Zero Mp System: Gclosed Pole : i Dmpig: 0.48 Overshoot (%): 6.4 Frequecy (rd/sec): 3 System: Gclosed Pole : - Dmpig: Overshoot (%): 0 Frequecy (rd/sec): Rel Axis Poles of the closed loop system, GCL(s) re: i i i i Ad the the step respose is give below

3 Amplitude 3.5 Step Respose K = Time (sec)

4 4 B- K = 3/9 3 Pole - Zero Mp K = 3/9 Imgiry Axis System: Gclosed Pole : i Dmpig: 0.65 Overshoot (%): 6.83 Frequecy (rd/sec):.3 System: Gclosed Pole : 9.6e i Dmpig: -3.55e-06 Overshoot (%): 00 Frequecy (rd/sec): Rel Axis Ru the code gi but with K = 3/9. Poles re: i i i i

5 Amplitude Step Respose K = 3/ Time (sec)

6 Imgiry Axis 6 C- K = 5 3 Pole-Zero Mp, K =5 0 System: Gclosed Pole : i Dmpig: Overshoot (%): 6. Frequecy (rd/sec):.7 System: Gclosed Pole : i Dmpig: -0.5 Overshoot (%): 44 Frequecy (rd/sec): Rel Axis The poles re: i i i i

7 Amplitude 7 Typo: The title should sy: Step respose for K = 5. 0 Step Respose K = 3/ Time (sec)

8 8 Observtio: The systems poles moved from Stble to mrgilly stble to ustble for slight chges i the forwrd gi, K. Questios: Is there lyticl wy of determiig the rge of K for which the system is stble?

9 9 Routh Hurwitz Test How is the RH test performed? First: Recll: - For closed loop system: G CL ( s) G( s) G( s) H( s) - The poles of the closed loop system re the roots of the deomitor of the trsfer fuctio, GCL(s): Defie: D( s) G( s) H( s) : This is referred to s the Chrcteristic Polyomil 3- Let us stdrdize form by expressig D(s) i polyomil form: Let s s... D( s) s s At this poit, it is time to explore the RH test. 0

10 0 Secod: The RH Test Steps: - Build the strtig RH rry S S Formed from the coefficiets of D(s) S - S -3 S S 0 b b b 3... c c c 3... d d d These coefficiets re clculted

11 - Clculte the remiig coefficiets of the RH rry Note: The first colum i the RH rry is the bsis vector. 3 b 5 4 b 3 b b b c 3 5 b b b c c b c b c d b b

12 3- Now the RH rry is formed, wht to do ext? All oe hs to do is look t the first colum i the RH rry. The umber of sig chges will be the umber of poles i the right hlf s-ple. (Thus ustble system if the umber is o-zero) Before explorig some specil cses of the RH rry, let us revisit the origil system of tody s lecture.

13 3 Exmple Determie the rge of K to hve stble system. The system: Correctio: H= /(s+) ot /(s+) 0K G ( s) s s s 0 H ( s) s G 0K( s ) CL s( s s 0)( s ) 0 K 4 3 D( s) s 3s s 0s 0K

14 4 - The RH rry S 4 0K S S 6/3 0K 0 Auxiliry Equtio S S K 0 0K 4 3 D( s) s 3s s 0s 0K

15 5 - Ivestigte the first colum.. Wheever possible, it is desired to hve o sig chges i the vector. b. Comig dow the bsis vector, the sigs re: +, +, +,?,? The lst two terms hve to be positive i order to hve stble system: 0K > 0, the K> 0 () K > 0 K < Therefore, it is cocluded tht the system is stble for 0 <K<3.556 Now it is obvious why whe K =5, the system resulted i ustble respose!!

16 6 For the ext lecture: - Ivestigte the uxiliry Equtio (ew ide) - Look t the specil cses.

17 7

18 8 Recp with i-clss Exmple: For the chrcteristic polyomil of the closed loop trsfer fuctio: D(s)=+G(s)H(s) = s 4 s 3 3s s 4 K () Determie the rge of K ecessry for stble respose (if y) () Determie the uxiliry equtio 6/3 S +0K = 0, K= 3.556, therefore, S = j. 58 rd /sec. (3) Determie the frequecy of oscilltio for mrgilly stble system.

19 9 Specil Cses Whe the RH tbultio rry termites premturely: - The first elemet i y oe row of the RH rry is zero while other elemets i the sme rry re NOT Exmple: D(s) = s 4 s 3 s s 3 0 The RH tbultio is: S 4 : 3 S 3 : 0 S : 0 3 Well, the remiig clculted elemets for s d s 0 will be ifiity hve problem. Solutio: Whe this hppes, substitute the zero with smll positive umber,, d cotiue buildig the tble. Which results i, S 4 : 3 S 3 : 0 S : 3 0 S : S 0 :

20 0 Two sig chges i the first colum Thus, there re two poles i the RHP. For fu: Solve for the roots of D(s), results i? (Note: the epsilo method my ot work ppropritely if D(s) hs pure imgiry roots.) - All the elemets i oe row re zeros Exmple: D(s) = S 5 + 4S 4 + 8S 3 + 8S + 7s + 4 = 0 The RH rry is: S 5 : 8 7 S 4 : S 3 : S : S : 0 0 The ext terms will be? Solutio: - Form uxiliry equtio with S A(s) = 4s + 4 = 0 - Tke the derivtive of A(s) w.r.t. s

21 d ( 4 s 4) 8s 0 ds Thus, the coefficiets of da re the coefficiets of: 8s + 0 =0 ds which re 8 d zero Use these to replce the elemets i the S row. Thus S : 8 0 S 0 : 4 No sig chges Stble Recp Exmple: Exmple Determie the rge of the vlues of K (if exists) for which the system is stble. I R(s) K s +4s+8 Y(s) Out G(s) H(s) 3 s +s+

22 Solutio G CL s K s 4s 8 K s s s K ( s s ) G CL ( s s )( s 4s 8) 3K 4 3 D(s) = s 6s 8s 64s ( 96 3K) RH Tbultio: S 4 : K S 3 : S : K 0 S : K 0 S 0 : 96+3K 96+3K>0 K> K>0 K<9.7 Thus, The rge of K for stble respose is: -3 < K < 9.7

23 3 Fil Vlue Theorem: Recll: Voltge cross cp i simple series RC circuit (E is the pplied DC voltge) Vc(t)= E e RC t lim t V c ( t) lim E e t t RC E This should mke sese to us: Vc fter log time [ t>5 ], the voltge cross the cp is equl to the Thevii s voltge. I this cse Vc = E. Let us visit this from LPlce poit of view: Cs V R s R s RC c ( s) R( s) ( ) ( ) R RCs s RC Cs Note: R(s) = E/s E is the mgitude of the DC voltge pplied. r(t) is the iput. Therefore,

24 4 E V RC c ( s) s s RC Accordig to the fil vlue theorem: lim c( t) lim s. C( s) c(t) is the output {if the limit exists} t s0 E lim sv s s RC c( ) lim E (grees with previous s 0 s 0 s s RC result.) Usig the fil vlue theorem, oe c predict, lyticlly, the stedy stte vlue without resortig to the time domi fuctio.