Krdeniz Technicl Univerity Deprtment of Electricl nd Electronic Engineering 6080 Trbzon, Turkey Chpter 8- nd Routh-Hurwitz Stbility Criterion Bu der notlrı dece bu deri ln öğrencilerin kullnımın çık olup, üçüncü hılr verilmei, herhngi bir yöntemle çoğltılıp bşk yerlerde kullnılmı, yyınlnmı Prof. Dr. İmil H. ALTAŞ ın yzılı iznine tbidir. Aki durumlrd yl işlem ypılcktır. Chpter 8-
It i fundmentl requirement of ny control ytem. It i tht property of ytem which enure boundedne of the repone of the ytem to et of input. The tbility of ytem i defined in three wy depending upon the type of pecified input. Definition A ytem i defined to be tble if ) it free repone (or zero input repone or nturl repone or initil condition repone) i bounded nd ) it pproche zero ymptoticlly. Chpter 8-3 Conider ytem hving trnfer function The output Y() i the um of repone Y forced () nd repone Y free () where Chpter 8-4
nd N I () depend upon initil condition. The ytem i tble if nd only if Exmple The free repone i found by etting u0 Tking Lplce trnform yield Chpter 8-5 Tking invere Lplce trnform yield The ytem i tble if nd only if 0 >0. Chpter 8-6
Exmple Bed the free repone behviour, i the following ytem tble? Solution Let u determine the differentil eqution relting the input, u(t) nd the output, y(t). To obtin the differentil eqution model conider the trnfer function model relting Y() nd U() which implie Chpter 8-7 Tking the invere Lplce trnform yield The free repone i obtined by etting u0: Tking Lplce trnform yield Hence the repone, Y free () Tking the invere Lplce trnform yield Clerly the ytem i tble ince y free (t) i bounded nd it i zero ymptoticlly. Chpter 8-8
Exmple The free repone i found by etting u0. The free repone The ytem i untble ince y free (t) lthough bounded doe not pproch zero t tend to infinity. Chpter 8-9 Comment An untble ytem will hve unbounded output even though the input i bounded (when the ytem h pole on the imginry xi) or even when there i no input (when the ytem h pole in the right hlf plne). Definition A ytem i id to be tble if ) it impule repone i bounded nd ) the impule repone pproche zero ymptoticlly. Chpter 8-0
Let the impule repone be h(t)l - {H()}. The ytem i tble if ) h(t) < nd ) limit h(t)0 t. Exmple It i tble ytem long 0 >0. Chpter 8- Exmple It i n untble ytem even though h(t) i bounded, Chpter 8-
Definition 3 A ytem i tble if nd only if the output i bounded for every bounded input. Tht i, the ytem i tble if nd only if Exmple Conider the ytem We find tht the output i bounded for ll bounded input except when u(t) i ome contnt. When u(t) i unit tep, the output y(t) i rmp. Thu the definition of tbility i violted, therefore the ytem i untble. Chpter 8-3 Exmple Apply bounded input nd nlyze the output: The output i unbounded for every bounded input. Hence it i untble. Chpter 8-4
Comment The tbility defined in the ene of definition 3 i generlly referred to bounded Input bounded Output (BIBO) tbility. In generl the phenomenon of reonnce occur whenever mode of the input ignl coincide with pole of the ytem. If pole of the ytem lie on the jω-xi nd the input ignl i inuoid hving frequency identicl to tht of the imginry pole loction, the output will be growing inuoid. In prctice mny phyicl ytem hve pole in the open left-hlf plne but cloer to jω xi. If the input contin inuoidl component whoe frequency coincide with the imginry prt d of the pole, ω d, then the output will be poorly dmped inuoid with lrge mplitude. Thi i to be voided by filtering out ignl which hve frequencie coincident with tht of `poorly dmped pole'. Chpter 8-5 Theorem A ytem i tble if nd only if ll the pole of the ytem re in the open left-hlf plne. Proof: We will ume for implicity tht the input u(t) h rtionl pectr, tht i the Lplce trnform U() i rtionl polynomil. The output Y() i Chpter 8-6
Conider the denomintor polynomil D()D u (). It root re the root of which re the pole of the plnt, which re the root of nd the root of which re the mode of the input ignl. The prtil frction expnion will hve the following term. A term contining the pole of the plnt A term contining the mode of the input ignl which re ditinct from the pole of the plnt. Chpter 8-7 When pole of the plnt coincide with mode of the input ignl, it cn, in generl, be clled reonnce, nd if the coincident root re on the imginry xi, the repone will be unbounded. If mode of input ignl coincide with zero of the plnt, then tht mode i bent. Conequently the output y(t) tke the following generl form c () t where -p i re the pole, p i, re the mode, c i (t) nd i re polynomil. The firt term on the right hnd ide of y(t) lo contin repone due to the mode of the input which re coincident with the pole. Since u(t) i bounded, tht i u(t) <, Chpter 8-8
The output y(t) will be bounded, tht i y(t) < for every u(t) < if nd only if Chpter 8-9 Note tht R e p i 0 will not enure boundedne of y(t) for every bounded input u(t) ince n input with mode coincident with p i will cue n unbounded output repone. Thu ytem i tble in the Bounded Input Bounded Output ene if nd only if ll it pole re in the open left-hlf plne. Comment: All the three definition of tbility re equivlent ince ytem i tble in the ene of ny of the definition if nd only if ll it pole re in the open left-hlf plne. Chpter 8-0
Definition I A ytem i tble if it free repone i bounded nd i zero ymptoticlly yfree () t < lim y ( t) 0 t free Definition II A ytem i tble if it impule repone i bounded nd i zero ymptoticlly ht () < lim ht ( ) 0 t All the pole re in the open left-heft hlf plne Definition III A ytem i tble if it output i bounded for every bounded input yt () < for ll u() t < Thu one cn chooe ny of definition of tbility depending upon the convenience of nlyi or deign. Chpter 8- ANALYSIS OF The tbility of ytem i nlyed by verifying whether the pole lie in the open left-hlf of the plne. For firt nd econd order ytem it i poible to obtin n nlyticl expreion for the pole of the ytem. Let ϕ() be the chrcteritic polynomil. For firt order ytem nd the pole i t -. For econd-order ytem nd the pole re t Chpter 8-
ANALYSIS OF nd For higher order ytem, tht i ytem with order 3 or more, there re no nlyticl expreion for obtining the pole nd one h to reort to numericl method. However, to verify the tbility of ytem exct pole-loction i not required. The informtion to whether ll the pole lie in the open left-hlf plne i ufficient. A imple necery but not ufficient condition i bed on verifying the ign of the coefficient of the chrcteritic polynomil. Conider the chrcteritic polynomil Chpter 8-3 ANALYSIS OF Let p i (i,,., n) be the root of the chrcteritic eqution. tht i -p i re the pole. The coefficient i relted to the pole -p i follow (i0,,,., n-), re Chpter 8-4
ANALYSIS OF Chpter 8-5 ANALYSIS OF where Σp i p j...p denote um of the product of ll combintion of negtive of the root tken y m t time with no repetition of product term. Chpter 8-6
ANALYSIS OF Theorem If the ytem i tble, then ll the coefficient of the chrcteritic polynomil ϕ() i trictly poitive tht i Proof Conider the reltion between the pole -p i nd the coefficient i, clerly if R e p i >0 for ll i then j > 0 for ll j. Comment If the ytem i tble, then ll coefficient of the chrcteritic polynomil re poitive nd none of the coefficient i miing. However the convere i not necerily true. The ytem could be untble even though ll the coefficient re poitive nd none i zero. Chpter 8-7 ANALYSIS OF Exmple ll the coefficient re trictly poitive. The ytem i untble ince the root of ϕ() re The ytem with i untble ince the root re -, 0±j. For econd order ytem, however the ign of the coefficient of the chrcteritic polynomil indicte tbility. Chpter 8-8
ANALYSIS OF Theorem A econd order ytem i tble if nd only if ll the coefficient of the chrcteritic polynomil re trictly poitive. Tht i, let ϕ() be the chrcteritic polynomil given by Then the ytem i tble if nd only if Chpter 8-9 ANALYSIS OF Proof Let p nd p be the root of Then Chpter 8-30
ANALYSIS OF clerly R e p > 0 nd R e p > 0. Thu econd-order ytem i tble if nd only if ll the coefficient of the chrcteritic polynomil i poitive nd none miing. Chpter 8-3 Definition I A ytem i tble if it free repone i bounded nd i zero ymptoticlly yfree () t < lim y ( t) 0 t free Definition II A ytem i tble if it impule repone i bounded nd i zero ymptoticlly ht () < lim ht ( ) 0 t All the pole re in the open left-heft hlf plne Definition III A ytem i tble if it output i bounded for every bounded input yt () < for ll introduction u() t < Thu one cn chooe ny of definition of tbility depending upon the convenience of nlyi or deign. Chpter 8-3
introduction The free repone The ytem i tble if nd only if 0 >0. The ytem i untble ince y free (t) lthough bounded doe not pproch zero t tend to infinity. Chpter 8-33 The Impule repone H() + 0 ht () e t 0 H() ht () introduction It i tble ytem long 0 >0. It i n untble ytem even though h(t) i bounded, lim ht ( ) 0 t Chpter 8-34
introduction BIBO u y bounded t u bounded y t y y u bounded u unbounded t t Chpter 8-35 introduction Stbility Iue Bounded Input Bounded Output Stbility: A ytem i BIBO tble if, for every bounded input, the output remin bounded with increing time (ll ytem pole mut lie in the left hlf of the - plne). Mrginl Stbility: A ytem i mrginlly tble if ome of the pole lie on the imginry xi, while ll other re in the LHS of the -plne. Some input my reult in the output becoming unbounded with time. Chpter 8-36
introduction Stbility Iue To tet the tbility of LTI ytem we need only exmine the pole of the ytem, i.e. the root of the chrcteritic eqution. Method re vilble for teting the root with poitive rel prt, which do not require the ctul olution of the chrcteritic eqution. Alo, method re vilble for teting the tbility of cloed-loop ytem bed only on the loop trnfer function chrcteritic. Chpter 8-37 ROUTH-HURWITZ CRITERION Routh-Hurwitz Stbility Criterion Aytemiuntble if ny coefficient of the chrcteritic polynomil i negtive or zero. However the ytem could be untble even though ll the coefficient re poitive nd none zero, except for econd order ytem for which the tbility i relted to the ign of the coefficient. No nlyticl procedure i vilble for computing the root of polynomil of degree 3. Chpter 8-38
ROUTH-HURWITZ CRITERION Routh-Hurwitz criterion give imple nd yet powerful nlyticl procedure for verifying the tbility of ytem. It give necery nd ufficient condition for tbility which re pplicble to ytem of ny order. Before we decribe the procedure, let u define the following: Chpter 8-39 ROUTH-HURWITZ CRITERION Hurwitz-polynomil: A polynomil i clled Hurwitz polynomil if ll of it root re in the open left-hlf plne. Thu ytem i tble if nd only if it chrcteritic polynomil i Hurwitz. Let Q()0 be the chrcteritic eqution. Tht i Q( ) n n + n n + L + + 0 Chpter 8-40
ROUTH-HURWITZ CRITERION Q( ) n n + n n + L + + 0 where 0 0. A necery (but not ufficient) condition for ll root to hve non-poitive rel prt i tht ll coefficient hve the me ign. For the necery nd ufficient condition, firt form the Routh rry. Chpter 8-4 ROUTH-HURWITZ CRITERION n n n n n n Q ( ) + + +... + + 0 n n n 4 ( ) n + n + n 4 +... + Q n n 3 ( ) n + n 3 +... + 0 Q Q () Q( + Q() Chpter 8-4
ROUTH-HURWITZ CRITERION n n n Q ( ) + +... + + 0 n The Routh Arry n n n- n-4 n-6 n- n- n-3 n-5 n-7 n- where b b b 3 b 4 n-3 c c c 3 c 4 n n b : : : : n k k 4 n n b l 0 n m etc. n n n 3 n 5 Chpter 8-43 ROUTH-HURWITZ CRITERION The Routh Arry n n n- n-4 n-6 n- n- n-3 n-5 n-7 n- b b b 3 b 4 n-3 c c c 3 c 4 : : : : k k l 0 m In imilr mnner, element in the 4th row, c, c, re clculted bed on the two previou row. c c b b n 3 n 5 b b n n The element in ll ubequent row re clculted in the me mnner. b b 3 Chpter 8-44
ROUTH-HURWITZ CRITERION b c.. k l n n m Necery nd ufficient condition: If ll element in the firt column of the Routh rry hve the me ign, then ll root of the chrcteritic eqution hve negtive rel prt. If there re ign chnge in thee element, then the number of root with non-negtive rel prt i equl to the number of ign chnge. Element in the firt column which re zero define pecil ce. Chpter 8-45 ROUTH-HURWITZ CRITERION b c.. k l n n m Necery nd ufficient condition: From the eqution for the clcultion of the element of the Routh rry, it cn be een tht the element of row cnnot be computed if the firt element of row bove i zero ince diviion by zero reult. Becue of thi poibility, we divide the ppliction of Routh Hurwitz criterion into 3 ce. Chpter 8-46
ROUTH-HURWITZ CRITERION Ce None of the element in the firt column of the Routh rry i zero. Thu no problem occur in the clcultion of the element of the rry. Exmple Conider n unity feedbck control ytem where G () 3 + 4 + 5+ r + K G () y Find K uch tht the tedy-tte error e 0% to tep input. Chpter 8-47 ROUTH-HURWITZ CRITERION SOLUTION r + e K G () Y () KG () R() + KG() Trnfer function from r to y Trnfer function from r to e E () R() + KG() E() R() + KG ( ) E() R() Y() KG() E() R() R() + KG ( ) + KG( ) KG ( ) E () R () + KG( ) E() R() + KG ( ) y Chpter 8-48
ROUTH-HURWITZ CRITERION SOLUTION The tedy tte error e i G () e lim E( ) 0 E() R() + KG ( ) For unit tep input, E () K + 3 + 4 + 5+ e r + e K 3 + 4 + 5+ E () 3 4 5 K + + + + 3 + 4 + 5+ lim 0 3 + 4 + 5+ + K e + K + K y Chpter 8-49 ROUTH-HURWITZ CRITERION SOLUTION e Since the tedy-tte error e 0% + K 0 e 00 0 + 0K + K 00 80 0K 4 K The ytem mut be tble for ome k 4. Chrcteritic polynomil i obtined by computing the trnfer function of the ytem T() Y () KG () R() + KG() Chpter 8-50
ROUTH-HURWITZ CRITERION SOLUTION The trnfer function of the ytem T(): Y() KG() R() + KG() K Y () 3 4 5 K + + + 3 R () K + + 4 + 5+ + K 3 + 4 + 5+ So, the chrcteritic eqution: 3 ( ) Q () + 4 + 5+ + K 0 Chpter 8-5 ROUTH-HURWITZ CRITERION SOLUTION The Routh rry 3 ( ) Q () + 4 + 5+ + K 0 3 5 0 4 +K 0 0 ( +K) 0 0 4 0 +K 0 0 Chpter 8-5
ROUTH-HURWITZ CRITERION SOLUTION The ytem i tble if 4 Tht men; 0 ( +K) 4 +K > 0 i tified. 0 ( +K) 4 +K >0 > 0 Chpter 8-53 ROUTH-HURWITZ CRITERION SOLUTION 0 ( + K) > 0 0 K > 0 9 > K K < 9 ( + K) > 0 + K > 0 K > K > < K < 9 Conidering tedy tte error 4< K < 9 Chpter 8-54
ROUTH-HURWITZ CRITERION Exmple r + e k + G k G y Find the rnge of k nd k o tht the cloed loop ytem i tble where G () 0 + 5 + 0 G () Chpter 8-55 ROUTH-HURWITZ CRITERION SOLUTION The trnfer function of the cloed loop ytem i r + e k + G k G y Y () 0k 3 R () + 5 + 0 + k + 0k ( ) The chrcteritic polynomil (or the denomintor polynomil) i 3 ( ) Q () + 5 + 0+ k + 0k 0 Chpter 8-56
ROUTH-HURWITZ CRITERION SOLUTION 3 ( ) Q () + 5 + 0+ k + 0k 0 3 0(+k ) 5 0k 00(+k )-0k 5 0 0k 0 0 Chpter 8-57 ROUTH-HURWITZ CRITERION SOLUTION For tbility 00( + k) 0k > 0 5 0k > 0 k > 0 0 + 0k 4k > 0 ( k k ) 45+ 5 > 0 5k > k 5 k > k 5 5 Chpter 8-58
ROUTH-HURWITZ CRITERION Exmple Where G () 3 Obtin the rnge of k P nd k I of PI controller r + + 4 + 5+ e C G () I P I () KP + C y K K + K The chrcteritic eqution ( it i Q()0 where Q() i the denomintor polynomil of the overll ytem relting the input to the output) i obtined from Q () + CG () () 0 Chpter 8-59 ROUTH-HURWITZ CRITERION Solution Q () + CG () () 0 KP+ KI Q () + 0 3 + 4 + 5+ 4 3 Q () + 4 + 5 + + K + K 0 4 3 P ( ) Q () + 4 + 5 + + K + K 0 Then, the Routh rry: P I I Chpter 8-60
ROUTH-HURWITZ CRITERION 4 3 ( ) Q () + 4 + 5 + + K + K 0 P I Solution Then, the Routh rry i: 4 5 K I 3 4 +K P 0 0 K P K 0 I 4 C 0 0 0 K 0 0 I 8 KP ( P) 8 4 + K K C 8 KP 4 9 KP ( + KP) 8KI C 9 KP C ( ) 9 K + K 8K 9 KP P P I I Chpter 8-6 ROUTH-HURWITZ CRITERION Solution C C ( )( ) ( )( ) 9 K + K 8K 9 K + K 6K 9 KP 9 KP P P I P P I ( ) 9+ 8K K 6K 9 K P P I P C P P I 8 + 6K K 6K 9 K P P P I 8 + 6K K 6K > 0 P P I 9+ 8K K 8K > 0 K I > 0 K > 0 K > 0 I I I < + P P 8K 9 8K K K I 9 8KP K < + 8 8 P Chpter 8-6
ROUTH-HURWITZ CRITERION Solution For K I 0; K I 9 8KP K < + 8 8 P 9+ 8K K 0 P P K + 8K + 9 0 K (,) P K P K P P b± b 4c P K I > 8± 64+ 36 (,) 8± 0 (,) K P() K P ( ) 0 9 K I > 0 Chpter 8-63 ROUTH-HURWITZ CRITERION Solution K I > 0 K I 9 8KP K < + 8 8 P K P( ) 9 K P() K I > 0 Chpter 8-64