Routh-Hurwitz Lecture Routh-Hurwitz Stability test

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1 ECE 35 Routh-Hurwitz Leture Routh-Hurwitz Staility test AStolp /3/6, //9, /6/ Denominator of transfer funtion or signal: s n s n s n 3 s n 3 a s a Usually of the Closed-loop transfer funtion denominator to test fo BIBO staility Test denominator for poles in CRHP (RHP inluding imaginary axis) For all poles to e in the LHP, all oeffiients must e > For a seond-order denominator, that is enough, skip the next step If all oeffiients are > & order >, then: Create Routh-Hurwitz array: s n a n 6 s n a n & divide s n n s 3 3 s z n n 3 a n n 3 n n 5 a 3 n n 5 n n 6 7 n 7 Look at first olumn: All positive All roots left of imaginary axis If any negative or, then there are poles on the Imaginary axis or in the RHP (Right-Half Plane) Count sign reversals down the first olumn Sign reversals numer of poles on the Imaginary axis or in the RHP (Right-Half Plane) an e replaed y ε to see if there are any other sign reversals ECE 35 Routh-Hurwitz Leture p

2 Example Uses ECE 35 Routh-Hurwitz Leture p The transfer funtions of C(s) and P(s) are given elow In eah ase determine if the steady-state error will go to zero and whether disturanes will e ompletely reeted Be sure to hek for losed-loop staility when needed a) C( s ) s s 5 s P( s ) s steady-state error? Reet Disturanes? e ss ( )? e ss ( ) s 5 no pole at zero no pole at zero for disturane? staility test needed to answer those questions ) C( s ) s s 3 P( s ) s Yes (Tentative answer) s s P( s) has pole at zero C( s) has no pole at zero Must test for staility: Closed loop transfer funtion C( s) P( s) C( s) P( s) N C ( s) D C ( s) D P ( s ) N C ( s) Closed loop denominator D C ( s) D P ( s ) N C ( s) C( s) P( s ) s s 3 s s s Closed loop denominator s s 3 s s ( )( s ) Routh-Hurwitz Staility test Test denominator for poles in CRHP (RHP inluding imaginary axis) All oeffiients must e > For a seond-order denominator, that is enough Create Routh-Hurwitz array: (RH Ex) Look at first olumn: D H ( s ) s 6 s 3 s D H ( s ) s 6 s 3 s s 5 s 3 6 s s s All positive, so All roots left of imaginary axis, so tentative answers aove are orret If any were negative or, then D H ( s ) would have poles on the Imaginary axis or in the RHP (Right-Half Plane) Alternatively, hek the atual roots Using your alulator, find the roots of: s 6 s 3 s Roots: ECE 35 Routh-Hurwitz Leture p roots all negative, stale So tentative answers aove are orret

3 ECE 35 Routh-Hurwitz Leture p3 More Routh-Hurwitz method examples RH Ex Given a loosed-loop denominator: D( s ) s s 3 3 s 5 s Are all the poles in the OLHP? RH Ex3 C( s ) 3 s 8 s 3 s s 3 s C( s) 8 P( s ) s 3 s s Routh-Hurwitz array: P( s ) s 3 s 3 s 3 5 s s s 3 3 Two sign reversals two prolem poles, in the RHP s 3 ECE 35 Routh-Hurwitz Leture p3 Atual roots: (tie that the Plant is not inherently stale) NO Two roots positive Closed loop denominator s 3 s s s 3 3 s 8 s 7 s 3 s s 6 7 s 6 s s 6 6 s 6 (-ε) Prolem, some root(s) in CRHP s ε 6 ε Consider this -ε & you get sign hanges, unstale poles Doesn't make sense to progress to the next row if all you want to know is staility, ut if you ount aove as -ε, this answer would ome out +, indiating two prolem poles Atual roots: First roots are on imaginary axis, unstale

4 ECE 35 Routh-Hurwitz Leture p Closed-loop transfer-funtion denominator Transfer funtion stale? a) D( s ) 3 s 8 s 3 3 s s The third oeffiient is negative, there must e root(s), & thus poles, in the losed right half plane ) D( s ) s 6 3 s 8 s 3 3 s s The oeffiient is zero, there must e root(s) in the losed right half plane ) D( s ) s s 3 s 3 s s The last oeffiient is zero, there must e root(s) in the losed right half plane d) D( s ) s s s Yes Neither fator has unstale poles so together they also have none Don't multiply and (Example in text) ompliate matters e) D( s ) s s s First fator has at least one unstale pole, so together they also have at least one Don't (Example in text) multiply and ompliate matters f) D( s ) s s 3 6 s s Can't tell without the full array RH Ex RH Ex5 Atual roots: s s 3 6 s s s Prolem, some root(s) in CRHP need to progress to the next row if all you want to 3855 know is staility, ut this extra steps an tell you 87 there are two prolem poles Last two roots are in the RHP t stale Use the Routh-Hurwitz method to determine the value range of that will produe a stale system D( s ) s s 3 s s Charateristi equation of a feedak sytem s s 3 s 5 s 5 5 s ( ) 5 > < 5 ECE 35 Routh-Hurwitz Leture p < < 5

5 ECE 35 Routh-Hurwitz Leture p5 RH Ex6 Use the Routh-Hurwitz method to determine the value range of that will produe a stale system D( s ) s s 3 s s s s 3 s 5 s s 5 5 > This ould have een seen from the original expression < 5 < < < 575 RH Ex7 Use the Routh-Hurwitz method to determine if all the poles are to the left of - 5 D( s ) s 3 s 3 s 68 Charateristi equation of a feedak sytem Replae all ouranes of s with (s - 5) ( ) 3 ( ) 3 ( ) 68 s 3 5 s 75 s 3 ( ) 68 s 3 5 s 75 s s 5 3 s s 3 9 s 5 s 3 RH Ex7 Are all the poles are to the left of -? Replae all ouranes of s with (s - ) ( s ) 3 ( s ) 3 ( s ) 68 s 3 s 8 s 6 s 8 s 6 3 ( s ) 68, this has egative oeffiient s 3 s 8 s 6 s 8 s 6 3 s 3 68 s 3 3 s 6 s 8 s 3 6 s 3 8 s s Look at first olumn: All positive, so all roots are indeed left of - Atual roots of: s 3 s 3 s Sure enough, all roots are left of -, and not all left of -5 ECE 35 Routh-Hurwitz Leture p5

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