EE3CL4: denominator coefficients Introduction to Linear Control - PowerPoint PPT Presentation

EE 3CL4, §4 1 / 55 Tim Davidson Stability Condition in terms of poles Condition in terms of EE3CL4: denominator coefficients Introduction to Linear Control Systems Routh Hurwitz condition Basics Section 4: Stability and Routh-Hurwitz Condition Disk drive example Dealing with zeros Zeros in first column Zero rows Using Routh Tim Davidson Hurwitz for design Turning control of a tracked vehicle McMaster University Winter 2020

EE 3CL4, §4 2 / 55 Outline Tim Davidson Stability Condition in terms of poles 1 Stability Condition in terms of denominator Condition in terms of poles coefficients Routh Hurwitz Condition in terms of denominator coefficients condition Basics Disk drive example Dealing with zeros Routh Hurwitz condition 2 Zeros in first column Zero rows Basics Using Routh Disk drive example Hurwitz for design Dealing with zeros Turning control of a tracked vehicle Zeros in first column Zero rows Using Routh Hurwitz for design 3 Turning control of a tracked vehicle

EE 3CL4, §4 4 / 55 Stability Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients Routh Hurwitz condition Basics Disk drive example A system is said to be stable if all bounded inputs r ( t ) give Dealing with zeros Zeros in first column rise to bounded outputs y ( t ) Zero rows Using Routh Hurwitz for Counterexamples design Turning control of a • Albert Collins, Jeff Beck (Yardbirds), tracked vehicle Pete Townshend (The Who), Jimi Hendrix, Tom Morello (Rage Against the Machine), Kurt Cobain (Nirvana) • Tacoma Narrows

EE 3CL4, §4 5 / 55 Conditions for stability Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients � ∞ Routh Hurwitz condition y ( t ) = g ( τ ) r ( t − τ ) d τ Basics −∞ Disk drive example Dealing with zeros Zeros in first column Let r ( t ) be such that | r ( t ) | ≤ ¯ r Zero rows Using Routh � ∞ Hurwitz for � � | y ( t ) | = g ( τ ) r ( t − τ ) d τ design � � � � Turning control of a −∞ tracked vehicle � ∞ � d τ � � ≤ � g ( τ ) r ( t − τ ) −∞ � ∞ � d τ ≤ ¯ � � r � g ( τ ) −∞ � ∞ � d τ is finite � � Using this: system G ( s ) is stable iff � g ( τ ) −∞

EE 3CL4, §4 6 / 55 Condition in terms of poles? Tim Davidson � ∞ � d τ to be finite � � We want � g ( τ ) Stability −∞ Condition in terms of Can we determine this from G ( s ) ? poles Condition in terms of denominator coefficients We can write a general rational transfer function in the form Routh Hurwitz condition K � i ( s + z i ) Basics G ( s ) = m ( s 2 + 2 α m s + ( α 2 Disk drive example s N � k ( s + σ k ) � m + ω 2 m )) Dealing with zeros Zeros in first column Zero rows Using Routh Poles: 0, − σ k , − α m ± j ω m Hurwitz for design Assuming N = 0 and no repeated roots, the impulse response is Turning control of a tracked vehicle zero for t < 0 and for t ≥ 0 it is A k e − σ k t + B m e − α m t sin( ω m t + θ m ) � � g ( t ) = m k � ∞ Stability requires −∞ | g ( t ) | dt to be bounded; that requires σ k > 0, α m > 0 In fact, system is stable iff poles have negative real parts

EE 3CL4, §4 7 / 55 Marginal stability Tim Davidson Stability Condition in terms of • Consider integrator: G ( s ) = 1 / s ; simple pole at origin poles Condition in terms of � t denominator • y ( t ) = −∞ r ( λ ) d λ coefficients • if r ( t ) = cos ( t ) , which is bounded, Routh Hurwitz condition then y ( t ) = sin ( t ) . Bounded Basics • If r ( t ) = u ( t ) , which is bounded, Disk drive example Dealing with zeros then y ( t ) = t . Not bounded Zeros in first column Zero rows Using Routh • Consider G ( s ) = 1 / ( s 2 + 1 ) , simple poles at s = ± j 1 Hurwitz for design • Unit step response: u ( t ) − cos( t ) . Bounded Turning control of a tracked vehicle • What if r ( t ) is a sinusoid of frequency 1 / ( 2 π ) Hz? Not bounded If G ( s ) has a pole with positive real part, or a repeated pole on j ω -axis output is always unbounded

EE 3CL4, §4 8 / 55 Routh-Hurwitz condition Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients Routh Hurwitz We have seen how to determine stability from the poles. condition Basics Disk drive example Much easier than having to find impulse response Dealing with zeros � ∞ Zeros in first column and then determining if −∞ | g ( τ ) | d τ < ∞ Zero rows Using Routh Can we determine stability without having to determine the Hurwitz for design poles? Turning control of a tracked vehicle Yes! Routh-Hurwitz condition

EE 3CL4, §4 9 / 55 Routh-Hurwitz condition Tim Davidson Let G ( s ) = p ( s ) Stability q ( s ) , where Condition in terms of poles Condition in terms of q ( s ) = a n s n + a n − 1 s n − 1 + . . . a 1 s + a 0 denominator coefficients = a n ( s − r 1 )( s − r 2 ) . . . ( s − r n ) Routh Hurwitz condition Basics where r i are the roots of q ( s ) = 0. Disk drive example Dealing with zeros Zeros in first column By multiplying out, q ( s ) = 0 can be written as Zero rows Using Routh q ( s ) = a n s n − a n ( r 1 + r 2 + · · · + r n ) s n − 1 Hurwitz for design Turning control of a + a n ( r 1 r 2 + r 2 r 3 + . . . ) s n − 2 tracked vehicle − a n ( r 1 r 2 r 3 + r 1 r 2 r 4 + . . . ) s n − 3 + · · · + ( − 1 ) n a n ( r 1 r 2 r 3 . . . r n ) = 0 If all r i are real and in left half plane, what is sign of coeffs of s k ? the same!

EE 3CL4, §4 10 / 55 Routh-Hurwitz condition Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients Routh Hurwitz condition That observation leads to a necessary condition. Basics Disk drive example Dealing with zeros Hence, not that useful for design Zeros in first column Zero rows A more sophisticated analysis leads to the Routh-Hurwitz Using Routh Hurwitz for condition, which is necessary and sufficient design Turning control of a tracked vehicle Hence, can be quite useful for design

EE 3CL4, §4 11 / 55 R-H cond: A first look Tim Davidson Consider G ( s ) = p ( s ) q ( s ) . Poles are solutions to q ( s ) = 0; i.e., Stability a n s n + a n − 1 s n − 1 + a n − 2 s n − 2 + · · · + a 1 s + a 0 = 0 Condition in terms of poles Condition in terms of denominator coefficients Routh Hurwitz Construct a table of the form condition Basics Row n a n a n − 2 a n − 4 . . . Disk drive example Dealing with zeros Row n − 1 a n − 1 a n − 3 a n − 5 . . . Zeros in first column Zero rows Row n − 2 b n − 1 b n − 3 b n − 5 . . . Using Routh Row n − 3 c n − 1 c n − 3 c n − 5 . . . Hurwitz for design . . . . . . . . . . . . . . . Turning control of a tracked vehicle Row 0 h n − 1 where b n − 1 = a n − 1 a n − 2 − a n a n − 3 = − 1 � � a n a n − 2 � � � � a n − 1 a n − 3 a n − 1 a n − 1 � � b n − 3 = − 1 � � c n − 1 = − 1 � � a n a n − 4 a n − 1 a n − 3 � � � � � � � � a n − 1 a n − 5 b n − 1 b n − 3 a n − 1 b n − 1 � � � �

EE 3CL4, §4 12 / 55 R-H cond: A first look Tim Davidson Stability Now consider the table that we have just constructed Condition in terms of poles Condition in terms of denominator Row n a n a n − 2 a n − 4 . . . coefficients Row n − 1 a n − 1 a n − 3 a n − 5 . . . Routh Hurwitz condition Row n − 2 b n − 1 b n − 3 b n − 5 . . . Basics Row n − 3 c n − 1 c n − 3 c n − 5 . . . Disk drive example Dealing with zeros . . . . . . . . Zeros in first column . . . . . . . Zero rows Row 0 h n − 1 Using Routh Hurwitz for design Turning control of a tracked vehicle Loosely speaking: • Number of roots in the right half plane is equal to the number of sign changes in the first column of the table • Stability iff no sign changes in the first column Now let’s move towards a more sophisticated statement

EE 3CL4, §4 14 / 55 Stability (Revision) Tim Davidson Stability Condition in terms of poles Condition in terms of denominator coefficients Routh Hurwitz condition Let G ( s ) = p ( s ) Basics q ( s ) , where Disk drive example Dealing with zeros Zeros in first column q ( s ) = a n s n + a n − 1 s n − 1 + . . . a 1 s + a 0 Zero rows Using Routh Hurwitz for design System is stable iff all poles of G ( s ) have negative real parts Turning control of a tracked vehicle Recall, poles are solutions to q ( s ) = 0 Can we find a necessary and sufficient condition that depends only on the set of coefficients { a k } n k = 0 so that we don’t have to solve q ( s ) = 0?