Lecture 7 Poles and Zeros Stability

Transfer Function Models General Representation wh where z i are the zeros p i are the poles n ≥ m to have a physically realizable system   The dynamic behavior of a transfer function model can be characterized by the numerical value of its poles and zeros

Transfer Function Models  Transfer Function can be expressed in gain/time constant form, i.e. factoring out b m and a n  Pole–Zero cancellation happens when the numerator and denominator terms cancels each other

Pole   The factor/s of the denominator of the transfer function   It is the value wherein the transfer function approaches infinity as the value of s approaches the pole

Example 1  Pole of order 1 or simple pole at s = 0  Pole of order 2 at s = 5 and pole of order 3 at s = – 7

Zero   The factor/s of the numerator of the transfer function   It is the value wherein the transfer function approaches 0 as the value of s approaches the zero

Complex Plane Plot Graphical representation of a rational transfer function in the complex plane which helps to convey certain properties of the system x–axis is real part y–axis is imaginary part

Pole–Zero Plot  Zero s = – 2 o in plot  Pole s = ±0.5i x in plot

Effects of Pole Location   Left Half Plane (LHP) results in a stable system, i.e. stable response   Right Half Plane (RHP) results in unstable system, i.e. unstable step response x x x Real axis Imaginary axis x x x Real axis Imaginary axis

Effects of Pole Location   Faster response when pole is farther from imaginary axis   Complex pole results to oscillatory response p = a + bj where j = √– 1 Real axis Imaginary axis x x x → complex poles

Effects of Pole Location   More oscillatory transient response when pole is farther from real axis   Pole at the origin, i.e. 1/s term in Transfer Function Model, results in an integrating element/process

Effects of Zero Location   Zeros have no effect on system stability   Zero in Right Half Plane (RHP) results in inverse response to a step change in the input xy t inverse response Real axis Imaginary axis

Effects of Zero Location   Zero in Left Half Plane (LHP) results in overshoot during a step response

Stability

Stability  Most industrial processes are stable without feedback control and are said to be open–loop stable or self–regulating  An open–loop stable process will return to the original steady state after a transient disturbance, i.e. one that is not sustained, occurs  By contrast there are a few processes, such as exothermic chemical reactors, that can be open- loop unstable

Stability Definition An unconstrained linear system is said to be stable if the output response is bounded for all bounded inputs or BIBO stability. Otherwise it is said to be unstable. Applicability –Any linear control system comprised of linear elements –Nonlinear systems operating near the point of linearization

Stability   Asymptotically stability is when the variables of the stable control system always decrease from their initial value and do not show permanent oscillations – –Permanent oscillations occur when a pole has a real part exactly equal to zero (in the continuous time case) or a modulus equal to one (in the discrete time case)   Marginally stability is when a simply stable system response neither decays nor grows over time, and has no oscillations – –System transfer function has non–repeated poles at complex plane origin, i.e. their real and complex component is zero in the continuous time case

Closed Loop System Transfer Function KmKm GmGm PY sp YmYm Y’ sp E – + G*dG*d GpGp GvGv GcGc U + + D YV

Closed Loop System   Characteristic Equation of Closed Loop System – –Is the denominator of the transfer function – –Can be simplified to 1 + G OL where G OL is the open loop transfer function – –Used to solve for poles by equation to zero, i.e. 1 + G OL = 0

Stability for Closed Loop System General Stability Criterion The feedback control system is stable if and only if all roots of the characteristic equation are negative or have negative real parts. Otherwise, the system is unstable. Assumptions –Set–point changes rather than disturbance changes were considered –Closed–loop transfer function was a ratio of polynomials –Poles are all distinct

Stability Region in Complex Plane A system is stable if the poles of the transfer function lie strictly in the closed left half of the complex plane, i.e. the real part of all the poles is less than zero

Closed Loop Response Stable System

Closed Loop Response Unstable System

Example 1 Determine whether system is stable or not given Solution: Solving for 1 + G OL Equating 1 + G OL = 0

Example 1 Solving for s Inference on system –System is stable since real part of pole/s is negative –Behavior is oscillatory due to the presence of imaginary term

Example 2 Determine whether system is stable or not given 1 + G OL = s + 0.2K c – 1 Solution: Equating 1 + G OL = 0 s + 0.2K c – 1 = 0 s = 1 – 0.2K c For system to be stable s should be less than 0, i.e. K c > 5

Example 3 Determine whether system is stable or not given Solution: Solving for 1 + G OL Equating 1 + G OL = 0

Example 3 Solving for s For system to be stable, s should be less than 0, i.e. 40(K c + 1) > 0 or K c > – 1

Stability Test   Direct Substitution Method   Routh Stability Criterion   Root Locus Diagram   Bode Stability Criterion   Nyquist Stability Criterion

Direct Substitution Method   Imaginary axis divides the complex plane into stable and unstable regions for the roots of characteristic equation   On the imaginary axis, the real part of s is zero, and thus we can write s = j . Substituting s = j  into the characteristic equation allows us to find a stability limit such as the maximum value of K c   As the gain K c is increased, the roots of the characteristic equation cross the imaginary axis when K c = K cm

Direct Substitution Method   Methodology – –s = j  is substituted in the characteristic equation – –K c is equated to K cm – –Both the real part and imaginary part is equated to 0 and value of  and K cm is computed – –Stability region is determined for K c

Example 4 Determine whether system is stable or not given 10s s 2 + 8s K c = 0 Solution: Substituting Substituting s = j  and K c = K cm in the characteristic equation – 10j 3 – j K cm = 0 – 10j  3 – 17  2 + 8j  K cm = 0or (1 + K cm – 17 2 ) + j (8 – 10 3 ) = 0 (1 + K cm – 17  2 ) + j (8  – 10  3 ) = 0

Example 4 Equating both real and imaginary part to zero 1 + K cm – 17 2 = K cm – 17  2 = 0 8 – 10 3 = 0 8  – 10  3 = 0 Solving 8 – 10 3 = 0 Solving  from 8  – 10  3 = 0 results to K cm = – 1  = 0 →K cm = – 1 K cm = 12.6  = ±0.894→ K cm = 12.6 Region of stability would be – 1 < K c < 12.6 – 1 < K c < 12.6

Example 5 Determine whether system is stable or not given Solution: Solving for 1 + G OL = 0

Example 5 s 3 + 6s s + 6(1 + K c ) = 0 Substituting Substituting s = j  and K c = K cm in the characteristic equation – j 3 – j + 6(1 + K cm ) = 0 – j  3 – 6  j  + 6(1 + K cm ) = 0or (6 + 6K cm – 6 2 ) + j (11 – 3 ) = 0 (6 + 6K cm – 6  2 ) + j (11  –  3 ) = 0

Example 5 Equating both real and imaginary part to zero 6 + 6K cm – 6 2 = K cm – 6  2 = 0 11 – 3 = 0 11  –  3 = 0 Solving – 3 = 0 Solving  from 11  –  3 = 0 results to K cm = – 1  = 0 →K cm = – 1 K cm = 10  = ±3.32→ K cm = 10 Region of stability would be – 1 < K c < 10 – 1 < K c < 10

Routh Stability Criterion   Developed by E J Routh in 1905   a.k.a. – –Routh–Hurwitz Stability Criterion – –Routh Test   Purely algebraic method   Used to establish stability in single input single output (SISO) linear time invariant control system

Routh Stability Criterion   Applied to systems with characteristic equation that has a polynomial form. Hence, it can not be used to systems with time delays or transport lag, i.e. e –  s term   For system with e –  s term, Padé approximation is done on the time delay term/s

Routh Stability Criterion   It is necessary (but not sufficient) that all the coefficients of the characteristic equation, i.e. a n, a n – 1,..., a 1 and a 0, be positive else the system is unstable. Hence, no need to perform the Routh Test

Flow Process for Performing Stability Analysis

Example 6 Determine the stability of system that has characteristic equation s 4 + 5s 3 + 3s = 0 Solution: Since the Since the s term is missing, its coefficient is zero Thus, the system is unstable. Recall that a necessary condition for stability is that all of the coefficients in the characteristic equation must be positive.

Routh Array Generation Given a polynomial of the form: a n s n + a n–1 s n–1 + ● ● ● + a 1 s + a 0 = 0 a n–1 anan a n–3 a n–2 z1z1 b1b1 Row 1 ● ● ● a n–5 a n–4 ● ● ● b3b3 b2b2 c1c1 c2c n + 1 ● ● ●

Theorems on Routh Test Theorem 1 A necessary and sufficient condition for all the roots of the characteristic equation to have a negative real parts (or stable system) is that all the elements of the first column in the Routh array be positive and nonzero Theorem 2 If some of the elements in the first column are negative, the number of roots with a positive real part, i.e. in the right hand plane, is equal to the number of sign changes in the first column

Theorems on Routh Test Theorem 3 If one pair of the roots is on the imaginary axis, equidistant from the origin, and all other roots are in the left half plane, all the elements of the n th row will vanish and none of the elements preceding row will vanish. The location of the pair of imaginary roots can be found by solving the equation Cs 2 + D = 0 where C and D are the elements of the (n–1) th row read left to right, respectively

Example 7 Determine the stability of system that has characteristic equation s 4 + 3s 3 + 5s 2 + 4s + 2 = 0 Solution: / 3 Row / System is stable since all terms in the 1 st column is positive (Theorem 1)

Example 8 Determine the stability of system that has characteristic equation s 6 + s 5 + 4s 4 + 3s 3 + 2s 2 + 4s + 2 = 0 Solution: –12 / 5 1 Row 1 2 – –74 /  System is unstable since not all terms in the 1 st column is positive  2 roots are in the right half plane due to 2 sign changes

Example 9 Determine value of K c to have a stable system Solution: Solving for 1 + G OL

Example 9 Characteristic equation s 3 + 6s s + 6(1 + K c ) = (1 + K c ) – K c Row 1 6(1 + K c ) For system to be stable, all terms in the 1 st column should be greater than zero (Theorem 1) 10 – K c > 0→ K c < 10 6 (1 + K c ) > 0→ K c > –1 Region of stability –1 < K c < 10

Example 9 When K c = 10, system is on the verge of instability. The Routh array becomes Row According to Theorem 3, the (n–1) th row is the coefficient C and D used in solving the imaginary roots. 6s = 0 s = ±√11

Example 10 Determine value of K c to have a stable system G OL = 5s +2K c e –s Solution: Solving for 1 + G OL 1 + G OL = 1 + 5s +2K c e –s Using 1/1 Padé approximation

Example 10 Characteristic equation 2.5s 2 + (5.5 – K c )s + (1 + 2K c ) = – K c K c Row For system to be stable, all terms in the 1 st column should be greater than zero (Theorem 1) 5.5 – K c > 0→ K c < K c > 0→ K c > –0.5 Region of stability –1 < K c < 10

Root Locus   The locus of the roots of the characteristic equation of the closed loop transfer function as the loop gain, K c, of the feedback system is increased from zero to infinty   It is a useful tool for analyzing the transient response, as well as the stability of a single input single output dynamic systems   A system is stable if all of its poles are in the left hand side of the s–plane   Graphical procedure for finding roots of the characteristic equation, 1 + G OL = 0

Root Locus  Methodology –Obtain characteristic equation –Vary value of K c –Solve for the roots of the equation –Plot the roots based on specific K c values –Connect the points based on increasing K c values

Example 11 Plot the Root Locus Diagram of the characteristic equation s 3 + 6s s + 6(1 + K c ) = 0 Solution: Rearranging equation to (s + 1) (s + 2) (s + 3) + K = 0 where K = 6K c

Example 11 K = 6K c Root 1Root 2Root 3 – – 3.61 – 2.60j– j –6 0– 3.00 – 1.41j– j 0– 3– 2– 1 0– 3– 2– – 3.1– 1.75 – – 3.16– 1.42– – 3.45– 1.28 – 0.75j– j 6.6– 4.11– 0.95 – 1.50j– j 6.6– 4.11– 0.95 – 1.50j– j 26.5– 5.10– 0.45 – 2.50j– j 60– 6.00 – 3.32j j 60– 6.00 – 3.32j j 100– – 4.00j j

Example 11 XXX X X K c = 0 K c = 10 X K c = 0.07 K c = –5 X K c = –1 X X

Example 12 Consider a feedback control system that has the open loop transfer function Plot the root locus for 0 ≤ K c ≤ 20 Solution: The characteristic equation 1 + G OL = 0 or (s + 1)(s + 2)(s + 3) + 4K c = 0 (s + 1)(s + 2)(s + 3) + 4K c = 0

Example 12 When K c = 0, the roots are merely the poles of the open loop transfer function, i.e. – 1, – 2 and – 3 X denotes an open loop pole Dots denote locations of the closed loop poles for different values of K c

References 1. Coughanowr, Donald R. Process Systems Analysis and Control. 2 nd ed. New York: McGraw–Hill, Inc, Seborg, Dale E. et al. Process Dynamics and Control. 2 nd ed. New York: John Wiley & Sons, Inc,