# Systems Analysis and Control

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1 Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 12:

2 Overview In this Lecture, you will learn: Review of Feedback Closing the Loop Pole Locations Changing the Gain Numerical Examples Pole Locations Routh-Hurwitz vs. A Review of Complex Numbers Polar Form Multiplication-Division M. Peet Lecture 12: Control Systems 2 / 24

3 The Effect of Feedback Feedback changes the open loop Ĝ(s) = n G(s) d G (s) ˆK(s) = n K(s) d K (s) u(s) + - G(s) K(s) y(s) to Ĝ(s) ˆK(s) 1 + Ĝ(s) ˆK(s) = n G (s)n K (s) d G (s)d K (s) + n G (s)n K (s) The pole locations are the roots of Objective: A closed loop denominator. d G (s)d K (s) + n G (s)n K (s) = 0 d(s) = (s p 1 ) (s p n ) Big Question: How to choose n K (s) and d K (s) so that d(s) = d G (s)d K (s) + n G (s)n K (s) M. Peet Lecture 12: Control Systems 3 / 24

4 The Effect of Feedback PD Control For a Second-Order System Ĝ(s) = 1 s 2 + as + b Im(s) Re(s) with PD feedback ˆK(s) = K (1 + T D s) We can achieve any denominator d(s) = s 2 + cs + d Question: What happens for more complicated systems: M. Peet Lecture 12: Control Systems 4 / 24

5 The Effect of Feedback Suspension Problem Open Loop: Closed Loop: s 2 + s + 1 s 4 + 2s 3 + 3s 2 + s + 1 K (1 + T D s) (s 2 + s + 1) s 4 + 2s 3 + 3s 2 + s K (1 + T D s) (s 2 + s + 1) K ( T D s 3 + (1 + T D )s 2 + (1 + T D )s + 1 ) = s 4 + (2 + KT D )s 3 + (3 + K + KT D )s 2 + (1 + K + KT D )s K Given a desired denominator d(s) = s 4 + as 3 + bs 2 + cs + d Which gives 4 equations and 2 unknowns a = 2 + KT D c = 1 + K + KT D There is No Solution!!!. b = 3 + K + KT D d = 1 + K M. Peet Lecture 12: Control Systems 5 / 24

6 The Effect of Feedback Solution We rarely need to achieve a precise set of poles. Performance Specifications Determine Regions of the Complex Plane. Stability Rise Time Settling time Overshoot Im(s) Re(s) New Question: What controller will ensure all roots of d G (s)d K (s) + n G (s)n K (s) lie in the desired region of the complex plane. M. Peet Lecture 12: Control Systems 6 / 24

7 The Effect of Feedback Proportional Feedback More fundamentally, how does changing n K (s) and d K (s) change the roots of The answer is complicated d G (s)d K (s) + n G (s)n K (s)? Must account for the effect of each term in n K and d K So simplify, lets consider a controller with only a single free parameter. Other options include: PD Control: ˆK(s) = 1 + TD s PI Control: ˆK(s) = T I s Question: How do the roots of change with k? ˆK(s) = k d G (s) + kn G (s)? M. Peet Lecture 12: Control Systems 7 / 24

8 Formal Definition Definition 1. Ĝ(s) = n G(s) d G (s) The of Ĝ(s) is the set of all poles of as k ranges from 0 to Alternatively: The roots of 1 + kĝ(s) for k 0 kĝ(s) 1 + kĝ(s) The roots of d G (s) + kn G (s) for k > 0 1 The solutions of Ĝ(s) = k for k 0 M. Peet Lecture 12: Control Systems 8 / 24

9 Video Surveillance System. We can estimate the root locus by finding the roots for several different values of k Example: Video Surveillance System. Pole at s = 0 to eliminate steady-state error. Open Loop: Closed Loop: Ĝ(s) = Pole Locations: 1 s(s + 10) k s s + k p 1,2 = 5 ± k M. Peet Lecture 12: Control Systems 9 / 24

10 Video Surveillance System p 1,2 = 5 ± k M. Peet Lecture 12: Control Systems 10 / 24

11 Video Surveillance System We can visualize the effect of changing k by plotting the poles on the complex plane. M. Peet Lecture 12: Control Systems 11 / 24

12 Video Surveillance System Plotting every possible value of k yields the root locus. Connect the dots. M. Peet Lecture 12: Control Systems 12 / 24

13 Example: Suspension System Imaginary Axis Real Axis From Routh Test: Stable for all k > 0. M. Peet Lecture 12: Control Systems 13 / 24

14 Example: Suspension System with Integral Feedback Now, if we add integral feedback: ˆK(s) = k 1 s Imaginary Axis Real Axis From Routh Test: Stable for all k <.1. M. Peet Lecture 12: Control Systems 14 / 24

15 Example: Inverted Pendulum Model Ĝ(s) = 1 s Imaginary Axis Real Axis M. Peet Lecture 12: Control Systems 15 / 24

16 Example: Inverted Pendulum Model Now an inverted pendulum with some derivative feedback: ˆK(s) = k(1 + s) Imaginary Axis M. Peet Real Axis Lecture 12: Control Systems 16 / 24

17 Complex Numbers Imaginary Axis Real Axis When Matlab calculates the root locus, it plots every point. Impractical for students Yields no intuition. is only one parameter. We must know how to manipulate the root locus by changes in controller type. Before we analyze the root locus, we begin with a review of Complex Numbers. M. Peet Lecture 12: Control Systems 17 / 24

18 Complex Numbers Polar Form Consider a Complex Number: Im(s) a s = a + b i s = a + bı b The Complex Plane is the a-b plane. Re(s) Im(s) s = r cos(θ) + r sin(θ) A complex number can also be represented in polar form r θ Re(s) Euler yields the more practical form: s = re θı s = r (cos θ + ı sin θ) Recall the Euler equation e ıθ = cos θ + ı sin θ M. Peet Lecture 12: Control Systems 18 / 24

19 Complex Numbers Polar Form Rectilinear s = a + bı Polar s = re θı Notation: r is called the Magnitude Denoted r = s θ = is called the Phase Denoted θ = s Im(s) r θ s = r cos(θ) + r sin(θ) Re(s) The relationship between Polar and Rectilinear coordinates is obvious ( ) b θ = tan 1 a = r cos θ a r = b = r sin θ a 2 + b 2 M. Peet Lecture 12: Control Systems 19 / 24

20 Complex Numbers Multiplication In polar form, Multiplying and Dividing complex numbers is cleaner. s 1 = r 1 e θ1ı s 2 = r 2 e θ2ı s 1 = a 1 + b 1 ı s 2 = a 2 + b 2 ı s 1 s 2 = r 1 e θ1ı r 2 e θ2ı = r 1 r 2 e θ1ı e θ2ı = r 1 r 2 e (θ1+θ2)ı s 1 s 2 = (a 1 + b 1 ı)(a 2 + b 2 ı) = (a 1 a 2 b 1 b 2 ) + (a 1 b 2 + a 2 b 1 )ı For multiplication magnitudes and phases decouple. magnitudes multiply phases add M. Peet Lecture 12: Control Systems 20 / 24

21 Complex Numbers Division For Division, the benefit is even greater. s 1 = r 1 e θ1ı s 2 = r 2 e θ2ı s 1 = a 1 + b 1 ı s 2 = a 2 + b 2 ı s 1 /s 2 = r 1 e θ1ı r 1 2 e θ2ı = = r 1 r 2 e (θ1 θ2)ı s 1 s 2 = a 1a 2 + b 1 b 2 a b2 2 + b 1a 2 a 1 b 2 a ı b2 2 For division in polar form, Again, magnitudes and phases decouple. magnitudes divide phases subtract M. Peet Lecture 12: Control Systems 21 / 24

22 Complex Numbers What does this mean for the root locus? Recall the root locus is the set of s such that In other words, In polar coordinates, this means 1 + kĝ(s) = 0 Ĝ(s) = 1 k Ĝ(s) = 1 k eπı Magnitude is 1/k Phase is πrad = 180 Since k can be anything greater than 0: Root locus is all point such that Ĝ(s) = 180 M. Peet Lecture 12: Control Systems 22 / 24

23 Complex Numbers Since Ĝ(s) = (s z 1) (s z m ) (s p 1 ) (s p n ) < s-p 1 Im(s) Then Ĝ(s) = (s z 1) + + (s z m ) (s p 1 ) (s p n ) m n = (s z i ) (s p i ) i=1 i=1 For a point on the root locus: < s-p 2 < s-z Re(s) m n (s z i ) (s p i ) = 180 i=1 i=1 M. Peet Lecture 12: Control Systems 23 / 24

24 Summary What have we learned today? Review of Feedback Closing the Loop Pole Locations The Effect of Changes in Gain Numerical Examples Pole Locations Routh-Hurwitz A Review of Complex Numbers Polar Form Multiplication-Division Next Lecture: Constructing the M. Peet Lecture 12: Control Systems 24 / 24

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