SECTION 5: ROOT LOCUS ANALYSIS
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1 SECTION 5: ROOT LOCUS ANALYSIS MAE 4421 Control of Aerospace & Mechanical Systems
2 2 Introduction
3 Introduction 3 Consider a general feedback system: Closed loop transfer function is 1 is the forward path transfer function May include controller and plant is the feedback path transfer function Each are, in general, rational polynomials in and
4 Introduction 4 So, the closed loop transfer function is 1 Closed loop zeros: Zeros of Poles of Closed loop poles: A function of gain, Consistent with what we ve already seen feedback moves poles
5 Closed Loop Poles vs. Gain 5 How do closed loop poles vary as a function of? Plot for 0, 0.5, 1, 2, 5, 10, 20 Trajectory of closed loop poles vs. gain (or some other parameter): root locus Graphical tool to help determine the controller gain that will put poles where we want them We ll learn techniques for sketching this locus by hand
6 Root Locus 6 An example of the type of root locus we ll learn to sketch by hand, as well as plot in MATLAB:
7 7 Evaluation of Complex Functions
8 Vector Interpretation of Complex Functions 8 Consider a function of a complex variable where are the zeros of the function, and are the poles of the function We can write the function as where is the # of zeros, and is the # of poles
9 Vector Interpretation of Complex Functions 9 At any value of, i.e. any point in the complex plane, evaluates to a complex number Another point in the complex plane with magnitude and phase where and
10 Vector Interpretation of Complex Functions 10 Each term represents a vector from to the point,, at which we re evaluating Each represents a vector from to For example: Zero at: Poles at:, 12 and 4 Evaluate at
11 Vector Interpretation of Complex Functions 11 First, evaluate the magnitude The resulting magnitude:
12 Vector Interpretation of Complex Functions 12 Next, evaluate the angle The result:
13 Finite vs. Infinite Poles and Zeros 13 Consider the following transfer function One finite zero: Three finite poles: 0, 3, and 10 But, as lim 0 This implies there must be a zero at All functions have an equal number of poles and zeros If has poles and zeros, where, then has zeros at is an infinite complex number infinite magnitude and some angle
14 14 The Root Locus
15 Root Locus Definition 15 Consider a general feedback system: Closed loop transfer function is 1 Closed loop poles are roots of 1 That is, the solutions to 1 0 Or, the values of for which 1 (1)
16 Root Locus Definition 16 Because and are complex functions, (1) is really two equations: that is, the angle is an odd multiple of 180, and 1 So, if a certain value of satisfies the angle criterion then that value of is a closed loop pole for some value of And, that value of is given by the magnitude criterion 1
17 Root Locus Definition 17 The root locus is the set of all points in the s plane that satisfy the angle criterion The set of all closed loop poles for We ll use the angle criterion to sketch the root locus We will derive rules for sketching the root locus Not necessary to test all possible s plane points
18 Angle Criterion Example 18 Determine if 32is on this system s root locus is on the root locus if it satisfies the angle criterion From the pole/zero diagram does not satisfy the angle criterion It is not on the root locus
19 Angle Criterion Example 19 Is on the root locus? Now we have is on the root locus What gain results in a closed loop pole at? Use the magnitude criterion to determine yields a closed loop pole at And at its complex conjugate, 2
20 20 Sketching the Root Locus
21 Real Axis Root Locus Segments 21 We ll first consider points on the real axis, and whether or not they are on the root locus Consider a system with the following open loop poles Is on the root locus? I.e., does it satisfy the angle criterion? Angle contributions from complex poles cancel Pole to the right of : 180 All poles/zeros to the left of : 0 satisfies the angle criterion,, so it is on the root locus
22 Real Axis Root Locus Segments 22 Now, determine if point is on the root locus Again angles from complex poles cancel Always true for real axis points Pole and zero to the left of contribute 0 Always true for real axis points Two poles to the right of : 360 Angle criterion is not satisfied is not on the root locus
23 Real Axis Root Locus Segments 23 From the preceding development, we can conclude the following concerning real axis segments of the root locus: All points on the real axis to the left of an odd number of open loop poles and/or zeros are on the root locus
24 Non Real Axis Root Locus Segments 24 Transfer functions of physically realizable systems are rational polynomials with real valued coefficients Complex poles/zeros come in complex conjugate pairs Root locus is symmetric about the real axis Root locus is a plot of closed loop poles as from varies Where does the locus start? Where does it end?
25 Non Real Axis Root Locus Segments 25 1 We ve seen that we can represent this closed loop transfer function as The closed loop poles are the roots of the closed loop characteristic polynomial As Δ Δ Closed loop poles approach the open loop poles Root locus starts at the open loop poles for 0
26 Non Real Axis Root Locus Segments 26 As Δ So, as, the closed loop poles approach the openloop zeros Root locus ends at the open loop zeros for Including the zeros at Previous example: 5poles, 1zero One pole goes to the finite zero Remaining poles go to the 4zeros at Where are those zeros? (what angles?) How do the poles get there as?
27 Non Real Axis Root Locus Segments 27 As, of the poles approach the finite zeros The remaining poles are at Looking back from, it appears that these poles all came from the same point on the real axis, Considering only these poles, the corresponding root locus equation is These poles travel from (approximately) to along asymptotes at angles of,
28 Asymptote Angles 28 To determine the angles of the consider a point,, very far from If is on the root locus, then That is, the multiple of asymptotes, angles from to sum to an odd, Therefore, the angles of the asymptotes are,
29 Asymptote Angles 29 For example poles and zeros poles go to as Poles approach along asymptotes at angles of, , If, 60,, 180,, 300
30 Asymptote Origin 30 The asymptotes come from a point,, on the real axis where is located? The root locus equation can be written 1 0 where According to a property of monic polynomials: Σ Σ where are the open loop poles, and are the open loop zeros
31 Asymptote Origin 31 The closed loop characteristic polynomial is If 1, i.e. at least two more poles than zeros, then Σ where are the closed loop poles The sum of the closed loop poles is: Independent of Equal to the sum of the open loop poles Σ Σ The equivalent open loop location for the poles going to infinity is These poles, similarly, have a constant sum:
32 Asymptote Origin 32 As, of the closed loop poles go to the open loop zeros Their sum is the sum of the open loop zeros The remainder of the poles go to Their sum is The sum of all closed loop poles is equal to the sum of the open loop poles The origin of the asymptotes is
33 Root Locus Asymptotes Example 33 Consider the following system open loop zero and open loop poles As : One pole approaches the open loop zero Four poles go to along asymptotes at angles of:,, 45,, 225,,
34 Root Locus Asymptotes Example 34 The origin of the asymptotes is Σ Σ As, four poles approach along four asymptotes emanating from at angles of,,, and
35 Root Locus Asymptotes Example 35
36 36 Refining the Root Locus
37 Refining the Root Locus 37 So far we ve learned how to accurately sketch: Real axis root locus segments Root locus segments heading toward, but only far from Root locus from previous example illustrates two additional characteristics we must address: Real axis breakaway/break in points Angles of departure/arrival at complex poles/zeros
38 Real Axis Breakaway/Break In Points 38 Consider the following system and its root locus Two finite poles approach two finite zeros as Where do they leave the real axis? Breakaway point Where do they re join the real axis? Break in point
39 Real Axis Breakaway Points 39 Breakaway point occurs somewhere between 1and 2 Breakaway angle: 180 where is the number of poles that come together here, 90 As gain increases, poles come together then leave the real axis Along the real axis segment, maximum gain occurs at the breakaway point To calculate the breakaway point: Determine an expression for gain,, as a function of Differentiate w.r.t. Find for / 0 to locate the maximum gain point
40 Real Axis Breakaway Points 40 All points on the root locus satisfy 1 On the segment containing the breakaway point,, so 1 The breakaway point is a maximum gain point, so Solving for 1 0 yields the breakaway point
41 Real Axis Breakaway Points 41 For our example, along the real axis Differentiating w.r.t Setting the derivative to zero , 3.37 The breakaway point occurs at
42 Real Axis Break In Points 42 The poles re join the real axis at a break in point A minimum gain point As gain increases, poles move apart Break in angles are the same as breakaway angles 180 As for the breakaway point, the break in point satisfies 1 In fact, this yields both breakaway and break in points For our example, we had Breakaway point: 1.63 Break in point:
43 Real Axis Breakaway/Break In Points
44 Angles of Departure/Arrival 44 Consider the following two systems Similar systems, with very different stability behavior Understanding how to determine angles of departure from complex poles and angles of arrival at complex zeros will allow us to predict this
45 Angle of Departure 45 To find the angle of departure from a pole, : Consider a test point,, very close to The angle from to is The angle from all other poles/zeros, /, to are approximated as the angle from or to Apply the angle criterion to find Solving for the departure angle, : In words: 180 Σ Σ 180
46 Angle of Departure 46 If we have complex conjugate open loop poles with multiplicity, then The different angles of departure from the multiple poles are where,
47 Angle of Arrival 47 Following the same procedure, we can derive an expression for the angle of arrival at a complex zero of multiplicity In summary, ,, Σ Σ Σ Σ
48 Departure/Arrival Angles Example 48 Angle of departure from Due to symmetry: Angle of arrival at , 183.8
49 Departure/Arrival Angles Example 49
50 Departure/Arrival Angles Example 50 Next, consider the other system Angle of departure from Angle of arrival at
51 Departure/Arrival Angles Example 51
52 Axis Crossing Points 52 To determine the location of a axis crossing Apply Routh Hurwitz Find value of that results in a row of zeros Marginal stability axis poles Roots of row preceding the zero row are axis crossing points Or, plot in MATLAB More on this later
53 Root Locus Sketching Procedure Summary Plot open loop poles and zeros in the s plane 2. Plot locus segments on the real axis to the left of an odd number of poles and/or zeros 3. For the poles going to, sketch asymptotes at angles,, centered at, where,
54 Root Locus Sketching Procedure Summary Calculate departure angles from complex poles of multiplicity 1 Σ Σ and arrival angles at complex zeros of multiplicity 1 Σ Σ Determine real axis breakaway/break in points as the solutions to 1 0 Breakaway/break in angles are 180 / to the real axis 6. If desired, apply Routh Hurwitz to determine axis crossings
55 Sketching the Root Locus Example 1 55 Consider a satellite, controlled by a proportionalderivative (PD) controller A example of a double integrator plant We ll learn about PD controllers in the next section Closed loop transfer function 1 Sketch the root locus Two open loop poles at the origin One open loop zero at 1
56 Sketching the Root Locus Example Plot open loop poles and zeros Two poles, one zero 2. Plot real axis segments To the left of the zero 3. Asymptotes to One pole goes to the finite zero One pole goes to at 180 along the real axis
57 Sketching the Root Locus Example Departure/arrival angles No complex poles or zeros 5. Breakaway/break in points Breakaway occurs at multiple roots at 0 2 Break in point: , 0
58 Sketching the Root Locus Example 2 58 Now consider the same satellite with a different controller A lead compensator more in the next section Closed loop transfer function Sketch the root locus 1 12
59 Sketching the Root Locus Example Plot open loop poles and zeros Now three open loop poles and one zero 2. Plot real axis segments Between the zero and the pole at Asymptotes to, ,
60 Sketching the Root Locus Example Departure/arrival angles No complex open loop poles or zeros 5. Breakaway/break in points , 2.31, Breakaway: 0, 5.19 Break in: 2.31
61 Sketching the Root Locus Example 3 61 Now move the controller s pole to Closed loop transfer function Sketch the root locus 1 9
62 Sketching the Root Locus Example Plot open loop poles and zeros Again, three open loop poles and one zero 2. Plot real axis segments Between the zero and the pole at 9 3. Asymptotes to, 90, Departure/arrival angles No complex open loop poles or zeros 4
63 Sketching the Root Locus Example Breakaway/break in points ,3,3 Breakaway: 0, 3 Break in: 3 Three poles converge/diverge at 3 Breakaway angles: 0, 120, 240 Break in angles: 60, 180, 300
64 64 Root Locus in MATLAB
65 feedback.m 65 sys = feedback(g,h,sign) G: forward path model tf, ss, zpk, etc. H: feedback path model sign: -1 for neg. feedback, +1 for pos. feedback optional default is -1 sys: closed loop system model object of the same type as G and H Generates a closed loop system model from forward path and feedback path models For unity feedback, H=1
66 feedback.m 66 For example: T=feedback(G,H); T=feedback(G,1); T=feedback(G1*G2,H);
67 rlocus.m 67 [r,k] = rlocus(g,k) G: open loop model tf, ss, zpk, etc. K: vector of gains at which to calculate the locus optional MATLAB will choose gains by default r: vector of closed loop pole locations K: gains corresponding to pole locations in r If no outputs are specified a root locus is plotted in the current (or new) figure window This is the most common use model, e.g.: rlocus(g,k)
68 68 Generalized Root Locus
69 Generalized Root Locus 69 We ve seen that we can plot the root locus as a function of controller gain, Can also plot the locus as a function of other parameters For example, open loop pole locations Consider the following system: Plot the root locus as a function of pole location, Closed loop transfer function is
70 Generalized Root Locus Want the denominator to be in the root locus form: 1 First, isolate in the denominator 1 1 Next, divide through by the remaining denominator terms
71 Generalized Root Locus The open loop transfer function term in this form is Sketch the root locus: 1. Plot poles and zeros 1 A zero at the origin and poles at 2. Plot real axis segments Entire negative real axis is left of a single zero
72 Generalized Root Locus Asymptote to Single asymptote along negative real axis 4. Departure angles Break in point ,1 1is not on the locus Break in point: 1
73 73 Design via Gain Adjustment
74 Design via Gain Adjustment 74 Root locus provides a graphical representation of closed loop pole locations vs. gain We have known relationships (some approx.) between pole locations and transient response These apply to 2 nd order systems with no zeros Often, we don t have a 2 nd order system with no zeros Would still like a link between pole locations and transient response Can sometimes approximate higher order systems as 2 nd order Valid only under certain conditions Always verify response through simulation
75 Second Order Approximation 75 A higher order system with a pair of second order poles can reasonably be approximated as second order if: 1) Any higher order closed loop poles are either: a) at much higher frequency ( ~5) than the dominant 2 nd order pair of poles, or b) nearly canceled by closed loop zeros 2) Closed loop zeros are either: a) at much higher frequency ( ~5) than the dominant 2 nd order pair of poles, or b) nearly canceled by closed loop poles
76 Design via Gain Adjustment Example 76 Determine for 10% overshoot Assuming a 2 nd order approximation applies: ln ln 0.59 Next, plot root locus in MATLAB Find gain corresponding to 2 nd order poles with 0.59 If possible often it is not Determine if a 2 nd order approximation is justified Verify transient response through simulation
77 Design via Gain Adjustment Example 77 Root locus shows that a pair of closed loop poles with exist for :, Where is the third closed loop pole?
78 Design via Gain Adjustment Example 78 Third pole is at Not high enough in frequency for its effect to be negligible But, it is in close proximity to a closedloop zero Is a 2 nd order approximation justified? Simulate
79 Design via Gain Adjustment Example 79 Step response compared to a true 2 nd order system No third pole, no zero Very similar response 11.14% overshoot 2 nd order approximation is valid Slight reduction in gain would yield 10% overshoot
80 Design via Gain Adjustment Example 80 Step response compared to systems with: No zero No third pole Quite different responses Partial pole/zero cancellation makes 2 nd order approximation valid, in this example
81 When Gain Adjustment Fails 81 Root loci do not go through every point in the s plane Can t always satisfy a single performance specification, e.g. overshoot or settling time Can satisfy two specifications, e.g. overshoot and settling time, even less often Also, gain adjustment affects steady state error performance In general, cannot simultaneously satisfy dynamic requirements and error requirements In those cases, we must add dynamics to the controller A compensator
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