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1 سیستمهای کنترل خطی پاییز 1389 بسم ا... الرحمن الرحيم دکتر حسين بلندي - دکتر سید مجید اسما عیل زاده

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Recap. Steady State Error Stability Routh Horowitz Method 2

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Root Locus The root locus is a plot of the roots of the characteristic equation of the closed-loop system as a function of the gain of the open- loop transfer function. This graphical approach yields a clear indication of the effect of gain adjustment. INTRODUCTION: 3

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Performance (stability and response) of a closed-loop system (feedback system) depends on its pole locations. In feedback control system design, it is usually necessary to adjust one or more parameters to achieve desired pole locations for the closed-loop system. Root locus is a plot of closed-loop system pole locations when the gain K is varied. A graphical technique exists for plotting the root locus which does not involve calculating the roots of a closed-loop polynomial. Root locus is a powerful graphical tool for analysis and design of feedback control systems. Root Locus 4

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i.e; A specific technique which shows how changes in one of a system’s parameter (usually the controller gain, K) will modify the location of the closed-loop poles in the s-domain. 5

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6 Graphical Evaluation )Hostetter): A rational function When evaluated at a specific value of the variable, is:

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On a pole-zero plot, suppose a directed line segment is drawn from the position of a pole, say to the value at which the function is to be evaluated. The segment has length and makes the angle With the real axis. As shown below, at a root s 1 7

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8 Thus:

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Consider the system: Calculation of Magnitude and Angles (Ogata) 9

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Definition The closed-loop poles of the negative feedback control: are the roots of the characteristic equation: The root locus is the locus of the closed-loop poles when a specific parameter (usually gain, K) is varied from 0 to infinity. 10

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Root Locus Method Foundations The values of s in the s-plane that make the loop gain KG(s)H(s) equal to -1 are the closed- loop poles (i.e. ) KG(s)H(s) = -1 can be split into two equations by equating the magnitudes and angles of both sides of the equation. 11

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Root Locus for Feedback Systems magnitude conditionsangle conditions 12

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Angle and Magnitude Conditions Independent of K 13

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14 APPLICATION OF THE MAGNITUDE AND ANGLE CONDITIONS: S Once the open-loop transfer function G(s)H(s) has been determined and put into the proper form, the poles and zeros of this function are plotted in the S plane. As an example, consider: with the damping ratio < 1, the complex-conjugate poles are:

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15 The plot of the poles and zeros is shown in bellow:

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17 magnitude conditions: angle conditions: All angles are considered positive, measured in the counter clock wise sense.

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PLOTTING ROOTS OF A CHARACTERISTIC EQUATION: Example1 : position-control system 18

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Example1 (Continue) : position-control system 19

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Example1 (Continue) : position-control system 20

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21 Example1 (Continue) : position-control system

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22 Example1 (Continue) : position-control system These curves are defined as the root-locus plot of

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23 Example1 (Continue) : position-control system Corresponding to the selected roots there is a required value of K that can be determined from the plot. When the roots have been selected, the time response can be obtained. Kpositive The value of K is normally considered to be positive. However, it is possible for K to be negative. For an increase in the gain K of the system, analysis of the root locus reveals the following characteristics:

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24 Example1 (Continue) : position-control system

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25 simple second-order system: A zero is added to the simple second-order system: QUALITATIVE ANALYSIS OF THE ROOT LOCUS: Compare

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26 If a pole, instead of a zero, is added to the simple second-order system, the resulting transfer function is: root locus of the closed-loop control system

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27 root-locus configurations for negative feedback control systems ?

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28 Example2: Determine the locus of all the closed-loop poles of C(s)/R(s), for

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29 >> sys = zpk( -4, [-1 -2 -5],1) Zero/pole/gain: (s+4) ---------------------- (s+1) (s+2) (s+5) >> rlocus(sys) Rlocus in MATLAB:

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30 GEOMETRICAL PROPERTIES (CONSTRUCTION RULES): To facilitate the application of the root-locus method, the following rules are established for K > 0. These rules are based upon the interpretation of the angle condition and an analysis of the characteristic equation. These rules can be extended for the case where K < 0. These rules provide checkpoints to ensure that the solution is correct. which provides a qualitative idea of achievable closed-loop system performance.

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31 Rule 1: Number of Branches of the Locus

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RL starts from n poles (k=0) RL ends at m zeroes (k=infinity) # of branches= max( n, m) 33

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37 Rule 2: Real-Axis Locus

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The locus includes all points along the real axis to the left of an odd number of poles plus zeros of GH 39

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44 Rule 3: Asymptotes of Locus as s Approaches Infinity

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45 Rule 3(continue):

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46 Rule 4: Real-Axis Intercept of the Asymptotes

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47 Rule 5: Breakaway Point on the Real Axis breakaway pointbreakin point

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48 The breakaway and break-in points can easily be calculated for an open-loop pole-zero combination for which the derivatives of W(s)=-K As an example, if then Multiplying the factors together gives

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49 by taking the derivative of this function and setting it equal to zero, the points can be determined or

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50 Rule 6: Complex Pole (or Zero): Angle of Departure

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51 In a similar manner the approach angle to a complex zero can be determined

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52 Rule 7: Imaginary-Axis Crossing Point

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53 Rule 7(continue):

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Learning by doing Example Sketch the root locus of the following system: 54

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#1 Assuming n poles and m zeros for G(s)H(s): The n branches of the root locus start at the n poles. m of these n branches end on the m zeros The n-m other branches terminate at infinity along asymptotes. First step: Draw the n poles and m zeros of G(s)H(s) using x and o respectively 55

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Draw the n poles and m zeros of G(s)H(s) using x and o respectively. 3 poles: p 1 = 0; p 2 = -1; p 3 = -2 No zeros 56

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Draw the n poles and m zeros of G(s)H(s) using x and o respectively. 3 poles: p 1 = 0; p 2 = -1; p 3 = -2 No zeros 57

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#2 The loci on the real axis are to the left of an ODD number of REAL poles and REAL zeros of G(s)H(s) Second step: Determine the loci on the real axis. Choose a arbitrary test point. If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus 58

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Determine the loci on the real axis: Choose a arbitrary test point. If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus 59

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Determine the loci on the real axis: Choose a arbitrary test point. If the TOTAL number of both real poles and zeros is to the RIGHT of this point is ODD, then this point is on the root locus 60

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#3 Assuming n poles and m zeros for G(s)H(s): The root loci for very large values of s must be asymptotic to straight lines originate on the real axis at point: radiating out from this point at angles: Third step: Determine the n - m asymptotes of the root loci. Locate s = α on the real axis. Compute and draw angles. Draw the asymptotes using dash lines. 61

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Determine the n - m asymptotes: Locate s = α on the real axis: Compute and draw angles: Draw the asymptotes using dash lines. 62

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Determine the n - m asymptotes: Locate s = α on the real axis: Compute and draw angles: Draw the asymptotes using dash lines. 63

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Breakpoint Definition The breakpoints are the points in the s- domain where multiples roots of the characteristic equation of the feedback control occur. These points correspond to intersection points on the root locus. 64

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#4 Given the characteristic equation is KG(s)H(s) = -1 The breakpoints are the closed-loop poles that satisfy: Fourth step: Find the breakpoints. Express K such as: Set dK/ds = 0 and solve for the poles. 65

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Find the breakpoints. Express K such as: Set dK/ds = 0 and solve for the poles. 66

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Find the breakpoints. Express K such as: Set dK/ds = 0 and solve for the poles. 67

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#5 Assuming n poles and m zeros for G(s)H(s): The n branches of the root locus start at the n poles. m of these n branches end on the m zeros The n-m other branches terminate at infinity along asymptotes. Last step: Draw the n-m branches that terminate at infinity along asymptotes 68

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Applying Last Step Draw the n-m branches that terminate at infinity along asymptotes 69

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Points on both root locus & imaginary axis? Points on imaginary axis satisfy: Points on root locus satisfy: Substitute s=jω into the characteristic equation and solve for ω. jω?jω? - jω 70

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71 EXAMPLE : ROOT LOCUS

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72 EXAMPLE (continue): ROOT LOCUS

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73 EXAMPLE (continue): ROOT LOCUS

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74 EXAMPLE (continue): ROOT LOCUS

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75 EXAMPLE (continue): ROOT LOCUS

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76 EXAMPLE (continue): ROOT LOCUS

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77 EXAMPLE (continue): ROOT LOCUS

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78 EXAMPLE (continue): ROOT LOCUS

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79 Example of an application of the root-locus method The systematic application of the for the construction of a root locus is shown in the following non-trivial example for the open-loop transfer function : The degree of the numerator polynomial is m=1. This means that the transfer function has one zero (s=-1). The degree of the denominator polynomial is n=4 and we have the four poles (s=0, s=-2 s=-6+2j, s=-6- 2j). First the poles (x) and the zeros (o) of the open loop are drawn on the S plane as

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80 Values of are in red

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81 We have (n-m=3) branches that go to infinity and the asymptotes of these three branches are lines which intercept the real axis according to rule. the crossing is at : the slopes of the asymptotes are: The asymptotes are shown in Figure as blue lines:

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83 Using Rule 2 it can be checked which points on the real axis are points on the root locus. The points -1~~
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84 The angle of departure of the root locus from the complex pole at can be determined from

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Breakaway Point Example Consider the following loop transfer function. Real axis loci exist for the full negative axis. Asymptotes: angles jjjj X X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j +1+1+1+1 60 ° asymptotes

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Breakaway Point Example Determine the breakaway points from then jjjj X X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j +1+1+1+1 3,1 0)3)(1(34 2 s ssss 0 )96( )9123( 96)3( 223 2 232 sss ssK sss K ds d ss K ds d

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Root Locus Plot Example Loop Transfer function: Roots: s = 0, s = –4, s = –2 4j Real axis segments: between 0 and –4. Asymptotes: angles = 2 4 )0224( a 4 7, 4 5, 4 3, 404 )12( kasymptotes jjjj X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j +1+1+1+1 45 ° 4j4j4j4j –4j–4j–4j–4j X X

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Root Locus Plot Example Breakaway points: Three points that breakaway at 90 °. solving, jjjj X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j +1+1+1+1 45 ° 4j4j4j4j –4j–4j–4j–4j X X 2

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Root Locus Plot Example The imaginary axis crossings: Characteristic equation Routh table s 4 1 36 K s 3 8 80 0 s 2 26 K 0 s 80-8K/26 0 0 s 0 K 0 0 Condition for critical stability 80-8K/26 > 0 or K 0 or K<260 The auxiliary equation 26 s 2 + 260 = 0 solving 080368 234=++++Kssss jjs16.3 10±=±=

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The final plot is shown on the right. What is the value of the gain K corresponding to the breakaway point at Root Locus Plot Example X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j 4j4j4j4j –4j–4j–4j–4j X X 3.16j jjjj

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From the general magnitude condition the gain corresponding to the point s 1 on the loci is For the point s 1 = –2 + 2.45j K = |–2 + 2.45j| ·|–2 + 2.45j + 4| ·|–2 + 2.45j + 2 + 4j| ·|–2 + 2.45j + 2 – 4j| / 1.0 = 3.163 · 3.163 · 6.45 · 1.55 = 100.0 = 3.163 · 3.163 · 6.45 · 1.55 = 100.0 Root Locus Plot Example Gain Calculations m i i n i i zspsK 1 1 1 1

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Is there a gain corresponding to a damping ratio of 0.707 or more for all system modes? = 0.707 = cos( ) = 45 ° Root Locus Plot Example X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j 4j4j4j4j –4j–4j–4j–4j X X 3.16j jjjj 45 °

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Examine the responses for the various gains and relate them to the location of the closed-loop roots. K = 64, roots are –2, –2, –2±3.46j K = 100, roots are –2±2.45j, –2±2.45j K = 260, roots are ±3.16j, –4±3.16j X –4–4–4–4 X –2–2–2–2 –2j–2j–2j–2j 2j2j2j2j 4j4j4j4j –4j–4j–4j–4j X X 3.16j jjjj K=64 K=100 K=260 Root Locus Plot Example Time Responses

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Time (sec.) Amplitude Step Response, K = 64 012345 0 0.2 0.4 0.6 0.8 1 s = –2, –2 s = –2±3.46j whole response

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Root Locus Plot Example Time Responses Time (sec.) Amplitude Step Response, K = 100 012345 0 0.2 0.4 0.6 0.8 1 whole response s = –2±2.45j (repeated)

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Root Locus Plot Example Time Responses Time (sec.) Amplitude Step Response K = 260 00.511.522.533.544.55 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 whole response s = ±3.16j s = –4±3.16j

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The root locus method focuses on the roots of the characteristic equation. There can be several different loop transfer functions that have the same closed-loop characteristic equation. To construct the root locus for a characteristic equation which has two parameters, we construct fictitious loop transfer functions and apply the normal methods.

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Consider the following characteristic equation. s 3 + s 2 + s + = 0. Write this in the general form of 1 + GH(s) = 0 with as a multiplying gain. This will allow the plotting of the root locus with respect to the gain for any given value of . The roots of the characteristic equation of GH´(s) define the starting points for the root loci. Consider the loci of these roots. 01 23 ss s GH´(s)

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The characteristic equation of GH´(s) is s 3 + s 2 + = 0, which may be written as Now construct the root locus of GH´´(s) in terms of the gain . 0 )1( 1 2 ss GH´´(s) jjjj X –2–2–2–2 –1–1–1–1 X X 1111 1111 2222 2222 1 = 0.3 s = –1.2, 0.1±0.5j 2 = 1.8 s = –1.66, 0.33±1.0j

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Now construct the root locus for where the open- loop poles correspond to the previous root locus for varying . Asymptotes: ± 90 ° a 5 a 5 a a 23 ss s GH´(s)

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Imaginary axis crossings: s 3 + s 2 + s + = 0 s 3 1 0 s 2 1 0 s 0 0 s 0 0 0 = s 2 + js jjjj –2–2–2–2 –1–1–1–1 11 11 22 22 O

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106 SUMMARY OF ROOT-LOCUS CONSTRUCTION RULES FOR NEGATIVE FEEDBACK

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Modern Control Systems (MCS) Dr. Imtiaz Hussain Assistant Professor URL :http://imtiazhussainkalwar.weebly.com/

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