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BME6360 Sp11 I P27-36 - 1.27 STABILITY OF CONTROL SYSTEM...

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Unformatted text preview: 1.27 STABILITY OF CONTROL SYSTEM The stability of a control system can be determined from the poles of its transfer function H(s). However, it is often difficult to do this and it proves easier to work with the Open Loop Transfer function G(s), or the experimental Bode Plot values for G(s). This much is known generally about the poles of H(s): F0(s) bO sm + + bm H(s) = , - ———-—-——————-————— Fi(5) 5n + a] s“—1 + + 21“,] s + an Chareq.: sn+a1 s“—1 + +an=O IMPULSE LOCATION RESPONSE s = i j(o jw axis undamped sinusoid (conditionally stable) = unstable 7 7 If the system is a standard feedback control system, it can be characterized by its Open Loop Transfer Function G(s), the feedback function H(s) and the Loop Gain G(s) H(s). + Fi(S) G(S FO(S) C—- FOG) G(s) ms) = 1 + Gm H(s) NYQUIST STABILITY CRITERION (more below) The system is unstable if either condition is met: Condition #1: IGQw) H003)! >l when 1(G(j(o) HOOD) =— 180° Condition #2 1(G003)H003)) <—180° when IGQm)HQm)t =1 =OdB I. 28 An example of an unstable system. Below are Bode plots of G(j(i)) H(ja)). Condition #1 is tested when the angle is ~180°, which occurs at approximately 0.22 Hz. The gain at this frequency is approximately 3dB or 1.414, which is greater than 1. Condition #2 is tested when the gain is 1 (linear scale) or 0 dB, which occurs at approximately 0.25 Hz. The phase shift is -l90°, which is usually described as ”more than 180°”. If the gain were reduced by more than 3 dB (a factor of 1.414), then the system would be stable. Gain in dB "5.2 0.22 0.24 0.26 0.28 Frequency in Hz -1 so _ ‘ -170 van-,5”.- : .............. ................... ...................... .......... .180 ........... .... 2 .................... ...................... .......... .1 go . ................ g ..................... ""'=-'::-=r:,_.,!:_ ,, 2 ..................... .......... Phase in Degrees .200 .................... ..................... .................... I........-...._:'!"“:?"5A5 ........ —21 ‘ ' ’ [0.2 0.22 0.24 0.26 0.28 Frequency In Hz A schematic example of a stable system is given below. It should be clear that when the gain is equal to 1 (linear) or 0 dB (log scale), that the phase angle is less negative than ~180° (Condition #2). Also, when the phase angle equals -180°, the gain is less than 1. The measures gain margin and phase margin are described graphically. They are used by control engineers to describe "how stable" a system is —— for instance, how much of an increase in feedback gain, perhaps due to heating of an amplifier, is permissible before the system goes unstable? gain 0 dB gain margin d> phase margin —180 1.29 NY UIST STABILITY CRITERION: a brief theoretical descri tion Let P(s) = $327375 = —(F3(—ss)) Then the Characteristic Equation is l + G(s) H(s) = 0 = F(s) If the roots of Rs) are in right half plane 2 unstable. Note that the roots of char. eq. = zeroes of F(s) = poles of P(s) 65 The Nyquist criterion says these > are equivalent questions: Are there any zeros 0fF(s) in 5 plane hatched area? radius 2 00 Is the system unstable? (5 To answer the question, we must use a procedure known as Conformal Mapping and will apply what is known as Cauchy's criterion below: ° point—to—point mapping from the s plane to the F(s) plane: - pick a point in s plane: s = +j(o ' compute complex point in Rs) plane: 130(1)) - pick a contour in 5 plane: 0 origin to +joo o +joo to —j<>o by large radius circle above - —joo to origin ' Compute a contour in F(s) plane 1.30 Example of Conformal Mapping: 1 F =1 G L : —————————————-—— (S) + (S)H(§) (s + 1) (5/10 +1) Bode Plot of . F(s) = l + G(s) H(s) —9O —l80 l 10 l 10 This Bode plot allows us to directly make this plot in the F(s) plane. to: —1 rad/sec A joa: —joo to origin ._'____ 0.4 jo): origin to +jo<> increasing je) for s = Reje R~—><>o 9: +90° —> —90° +jw / . . . does not Circle ongm Cauchy’s criterion: (simple form: assume F(s) has no poles) The # of zeroes inside contour in 5 plane: # encirclements of origin in the F(s) plane The 2nd order system above does not circle the origin in the F(s) plane. Therefore it has no r.h.p. zeroes and it is stable. 1.31 APPLICATIONS TO A FEEDBACK SYSTEM Consider that F(s) = 1 + G(s) H(s). Instead of worrying about encirclements of origin of F(s) plane, it is easier to consider encirclements of the point ~l in the F’(s) = G(s) H(s) plane. . . . . . 1 First consrder a stable example With a non—rational transfer function: G(s) = s + 1 H(s) = 6-5/5 Bode Plot l 0 phase -90 / \ / \S = 1 1 ~ m/ \ the point ——1 is 951 encircled :> F(s) has no r.h.p. zeroes :> P(s) has no r.h.p. poles :> stable system s i 1 H(s) = 10 e—S/5 Then the system will be unstable as the point -1 will be encircled. Notice that, since the transfer function is non—rational, it is very difficult to determine whether or not the system is stable by inspection of H(s) and G(s). However, the Nyquist criterion works graphically and is a great aid to understanding stability. Note, however, that you can determine stability directly from the Bode plots and do not need the Nyquist plot below. Now if we increase the feedback so that G(s) = Increased feedback fa K=10 / \ => F(s) has the point—1 r.h.p. zeroes is encircled => P(s) has r.h.p. poles = unstable system I. 32 EXAMPLES OF STABILITY, NYQUIST, CAUCHY These examples were added because students were curious about these stability criteria. These are best understood by example, but it takes some time to work through the numbers, drawings, and concepts. Y(s) T _ Y(s) _ G(s) (S) ‘ X(s) ‘i + kG(s) H(s) Poles of T(s) Zeroes of T(s) Stability l l l l _—_<s+1)<s+10> —-- 999880110 1 l The first question asked was: Why not use F=l+KGH? Isn‘t that just as easy to analyze? As suggested by this table it is not. Below you will see the Bode and Nyquist plots and see that this observation is true. F'=KGH F=l+KGH _____1__ 52+ 115+ ll (s+l)(s+10) ”52+ lls: 10 ______1____ s3+11152+11105+1001 S3+l1152+llIOS+1000 S3+11152+11105+1000 _____._1QQ_®90____ s3+iiis2+iii0s+1001000 s3 +11152 +1110s +1000 ['11 t» N >——* >< :tt s3+lllsz+11105+1000 The next question followed when a student factored T(s) and asked, "Why not apply the Nyquist criterion, conformal mapping, etc. to the denominator? It looks pretty easy to analyze." As you'll see below, it is more difficult than would be guessed from the simple form off: . Ex # T=G/(l+KGH) 1’:— (denom of T) 5+ 10 1 52+115+11 $2+1IS+11 5+ 100 2 S3+11132+11103+1001 93+llls2+11105+1001 5+ 100 s3 +11152 +1110. +1001,000 S3+11152+11105+1001000 I For each example the following analyses were done: -A. Bode Plot of T(s), the overall transfer function Polar Plot of T(s) — not useful for stability -B. Bode Plot of the denominator F(s) = l + KG(s) H(s) Polar Plot of F(s), which can be used for stability, although it is not usually used. -C. Bode Plot of KG(s)H(s). This is the most popular as one can easily find the gain and phase margins. Nyquist Plot — this is often used for stability — look for encirclement of the point -l. 0D. Bode and Polar Plots of PG) , the simplified denominator of T(s). For Example 1 this is 1.33 1/s+i s+10 __s+10. I .___1_:sz+11s+11—A s+1 s+10 HS) A 0 HS) ) = se+1ls+11 T(s) = 1 + a When looking at an example, students usually ask, “why not apply the Cauchy criterion to the polynomial E(s) ?” Answer: it’s usually amess. In these examples, we’ll see that it works, but is not easy to use. Why is the KGH form used and not the others? — because most systems directly specify them — if you add 1 (e.g., 1 + KGH), the Bode plots and contours become surprisingly difficult to plot! — if you have irrational transfer functions, the plots and roots quickly become very difficult to find, while KGH is still easy to determine. NOTE: To follow Contours, you have to know your insides from your outsides! at r = °° When going forward 1 flu) inside is to your right —> and outside is to your left.<—— outside _j(D Applying the Cauchy Criterion with Sample Contours Contour in s—plane First Example of Contour in F(s) plane G) 6 : —90° ——> 90° s=-j(n @ Origin Circled Twice #Zeroes= Z+#Poles;Z=N+P;N=2 Second Example of Contour phase shift of 540° (1 1/2 circles) during GD The origin is OUTSIDE! N = 0 Z = P = usually 0. I. 34 1 1 s+1’H=s+10 EXAMPLE 1 K = 1, G = T"-Y(‘S) _ 5+ 10 zeroat—lO 1” ’ X(s) ‘82 + 11—5 + 11 poles at —9.8875,—1.1125 , o Bode Plot % . 0 Polar Plot : —20 dB : decade —90° b l 1.11 s2 + 115 + 11 zeroes Ext—9.8875, —1.1125 52+ 11s: 10 poles at—1,—10 Bode 00 Polar No encirclements of origin. Stable. No zeroes in R.H.P_ for 1 + KGH F(s) = 1 + KGH= . . . 1 1 . Nqust Criterion. KGH—S + 1 s + 10, poles at -1, -10; no zeroes. Actual Bode 0 dB 0 / —90° f +1 ~180° stylized l 10 never uite reaches —180° infinite ain and base marOins does not encircle —1 A F(s) = 52 +115 +11; zeroes at —9.8875, —1.1125) @ corresponds to s = rele; r—><><> Bode gain 1 1 111 1 9.89 180° 90" 0 Does not encircle origin 0 No RHP zeroes Stable 1.35 1 1 _ _ 1 EXAMPLE#2 K=1*G=§TT'S+10’H‘5+100 1111,.) _ Y(S) g G _ poles: —100.001, —9.9988, — 1" HX(s) 1 + KGH 1.0011 s + 100 . zeroes: —100 53 +11152 + 1110s + 1001’ Bode Plot ‘1 I I Polar 0 .1 l l | ' ' —180 1 | I | 1 10 s3+11132+ 11105+ 1001 polesat— 41—10—100 F s =————-——————— =1+ KGH _ _ _ ( 1 $3 +11152+11105 +1000 zeroes at 1.0011, 99988, 100.001 origin Bode Plot 1 001 i stable Nyquist Criterion: KGH: poles at—1,*10,—100 ’1', 50x1 .001 does not encircle 53+ 11152+1111Os+ 1000 C1) —90 —180 —270 1 10 100 Does not encircle —1 Stable Gain Margin = 100 dB Phase Margin = infinite Nyquist Plot = 105 (gain always < 1) fi‘(s) = S3 + 11152 + 11105 +1001 ZGI'OCS 31—100.001, -9.9988, —1.0011 Bode Plot 1001 +100. 001 +1.0011 +9.9988 270° 180° 90° Crazy as it may seem the origin is outside the area 0 encircled. Therefore, the number of poles and zeroes is the same (both are zero). l l l 1.36 ' H=s+100 EXAMPLE #3 K = 106 G = unstable 5 + 1 s + 10 YlS) _ G 5+100 poles: —1.4877; +0.1888ij0.7982 T(S) = — —’—" = ——«,—_""“‘— 7Prn Q’ -100 X(s) 1+KGH :3 +llls‘ +11105+l,001,000 “We“ l POM k +180° / —40 dB unusual R z 7 ~ ' s” +11 13- + 1 1 103+ 1001000 zeroes alt—1.4877, + 0.1888 i} 0.7982 F(S)=1+KGH=——————-———-—— poles at—l,—lO,—l00 s3+111s2+11105+1000 Polar Plot Bode Plot 00 l 60 dB I N + I ' @ Cb) i l \\ ’/ . - —360 \-_,k 1 100 © The oriOin is encircled twice. There are two more zeroes than oles. . . . l 0 l t—l,—10,—100 Nqust Criterion: KGH = __________Q_QQQQ____ p0 es a s3 + llls2 + 11105 +1000 We “0‘ Nyquist Plot 103 0 dB at 100 rad encircles ml twice Negative gain margin two r.h.p. zeroes 1 ‘0 100 in KGH and 2 poles 0 in total _90 ‘l 80” at 30 radians ’ ‘180 Negative phase ‘270 margin f;(s) = S3 + 11182 + 11105 + 1001000 ZCFOCS at —l.4877, +0.1888 ij0.7982 Bode *1” Plot 120 dB i I 6:) 100 OD -__\/— —90° -------- dOCS DOt encircles origin me go to —180°! —i~ ...
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