Note#22 Root Locus - Note 22. Sketching Root Locus Diagrams...

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Unformatted text preview: Note 22. Sketching Root Locus Diagrams ECSE—2410 Signals & Systems (Wozny) W Consider the Feedback system X(S) +- H(S) F4476?) X(s) JQAKH“) ms) or where T(s) 2 {LS—)- — ___H(“:)____ m T“) : [01(3) X(s) T 1+ KG(s)H(s) 3+KG(s)H(s)' The characteristic equation for either system is 1+ K G(s)H (s) = 0. The closed loop poles of either feedback system area the roots of the characteristic equation. The loop poles and zeros are defined as the poles and zeros of K G(s)H (s). The root locus is a plot of the roots of the characteristic equation (i.e., the closed loop poles) when the parameter K is varied from zero to infinity. Thus the defining equation of the root locus is 1+ KG(s)H(s) 2 0 or l KG(s)H(s) = ——1, or equivalently A(KG(S)H(S)) 2 ~71 n, 21 odd and IKG(s)H(s)i = 1 or K 2 ~—-———-—. 1G“ )H (S )l Z(KG(S)H(S)) = —7r n, n odd is known as the angle criterion, and K = ————1———— as the magnitude value. IG(S)H(S)i Example. The root locus of ~>Y(S) is defined by the system characteristic equation 1+ KG(s)H(s) = 0 or 52 + 23 + K = 0 for 0 S K _<_ oo. Factoring the quadratic gives 5 = —l i «11- K . Now plot root locations (closed loop poles) for various values of K: K i Closed Loop Poles 0 s = 0, s: ——2 0.5 s = —l.707, s = -O.293 l s = —-1, S: ——l 2 s = -1ij 5 1 S 2 ~1ij2 Another, general solution is to graphically find points in the s—plane that satisfy the angle criterion, A(KG(S)H(S)) = 4 ( K 2)] = -7t 11, n odd. Choosing the points shown on plot below: S S + First column A 1 test pomt S(S + 2) Sz’Jz —————1-——-—- = 90° + 45° 2135” (—JX—J + 2) -—-~ Second column test 4 oint . (-1“12)(1~12) (—1-!)(1-1) 3'1“ ————.1——.=—135°—45° :——180° (—1+J)(1+J) =-1+j2 1 (—1 + j2)(l + j2) max—12 + 2) = ~90° - 45° = —l35° (J'ZXJ'Z + 2) = —(180“ ~tan'12)-tan'12 = ~180° Third column 1 t t ' t 1 es pom (“Sf 2) 5:212 __1__ 2 135° + 90° 2 225° . (-2 — j2)(~j2) — — J —1 ("53—75 = —(180° ~ tan—1%)“ 90" = -243~4° - - J J s=-2+J2 __1_ 2 435° —- 90° 2 ~225° (—2 + j2)(12) Fourth column test A oint l s(s +2)) ______1_._._ (-3 ~ j2)(-1- 1'2) 1 (-3-j)(—1*-J‘) (—3 ‘1)(“1— j) : (180° : tan-"1 -§-) + (180 -— mi1 2) : 262.9“ 2 ~(180° — tan"1 %)~135° = ~296.6° : (180° -tan"1—1—)+135° : 2966" 3 If a point so satisfies the angle criterion, then the value of K at that point is K 2 point satisfying angle criterion 1 K = —--—-————= 2 p(s>H<s>| M” )' s=—1——j2 s=—1—j s=—1 s=-—1+j 52-1+j2 K: K—i—j2x1—j2) 1:5 K= k~1~j><1—j>l=2 K: [-1)(+1)|:1 K: K—1+j)(l+j)]= 2 K: k—l+j2)(1+j2)|=5 l lG(s)H<s)l 2150(50 + 2)}. Instead of plotting root locus diagrams by the above method, let’s use the angle criterion to create a set of sketching rules which will give us more insight into feedback system behavior. 1. The root locus starts (K =0) at the poles of the loop transfer function KH(S)G(S), and ends (K -—a 00) at the zeros of the loop transfer function KH(S)G(s), including zeros at infinity. 2. To find the root locus along the real axis, start at +00 and move left along the axis toward minus infinity. There will be a root locus along the real axis whenever you pass an odd number of roots (poles or zeros) of the loop transfer function KH(s)G(s). The locus alternates along the real axis: no, yes, no, yes, no, yes, Ks(s + 3)(s + 4) (s «1)(5 +1)(s + 2) yes yes yes 1 1 Example. KG(S)H (S) = K 1 2 Example. Two co-incident roots. K 0(5) H ( S) 2 % S .- ' yes yes (2) (2) Think of l ‘ l @ l X it: % fix -1 | +1 T _1 I I +1 T l “ID no no K Example. Three co-incident poles. K G(S)H (s) = 3 (s + 1) yes yes (3) Think of Note #22. Sketching Root Locus Diagrams - p2 3. T119 fao‘f‘ INKS ulfi/efius mew‘Fle. ffieyguos Aj' egfimf akajes y?! r Bangs. 01:2.) 4_ auntielnfl) Emufijim‘f) - K = q [(55) H49 = C52? ‘ 146(5) HQ - Ems (54—!) t2.) {’9' m 50' , “I " N046 0Jng samme abmd" We, Pea/0 @2455, 4, Po‘e*%evu E11293: CPfiE) Pa; = #ane. affix [our Fatal -— it {mat are“ 1“? 8W kegw=fléfll . P35=3~1=Z s‘(s+3) , S, 2:, pm {240%. is a shred i- (Me +0 :0 ‘14:} a bramfaé flax“ flows ‘1‘de wwL 6. Asfimfiok Anga (AA) —.-. :11?“ m «4:3 P85 J L. K E e.} EmuF‘e. kexs) =§T Emfierwg-E Mi MB we“) AA: ~yfx=é+1rm=4 135:2. Note #22. Sketching Root Locus Diagrams — p.3 7 LOWHM (AL) : Enamels. kécs) Ha) = K $($+ZX,S+L’ )/ ) Eugen 10°F refs, (ocdm: -ZOfen ’00? gm loadv‘ms PbE 175—823) fimflwzs AA; :60", 1%0° AL= 0‘23"“ = ~2— Note #12. Sketching Root Locus Diagrams ~p.4 Ci. Break-mag rmkh LEAF). 712. {Egg beam §MH11§10°€ 4+ breakraufi Fern-h; m e954 . 1!. Catawba. EM” ifBIF. Sake—5%, K mfle. 6%N K+KGLQHL§=O => K= - d3 33 6mm Pa‘es ma }1‘ braid Wag/(3W “(LL “we ad- 11 Max W M . s-flmc 31? 3A? .__L._. Gles’ Thw‘HR éE=_d(__L_.)=O Wcal%{,;1Mwu0W 0+5. Fnr'rEQMmr‘e. K M M"; “OHM”. AS K inWoJfiSJTLL locus Mow-£5 02:” K Mu‘!‘ s£s+23 ' “BA? new: «r s=-x “ax Note #22. Sketching Root Locus Diagrams «p5 II. defia‘h W? 37—? (Cmfimei), Emea, Maw: K(s+‘+1(s+£) Sowing) 35‘”) . _ 1 : _s($+_z‘) 1” 85—25 k t 2&9 ($440515) s’q—xos-t 2‘? (g Eaxfios-yz‘insu) ~ (@2512; Haj : D As ‘ ’ (s‘wsn‘rf $4“. 5L+é$+gzo fi 32—3163: g = BAP: .314; 4.267? 52... 53?: -343 = -—‘h7;a 12.. 200+ Loci (nosing +52 gnaw/(us. (be. firm Wé 1Q Mam/‘3 (Bauxfio‘k K Jkrdqafl WSSI‘bg"? KGB) H6): s5+t>£$+i> sly—lbw F51,“ Radix M 5‘ ’ z .5“ 3 K 3‘ Egg 4—— FELL 3‘ raw *3 312m} 43:6. 1211“ figs—=0 ’7 K36. 5" K ‘ ‘ 2' abotb‘l'f’ 1 1 o 1; a” M‘MSJKM m8 Nu) ‘ ($31: 5 6k J2: P065 GAL roo‘l‘S 4.1% original Folahoml. TfiaAfl £31,515 $‘+K§;=o =7 Sat-2:0 =7 sag 2. Note #12. Sketching Root Locus Diagrams ~p‘6 (/ :3. Ayah; _¢% Pefmwefi Court“ EMECAD) R‘dc. a. Falh+amdflmo+ c? 50va fl! Q? KéLS)WS)=-’-'l :{h M61813. 463mg) 3-1M“) MD EmelE. |< chm250 K ’12P?) 5+2 Peck S. m -— ._ J dam-ha Hle ‘ Mu) " T“) “9d Note #22. Sketching Root Locus Diagrams — 13.7 H. Age 4: Arr?wa 19L CemFW rook. ‘ F; [om glefs in Sec. 13 QLOU-e.. FALMTE. KG“) H19: K “5+0?! 1M?" To—gx‘vd BM): ,i<.=~—L—-—— g. .. £3115... GNH’S slur-2.5+?— ¢ iE=_E (32‘*25+2-X2$+¥)“(5z+%)(a$+zfizo 2) sz-zs—‘r=o 2. Note #22. Sketching Root Locus Diagrams - 13.8 Let’s use MATLAB to plot the root loci of some of the previous examples. But first some key MATLAB routines: POLY If V is a vector, POLY(V) is a vector whose elements are the coefficients of the polynomial whose roots are the elements of V . For vectors, ROOTS and POLY are inverse functions of each other, up to ordering, scaling, and roundoff error. Example: Show the roots (s)(s + l)(s + 2) expand to the polynomial s3 + 332 + 23. EDU>>S=[O -l -2]; EDU»pOl)’(S) ans = 1320 ROOTS Find polynomial roots. ROOTS(C) computes the roots of the polynomial whose coefficients are the elements of the vector C. If C has N+l components, the polynomial is C(1)*XAN + + C(N)*X + C(N+l). Example: Show the polynomial s3 + 332 + 25 factors to the roots (s)(s + l)(s + 2). EDU»C=[1 3 2 0]; EDU>>roots(c) ans = 0 —2 ~] RLOCUS Evans root locus. RLOCUS(NUM,DEN) calculates and plots the locus of the roots of: NUM(s) H(s) = 1 + k «mm» = 0 DEN(s) for a set of gains K which are adaptively calculated to produce a smooth plot. Alternatively the vector K can be specified with an optional right~hand argument RLOCUS(NUM,DEN,K). Vectors NUM and DEN must contain the numerator and denominator coefficients in descending powers of s or 2. When invoked with left hand arguments _ R = RLOCUS(NUM,DEN,K) or [R,K] = RLOCUS(NUM,DEN) returns the matrix R with LENGTH(K) rows and (LENGTH(DEN(-l) columns containing the complex root locations. Each row of the matrix corresponds to a gain from vector K. If a second left hand argument is included, the gains are returned in K. J... W?» Note #12. Sketching Root Locus Diagrams . p.9 Example in Sec. 7, p.3. 1 Find root locus for G(S)H(S) = -—————-———-. s(s + 2)(s + 4) EDU>>S=[0 —2 ~41; EDU>>den=poly(s) den: 1 6 8 0 EDU»num=[1]; EDU>>den=[1 6 8 0]; EDU>>rlocus(num,den) 8 lmag Axis Note $9.2. Sketching Root Locus Diagrams - p. 10 Example in Sec. 7, p.3. (5+4) Find root locus for G(s)H(s) = -—--—~———-—————. 5(5 + 2)(s + 6) EDU>>S=[0 '2 —6]; EDU»den=poly(s); EDU>>num=[l 4]; EDU>>rlocus(num,den) Warning: Divide by zero Warning: Divide by zero EDU>>hold Current plot held EDU»axis([-7 2 ~10 10]) mTflfi -7 -6 -5 -4 -3 -2 -1 Real Axis f”“\ Note #22. Sketching Root Locus Diagrams - p.11 Example in Sec. 10, p.4. (s + 3)(s + 5) 5(5 + 2) ‘ Find root locus for G(s)H(s) = EDU»s=[-3 ~5]; EDU>>num=poly(s); EDU>>p=[0 —2]; EDU>>den=poly(p); EDU»rlocus(num,den) EDU>>hOId Current plot held EDU>>axis([-6 1 ~2 2}) Imag Axis Note #2. Sketching Root Locus Diagrams ~ p. 12 Example in Sec. 11, p.5. (5 + 4)(s + 6) Find root locus for G(S)H(s) = s(s + 2) EDU»S=[-4 ~6}; EDU>>num=poly(s); EDU»p=:[O -2]; EDU>>den=poly(p); EDU>>rlocus(num,den) EDU>>hold Current plot held EDU>>axis([-6.5 .5 -2 2]) 2 I T . I W 1.5 lmag Axis O -6 -5 -4 -3 -2 v1 0 Real Axis om , I”, Note #22. Sketching Root Locus Diagrams - p.13 Example in Sec. 13, p.6. 1 s((s + 2f + 22)' Find root locus for G(s)H(s) = EDU>>num=[1]; EDU>>p=[O ~2+2i -2-2i]; EDU>>den=poly(p); EDU>>rlocus(num,den) EDU>>hOId Current plot held EDU>>axis({-4 2 —4 4]) Note #22. Sketching Root Locus Diagrams - p.14 Example in Sec. 14, p.7. ((5 +1)2 +1) Find root locus for G(S)H(S) = s(s + 4) EDU»S=[-l+i -1~i]; EDU>>num=poly(s); EDU»p=[O 4}; EDU>>den=poly(p); EDU>>rlocus(num,den) EDU>>hOId Current plot held EDU>>axis([-4.5 .2 4.5 15]) imag Axis 14.5 -4 -3.5 -3 ~25 ~2 -1 .5 -1 -O.5 0 ' Real Axis ...
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Note#22 Root Locus - Note 22. Sketching Root Locus Diagrams...

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