hw5 - Introduction: Bode plots consists of two graphs, the...

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Unformatted text preview: Introduction: Bode plots consists of two graphs, the magnitude of H (w) and the phase angle of H ((0), both plotted as a function of frequency w. The db magnitude vs. logw is called the Bode magnitude plot, or log~modulus plot. The phase angle vs. logw is called the Bode phase angle plot. Logarithmic scales are used for Bode plots because they considerably simplify their construction, manipulation, and interpolation. Bode plots clearly illustrate the relative stability of a system. Gain and phase margins are often defined in terms of Bode plots. Bode plot rules: Considering the following transfer function, 5 + s + + Haw) ___ K (1 21X 22) (5' Zn) (3) (S + MKS + 292) (S + pm) we first write it in the so-called Bode form (3 = jw) as below: How) _ [((Hfllzi) (jw/zl + DOE/22 + 1) (jw/zm + 1) £le (inLUw/pl + ooze/p2 + 1) Uw/pn + 1) Where I is a nonnegative integer. The Bode gain K3 is defined as KB K<Hi7=lei) n i=1pi UP down} by ZOdb/dec at { The Magnitude plot breaks { The Phase plot: breaks { up } by 45°/dec one decade before { zeros polesi' zeros poles} zeros poles} down breaks { darn} by 45°/dec one decade after { Initial magfiitude plot: - nit; Othen lHUwNIMSD in db, so Q Initial magnitude= ZOZongQ'OM 0 Hit ¢ 0 then a) = (KE)1/l with an initial slope of—ZOdb/dec. Initial Phase plot: Initial 4 = Azeros — Lpoles /_ Final 4 = —90(RD) Where RD is the relative degree of the system= deg[den]—deg[num]. Relative stability, gain margin and phase margin: he basic goal of control system design is meeting performance specifications which can be stated as frequency-domain specifications (frequency response) or time—domain specifications (time response). Frequency—domain specifications for continuous and discrete time systems are often stated in one or more of the seven ways, Gain margin, Phase Margin, Delay Time, Bandwidth, Cutoff Rate, Resonance Peak and Resonance frequency. As a measure of stability, Gain Margin, is defined as the magnitude of the reciprocal of the open-loop transfer function, evaluated at the frequency (on at which the phase angle is —180°. That is: 1 l H Cw”)! a)” is called the phase crossover frequency. Since 0 db corresponds to a magnitude of 1, the gain margin is the number of decibels that [H is below 0 db at the phase CtQSanet_frenupn oJr__._._. ~47 W - _ 2 Phase Margin, ¢pM is a measure of relative stability and is defined as 180° plus the phase angle (1)1 of the open-loop transfer firnction at unity gain. That is: ¢PM = [180 + QTHH(0)1)]° Where lHCwfll = 1 and col is called the gain crossover frequency. [1]. SCHAUM’S outlines, Feedback and Control System 1. Draw the asymptotic bode plot for the following open-loop transfer functions. Solution: The following Matlab code was written to define a function SemiLog in order to draw semilog paper for Bode plot hand draw. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% 53; Function Se IiLog to draw Magnitude- 52; and Phase plot axis for Homework #5 its ‘2: EML 4312, Dr. Dixon, Control 35 *2; Nov.22.2009, Ramin Shamshiri $5 ramiri.sh@ufl .eclu is r 1 r 0") t Function SemiLog ( wMin , wMax, ' 630M121, dbMax, Phasettin , Phaselviax) figurel = figure; %=====Create Magnitude plot axis=========== deick=dbMin:20:dbMaX; ‘XMinerGrid',’on',... ‘XMinorTick','on',... ‘XScale’,'log',... ‘YGrid‘,'en’,... ’Y‘ick‘,deick,... 'Parent‘,figurel); - axis(axesl,[wmin wMaX dbMin dbMaXJ); ylabel(axesl,'magnitude (dBl’); %=====Create Phase plot axis=========== PhaseTick=PhaseHin:45:PhaseMax; axesZ = axes(... 'Fontsize',8,... 'Position',[0.l 0.1 0.8 0.41,... 'XGrid',’on',... 'XMiuorGrid','on',... ‘XMinorTick’,'on',.. ‘XScale','log’,... 'YGrid‘,’on',. ’YTick‘,PhaseTick,... ‘Parent‘,figurel): axis(axesE,[wMin wMax PhaseMin PhaseMaXJ); ylabel(axe52,'Phase (degl'); 5+2 a. H (s) = (5+1)2 Writing the given transfer function in bode form: 2 'w 2 + 1 :. (—) 1 (1w+1)(}w+1). Zeros = 2 {Poles = 1,1 KB : 2 Magnitude Plot: We have a zero at w = 2, so the magnitude plot breaks up by 20db/ dec. We also have a pole with multiplicity of 2 at a) = 1, so magnitude plot breaks down by 2 X 20 = 40db/dec. Initial magnitude plot: with l = 0, which means we don’t have integrator (1 / S) in the transfer function, thus we begin at: zozog|H0an|1w=0 = 2010911100): = ZOIOgIZI = 6.02 Phase Plot: It first breaks down by 2 X 45 = 90°/dec at a) = 0.1 (one decade before where we have pole) then breaks up at a) = 10 by 2 X 45 = 90° / dec (one decade after Where we have pole). Also should breaks up by 45°/dec at a) = 0.2 (one decade before Where we have zero) and then breaks down 45°/dec at w = 20 (one decade after where we have zero). Initial angle of phase plot: Lzeros — Lpoles = 0" — 0" = 0" Final angle of phase plot: RD (—90) = (1)(~90°) The Bode plot for this problem is drawn and shown in figure.1. l M o m 0 <2!“ Magnkude (as) L) O 5. o to: 420 _‘ j " ", 10" w“ 10“ _ 10 10 Frequency (moisten) Figure 1. Bode plots asymptotes for La. created using function: SemiLog(0.0:l,100,—60,20, w180,0) Phase (deg') - b. HCS) = ~53- Transfer function in Bode form: (jw + 1) H ' = —- 0‘”) 0am Zeros = 1 Poles = 0,0 KB = 1 Magnitude Plot: Should breaks up at a) = l by ZOdb/dec. Initial magnitude, l = 2, so a) = (1)“2 = 1. This is the w—intercept. Because the minimum value on the frequency axis is 102, we need the amplitude for H (1'10‘2). We are only concerned with the amplitude and not the phase, so we can ignore j and do the following calculation to determine the value of initial magnitude: ZOZOQIHmem =ZOZog[H(1O‘2)| =0.01 0“2 1 = 20log -(1——-:i—) = 2010910100 E 80 —2 ..—,¢_..;c (10 ) " Phase Plot: Should breaks up by 45°/dec atw = 0.1 and then breaks down by 45°/dec at a) = 10. ,Initial angle of phase plot: Azeros — Lpoles = 0° — 90° — 90° = —180° Final angle of phase plot: RD(——90) = (1)(——90°) The Bode plot for this problem is drawn and shown in figure.2. ii» 100 so so 40 Magnitude (dB) .135 -1 80 I -225 10* 1o“ 10“ 10 10 Frequency (maisec) Figure 2. Bode plots asymiptotes for 1.13. Axis created using function: SemiLogKL01,100,»40,100, 1 c' H“) = s(2s+1)(s+20)(s+300) Transfer function in Bode form: HUw) = jw<e+1><a+ma+u Zeros = dont have Poles = 0,0.5,20,300 K_ 1 3—6000 Magnitude Plot: Should breaks down at a) = 0.5, 20, 300 each by 20db/dec. Initial magnitude, withl = 1, so to = (Kiwi/1 = 1.66 x 10—4. This is the w-intercept. To determine the value of initial magnitude: (since the minimum value on the w is 0.01) 20long(jw)||w =zozog|H(1o-Z): =0.01 I 1 _ 20109 (W) 0,01 0—951 + 1) (9% + 1) (93%)51+ 1) 0 = 201091.66 X 10"4 E —35.7 Phase Plot: Should breaks down by 45°/dec at w = 0.05, 2, 30 and then breaks up by 45°/dec atw = 5,200,3000 . Initial angle of phase plot: Azeros — Apoles = 0° — (90° + 0° + 0° + 0°) = —90° Final angle of phase plot: RD(—90) = (4)(—90°) = —360° The Bode plot for this problem is drawn and shown in figure.3. i: l; .l, :1 ,. -‘l 00 420 440 .160 -‘l BO -200 420 440 £60 ‘ - 3130 Magnitude (as) Jilly—l 350 ~ ' 1o“ 10' J A i 2 ° 10 10" _ 10 Frequency [rad/sec) Figure '3. Bode plots asymptotes for 1.0 Axis created using function: SemiLog(0 , 01,10000, —34-0, 0, —-360, 0) 10 10 20(s+10) (5+1) (s+2) (5+100) d. H(s) = Transfer function in Bode form: How) = (20 X 10) ZOQw/lo + 1') 2 X 100 (jw+1)(J—29+1)(~11%’—6+ 1) Zeros = 10 {Poles = 1,2,100 KB = 1 Magnitude Plot: Should breaks down at a) = 1, 2, 100 each by ZOdb/dec and breaks up at co = 10 by 20db/dec. Initial magnitude plot: with l = 0, we begin at: ZOlogIHUw)]|w=o = 2010gIH(iO)| = 2010911! = 0 Phase Plot: breaks { dg‘i’m} by 45°/dec at { breaks { dign} by 45°/deoc gfi' a) = 1 w = 0.1, 0.2, 10} = o ‘lw = (1,0,231000} Initial angle of phase plot: Lzeros — Apoles = 0" — (0" + 0° + 0°) = 0° Final angle of phase plot: RD(—90) = (2)(—90°) = —180" The Bode plot for this problem is drawn and shown in figure.4. g) j’ 1“ u. Magnitude (:18) Phase (deg) —1 35‘ ~ -’l 80 10'2 410' 1o 10‘ 102 to“ 10‘ Frequency (rad/sec) 1 [I Figure 4. Bode plots asymptotes for l Axis created ' using function: SemiLogtO . 01.10000 , —150, O , ~180 , 0) 2. Do the bode plot for the above examples using MATLAB, turn in your plots for each example. Solution: numl=[1 2];denl=[l 2 l]; Iiun12=[l 1];den2=[l 0 O]; num3=[l];den3=conv(conv(conv([2 l],[1 20]),[1 300]):[1 0]); num4=20*[l 10];dené=conv(conv([l ll, [1 2]),[1 100]); margin (numl , Eleni) ; figure; margin (mun? , denZ) ; figure.- margin (num3 , den3) ; figure,- margin (numé , (18114) ; Ema Diagram Gm = Inl. Pm = 112 deg (al1.14 rad/sec) a s a B 'E 3 " 41) 15 .23 E 3 3 «15 m g a. .20 ' L—u-u—l 4' .1 5 1o 10 1a Fremency lmdlsecl Figure 5. Bode plot drawn by Matlab for Prob La. Bode Diagram Gm = Inf. Pm = z1.8 deg (ax 127 mulsae) 150 mu - Maglilmie (as) g / fifibm'ml l:- ZJWMW§u ma Phase (deg) '3 m l -1BD m" 10" w" Frequency (radrsec) Figure 6 . 1 @an ‘ ~- . . . l . . l . . . . . , . . . . . . . . , .5.......,...l,.. Hour“... ....,.... we qu-Iw' elm Bode pth draxm by Matlab for Prob 1.13. 10 1n2 Bode Dlagnm Em =10105(313.06 radlsec). Fm = 90 den (:1! 0000167 rad/sec) , i, . . 50 \ .100 - _ . .150.— \ - a Magn‘ltde (d8) A.» 3 1 I h. Ln :3 l u WEIJIEV‘ ,i Mfifim .agJuA a .uui 0“ .'~ H1 m‘ m“ m. m Flequency (mdlsec) Figure I Bode plot drawn by Iviatlab for Prob l.c. BodeDiaaram Gm = In! 63 (al 1m ndlsec). Fm = 480 deg (all) mdlsec) u Wax awwwwq—v—r—s—rnwr—r—me—w—w \ \\ .5u._ \ - a \ :1 u E i E ‘ 400— _ 453 1 'r ‘ 'uéneéur‘ ' _thI=.‘{ br‘nfi-xl-é‘fi‘ei a. "45:; u- ' .15” \ . a i 3 -9m— -‘ S E w «135 — . \ — \\ «NEW v1.4 r" vI-Luu AllllmL—A w: . i . 4 1o2 10‘ 10° 10‘ m‘ in“ 1:: Frequency (mdlsec) Figure 8. Bode plot. drawn by Matlab for Prob l.d. El 3. Determine the phase and gain margins for each plant in the above examples. Solution: GM = IH(wn)| ¢PM = [180 + argH(w1)]° 1.a. GM = oo db, (pm = (180°) + (—68°) = 112° 1.b. GM = oo db, (pm = (180°) + (—130°) = 50° Lo. GM = 100 db, (pm = (180°) + (—90°) = 90° 1.d. GM = oo db, (pm = 180°+ (—360") = —180° 5. For 1(d), find k so that the phase margin is 45 deg. Solution: Finding the angle that results 30 deg: ¢PM = (180°) + (95°) = 45° x° = ~135° 4. For 1(c), find k so that the phase margin is 30 deg. Solution: Finding the angle that results 30 deg: ¢PM = (180°) + (95°) = 30° x° = —150° Using magnitude condition: 20logK = —ZOZog|H(jw)| Now, our job is to find the frequency, a), corresponding to —135°. From Matlabbode plot, this frequency is approximately 0.642 and the corresponding magnitude which is: ZOZogIHOwjl = ~1.91 Therefore ZOZOgK = 1.91 1.91 OrK = 107 = 1.245 Using magnitude condition: K [H (jw)| = 1 20logKIHO'w)I = ZOlogl 20logK + 20logIH(jw)l = O 20logK = —ZOIog|H(jw)l Now, our job is to find the frequency, a), corresponding to —150°. From Matlab bode plot, we can precisely determine this frequency which is 0.777 and the corresponding magnitude which is: 20109|H(jw)[ = ~78.7 Therefore 20logK = 78.7 78.7 OrK = 1075‘ = 8609.93 Bode Diagmm Bode Diagram Gm = dB [31 3115 nausea) . P 90 deg (:1! 0.000157 mdlsec) em. lequenzy lmdlsec): 0:777 continue ‘ Magrilme (dB) I SysIern: unIillsm fluency (radlsscl' 0.777 tern. u requency (muses): 0542 Phase (deg) Phase (deg) in" m 10 1o 1&72 1a 1a " m Frequency (mdlsec) Frequency (msec) ...
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This note was uploaded on 09/05/2011 for the course EML 4312 taught by Professor Dixon during the Fall '07 term at University of Florida.

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hw5 - Introduction: Bode plots consists of two graphs, the...

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