Chapter 2 Feedback Control Theory Continued

Transcription

1 Chapter Feedback Cotrol Theor Cotiued. Itroductio I the previous chapter, the respose characteristic of simple first ad secod order trasfer fuctios were studied. It was show that first order trasfer fuctio, sometimes called first order lag, has a overdamped respose ad the output lags iput as it was show i the frequec respose. The secod order trasfer fuctio as was show i the previous chapter ca have overdamped respose for ζ > ad ca be oscillator for ζ <. It becomes clear that trasfer fuctio of higher order ma become ustable. Stabilit of cotrol sstem is a major cosideratio ad must be studied. If the sstem is stable, it must be studied how oscillator the sstem is.. Routh Hurwitz Stabilit Criteria For a sstem with trasfer fuctio of Y() s = X() s as + a s + a 0 (.) it is better to stud stabilit i time domai. Covertig Eq. (.) i differetial form ields d d a + a 0 : () + + a = xt d dt (.) For this kid of differetial equatios, there are two solutios. Oe is the trasiet respose ad the other oe is stead state solutio. The stead state solutio is obtaied b assumig a solutio i the form of the iput x( t) with ukow coefficiets. The the solutio is substituted i the differetial equatios ad the ukow coefficiets are obtaied b equatig the coefficiet of the same order i s. Usuall R. Firoozia, Servo Motors ad Idustrial Cotrol Theor, DOI 0.007/ _, Spriger Iteratioal Publishig Switzerlad 04 7

2 8 Feedback Cotrol Theor Cotiued the solutios i cotrol are obtaied for step, ramp, acceleratio, ad siusoidal iputs. More iterestig is the trasiet respose, which determies stabilit ad how oscillator the output is. To obtai the trasiet respose the right had side of Eq. (.) is equated to zero ad a solutio i the form of t () = e st (.3) is assumed. Substitutig Eq. (.3) i Eq. (.) ad after some algebraic maipulatio ields as + a s + as + a0 = 0 (.4) Equatio (.4) is kow as the characteristic equatio ad the roots of the characteristic equatio determie the trasiet respose. For stabilit all real parts of the roots must be egative. For real phsical sstem, the complex roots appear i cojugate. This meas that if Eq. (.4) cotais complex cojugate the respose could be overdamped or oscillator. Routh Hurwitz method is a quick wa of establishig the stabilit of the sstem. Ufortuatel, this method does ot idicate how oscillator the sstem is. To determie stabilit a arra i the followig form is costructed s : a a a s a a 3 a 5 s b b b3 3 s c c c3... s s 0 :...0 :...0 :...0 : g 0 : h 0 where b = ( a a aa ) 3 a b = ( a a aa ) 4 5 a b = ( a a aa ) a

3 .3 Root Locus Method 9 Fig.. Block diagram of a simple positio cotrol sstem x + e K s(0.5 s + ) c = ( ba ba ) 3 b c = ( ba ba ) 5 3 b The calculatio for the parameters i the above arra ma be writte i determet form. For stabilit, all the coefficiets of the characteristic equatio must be positive ad there must ot be sig chages i the first colum of the arra. A sig chage idicates that there is a root with positive real part. If a elemet i the first colum is zero, a small positive ε is assumed ad the sig chage is determied whe ε teds to zero. If all elemets i a row are zero, there is a root with positive real part or zero real part. There are computer programs that calculate the roots of the characteristic equatio. I this case, the roots ca be plotted o a complex plae. This brigs us to a powerful aalsis of stabilit kow as Root Locus method..3 Root Locus Method It was show that the stabilit of cotrol sstem could be studied b the roots of characteristic equatio. I this sectio, the Root Locus will be studied for a secod order sstem. For higher order sstem there, is a aaltical approach that ca be used to plot the Root Locus from the ope loop trasfer equatio. The method is tedious ad the loci are plotted from the zeros ad poles of the ope loop trasfer fuctio. It should be metioed that the umber of loci are equal of the order of the characteristic equatio. The loci will ed to the zeros or ifiit as the gai of the sstem is icreased. The block diagram of a egative feedback of a simple servo positio cotrol is show i Fig... The itegrator shows the fact that the positio is obtaied from velocit ad the first order lag shows that because of the iertia there is a time lag. The closed loop trasfer fuctio ca be obtaied b usig the block diagram algebra. With some maipulatio it ca be show that the closed loop trasfer fuctio becomes

4 0 Feedback Cotrol Theor Cotiued k = x s(0.5s + ) + k (.5) The characteristic equatio is, therefore, The roots are 0.5 s + s + k: = 0 s, = ± k (.6) (.7) Now the root locus ca be obtaied b varig k from zero to ifiit. Some importat poits are the roots whe k = 0, k = 0.5 ad k > 0.5. For k < 0.5, the roots are egative startig from the poits 0 ad. Ad there is a breakawa poit at poit k = 0.5 where the roots becomes complex ad as k is icreased, the two complex roots move towards ifiit parallel to the imagiar axis. B comparig the trasfer fuctio with the secod order trasfer fuctio studied i previous chapter, it ca be show that ad Ad it ca be show that ω = ζ = k k cos θ= ζ The Root Locus for the above secod order sstem is show i Fig... X ad Y cotai the real ad imagiar part of the root for values of Gai ( K). I this case, there are two loci, which ed at ifiit. The roots are show b crosses. Figure.3 shows a simple third order trasfer fuctio which could represet a positio cotrol sstem with DC motor. This model cotais the effect of iductace i the sstem. I the ope loop trasfer fuctio, there is a itegrator ad two first order lags. Therefore, there are three poles at s = 0, s =, s =. The loci start at these three poles ad ed to ifiit as the gai K is icreased. It should be oted that there is o zeros, which makes the umerator of the ope loop trasfer fuctio equal to zero. The closed loop trasfer fuctio ma be calculated as K : = x s + s + s + k (.8)

5 .3 Root Locus Method Fig.. Root Locus for a secod order sstem x: = : = z x Fig..3 Block diagram of a third order sstem with gai K x K s(0.5s + )(s + ) With the characteristic equatio of s +.5 s + s + k: = 0 (.9) The MathCAD Polroots expressio ca be used to calculate the roots of characteristic equatio for various K. The root locus for this sstem is show i Fig..4. At K = 0, there are three egative real roots. As K is icreased the two real roots move towards each other ad the third real root moves towards ifiit. The two real roots break awa from the real axis ad become complex. As k is icreased, the real part of complex roots becomes positive showig that the sstem becomes ustable. The root locus i this case eter the right-had side of the s-plae.

6 Feedback Cotrol Theor Cotiued Fig..4 A tpical root locus for a third order characteristic equatio 4 4 imagiar x 4 real The root locus for this sstem is show i Fig..4. Higher order sstems have more roots ad the root locus becomes more complicated. There are also computer programs, which presets the root locus from the ope loop trasfer fuctio removig the eed to calculate the closed loop trasfer fuctio. I this book, the MathCAD computer program is used throughout to plot the root locus. There are also methods of calculatig the gai for each locatio of the roots. With MathCAD polroots facilit, the roots ca be calculated for each gai or parameter of iterest separatel. The correct gai or parameter of iterest ca be obtaied b ispectio of the table of roots ad there is o eed to go ito details of the graphical methods. There are computer programs that ca calculate the roots of characteristic equatio of a order. MatLab ad MathCAD are two of the most commol used computer programs. The Root Locus ca the be plotted. The dampig ratio idicated b the real part ad the frequec of oscillatio is determied b the imagiar part. For most cotrol sstems, a dampig ratio of 0.7 is preferred. The dampig ratio of 0.7 will result i small overshoot ad the dampig ratio of results i o overshoot. This meas that all roots must lie betwee the ± 45 lies o the left side of the s-plae. For stabilit all roots must lie i the left had side of the s-plae ad the imagiar part show i fact the frequecies of oscillatio. The further awa from the origi the faster the respose to a step iput. For a good respose to step iput all roots must lie i the 45 ad the egative real axis. For complicated sstem which have ma roots the root earest to the imagiar axis domiates the step iput respose.

7 .4 Importat Features of Root Locus 3 Fig..5 A simple uit feedback sstem x i + K (s + 50) 0 (s + 0 s + 04) (s + 5).4 Importat Features of Root Locus Usuall the trasfer fuctio of elemets of a cotrol sstem is i form of first or secod order lag form ad the closed loop sstem is i the form of several cascaded of these trasfer fuctios. This ca be used to sketch the root locus mauall from the ope loop trasfer fuctios. There are importat features o the Root Locus that the loci ca be sketched mauall ad the are useful to kow. It ca be used for simple sstem ad it is ver complicated for complex sstems. Without loss of geeralit cosider the simple sstem show i Fig..5. Where x i is the iput variable ad ( o ) is the output variable to be cotrolled. The secod order term i the deomiator could be as a mass-sprig-damper trasfer equatio. The other term i the umerator ad i the deomiator ma represet a lead lag compesatio etwork. The K is the parameter of iterest that should be adjusted to obtai a stable ad fast respodig sstem. The closed loop trasfer fuctio becomes, o : i ( ) [ K (s + 50) ] = x s + 0s + 04 (s + 5) + K (s + 50) Therefore the characteristic equatio becomes, ( s ) s (s + 5) + K (s + 50) : = 0 Dividig both side of the above equatio gives the characteristic equatio i stadard for as, ( ) s + 0s + 04 (s + 5) : = K (s + 50) (.0) The agle Law states that the agle of left side of Eq..0, which is a complex umber, should be ± 80 measured from the real axis couter clockwise. The magitude law states that the magitude of the left side of Eq. (.0) should be. These two laws ad some feature of the loci help us to draw the root locus mauall. The order of the deomiator is = 3 ad the order the umerator is m = for the ope loop trasfer fuctio. From the characteristic equatio is clear that there

8 4 Feedback Cotrol Theor Cotiued Fig..6 Sketch of the root locus for the example cosidered i this sectio 45 degree A Imag.. Asmptote +50 K=83. s=3.9 B A.Real A 3 50 K Icreasig are three loci. Whe K = 0 the loci start at the pole of the trasfer equatio ad ed to the zero of the umerator ad the remaiig poles ed to ifiit. For this example the pole whe K = 0 are. ( ) s + 0 s+ 04 (s+ 5) + K (s + 50) := 0 s : = 5 04 F := 0 s : = polroots (F) 0 i s,3 : = 0 + i (.) It ca be see that eve for simple secod order equatio the Mathcad software ca be used to determie the roots. The poles are show o the graph paper b crosses ad zeros are show b a circle. I this example there is ol oe zero ad it is, Z = 50 The poles ad zeros are show o Fig..6. Those loci which go to ifiit have asmptotes determied b the followig equatio, [( + ) 80] θ : = m [( + ) 80] θ : = m

9 .4 Importat Features of Root Locus 5 is a iteger umber. For real sstems the complex roots appears i cojugate so for this complex cojugate there is oe positive ad oe egative asmptote. Selectig oe for the iteger l = the agle of asmptotes becomes, θ = 90 θ = 90 All asmptotes itersect the real axis at a sigle poit ad is give b, The sum of all ope loop poles the sum of ope loop zeros d = m Where is the umber of ope loop poles ad m is the umber of ope loop zeros. It should be oted because all the complex roots appear i cojugate the distace d will be a real umber. Therefore the distace d ca be calculated as, d: = ( 0 i 0 + i ) d : =.5 It is show i Fig..6. The loci depart from complex ope loop poles ad arrive to the complex ope loop zeros at agle that satisf the agle law. For the example i had the departure agle cosiderig a poit ver ear to the complex root poles becomes, α=± (l + )80 So α = ±0. This is show i Fig..6. Aother importat poit is the itersectio of the loci with Imagiar axes. The Routh Hurwitz arra is costructed for the sstem costructed above. The characteristic equatio is give as below, 3 s + 5 s + (K + 04) s k : = 0 Costructig the Routh Hurwitz arra gives, 3 s 04 + K 0 s K 0 s 83. K K 0 s ( K 50 K ) 0 0 The value of k must be as such that the first colum must be all positive for the sstem to remai stable. Doig this the value of K where the loci crosses the imagiar

10 6 Feedback Cotrol Theor Cotiued axis ca be foud. The first colum of third row idicates that the value of K must be less tha 83.. The first colum of forth row is more complicated because there is the K to power of ad the deomiator has also the coefficiet K. It is clear that for k = 83. the first colum fourth row for the above K is positive. To fid the itersectios poits o the imagiar axis the auxiliar equatio is formed as, K : = s K : = 0 so, s : = ( K) 50 s = 3.94i s : = 3.9i s : = 3.9i These poits are show o the imagiar axes b crosses. The reader is ecourage to do the Routh Hurwitz algebra themselves to show the data preseted here is correct. For desig purpose the value of K must be selected to have a dampig ratio 0.7. This dampig ratio is achievable b drawig a 45 lie from the origi ad its itersectio with the loci gives a dampig ratio of 0.7. The followig procedures remai as before ol the magitude chages. This is show b poit A o the loci. The the value of gai is obtaied b magitude law as, K: = ( A A ) These measuremet is show i Fig..6, so, B B (.5.5.3) K: = 6 K : = 0.83 B rememberig these importat poits the root locus ca be sketched mauall o a graph paper as show i Fig..6. This is just a approximate of the root locus ad for complicated sstem, the procedures becomes complex. It is useful to kow these poits ad whe the root locus is plotted usig a software like Mathcad these poit

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