ECE 3510 Root Locus Design Examples. PI To eliminate steady-state error (for constant inputs) & perfect rejection of constant disturbances

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1 ECE 350 Root Locu Deign Example Recall the imple crude ervo from lab G( ) σ = = PI To eliminate teady-tate error (for contant input) & perfect reection of contant diturbance Note: The DC motor ha a pole at zero and hould do zero the teadytate error by itelf, but nonlinearitie prevent it from doing it well. G( ). 0. Add pole at 0 and zero at -0. C( ) = =. p LAG An alternative i a Lag Compenator, here with a pole at -0. and a zero at -0.5 G( ) = Thi wor very much lie the PI controller, but without the need for active component. The area near the origin Root Locu Deign Example p.

2 Root Locu Deign Example p.2 Let' eep the pole at 0 and zero at -0. for elimination of teady-tate error and reection of diturbance CL pole at p and = = G( ) 3.44 At gain of = deg 8.6 o 36.4 o 35 o Thi i a point in the root locu becaue: 8.6. deg deg 35. deg 35. deg 35. deg = 80 deg 6.64 = deg PD or PID To Improve the dynamic repone Want to double the peed Want pole to move to: p = deg 9.4 o 79.3 o 35 o Unfortunately, thi point in NOT on the root locu = deg deg = deg Maybe we could add a zero o that it' angle i: θ. z deg 80. deg θ z = 53.7 deg x =. = 0.28 z Re( p ). z = G( ).( 0. ).( 24.28) = = 0.48 i the required gain G( 4 4. ) Root Locu Deign Example p o 79.3 o x 35 o

3 Root Locu Deign Example p.3 We have deigned a our compenation with the following: A pole at the origin A zero at -0. A zero at Gain of 0.48 Find the,, & d of a PID controller.. d C( ) = =. d =.. i 2 d.. 2 d 2 p. =. d d d ( 0. ).( ) = gain = d 0.48 = 2. d d = d 2.43 =.06 d = i p d = 0.9 Notice: = 0. ~ 0. d Notice that the proportional gain i actually almot 3 time higher than it wa before = 0.32 LEAD An alternative to the differentiator i a Lead Compenator. Intead of a ingle zero with: θ z = 53.7 deg How about a zero with θ z. 70 deg And a pole with θ p. 70 deg deg x =. = θ p = 6.29 deg z Re( p ). z = xp =. = tan θ p 6.3 o 70 o p Re( p ). p = tan θ p Thi example i actually a PI-Lead controller Root Locu Deign Example p.3

4 Problem with the differentiator Root Locu Deign Example p.4. Trie to differentiate a tep input into an impule -- not liely. You'll have to conider how your differentiator will actually handle a tep input and how your amplifier will aturate. If the differentiator and amplifier aturate in uch a way the the "area under the curve" approximate the impule "area under the curve", then thi may not be uch a problem. It may not be a fat a predicted from the linear model, but it may be a fat a the ytem limit allow. (Pedal-to-the-metal.) 2. It' a high-pa filter and can accentuate noie. Thi i actually common to all compenator that peed up the repone. 3. Require active component and a power upply to build. Uually no big deal ince your amplifier (ource of gain) doe too. 4. I never perfect (alway ha higher-order pole), but then neither i anything ele. Epecially in mechanical ytem, thee pole uually are well beyond where they could caue problem. Alternative:. Lag-Lead or PI-Lead compenation. Thi eliminate the differentiator, but it i till a high-pa filter that can be a noie problem and it could till aturate the amplifier if the input change too rapidly. Be ure to chec for aturation problem. 2. Place the differentiator in the feedbac loop. The output of the plant i much le liely to be a tep or to change o rapidly that it caue problem. Differentiation in the feedbac P( ) Find the,, & d of thi controller.. d Note: The differential ignal i often taen from a motor tachometer when the output i a poition. Then you don't need a eparate differentiator circuit, ut a eparate gain for that ignal. F( ). C( ) =. p C( ) F( ). d =.. p. p d In thi cae the open-loop zero in the feedbac loop IS NOT in the cloed-loop. Thi turn out to mae the tep repone lower than predicted by the econd-order approximation, but try a imulation, you may be able to ue ignificantly more gain with no more overhoot. The differentiator in thi poition inhibit overhoot. =.. p d d For our example: = ( 0. ).( 24.28) d d d = = 0.9 d. i 0. =.09 Root Locu Deign Example p.4

5 PI and PID Deign Example p.5 Ex.2, from S6 Exam 3 Conider the tranfer function: G( ) a) Find the departure angle from a complex pole. Angle: from pole at - θ p. 80 deg 4 θ p = deg 5 ( ) from pole at -2-4 θ p2 90. deg θ p2 = 90 deg from zero at -5 θ z 4 3 θ z = 53.3 deg deg θ = deg 90. deg deg 80. deg = deg b) Draw a root locu plot. Calculate the centroid and accurately draw the departure angle σ σ = 0 2 c) I there any decent place to locate the cloed-loop pole? NO deg d) You would lie to place your cloed-loop pole to get a ettling time of /2 ec and 0.656% overhoot. Add the implet poible compenator to accomplih thi and calculate what the compenator hould be. 2% ettling time: T = 4 a Overhoot: a b = ln( OS) π OS = e. π a b ln( ) = = π 4 a = = 2 8 %OS =. 00% e 8.6 b = = 5.6. π a b Pole hould be at Angle: from pole at deg 5 = deg 7 from pole at deg = deg deg from pole at deg 9 = deg 6 from zero at deg 5 = deg deg deg deg deg = deg θ. z deg 80. deg θ z = deg tan( deg 90. deg) =. = x 5 x x = 2.5 C( ) = 2.5 G c ( ) ( 5 ).( 2.5) ( ) PI and PID Deign Example p ( 5 ).( 2.5 ) Chec: arg = 80 deg ( )

6 PI and PID Deign Example p.6 e) What i the gain? G c ( ) ( 8 5. ). ( 8 5. ) 2 4.( 8 5. ) 20 = = ( ).( ) f) What i the teady-tate error for a unit-tep input? G c ( ) ( 5 ).( 2.5) ( ) G c ( 0 ) = G c ( 0) = e tep = = 0.9 % ( 0 5 ).( 0 2.5) ( 0 ) ( 5 ).( 2.5) = = ( ).( 20) g) If thi teady-tate error wa a little too big, what would be the very implet way to reduce it? turn up the gain Ex.3, from S6 Exam 3 a) Setch the root locu plot of, G( ) σ C = = n m The gain i et at 452, o that one of the cloed-loop pole i at, Further calculation yield: Settling time: ec % overhoot: 5.92.% Steady-tate error to a unit-tep input: 60.8% 00 m 0 ( 25 ).( ).( 70) n 3 45 n m = 3 o aymptote are at + 60 o & 80 o b) You wih to increae the frequency of ringing to rad/ec without changing the % overhoot at all. Where hould the cloed-loop pole be located? a b = = 0.9 new b new a = 0.9. b = 36 New location: 36. c) Add a LEAD compenator o that you will be able to place the cloed-loop pole at the location found in b). Add the new zero at -30. Find the location of the new pole. Angle: from pole at -25 θ deg from pole at - θ 36 from pole at -70 θ from new zero at -30 θ deg θ 25 = deg θ = deg θ 70 = deg θ 30 = deg PI and PID Deign Example p.6

7 θ 25 θ θ 70 θ 30 θ p =. 80 deg PI and PID Deign Example p.7 θ p. 80 deg θ 25 θ θ 70 θ 30 θ p = deg p 36 p = = tan θ p 85 G c ( ) 00.( 30) ( 25 ).( ).( 70 ).( 85) 00.( 30) Chec: arg = ( 25 ).( ).( 70 ).( 85) deg d) With the compenator in place and a cloed-loop pole at the location deired in part b) i) What i the gain? G c ( ) = 369 ii) What i the 2% ettling time? 4 T = = 0. ec 36 Ue the econd-order approximation. iii) What i the teady-tate error to a unit-tep input? 00.( 0 30) G c ( 0 ) = = ( 0 25 ).( 0 ).( 0 70 ).( 0 85) e) Add another compenator: C 2 ( ) e tep = =. G c ( 0) and maintain the gain of part d) i) What i thi type of compenator called and what i it purpoe? PI, ued to eliminate teady-tate error 59.6 % ii) Calculate what you need to to how that thi compenator achieved it purpoe. 00.( 30) G c ( ). ( 2) ( 25 ).( ).( 70 ).( 85) G c ( 0 ) = e tep =. = 0.% f) With both compenator in place, i there poibility for improvement (quicer ettling time peed and/or lower ringing)? If ye, what would be the implet thing to do? Jutify your anwer. A quic etch of the new root-locu how that imply decreaing the gain would improve the ytem PI and PID Deign Example p.7 move down here