SKEE 3143 CONTROL SYSTEM DESIGN CHAPTER 3 Compenator Deign Uing the Bode Plot 1
Chapter Outline 3.1 Introduc4on Re- viit to Frequency Repone, ploang frequency repone, bode plot tability analyi. 3.2 Gain Adjutment Compena4on 3.3 Lag Compena4on 3.4 Lead Compena4on 3.5 Lead- Lag Compena4on 3.6 Deign the Compenator uing MATLAB 2
3 The log- magnitude and phae frequency repone curve a a func4on of log ω i called Bode plot. Bode plot i a technique for analye and deign of control ytem. Conider a tranfer func4on The magnitude frequency repone Conver4ng into db ) ( ) )( ( ) ( ) )( ( ) ( 2 1 2 1 k m k z z z z z z K G!! ω ω j n m k p p p z z z K j G ) ( ) ( ) ( ) ( ) ( ) ( ) ( 2 1 2 1!!!! ) 20log( ) 20log( ) 20log( ) 20log( ) 20log( 20log ) ( 20log 2 1 2 1 p j p j j z j z j K j G m ω ω ω ω ω ω 3.1 IntroducGon Re- viit: The Bode Plot
3.1 IntroducGon Re- viit: The Bode Plot The phae frequency repone G( jω) K ( jω z1) ( jω z2)! ( jω p1) ( jω p2)! If we know the magnitude and phae repone of each term, total frequency repone can be obtained by algebraic um of each term. The frequency repone can be implified by u4lizing traight- line approxima4on. Therefore, total frequency repone can be obtained by graphic addi4on. 4
3.1 IntroducGon Re- viit: The Bode Plot G() K G( ) Gjω) K; G( jω) db G( jω) 20log K 0 0 K 5
3.1 IntroducGon Re- viit: The Bode Plot G() G( ) Gjω) ; G( jω) db G( jω) 20logω 0 90 jω 20 db/decade At ω 1, gain 0 db. 6
3.1 IntroducGon Re- viit: The Bode Plot G() 1/ G( ) 1 ; G( jω) 1 Gjω) log(1/ ω) Gjω) db G( jω) 90 At ω 1, gain 0 db. 20log(1/ ω) 0 jω 20logω - 20 db/decade 7
3.1 IntroducGon Re- viit: The Bode Plot For At ω << a, At ω a, At ω >> a, G( ) 1 a Gjω) G( jω) Gjω) db db 20 log1 G( jω) tan tan 1 20log tan 1 0 1 1 ω 0dB 0 0 2 3.01dB 45 0 90 0 G( ) Gjω) ( G( jω) Gjω) 20logω db 20 db/decade G( jω) a 1); G( jω) ω a tan 2 1 1 ω a jω 1 a 8
3.1 IntroducGon Re- viit: The Bode Plot The low- frequency approxima4on i called the low- frequency aymptote. The high- frequency approxima4on i called the high- frequency aymptote. The frequency, ω a i known a break frequency becaue it i the break between the low- and high- frequency aymptote. 9
3.1 IntroducGon Re- viit: The Bode Plot high- frequency aymptote low- frequency aymptote break- frequency 10
3.1 IntroducGon Re- viit: The Bode Plot high- frequency aymptote low- frequency aymptote aympto4c approxima4on 11
3.1 IntroducGon Re- viit: The Bode Plot 12
3.1 IntroducGon Re- viit: The Bode Plot 13
For G( ) At ω << a, At ω a, At ω >> a, a 1 1 Gjω) db G( jω) Gjω) 20 log1 tan 1 G( jω) tan Gjω) db db 0 20 log(1 1 1 20log ω G( jω) tan G( ) 0 Gjω) 0dB 0 1 45 1 ω 0 1 1 ; G( jω) a 1 jω a 1 1 1 ; G( jω) tan 2 ω 1 a 2) 3.01dB 90 0 3.1 IntroducGon Re- viit: The Bode Plot - 20 db/decade ω a 14
3.1 IntroducGon Re- viit: The Bode Plot high- frequency aymptote low- frequency aymptote break- frequency 15
3.1 IntroducGon Re- viit: The Bode Plot high- frequency aymptote low- frequency aymptote aympto4c approxima4on 16
Example 1 3.1 IntroducGon Re- viit: The Bode Plot Sketch the Bode plot for the ytem hown where G( ) K( 3) ( 1)( 2) Compare your ketch with the bode plot obtained from MATLAB 17
18 For At low frequency, ω << ω n, At high frequency, ω >> ω n, 2 2 2 2 ) ( n n n G ω ζω ω n n n n n n n n n j j j j G G ω ζω ω ω ω ζω ω ω ω ω ζ ω ω ζω ω 2 1 1 1 2 ) ( 1 ) ( 1 2 1 2 ) ( 2 2 2 2 2 2 2 2 2 0 0 ) ( ; 0 20log1 ) ω ω j G db Gj db 1 ) ( jω G 0 2 180 ) ( ; 40log 20log ) ω ω ω ω j G Gj db 2 1 ) ( ω ω j G 3.1 IntroducGon Re- viit: The Bode Plot econd order ytem
3.1 IntroducGon Re- viit: The Bode Plot econd order ytem high- frequency aymptote low- frequency aymptote break- frequency 19
3.1 IntroducGon Re- viit: The Bode Plot econd order ytem high- frequency aymptote low- frequency aymptote aympto4c approxima4on 20
At the break frequency, ω ω n, G( jω) 20log(2ζ ); G( jω) 90 db 3.1 IntroducGon Re- viit: The Bode Plot econd order ytem G ( jω) 1 j2ζ The magnitude depend on ζ. For example: 0 ζ lg(jω)l db 0.1-20log (0.2) 13.9 db 0.7-20log (1.4) - 2.92 db 1-20log (2) - 6.02 db 21
3.1 IntroducGon Re- viit: The Bode Plot econd order ytem 22
3.1 IntroducGon Re- viit: The Bode Plot econd order ytem 23
3.1 IntroducGon Re- viit: The Bode Plot econd order ytem Example 2 Sketch the Bode plot for G() for the unity feedback ytem hown below where G() ( 3)/[( 2)( 2 2 25)]. Compare your ketch with the one obtained uing MATLAB 24
For 1/( 2 2 25)], ω n 5 rad/; ζ 0.2 Normalie: G 3.1 IntroducGon Re- viit: The Bode Plot econd order ytem SoluGon (3 50) 1 3 ) 2 1 1 2 25 25 ( 2 25
SoluGon 3.1 IntroducGon Re- viit: The Bode Plot econd order ytem 26
SoluGon 3.1 IntroducGon Re- viit: The Bode Plot econd order ytem 27
3.1 IntroducGon Re- viit: The Bode Plot Gain & Phae Margin 180 28
3.1 IntroducGon Re- viit: The Bode Plot Gain & Phae Margin Example 4 Conider a unity feedback ytem with G() 200/[(2) (4)(5)]. By uing MATLAB, find the gain margin and the phae margin. SOLUTION: ω GM 7 rad/. G M 6.02 db. ω ΦM 5.5 rad/. Φ M i 180 0 165 0 15 0. 29
3.1 IntroducGon Re- viit: The Bode Plot Damping RaCo & Phae Margin Rela4onhip of phae margin and damping ra4o: Φ M tan 1 2ζ 2 2ζ 1 4ζ 4 30
3.1 IntroducGon Re- viit: The Bode Plot Steady- tate Error Steady- tate error can alo be obtained from the open- loop Bode plot. Sytem type 0 Sytem type 1 31
3.1 IntroducGon Re- viit: The Bode Plot Steady- tate Error Sytem type 2 32
3.1 IntroducGon Re- viit: The Bode Plot Stability with GM and PM 33
3.2 Gain Adjutment The Deign of Controller For deign in frequency repone, the concept of tability, tranient repone and teady- tate error are ued. A cloed- loop ytem i table if the open- loop magnitude frequency repone ha a gain le than 0 db at the phae frequency repone - 180 0. Percentage OS i reduced by increaing the phae margin. Speed of repone i increaed by increaing the bandwidth. The teady- tate error i improved by increaing the lower frequency magnitude repone. 34
3.2 Gain Adjutment The Deign of Controller The deired OS can be obtained by deigning the phae margin. If we deire a phae margin, Φ M repreented by CD, we have to raie the magnitude curve by AB. A imple gain adjutment can be ued to deign the phae margin and hence, the OS. 35
Example 3 3.2 Gain Adjutment The Deign of Controller For the poi4on control ytem, find the value of the preamplifier, K to yield a 9.5% OS in the tranient repone for a tep input. Ue only the frequency repone method. 36
SoluGon 3.2 Gain Adjutment The Deign of Controller K 3.6 ω 14.8 rad/ 37
SoluGon OS 9.5%, ζ 0.6, Φ M 59.2 0. From the plot, to obtain Φ M 59.2 0, ω ΦM 14.8 rad/. At ω 14.8 rad/, the gain - 44.2 db. The magnitude ha to be raied to 0 db to yield the required phae margin. K 1 162.2 K 3.6K 1 162.2 x 3.6 583.9 and e 0.062 (for ramp input) The gain- adjuted open- loop tranfer func4on: G( ) 58390 ( 36)( 100) 3.2 Gain Adjutment The Deign of Controller 38
Example 4 For a unity feedback ytem with a forward tranfer func4on G( ) ( K 50)( 3.2 Gain Adjutment The Deign of Controller 120) ue frequency repone technique to find the value of gain, K to yield a cloed- loop tep repone with 20% OS. ANSWER: K 194200 39
3.3 Lag CompenaGon The Lag Compenator Lag compenator i ued to improve teady- tate error without affec4ng the tranient repone. With the Bode diagram, the lag compenator: Improve the ta4c error contant by increaing only the low frequency gain. Increae the phae margin to yield the deired tranient repone. The tranfer func4on of the lag compenator: G c ( ) 1 T 1 αt which can be implified to G c ( ) z p c c 1 1 where z c > p c 40
The Bode plot for lag compenator i: 3.3 Lag CompenaGon The Lag Compenator 41
3.3 Lag CompenaGon The Lag Compenator 42
3.3 Lag CompenaGon The Lag Compenator The effect of lag compenator: The uncompenated ytem i untable to achieve deired ta4c error contant. The lag compenator reduce the high- frequency gain, while not changing the low- frequency gain. Thu, the low- frequency gain of the ytem can be made high to yield a large K v without crea4ng intability. 43
3.3 Lag CompenaGon The Lag Compenator Step to deign: i Obtain the value for K which a4fie the e and plot the correponding Bode Plot. ii Find the gain croover freq o that the deired phae margin i obtained. iii Read the gain of the uncompenated ytem at thi freq. The compenator gain needed i the amount needed to change the compenated gain to 0 db. iv Hence, the higher break freq (z c ) for the compenator i one decade below the gain croover freq. v Then, draw a - 20 db/decade lope backward un4l it meet the 0 db line. Thi i the lower break freq (p c ) for the compenator. 44
Example 5 3.3 Lag CompenaGon The Lag Compenator Given the ytem below (Example 3), ue Bode diagram to deign a lag compenator to yield a tenfold improvement in teady- tate error over the gain compenated ytem while keeping the percent OS at 9.5%. 45
SoluGon From example 3, OS 9.5%, K 583.9; K v 16.22 Deired K v 16.22 (10) 162.2 Therefore, K 5839 to achieve K v 162.2. Bode plot with K 5839. 3.3 Lag CompenaGon The Lag Compenator OLTF, G( ) 583900 ( 36)( 100) 46
SoluGon 3.3 Lag CompenaGon The Lag Compenator K 5839 ω 0.062 20 db/dec ω 0.98 110 47
SoluGon 3.3 Lag CompenaGon The Lag Compenator Deired OS 9.5%, ζ 0.6, Φ M 59.2 0. To compenate (uing lag compenator), Φ M 59.2 0 10 0 69.2 0 ω 9.8 rad/, magnitude 24 db. Draw a lag compenator: high break frequency 0.98 rad/, low break frequency 0.062 rad/. [20 db/dec line] 0.063( 0.98) The tranfer func4on: G c ( ) ( 0.062) The compenated ytem: G c ( ) 36786( 0.98) ( 36)( 100)( 0.062) 48
Example 6 3.3 Lag CompenaGon The Lag Compenator For a unity feedback ytem with a forward tranfer func4on G( ) K ( 50)( 120) ue Bode diagram to deign a lag compenator for the ytem that will improve the teady- tate error tenfold, while 4ll opera4ng with 20% OS. ANSWER: G lag ( ) 0.0691( 2.04) ( 0.141) 49
3.4 Lead CompenaGon The Lead Compenator Lead compenator i ued to improve the tranient repone. With the Bode diagram, the lead compenator: Increae the phae margin to reduce the percent OS. Increae the gain croover and bandwidth to realize a fater tranient repone. 50
3.4 Lead CompenaGon The Lead Compenator The tranfer func4on of the lead compenator: G lead ( ) 1 1 T β 1 βt which can be implified to G c ( ) zc p c 1 1 where, β <1 where z c < β p z p c c c < 1 51
The Bode plot for lead compenator i: 3.4 Lead CompenaGon The Lead Compenator 52
3.4 Lead CompenaGon The Lead Compenator 53
3.4 Lead CompenaGon The Lead Compenator The uncompenated ytem ha a mall phae margin (B) and a low phae- margin frequency (A). Uing the lead compenator, the gain croover frequency i increaed from A to C. Thi yield a larger phae margin (D), a higher phae- margin frequency (C) and a larger bandwidth. 54
The Bode plot for lead compenator i: 3.4 Lead CompenaGon The Lead Compenator The peak and phae curve vary in maximum angle at the frequency where the maximum occur. 55
3.4 Lead CompenaGon The Lead Compenator The frequency at the maximum phae angle: zc ωmax [1] β The maximum phae angle: φ max 1 β in 1 1 β [2] The compenator magnitude at ω max : G c ( max) jω 1 β [3] 56
3.4 Lead CompenaGon The Lead Compenator Step to deign: i Obtain the value for K which a4fie the e and plot the correponding Bode Plot. ii Find the Phae margin of the uncompenated ytem. iii Calculate the addi4onal phae needed by the lead compenator o that it a4fie the ytem requirement (addi4onal correc4on factor might be needed). iv Calculate the β needed. v Calculate the compenator magnitude to obtain β above. vi Obtain the new gain croover freq ω G. t. the magnitude of uncompenated ytem i equal but nega4ve to that of the compenator. Hence, calculate for z c and p c. vii Draw the bode plot and check the ytem performance. Repeat the deign tep with different correc4on factor if the ytem requirement i ill not achieved. 57
Example 7 3.4 Lead CompenaGon The Lead Compenator Given the ytem below, ue Bode diagram to deign a lead compenator to a 20% OS and K v 40, with a peak 4me of 0.1 econd. 58
SoluGon 3.4 Lead CompenaGon The Lead Compenator For K v 40, K 1440. Plot Bode diagram with K 1440. Deired OS 20%; Φ M 48.1 0. 59
SoluGon 3.4 Lead CompenaGon The Lead Compenator From the Bode diagram, Φ M 34 0 at ω 29.6 rad/. Therefore the required phae margin from the lead compenator i 48.1 0 34 0 10 0 24.1 0 60
SoluGon 3.4 Lead CompenaGon The Lead Compenator 3.76 db Uing equa4on [2], for φ max 24.1 0, β 0.42. Uing equa4on [3], the lead compenator magnitude 3.76 db at ω max 61
To elect a the new phae- margin frequency, find the frequency (ω max ) at which the uncompenate d ytem magnitude i - 3.76 db. From the Bode diagram, ω max 39 rad/. SoluGon 3.4 Lead CompenaGon The Lead Compenator 39 rad/ 62
SoluGon 3.4 Lead CompenaGon The Lead Compenator Uing equa4on [1], z c 25.3 and p c 25.3/0.42 60.2 Hence, the lead compenator i G c ( ) 2.38 25.3 60.2 The final compenated open- loop tranfer func4on: G c ( ) ( 342600( 25.3) 36)( 100)( 60.2) 63
SoluGon 3.4 Lead CompenaGon The Lead Compenator 64
Example 8 3.4 Lead CompenaGon The Lead Compenator For a unity feedback ytem with a forward tranfer func4on K G( ) ( 50)( 120) deign a lead compenator for the ytem to yield 20% OS and K v 50. ANSWER: G lead ( ) 2.27( 33.2) ( 75.4) 65
3.5 Lead- Lag CompenaGon The Lead- Lag Compenator Lead- lag compenator i ued to improve both the tranient repone and the teady tate error. It i the combina4on of lead compenator and the lag compenator. With the Bode diagram, the lead- lag compenator: Improve the performance at low frequency range via the effect from the lag compenator Improve the performance at high frequency range via the effect from the lead compenator. 66
67 The tranfer func4on of the lead- lag compenator: where the condi4on for the zero and pole are ame a being et earlier for the individual compenator. Further detail example can be found in the textbook for thi part. 1 1 1 1 ) ( 2 1 2 1 c c c c c p p z z G 3.5 Lead- Lag CompenaGon The Lead- Lag Compenator
Concluion We have covered the deign of ytem performance uing Bode Plot by u4liing: ü The gain adjutment ü The lag compenator, ü The lead compenator. ü The lead- lag compenator. THE END 68