tài liệu bài giảng 4.2 bode diagram (cont )

The slides used the source PowerPoint files of Anthony Rossiter • What is frequency response? • How I compute this efficiently? • How I represent frequency response information in a helpful fashion? • Why is this relevant to feedback loop analysis and design? • Previous topic showed how we can determine expressions for frequency response gain and phase based on a transfer function • These formulae are not insightful on their own • Humans often relate better to pictures and graphs • Hence it is useful to sketch the gain and phase in graphical form and thus to judge whether such a sketch would be helpful a transfer function object G a frequency ‘w’ Then [gain,phase]=bode(G,w); 3 1 w st take GTo ( s )begin,  ; Ga( simple jw )  order;example G ( jw )  tan 2 s2 w 2 1.5 -20 Gain Phase -40 -60 0.5 -80 0 20 40 60 Frequency 80 100 -100 20 40 60 Frequency 80 100 Gain For low frequencies gain is 1.5 As frequency increases, gain drops to zero For low frequencies phases is As frequency increases, phase tends to -90 0.5 0 20 40 60 80 100 Frequency -20 Phase k G( s)  sa 1.5 -40 -60 -80 -100 20 40 60 Frequency 80 100 G ( s)  ; s  5s  G ( jw)   tan 1 G ( jw)  ( w  )(w  ) 2 2 ; w w  tan 1 0.7 0.6 -50 Phase Gain 0.5 0.4 0.3 0.2 -100 -150 0.1 0 20 40 60 Frequency 80 100 -200 20 40 60 Frequency 80 100 0.7 0.6 G( s)  s  5s  Gain 0.4 0.3 0.2 0.1 0 20 40 60 80 100 Frequency -50 Phase For low frequencies gain is 0.67 and as frequency increases, gain drops to zero For low frequencies phases is and as frequency increases, phase tends to -180 0.5 -100 -150 -200 20 40 60 Frequency 80 100 s4 G( s)  ; s  2s  20 G ( jw)  w2  (20  w2 )  4w2 ; 2w G ( jw)   tan 20  w2 1 0.7 20 0.6 0.5 Phase Gain -20 0.4 0.3 -60 0.2 -80 0.1 0 -40 20 40 60 Frequency 80 100 -100 20 40 60 Frequency 80 100   a /(10)  G  0;   a /(10)  G   tan 1 (0.1)  6o ;   a  G  45o ;   10a  G   tan 1 (10)  84o   10a  G  90o ; a  G a a  G  a  or [20 log 10 a]dB or [20 log 10 a  3]dB a G  w or [20 log 10 w]dB G sa We will make use of these insights in combination with the additive property • Break the frequency range into domains based on approximations for each factor  1  1  , jw  1 1  jw  5 or G  [20 log 10 ]dB  4.5dB 1 1 1     ,  jw  w jw  5 or  5  or 3 ]dB  [20 log 10  20 log 10 w] 2w 1 1  ,  jw  w jw  w G  [20 log 10 G  [20 log 10 ]dB or [20 log 10  40 log 10 w]dB w G ( s  1)( s  5) SLOPE OF MINUS 20dB/DECADE SLOPE OF MINUS 40dB/DECADE Use the asymptotic information to form a sketch Bode magnitude asymptotes Asymptotes Exact -10 Decibels -20 -30 G ( s  1)( s  5) -40 -50 -60 -1 10 10 10 Freq rad/s • Break the frequency range into domains based on approximations for each factor 1 1  1   ,  G jw  1 jw  5 ( s  1)( s  5) or G  1    1  , jw  jw 1  jw  5 or G  90o  5  1  , jw  jw or G  180o 1  jw  jw Correction of about 45o at corner frequency Bode phase asymptotes Asymptotes Exact -20 -40 G ( s  1)( s  5) Phase (degrees) -60 -80 -100 G ( j1)  45  11 -120 G ( j 5)  79  45 -140 -160 -180 -1 10 10 w Frequency (rad/s) G   tan w  tan 1 1 10 10 jw+1 jw+2 jw+3 G(jw) w[...]... actual w which is helpful) 3 y-axis plots are either: a 20log10|G(jw)| b Arg(G(jw )) (denoted decibels) (usually in degrees) Bode Diagram 10 Magnitude (dB) 0 -10 -20 Note: use of decibels for gain -30 Phase (deg) -40 0 Note: frequency on a log10 scale -45 -90 -1 10 10 0 1 10 Frequency (rad/s) (rad/s) 2 10 Bode Diagram 0 Magnitude (dB) -10 -20 -30 s4 G 2 s  2s  20 -40 45 Phase (deg) 0 -45 -90 -135 -1... (dB) 20 0 -20 -40 91 Phase (deg) 90.5 90 89.5 89 -2 10 -1 10 0 10 Frequency (rad/s) (rad/s) 1 10 2 10 Substitute s=jw and use the definition of the Bode diagram 20 log 10 1 jw against log 10 w 1  against log 10 w jw Bode Diagram 40 Magnitude (dB) 20 0 -20 -40 -89 Phase (deg) -89.5 -90 -90.5 -91 -2 10 -1 10 0 10 Frequency (rad/s) (rad/s) 1 10 2 10 Substitute s=jw and use the definition of the Bode diagram. .. 2 w2 (2  jw) against log 10 w w  2 w  2 w2 Bode Diagram 50 Magnitude (dB) 40 30 20 10 Phase (deg) 0 90 45 0 -1 10 0 10 10 1 Frequency (rad/s) (rad/s) 10 2 Substitute s=jw and use the definition of the Bode diagram 20 log 10 1 4  jw against log 10 w w  4 w  4 w4 1 ) against log 10 w 4  jw w  4 ( w  4 w4 Bode Diagram -10 Magnitude (dB) -15 -20 -25 -30 -35 Phase (deg) -40 0 -45 -90... divide by 1 0) 2 A change in gain of 3dB is equivalent to a change in gain of a factor of √2 (approx) Similarly one can see that 6dB is equivalent to a factor of 2 (approx .) It can be helpful, for mental arithmetic and insight, to know logarithms for several key integers log10( 1) = 0 log10( 2) ≈ 0.3 log10(1 0) =1 log10( 3) ≈ 0.48 (or 9.6dB) log10(10 0) = 2 log10(10n) =n log10( 4) ≈ 0.6 (or 12dB) log10( 5) ≈ 0.7... to sketch Bode diagrams for: 1 1 G  s, , ( s  a), s sa Comprises 2 plots 1 Both plots use log10w or w on a log axis 2 The other axis plots are either: a 20log10|G(jw)| b Arg(G(jw )) (denoted decibels) (usually in degrees) Next, apply these definitions to each factor in turn • Substitute s=jw and use the definition of the Bode diagram 20 log 10 jw against log 10 w jw against log 10 w Bode Diagram. .. ≈ 0.7 (or 14dB) log10(0. 1) =-1 log10( 6) ≈ 0.78 log (0.0 1) =-2 10 log10( 8) ≈ 0.9 (or 18dB) • The previous topic showed that MATLAB can be used for forming exact Bode diagrams and reminded students of core properties of logarithms • However, it is recognised that the ability to sketch is core for developing insight and for use in design • This topic will use the definition of the Bode diagram, and properties... to shapes in the Bode plot • This understanding will allow useful insight, especially for control design which comes later • As will become apparent, we need to be comfortable with rules of logarithms log(ab)=log(a)+log(b) log(a/b)=log a- log(b) How does a change in the decibel scale relate to changes in the underlying gain? 20log10(10a)= 20log10 (√2 a)= 20log10 (a/√2 )= 20log10 (a/1 0)= 1 A change of... Bode Diagram -10 Magnitude (dB) -15 -20 -25 -30 -35 Phase (deg) -40 0 -45 -90 -1 10 0 10 10 1 Frequency (rad/s) (rad/s) 10 2 Generic observations can be given for simple factors:   a /(1 0)  G  0; G  ( s  a)   a /(1 0)  G  tan 1 (0. 1)  6o ;   a  G  45o ;   10a  G  tan 1 (1 0)  84o   10a  G  90o ;   a  G  a or [20 log 10 a]dB a  G a 2 a  G  w or [20 log 10 w]dB or... Generic observations can be given for simple factors:   a /(1 0)  G  0; 1   a /(1 0)  G   tan 1 (0. 1)  6o ;   a  G  45 ; o   10a  G   tan 1 (1 0)  84o   10a  G  90o ; a  G  a  G  a  1 a or [20 log 10 a]dB 1 or [20 log 10 a  3]dB a 2 G  w or [20 log 10 w]dB G sa • This video has presented Bode diagrams for simple factors • Shown that an ‘accurate’ sketch... 10 10 0 1 10 Frequency (rad/s) (rad/s) 2 10 Bode Diagram 0 Magnitude (dB) -10 -20 -30 s4 G 2 s  2s  20 -40 45 Phase (deg) 0 -45 -90 -135 -1 10 0 10 1 10 Frequency (rad/s) (rad/s) 2 10 It is straightforward to sketch the bode diagrams using MATLAB and hence to get an overview of how gain and phase change over a large range of frequencies The use of log axis for frequencies enables us to focus on ... log10( 1) = log10( 2) ≈ 0.3 log10(1 0) =1 log10( 3) ≈ 0.48 (or 9.6dB) log10(10 0) = log10(10n) =n log10( 4) ≈ 0.6 (or 12dB) log10( 5) ≈ 0.7 (or 14dB) log10(0. 1) =-1 log10( 6) ≈ 0.78 log (0.0 1) =-2 10... (rad/s) (rad/s) 10 Bode Diagram Magnitude (dB) -10 -20 -30 s4 G s  2s  20 -40 45 Phase (deg) -45 -90 -135 -1 10 10 10 Frequency (rad/s) (rad/s) 10 It is straightforward to sketch the bode diagrams... helpful) y-axis plots are either: a 20log10|G(jw)| b Arg(G(jw )) (denoted decibels) (usually in degrees) Bode Diagram 10 Magnitude (dB) -10 -20 Note: use of decibels for gain -30 Phase (deg) -40