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Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 19: Drawing Bode Plots, Part 1

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Overview In this Lecture, you will learn: Drawing Bode Plots Drawing Rules Simple Plots Constants Real Zeros M. Peet Lecture 19: Control Systems 2 / 30

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Review Recall from last lecture: Frequency Response Input: ) = sin( ωt Output: Magnitude and Phase Shift ) = sin( ωt 10 12 14 16 18 20 −2 −1.5 −1 −0.5 0.5 1.5 2.5 Linear Simulation Results Time (sec) Amplitude Frequency Response to sin ωt is given by M. Peet Lecture 19: Control Systems 3 / 30

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Bode Plots We know determines the frequency response. How to plot this information? 1 independent Variable: 2 Dependent Variables: Re )) and Im )) Im G(iω) Re G(iω) Figure: The Obvious Choice Really 2 plots put together. M. Peet Lecture 19: Control Systems 4 / 30

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Bode Plots An Alternative is to plot Polar Variables 1 independent Variable: 2 Dependent Variables: and |G(iω)| < G(iω) Advantage: All Information corresponds to physical data. Can be found directly using a frequency sweep. M. Peet Lecture 19: Control Systems 5 / 30

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Bode Plots If we only want a single plot we can use as a parameter −0.6 −0.4 −0.2 0.2 0.4 0.6 −0.6 −0.4 −0.2 0.2 0.4 0.6 Nyquist Diagram Real Axis Imaginary Axis A plot of Re )) vs. Im )) as a function of Advantage: All Information in a single plot. AKA: Nyquist Plot M. Peet Lecture 19: Control Systems 6 / 30

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Bode Plots We focus on Option 2 Deﬁnition 1. The Bode Plot is a pair of log log and semi- log plots: 1. Magnitude Plot: 20 log 10 vs. log 10 2. Phase Plot: vs. log 10 20 log 10 is units of Decibels (dB) Used in Power and Circuits. 10 log 10 |·| in other ﬁelds. Note that by log , we mean log base 10 ( log 10 In Matlab, log means natural logarithm. M. Peet Lecture 19: Control Systems 7 / 30

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Bode Plots Example Lets do a simple pole ) = + 1 We need Magnitude of Phase of Im(s) Re(s) ____ √1+ Recall that |···| |···| So that + 1 1 + M. Peet Lecture 19: Control Systems 8 / 30

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Bode Plots Example How to Plot 1+ We are actually want to plot 20 log = 20 log 1 + = 20 log(1 + 10 log(1 + Three Cases: Case 1: ω<< Approximate 1 + 20 log 10 log(1 + 10 log 1 = 0 Case 2: = 1 20 log 10 log(1 + 10 log(1 + 01 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram M. Peet Lecture 19: Control Systems 9 / 30

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Bode Plots Example Case 3: ω>> Approximate: 1 + 20 log 10 log(1 + 10 log 20 log -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram But we use a log log plot. -axis is = log -axis is = 20 log 20 log 20 Conclusion: On the log log plot, when ω>> Plot is Linear Slope is -20 dB/Decade! M. Peet Lecture 19: Control Systems 10 / 30

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Bode Plots Example Of course, we need to connect the dots. -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram Compare to the Real Thing: -35 -30 -25 -20 -15 -10 -5 Magnitude (dB) Bode Diagram M. Peet Lecture 19: Control Systems 11 / 30

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Bode Plots Example: Phase Now lets do the phase. Recall: ) = =1 =1 In this case, ) = + 1) tan Again, 3 cases: Case 1: ω<< tan( )) tan( )) Im(s) Re(s) <(i +1) 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) M. Peet Lecture 19: Control Systems 12 / 30

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Bode Plots Example: Phase Case 2: = 1 tan( )) = 1 45 Case 3: ω>> tan( )) 90 Fixed at 90 for large Im(s) Re(s) <(i +1) 10 -2 10 -1 10 10 10 -90 -45 ω<<1 Phase (deg) Frequency (rad/sec) ω>>1 ω=1 M. Peet Lecture 19: Control Systems 13 / 30

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Bode Plots Example We need to connect the dots somehow. 10 -2 10 -1 10 10 10 -90 -45 ω<<1 Phase (deg) Frequency (rad/sec) ω>>1 ω=1 Compare to the real thing: 10 -2 10 -1 10 10 10 -90 -45 Phase (deg) Frequency (rad/sec) M. Peet Lecture 19: Control Systems 14 / 30

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Bode Plots Methodology So far, drawing Bode Plots seems pretty intimidating. Solving tan dB and log -plots Lots of trig The process can be Greatly Simpliﬁed Use a few simple rules. Example: Suppose we have ) = Then || and log = log + log M. Peet Lecture 19: Control Systems 15 / 30

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Bode Plots Rule # 1 Rule # 1: Magnitude Plots Add in log -space. For ) = 20 log = 20 log + 20 log Decompose into bite-size chunks: ) = + 3 + 1) + 3 + 1 M. Peet Lecture 19: Control Systems 16 / 30

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Bode Plots Rule #2 Rule # 2: Phase Plots Add For ) = ) = ) + M. Peet Lecture 19: Control Systems 17 / 30

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Bode Plots Approach Our Approach is to Decompose into simpler pieces. Plot the phase and magnitude of each component. Add up the plots. Step 1: Decompose into all its poles and zeros ) = ··· ··· Then for magnitude 20 log 20 log 20 log 20 log | 20 log And for phase: ) = But how to plot and 20 log M. Peet Lecture 19: Control Systems 18 / 30

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Plotting Simple Terms The Constant Before rushing in, lets make sure we don’t forget the constant term. If ) = ··· ··· Magnitude: ) = 20 log = 20 log -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram Frequency (log ω) 20 log |c| Conclusion: Magnitude is Constant for all M. Peet Lecture 19: Control Systems 19 / 30

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Plotting Simple Terms The Constant Phase: ) = ) = c> 180 c< 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) c > 0 c < 0 Conclusion: phase is if c> , otherwise 180 M. Peet Lecture 19: Control Systems 20 / 30

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Plotting Simple Terms A “Pure” Zero Lets start with a zero at the origin: ) = Magnitude: ) = 20 log = 20 log Our x-axis is log Plot is Linear for all Slope is +20 dB/Decade! Need a point: = 1 20 log || =1 = 20 log 1 = 0 Passes through dB at = 1 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram ω=1 High Gain at High Frequency A pure zero means The faster the input, The larger the output M. Peet Lecture 19: Control Systems 21 / 30

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Plotting Simple Terms A “Pure” Zero: Phase Phase: ) = ) = = 90 Always 90 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) Always 90 out of phase. Why? M. Peet Lecture 19: Control Systems 22 / 30

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Plotting Simple Terms A “Pure” Zero: Multiple Zeros What happens if there are multiple pure zeros Just what you would expect. Magnitude: ) = 20 log = 20 log = 20 log Slope is +20k dB/Decade! Need a Point At = 1 20 log || =1 = 20 log 1 = 0 Still Passes through dB at = 1 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram ω=1 k = 2 k = 1 k = 3 k = 4 pure zeros added together. M. Peet Lecture 19: Control Systems 23 / 30

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Plotting Simple Terms A “Pure” Zero: Multiple Zeros And phase for multiple pure zeros? Phase: ) = ) = = 90 Always 90 10 -2 10 -1 10 10 10 -45 45 90 135 180 225 270 315 360 405 Phase (deg) Frequency (rad/sec) k = 2 k = 1 k = 3 k = 4 pure zeros added together. M. Peet Lecture 19: Control Systems 24 / 30

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Plotting Simple Terms Plotting Normal Zeros A zero at the origin is a line with slope +20 /Decade. What if the zero is not at the origin? We did one example already ( +1 ). Change of Format : to simplify steady-state response, we use ) = ( τs + 1) Pole is at Also put poles in this form Rewrite + 1) ) = ··· ··· ··· ··· + 1) ··· + 1) + 1) ··· + 1) + 1) ··· zm + 1) + 1) ··· pn + 1) Where zi pi ··· ··· Assume and are Real M. Peet Lecture 19: Control Systems 25 / 30

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Plotting Simple Terms Plotting Normal Zeros ) = + 1) ··· zm + 1) + 1) ··· pn + 1) The advantage of this form is that steady-state response to a step is ss = lim ) = (0) = 10 -2 10 -1 10 10 10 -90 -45 Phase (deg) Frequency (rad/sec) Low Frequency Response is given by the constant term, M. Peet Lecture 19: Control Systems 26 / 30

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Plotting Simple Terms Plotting Normal Zeros ) = ( τs + 1) + 1 1 + Magnitude: 20 log = 20 log(1+ = 10 log(1+ Im(s) Re(s) ______ √1+ 2 Case 1: ωτ << Approximate 1 + 20 log = 20 log(1 + 20 log 1 = 0 Case 2: = 1 20 log = 10 log(1 + = 10 log 2 = 3 01 10 15 20 25 30 35 40 Magnitude (dB) Bode Diagram ω = -1 10 -2 10 -1 10 10 10 Frequency (rad/sec) M. Peet Lecture 19: Control Systems 27 / 30

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Bode Plots Example Case 3: ωτ >> Approximate 1 + 20 log = 20 log 1 + 10 log = 20 log = 20 log + 20 log 10 15 20 25 30 35 40 Magnitude (dB) Bode Diagram ω = -1 +20 dB / decade 10 -2 10 -1 10 10 10 Frequency (rad/sec) Conclusion: When ω>> Plot is Linear Slope is +20 dB/Decade! inﬂection at M. Peet Lecture 19: Control Systems 28 / 30

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Plotting Simple Terms Plotting Normal Zeros Compare this to the magnitude plot of ) = ω << -1 ω >> -1 This is why we use the format ) = τs + 1 We want dB (no gain) at low frequency. M. Peet Lecture 19: Control Systems 29 / 30

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Summary What have we learned today? Drawing Bode Plots Drawing Rules Simple Plots Constants Real Zeros Next Lecture: More Bode Plotting M. Peet Lecture 19: Control Systems 30 / 30

Peet Illinois Institute of Technology Lecture 19 Drawing Bode Plots Part 1 brPage 2br Overview In this Lecture you will learn Drawing Bode Plots Drawing Rules Simple Plots Constants Real Zeros M Peet Lecture 19 Control Systems 2 30 brPa ID: 22588

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Page 1

Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 19: Drawing Bode Plots, Part 1

Page 2

Overview In this Lecture, you will learn: Drawing Bode Plots Drawing Rules Simple Plots Constants Real Zeros M. Peet Lecture 19: Control Systems 2 / 30

Page 3

Review Recall from last lecture: Frequency Response Input: ) = sin( ωt Output: Magnitude and Phase Shift ) = sin( ωt 10 12 14 16 18 20 −2 −1.5 −1 −0.5 0.5 1.5 2.5 Linear Simulation Results Time (sec) Amplitude Frequency Response to sin ωt is given by M. Peet Lecture 19: Control Systems 3 / 30

Page 4

Bode Plots We know determines the frequency response. How to plot this information? 1 independent Variable: 2 Dependent Variables: Re )) and Im )) Im G(iω) Re G(iω) Figure: The Obvious Choice Really 2 plots put together. M. Peet Lecture 19: Control Systems 4 / 30

Page 5

Bode Plots An Alternative is to plot Polar Variables 1 independent Variable: 2 Dependent Variables: and |G(iω)| < G(iω) Advantage: All Information corresponds to physical data. Can be found directly using a frequency sweep. M. Peet Lecture 19: Control Systems 5 / 30

Page 6

Bode Plots If we only want a single plot we can use as a parameter −0.6 −0.4 −0.2 0.2 0.4 0.6 −0.6 −0.4 −0.2 0.2 0.4 0.6 Nyquist Diagram Real Axis Imaginary Axis A plot of Re )) vs. Im )) as a function of Advantage: All Information in a single plot. AKA: Nyquist Plot M. Peet Lecture 19: Control Systems 6 / 30

Page 7

Bode Plots We focus on Option 2 Deﬁnition 1. The Bode Plot is a pair of log log and semi- log plots: 1. Magnitude Plot: 20 log 10 vs. log 10 2. Phase Plot: vs. log 10 20 log 10 is units of Decibels (dB) Used in Power and Circuits. 10 log 10 |·| in other ﬁelds. Note that by log , we mean log base 10 ( log 10 In Matlab, log means natural logarithm. M. Peet Lecture 19: Control Systems 7 / 30

Page 8

Bode Plots Example Lets do a simple pole ) = + 1 We need Magnitude of Phase of Im(s) Re(s) ____ √1+ Recall that |···| |···| So that + 1 1 + M. Peet Lecture 19: Control Systems 8 / 30

Page 9

Bode Plots Example How to Plot 1+ We are actually want to plot 20 log = 20 log 1 + = 20 log(1 + 10 log(1 + Three Cases: Case 1: ω<< Approximate 1 + 20 log 10 log(1 + 10 log 1 = 0 Case 2: = 1 20 log 10 log(1 + 10 log(1 + 01 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram M. Peet Lecture 19: Control Systems 9 / 30

Page 10

Bode Plots Example Case 3: ω>> Approximate: 1 + 20 log 10 log(1 + 10 log 20 log -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram But we use a log log plot. -axis is = log -axis is = 20 log 20 log 20 Conclusion: On the log log plot, when ω>> Plot is Linear Slope is -20 dB/Decade! M. Peet Lecture 19: Control Systems 10 / 30

Page 11

Bode Plots Example Of course, we need to connect the dots. -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram Compare to the Real Thing: -35 -30 -25 -20 -15 -10 -5 Magnitude (dB) Bode Diagram M. Peet Lecture 19: Control Systems 11 / 30

Page 12

Bode Plots Example: Phase Now lets do the phase. Recall: ) = =1 =1 In this case, ) = + 1) tan Again, 3 cases: Case 1: ω<< tan( )) tan( )) Im(s) Re(s) <(i +1) 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) M. Peet Lecture 19: Control Systems 12 / 30

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Bode Plots Example: Phase Case 2: = 1 tan( )) = 1 45 Case 3: ω>> tan( )) 90 Fixed at 90 for large Im(s) Re(s) <(i +1) 10 -2 10 -1 10 10 10 -90 -45 ω<<1 Phase (deg) Frequency (rad/sec) ω>>1 ω=1 M. Peet Lecture 19: Control Systems 13 / 30

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Bode Plots Example We need to connect the dots somehow. 10 -2 10 -1 10 10 10 -90 -45 ω<<1 Phase (deg) Frequency (rad/sec) ω>>1 ω=1 Compare to the real thing: 10 -2 10 -1 10 10 10 -90 -45 Phase (deg) Frequency (rad/sec) M. Peet Lecture 19: Control Systems 14 / 30

Page 15

Bode Plots Methodology So far, drawing Bode Plots seems pretty intimidating. Solving tan dB and log -plots Lots of trig The process can be Greatly Simpliﬁed Use a few simple rules. Example: Suppose we have ) = Then || and log = log + log M. Peet Lecture 19: Control Systems 15 / 30

Page 16

Bode Plots Rule # 1 Rule # 1: Magnitude Plots Add in log -space. For ) = 20 log = 20 log + 20 log Decompose into bite-size chunks: ) = + 3 + 1) + 3 + 1 M. Peet Lecture 19: Control Systems 16 / 30

Page 17

Bode Plots Rule #2 Rule # 2: Phase Plots Add For ) = ) = ) + M. Peet Lecture 19: Control Systems 17 / 30

Page 18

Bode Plots Approach Our Approach is to Decompose into simpler pieces. Plot the phase and magnitude of each component. Add up the plots. Step 1: Decompose into all its poles and zeros ) = ··· ··· Then for magnitude 20 log 20 log 20 log 20 log | 20 log And for phase: ) = But how to plot and 20 log M. Peet Lecture 19: Control Systems 18 / 30

Page 19

Plotting Simple Terms The Constant Before rushing in, lets make sure we don’t forget the constant term. If ) = ··· ··· Magnitude: ) = 20 log = 20 log -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram Frequency (log ω) 20 log |c| Conclusion: Magnitude is Constant for all M. Peet Lecture 19: Control Systems 19 / 30

Page 20

Plotting Simple Terms The Constant Phase: ) = ) = c> 180 c< 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) c > 0 c < 0 Conclusion: phase is if c> , otherwise 180 M. Peet Lecture 19: Control Systems 20 / 30

Page 21

Plotting Simple Terms A “Pure” Zero Lets start with a zero at the origin: ) = Magnitude: ) = 20 log = 20 log Our x-axis is log Plot is Linear for all Slope is +20 dB/Decade! Need a point: = 1 20 log || =1 = 20 log 1 = 0 Passes through dB at = 1 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram ω=1 High Gain at High Frequency A pure zero means The faster the input, The larger the output M. Peet Lecture 19: Control Systems 21 / 30

Page 22

Plotting Simple Terms A “Pure” Zero: Phase Phase: ) = ) = = 90 Always 90 10 -2 10 -1 10 10 10 -225 -180 -135 -90 -45 45 90 135 180 225 Phase (deg) Frequency (rad/sec) Always 90 out of phase. Why? M. Peet Lecture 19: Control Systems 22 / 30

Page 23

Plotting Simple Terms A “Pure” Zero: Multiple Zeros What happens if there are multiple pure zeros Just what you would expect. Magnitude: ) = 20 log = 20 log = 20 log Slope is +20k dB/Decade! Need a Point At = 1 20 log || =1 = 20 log 1 = 0 Still Passes through dB at = 1 -35 -30 -25 -20 -15 -10 -5 10 Magnitude (dB) Bode Diagram ω=1 k = 2 k = 1 k = 3 k = 4 pure zeros added together. M. Peet Lecture 19: Control Systems 23 / 30

Page 24

Plotting Simple Terms A “Pure” Zero: Multiple Zeros And phase for multiple pure zeros? Phase: ) = ) = = 90 Always 90 10 -2 10 -1 10 10 10 -45 45 90 135 180 225 270 315 360 405 Phase (deg) Frequency (rad/sec) k = 2 k = 1 k = 3 k = 4 pure zeros added together. M. Peet Lecture 19: Control Systems 24 / 30

Page 25

Plotting Simple Terms Plotting Normal Zeros A zero at the origin is a line with slope +20 /Decade. What if the zero is not at the origin? We did one example already ( +1 ). Change of Format : to simplify steady-state response, we use ) = ( τs + 1) Pole is at Also put poles in this form Rewrite + 1) ) = ··· ··· ··· ··· + 1) ··· + 1) + 1) ··· + 1) + 1) ··· zm + 1) + 1) ··· pn + 1) Where zi pi ··· ··· Assume and are Real M. Peet Lecture 19: Control Systems 25 / 30

Page 26

Plotting Simple Terms Plotting Normal Zeros ) = + 1) ··· zm + 1) + 1) ··· pn + 1) The advantage of this form is that steady-state response to a step is ss = lim ) = (0) = 10 -2 10 -1 10 10 10 -90 -45 Phase (deg) Frequency (rad/sec) Low Frequency Response is given by the constant term, M. Peet Lecture 19: Control Systems 26 / 30

Page 27

Plotting Simple Terms Plotting Normal Zeros ) = ( τs + 1) + 1 1 + Magnitude: 20 log = 20 log(1+ = 10 log(1+ Im(s) Re(s) ______ √1+ 2 Case 1: ωτ << Approximate 1 + 20 log = 20 log(1 + 20 log 1 = 0 Case 2: = 1 20 log = 10 log(1 + = 10 log 2 = 3 01 10 15 20 25 30 35 40 Magnitude (dB) Bode Diagram ω = -1 10 -2 10 -1 10 10 10 Frequency (rad/sec) M. Peet Lecture 19: Control Systems 27 / 30

Page 28

Bode Plots Example Case 3: ωτ >> Approximate 1 + 20 log = 20 log 1 + 10 log = 20 log = 20 log + 20 log 10 15 20 25 30 35 40 Magnitude (dB) Bode Diagram ω = -1 +20 dB / decade 10 -2 10 -1 10 10 10 Frequency (rad/sec) Conclusion: When ω>> Plot is Linear Slope is +20 dB/Decade! inﬂection at M. Peet Lecture 19: Control Systems 28 / 30

Page 29

Plotting Simple Terms Plotting Normal Zeros Compare this to the magnitude plot of ) = ω << -1 ω >> -1 This is why we use the format ) = τs + 1 We want dB (no gain) at low frequency. M. Peet Lecture 19: Control Systems 29 / 30

Page 30

Summary What have we learned today? Drawing Bode Plots Drawing Rules Simple Plots Constants Real Zeros Next Lecture: More Bode Plotting M. Peet Lecture 19: Control Systems 30 / 30

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