443 Lecture 19

Systems Analysis and Control

Matthew M. PeetIllinois Institute of Technology

Lecture 19: Drawing Bode Plots, Part 1

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

Review

Recall from last lecture: Frequency Response

Input:

u(t) =M sin(t+ )

Output: Magnitude and Phase Shift

y(t) = |G()|M sin(t+ + G())

0 2 4 6 8 10 12 14 16 18 202

1.5

1

0.5

0

0.5

1

1.5

2

2.5

Linear Simulation Results

Time (sec)Am

plitu

de

Frequency Response to sint is given by G()


Bode Plots

We know G() determines the frequency response.How to plot this information?

1 independent Variable: 2 Dependent Variables: Re(G()) and Im(G())

Im G(i)

Re G(i)

Figure: The Obvious Choice

Really 2 plots put together.M. Peet Lecture 19: Control Systems 4 / 30

Bode Plots

An Alternative is to plot Polar Variables 1 independent Variable: 2 Dependent Variables: G() and |G()|

|G(i)|

< G(i)

Advantage: All Information corresponds to physical data.I Can be found directly using a frequency sweep.


Bode Plots

If we only want a single plot we can use as a parameter.

0.6 0.4 0.2 0 0.2 0.4 0.6

0.6

0.4

0.2

0

0.2

0.4

0.6

Nyquist Diagram

Real Axis

Imag

inar

y Ax

is

A plot of Re(G()) vs. Im(G()) as a function of .

Advantage: All Information in a single plot. AKA: Nyquist Plot


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 log10 |G()| vs. log10 2. Phase Plot: G() vs. log10

20 log10 |G()| is units of Decibels (dB) Used in Power and Circuits. 10 log10 | | in other fields.

Note that by log, we mean log base 10 (log10)

In Matlab, log means natural logarithm.


Bode PlotsExample

Lets do a simple pole

G(s) =1

s+ 1

We need

Magnitude of G() Phase of G()

Im(s)

Re(s)

1

} } ____

1+2

Recall that

|G(s)| = |s z1| |s zm||s p1| |s pn|So that

|G()| = 1| + 1| =1

1 + 2


Bode PlotsExample

How to Plot |G()| = 11+2

?

We are actually want to plot

20 log |G()| = 20 log 11 + 2

= 20 log(1 + 2)12 = 10 log(1 + 2)

Three Cases:

Case 1:

Bode PlotsExample

Case 3: >> 1

Approximate: 1 + 2 = 220 log |G()| = 10 log(1 + 2)

= 10 log2= 20 log

-35

-30

-25

-20

-15

-10

-5

0

5

10

Mag

nitu

de (d

B)

Bode Diagram

But we use a log log plot. x-axis is x = log y-axis is y = 20 log |G()| = 20 log = 20x

Conclusion: On the log-log plot, when >> 1,

Plot is Linear Slope is -20 dB/Decade!


Bode PlotsExample

Of course, we need to connect the dots.

-35

-30

-25

-20

-15

-10

-5

0

5

10

Mag

nitu

de (d

B)

Bode Diagram

Compare to the Real Thing:

-35

-30

-25

-20

-15

-10

-5

0

Mag

nitu

de (d

B)

Bode Diagram


Bode PlotsExample: Phase

Now lets do the phase. Recall:

G(s) =mi=1

(s zi)ni=1

(s pi)

In this case,

G() = ( + 1)= tan1()

Again, 3 cases:Case 1:

Bode PlotsExample: Phase

Case 2: = 1

tan(G()) = 1 G() = 45

Case 3: >> 1

tan(G()) = 10 G() = 90 Fixed at 90 for large !

Im(s)

Re(s)

1

} }

Bode PlotsExample

We need to connect the dots somehow.

10 -2 10 -1 100 101 102-90

-45

01

=1

Compare to the real thing:

10 -2 10 -1 100 101 102-90

-45

0

Phas

e (de

g)

Frequency (rad/sec)


Bode PlotsMethodology

So far, drawing Bode Plots seems pretty intimidating.

Solving tan1 dB and log-plots Lots of trig

The process can be Greatly Simplified:

Use a few simple rules.

Example: Suppose we have

G(s) = G1(s)G2(s)

Then|G()| = |G1()||G2()|

andlog |G()| = log |G1()|+ log |G2()|


Bode PlotsRule # 1

Rule # 1: Magnitude Plots Add in log-space.For G(s) = G1(s)G2(s),

20 log |G()| = 20 log |G1()|+ 20 log |G2()|

Decompose G into bite-size chunks:

G(s) =1

s+ 3(s+ 1)

1

s2 + 3s+ 1= G1(s)G2(s)G3(s)


Bode PlotsRule #2

Rule # 2: Phase Plots Add.For G(s) = G1(s)G2(s),

G() = G1() + G2()


Bode PlotsApproach

Our Approach is to Decompose G(s) into simpler pieces. Plot the phase and magnitude of each component. Add up the plots.

Step 1: Decompose G into all its poles and zeros

G(s) =(s z1) (s zm)(s p1) (s pn)

Then for magnitude

20 log |G()| =i

20 log | zi|+i

20 log1

| pi|=i

20 log | zi| i

20 log | pi|

And for phase:G() =

i

( zi)i

( pi)

But how to plot ( zi) and 20 log | zi|?M. Peet Lecture 19: Control Systems 18 / 30

Plotting Simple TermsThe Constant

Before rushing in, lets make sure we dont forget the constant term. If

G(s) = c(s z1) (s zm)(s p1) (s pn)

Magnitude: G1(s) = c |G1()| = |c| 20 log |G1()| = 20 log |c|

-35

-30

-25

-20

-15

-10

-5

0

5

10

Mag

nitu

de (d

B)

Bode Diagram

Frequency (log )

20 log |c|

Conclusion: Magnitude is Constant for all M. Peet Lecture 19: Control Systems 19 / 30

Plotting Simple TermsThe Constant

Phase: G1(s) = c

G1() = c ={0 c > 0180 c < 0

10 -2 10 -1 100 101 102-225

-180

-135

-90

-45

0

45

90

135

180

225

Phas

e (de

g)

Frequency (rad/sec)

c > 0

c < 0

Conclusion: phase is 0 if c > 0, otherwise 180.


Plotting Simple TermsA Pure Zero

Lets start with a zero at the origin: G1(s) = s.

Magnitude: G1(s) = s

|G1()| = || = || 20 log |G1()| = 20 log ||

Our x-axis is log.

Plot is Linear for all Slope is +20 dB/Decade! Need a point: = 1

20 log |G1()||=1 = 20 log 1 = 0 Passes through 0dB at = 1

-35

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-5

0

5

10

Mag

nitu

de (d

B)

Bode Diagram

=1

High Gain at High Frequency

A pure zero means u(t) The faster the input, The larger

the output


Plotting Simple TermsA Pure Zero: Phase

Phase: G1(s) = s

G1() = = 90 Always 90!

10 -2 10 -1 100 101 102-225

-180

-135

-90

-45

0

45

90

135

180

225

Phas

e (de

g)

Frequency (rad/sec)

Always 90 out of phase. Why?


Plotting Simple TermsA Pure Zero: Multiple Zeros

What happens if there are multiple pure zeros

Just what you would expect.Magnitude: G1(s) = sk

|G1()| = ||k = ||k

20 log |G1()| = 20 log ||k= 20k log ||

Slope is +20k dB/Decade!Need a Point

At = 1:

20 log |G1()||=1 = 20k log 1 = 0 Still Passes through 0dB at = 1

-35

-30

-25

-20

-15

-10

-5

0

5

10

Mag

nitu

de (d

B)

Bode Diagram

=1

k = 2

k = 1

k = 3

k = 4

k pure zeros added together.


Plotting Simple TermsA Pure Zero: Multiple Zeros

And phase for multiple pure zeros?Phase: G1(s) = sk

G1() = ()k = k = 90k Always 90k

10 -2 10 -1 100 101 102-45

0

45

90

135

180

225

270

315

360

405

Phas

e (de

g)

Frequency (rad/sec)

k = 2

k = 1

k = 3

k = 4

k pure zeros added together.


Plotting Simple TermsPlotting Normal Zeros

A zero at the origin is a line with slope +20/Decade. What if the zero is not at the origin?

I We did one example already ( 1s+1

).

Change of Format: to simplify steady-state response, we use

G1(s) = (s+ 1) Pole is at s = 1 Also put poles in this form

Rewrite G(s): (s+ p) p( 1ps+ 1).

G(s) = k(s+ z1) (s+ zm)(s+ p1) (s+ pn)

= kz1 zmp1 pn

( 1z1 s+ 1) ( 1zm s+ 1)( 1p1 s+ 1) ( 1pn s+ 1)

= c(z1s+ 1) (zms+ 1)(p1s+ 1) (pns+ 1)

Where

zi = 1zi pi = 1pi c = k z1zmp1pn

Assume zi and pi are Real.



G(s) = c(z1s+ 1) (zms+ 1)(p1s+ 1) (pns+ 1)

The advantage of this form is that steady-state response to a step is

yss = lims0

G(s) = G(0) = c

10 -2 10 -1 100 101 102-90

-45

0

Phas

e (de

g)

Frequency (rad/sec)

Low Frequency Response is given by the constant term, c.



G1(s) = (s+ 1)

|G1()| = | + 1| =1 + 22

Magnitude:

20 log |G1()| = 20 log(1+22) 12 = 10 log(1+22)

Im(s)

Re(s)

1

} } ______

1+2 2

Case 1:

Bode PlotsExample

Case 3: >> 1

Approximate 1 + 22 = 2220 log |G()| = 20 log

1 + 22

= 10 log22= 20 log

= 20 log + 20 log

0

5

10

15

20

25

30

35

40

Mag

nitu

de (d

B)

Bode Diagram

= -1

+20 dB / decade

10 -2 10 -1 100 101 102

Frequency (rad/sec)

Conclusion: When >> 1,

Plot is Linear Slope is +20 dB/Decade! inflection at = 1



Compare this to the magnitude plot of

G1(s) = s+ a

> -1

This is why we use the format G1(s) = s+ 1

We want 0dB (no gain) at low frequency.M. Peet Lecture 19: Control Systems 29 / 30

Summary

What have we learned today?

Drawing Bode Plots

Drawing Rules

Simple Plots

Constants Real Zeros

Next Lecture: More Bode Plotting


Control Systems

Documents

443 Lecture 19