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443 Lecture 19
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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()
M. Peet Lecture 19: Control Systems 3 / 30
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.
M. Peet Lecture 19: Control Systems 5 / 30
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
M. Peet Lecture 19: Control Systems 6 / 30
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.
M. Peet Lecture 19: Control Systems 7 / 30
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
M. Peet Lecture 19: Control Systems 8 / 30
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!
M. Peet Lecture 19: Control Systems 10 / 30
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
M. Peet Lecture 19: Control Systems 11 / 30
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)
M. Peet Lecture 19: Control Systems 14 / 30
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()|
M. Peet Lecture 19: Control Systems 15 / 30
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)
M. Peet Lecture 19: Control Systems 16 / 30
Bode PlotsRule #2
Rule # 2: Phase Plots Add.For G(s) = G1(s)G2(s),
G() = G1() + G2()
M. Peet Lecture 19: Control Systems 17 / 30
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.
M. Peet Lecture 19: Control Systems 20 / 30
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
-30
-25
-20
-15
-10
-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
M. Peet Lecture 19: Control Systems 21 / 30
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?
M. Peet Lecture 19: Control Systems 22 / 30
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.
M. Peet Lecture 19: Control Systems 23 / 30
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.
M. Peet Lecture 19: Control Systems 24 / 30
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.
M. Peet Lecture 19: Control Systems 25 / 30
Plotting Simple TermsPlotting Normal Zeros
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.
M. Peet Lecture 19: Control Systems 26 / 30
Plotting Simple TermsPlotting Normal Zeros
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
M. Peet Lecture 19: Control Systems 28 / 30
Plotting Simple TermsPlotting Normal Zeros
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
M. Peet Lecture 19: Control Systems 30 / 30
Control Systems
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