Systems Analysis and Control
Matthew M. Peet
Arizona State University
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
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Review
Recall from last lecture: Frequency Response
Input:
Linear Simulation Results
u(t) = M sin(ωt + φ)
2.5
2
1.5
Output: Magnitude and Phase Shift
y(t) = |G(ıω)|M sin(ωt + φ + ∠G(ıω))
Amplitude
1
0.5
0
−0.5
−1
−1.5
−2
0
2
4
6
8
10
Time (sec)
12
14
16
18
20
Frequency Response to sin ωt is given by G(ıω)
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Lecture 19: Control Systems
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Bode Plots
We know G(ıω) determines the frequency response.
How to plot this information?
• 1 independent Variable: ω
• 2 Dependent Variables: Re(G(ıω)) and Im(G(ıω))
Im G(iω)
Re G(iω)
ω
Figure: The Obvious Choice
Really 2 plots put together.
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Lecture 19: Control Systems
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Bode Plots
An Alternative is to plot Polar Variables
• 1 independent Variable: ω
• 2 Dependent Variables: ∠G(ıω) and |G(ıω)|
|G(iω)|
< G(iω)
ω
• Advantage: All Information corresponds to physical data.
I Can be found directly using a frequency sweep.
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Lecture 19: Control Systems
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Bode Plots
If we only want a single plot we can use ω as a parameter.
Nyquist Diagram
0.6
0.4
Imaginary Axis
0.2
0
−0.2
−0.4
−0.6
−0.6
−0.4
−0.2
0
Real Axis
0.2
0.4
0.6
A plot of Re(G(ıω)) vs. Im(G(ıω)) as a function of ω.
• Advantage: All Information in a single plot.
• AKA: Nyquist Plot
M. Peet
Lecture 19: Control Systems
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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 log10 |G(ıω)| vs. log10 ω
2. Phase Plot: ∠G(ıω) vs. log10 ω
20 log10 |G(ıω)| 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 Plots
Example
Lets do a simple pole
Im(s)
1
G(s) =
s+1
}
____
√1+ω2
We need
• Magnitude of G(ıω)
• Phase of G(ıω)
ω
}
Re(s)
1
Recall that
|G(s)| =
|s − z1 | · · · |s − zm |
|s − p1 | · · · |s − pn |
|G(ıω)| =
1
1
=√
|ıω + 1|
1 + ω2
So that
M. Peet
Lecture 19: Control Systems
8 / 30
Bode Plots
Example
How to Plot |G(ıω)| =
√ 1
?
1+ω 2
We actually want to plot it in dB, so ...
1
1
20 log |G(ıω)| = 20 log √
= 20 log(1 + ω 2 )− 2 = −10 log(1 + ω 2 )
2
1+ω
Three Cases:
Case 1: ω << 1
Bode Diagram
• Approximate 1 + ω 2 ∼
=1
5
0
-5
Magnitude (dB)
20 log |G(ıω)| = −10 log(1 + ω 2 )
∼
= −10 log 1 = 0
10
-10
-15
-20
-25
-30
Case 2: ω = 1
-35
20 log |G(ıω)| = −10 log(1 + ω 2 )
= −10 log(1 + ω 2 )
= −3.01
M. Peet
Lecture 19: Control Systems
9 / 30
Bode Plots
Example
Case 3: ω >> 1
Bode Diagram
• Approximate: 1 + ω 2 ∼
= ω2
10
5
= −20 log ω
0
-5
Magnitude (dB)
20 log |G(ıω)| = −10 log(1 + ω 2 )
∼
= −10 log ω 2
-10
-15
-20
-25
-30
-35
w
But we use a log − log plot.
• x-axis is x = log ω
• y-axis is y = 20 log |G(ıω)| = −20 log ω = −20x
Conclusion: On the log-log plot, when ω >> 1,
• Plot is Linear
• Slope is -20 dB/Decade!
M. Peet
Lecture 19: Control Systems
10 / 30
Bode Plots
Example
Of course, we need to connect the dots.
Bode Diagram
10
5
0
Magnitude (dB)
-5
-10
-15
-20
-25
-30
-35
Compare to the Real Thing:
Bode Diagram
0
-5
Magnitude (dB)
-10
-15
-20
-25
-30
-35
M. Peet
Lecture 19: Control Systems
11 / 30
Bode Plots
Example: Phase
Now lets do the phase. Recall:
∠G(s) =
m
X
∠(s − zi ) −
i=1
n
X
Im(s)
∠(s − pi )
i=1
In this case,
<(iω +1)
}
∠G(ıω) = −∠(ıω + 1)
}
ω
Re(s)
1
= − tan−1 (ω)
225
Again, 3 cases:
Case 1: ω << 1
• tan(∠G(ıω)) ∼
= ∠G(ıω) ∼
=0
135
90
Phase (deg)
• tan(∠G(ıω)) ∼
=0
180
45
0
-45
-90
-135
-180
-225
10
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10
-1
Lecture 19: Control Systems
0
10
Frequency (rad/sec)
10
1
10
2
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Bode Plots
Example: Phase
Case 2: ω = 1
Im(s)
• tan(∠G(ıω)) = 1
• ∠G(ıω) ∼
= 45◦
Between ω = .1 and ω = 10:
• Approximate Slope:
}
I
<(iω +1)
−45◦ /Decade
}
ω
Re(s)
1
Case 3: ω >> 1
• tan(∠G(ıω)) ∼
=
1
0
0
• ∠G(ıω) ∼
= −90
◦
Phase (deg)
• Fixed at −90◦ for large ω!
ω<<1
ω=1
-45
ω>>1
-90
10
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-2
10
-1
0
10
Frequency (rad/sec)
Lecture 19: Control Systems
10
1
10
2
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Bode Plots
Example
We need to connect the dots somehow.
0
Phase (deg)
ω<<1
ω=1
-45
ω>>1
-90
10
-2
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
Compare to the real thing:
Phase (deg)
0
-45
-90
10
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10
-1
0
10
Frequency (rad/sec)
10
1
Lecture 19: Control Systems
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2
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Bode Plots
Methodology
So far, drawing Bode Plots seems pretty intimidating.
• Solving tan−1
• 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(ıω)| = |G1 (ıω)||G2 (ıω)|
and
log |G(ıω)| = log |G1 (ıω)| + log |G2 (ıω)|
M. Peet
Lecture 19: Control Systems
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Bode Plots
Rule # 1
Rule # 1: Magnitude Plots Add in log-space.
For G(s) = G1 (s)G2 (s),
20 log |G(ıω)| = 20 log |G1 (ıω)| + 20 log |G2 (ıω)|
Decompose G into bite-size chunks:
G(s) =
M. Peet
1
1
(s + 1) 2
= G1 (s)G2 (s)G3 (s)
s+3
s + 3s + 1
Lecture 19: Control Systems
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Bode Plots
Rule #2
Rule # 2: Phase Plots Add.
For G(s) = G1 (s)G2 (s),
∠G(ıω) = ∠G1 (ıω) + ∠G2 (ıω)
M. Peet
Lecture 19: Control Systems
17 / 30
Bode Plots
Approach
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(ıω)| =
X
20 log |ıω − zi | +
X
i
i
X
20 log |ıω − zi | −
X
=
i
And for phase:
∠G(ıω) =
20 log
1
|ıω − pi |
20 log |ıω − pi |
i
X
i
∠(ıω − zi ) −
X
∠(ıω − pi )
i
But how to plot ∠(ıω − zi ) and 20 log |ıω − zi |?
M. Peet
Lecture 19: Control Systems
18 / 30
Plotting Simple Terms
The Constant
Before rushing in, lets make sure we don’t forget the constant term. If
G(s) = c
( zs1 − 1) · · · ( zsm − 1)
( ps1 − 1) · · · ( psn − 1)
Magnitude: G1 (s) = c
• |G1 (ıω)| = |c|
• 20 log |G1 (ıω)| = 20 log |c|
Bode Diagram
10
5
0
Magnitude (dB)
-5
20 log |c|
-10
-15
-20
-25
-30
-35
Frequency (log ω)
Conclusion: Magnitude is Constant for all ω
M. Peet
Lecture 19: Control Systems
19 / 30
Plotting Simple Terms
The Constant
Phase: G1 (s) = c
(
0◦
∠G1 (ıω) = ∠c =
180◦
c>0
c<0
225
180
c<0
135
Phase (deg)
90
45
0
c>0
-45
-90
-135
-180
-225
10
-2
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
Conclusion: phase is 0◦ if c > 0, otherwise 180◦ .
M. Peet
Lecture 19: Control Systems
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Plotting Simple Terms
A “Pure” Zero
Lets start with a zero at the origin: G1 (s) = s.
Magnitude: G1 (s) = s
Bode Diagram
• |G1 (ıω)| = |ıω| = |ω|
10
5
• 20 log |G1 (ıω)| = 20 log |ω|
0
Our x-axis is log ω.
• Plot is Linear for all ω
• Slope is +20 dB/Decade!
Magnitude (dB)
-5
-10
-15
-20
-25
-30
-35
• Need a point: ω = 1
20 log |G1 (ıω)||ω=1 = 20 log 1 = 0
• Passes through 0dB at ω = 1
ω=1
High Gain at High Frequency
• A pure zero means u0 (t)
• The faster the input, The larger
the output
M. Peet
Lecture 19: Control Systems
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Plotting Simple Terms
A “Pure” Zero: Phase
Phase: G1 (s) = s
• ∠G1 (ıω) = ∠ıω = 90◦
• Always 90◦ !
225
180
135
Phase (deg)
90
45
0
-45
-90
-135
-180
-225
10
-2
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
Always 90◦ out of phase. Why?
M. Peet
Lecture 19: Control Systems
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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: G1 (s) = sk
Bode Diagram
• |G1 (ıω)| = |ıω|k = |ω|k
10
5
0
k
20 log |G1 (ıω)| = 20 log |ω|
• Slope is +20k dB/Decade!
Need a Point
Magnitude (dB)
= 20k log |ω|
-5
-10
k=1
-15
k=4
-20
-25
k=2
-30
-35
k = 3 ω=1
• At ω = 1:
20 log |G1 (ıω)||ω=1 = 20k log 1 = 0
k pure zeros added together.
• Still Passes through 0dB at ω = 1
M. Peet
Lecture 19: Control Systems
23 / 30
Plotting Simple Terms
A “Pure” Zero: Multiple Zeros
And phase for multiple pure zeros?
Phase: G1 (s) = sk
• ∠G1 (ıω) = ∠(ıω)k = k∠ıω = 90◦ k
• Always 90◦ k
405
k=4
360
315
k=3
Phase (deg)
270
225
k=2
180
135
k=1
90
45
0
-45
-2
10
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
k pure zeros added together.
M. Peet
Lecture 19: Control Systems
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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?
I
1
).
We did one example already ( s+1
Change of Format: to simplify steady-state response, we use
G1 (s) = (τ s + 1)
• Pole is at s = − τ1
• Also put poles in this form
Rewrite G(s): (s + p) → p( p1 s + 1).
Where
(s + z1 ) · · · (s + zm )
G(s) = k
(s + p1 ) · · · (s + pn )
1
1
z1 · · · zm ( z1 s + 1) · · · ( zm s + 1)
=k
p1 · · · pn ( p11 s + 1) · · · ( p1n s + 1)
=c
(τz1 s + 1) · · · (τzm s + 1)
(τp1 s + 1) · · · (τpn s + 1)
M. Peet
1
zi
τpi = p1i
m
c = k zp11···z
···pn
• τzi =
•
•
Assume zi and pi are Real.
Lecture 19: Control Systems
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Plotting Simple Terms
Plotting Normal Zeros
G(s) = c
(τz1 s + 1) · · · (τzm s + 1)
(τp1 s + 1) · · · (τpn s + 1)
The advantage of this form is that steady-state response to a step is
yss = lim G(s) = G(0) = c
s→0
Phase (deg)
0
-45
-90
10
-2
10
-1
0
10
Frequency (rad/sec)
10
1
10
2
Low Frequency Response is given by the constant term, c.
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Lecture 19: Control Systems
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Plotting Simple Terms
Plotting Normal Zeros
G1 (s) = (τ s + 1)
p
|G1 (ıω)| = |ıωτ + 1| = 1 + τ 2 ω 2
Im(s)
}
______
√1+ω2 τ2
Magnitude:
1
20 log |G1 (ıω)| = 20 log(1+ω 2 τ 2 ) 2 = 10 log(1+ω 2 τ 2 )
}
Bode Diagram
• Approximate 1 + ω 2 τ 2 ∼
=1
= 10 log 2 = 3.01
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40
35
30
Magnitude (dB)
Case 2: ωτ = 1
20 log |G(ıω)| = 10 log(1 + ω 2 τ 2 )
Re(s)
1
Case 1: ωτ << 1
20 log |G(ıω)| = 20 log(1 + ω 2 τ 2 )
∼
= 20 log 1 = 0
ωτ
25
20
15
10
ω = τ -1
5
0
-2
10
10
-1
Lecture 19: Control Systems
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10
Frequency (rad/sec)
10
1
10
2
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Bode Plots
Example
Case 3: ωτ >> 1
Bode Diagram
• Approximate 1 + ω 2 τ 2 ∼
= ω2 τ 2
= 20 log ωτ
= 20 log ω + 20 log τ
35
30
Magnitude (dB)
p
20 log |G(ıω)| = 20 log 1 + ω 2 τ 2
∼
= 10 log ω 2 τ 2
40
25
20
15
10
5
0
-2
10
-1
10
0
10
Frequency (rad/sec)
1
10
2
10
Conclusion: When ωτ >> 1,
• Plot is Linear
• Slope is +20 dB/Decade!
• inflection at ω =
M. Peet
1
τ
Lecture 19: Control Systems
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Plotting Simple Terms
Plotting Normal Zeros
Compare this to the magnitude plot of
G1 (s) = s + a
ω >> τ -1
ω << τ -1
This is why we use the format G1 (s) = τ s + 1
• We want 0dB (no gain) at low frequency.
M. Peet
Lecture 19: Control Systems
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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
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Lecture 19: Control Systems
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