Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Bode Phase Plots
Outline of Today’s Lecture
 Review
 Frequency Response
 Reading the Bode Plot
 Computing Logarithms of |G(s)|
 Bode Magnitude Plot Construction
 Phase
 Phase Computations
 Full Bode Plot
 System Identification
 Using Bode Plots for System Identification
Frequency Response
General form of linear time invariant (LTI) system was previously expressed as
(n)
s
 n 1
 n 2 
m
 m 1
 m 2 
y  a1 y  a2 y  ...  an 2 y  an 1 y  an y  b0 u  b1 u  b2 u  ...  bn 2u  bn 1u  bnu
n
 a1s n 1  a2 s n 2  ...  an 2 s 2  an 1 2  an  y0e st   b0 s m  b1s m1  b2 s m2  ...  bn 2 s 2  bn 1s  bn  e st
b s bs


s  a s
m
y (t )  y (0)e
st
0
m 1
1
n 1
n
1
 b2 s m2  ...  bn 2 s 2  bn 1s  bn 
 a2 s
n 2
 ...  an 2 s  an 1 2  an 
2
e st
The transfer function form is then
b0 s m  b1s m1  b2 s m2  ...  bn 2 s 2  bn 1s  bn b( s )
y ( t )  G ( s )u ( t )  G ( s )  n

s  a1s n 1  a2 s n 2  ...  an 2 s 2  an 1 2  an
a(s)
We now want to examine the case where the input is sinusoidal. The
response of the system is termed its frequency response.
s  i  u(t )  eit  cos  i   i sin  i 
y (t )  G (i )eit  Me
i  t
where M  G (i ) is the magnitude or gain of G  i  and
 Im  G  i   
  tan 
 is the phase angle or argument of G  i 
 Re  G  i   


1
Reading the Bode Plot
Amplifies
Attenuates
Input
Response
The magnitude is in decibels
decade
also, cycle
Note: The scale for  is logarithmic
What is a decibel?
 The decibel (dB) is a logarithmic unit that indicates the ratio of a
physical quantity relative to a specified or implied reference level.
A ratio in decibels is ten times the logarithm to base 10 of the
ratio of two power quantities.
(IEEE Standard 100 Dictionary of IEEE Standards Terms, Seventh Edition, The Institute of Electrical and
Electronics Engineering, New York, 2000; ISBN 0-7381-2601-2; page 288)
Because decibels is traditionally used measure of power, the decibel
value of a magnitude, M, is expressed as 20 Log10(M)
 20 Log10(1)=0 … implies there is neither amplification or
attenuation
 20 Log10(100)= 40 decibels … implies that the signal has been
amplified 100 times from its original value
 20 Log10(0.01)= -40 decibels … implies that the signal has
been attenuated to 1/100 of its original value
Computing Logarithms of G(s)
20Log10 K  constant
Since this does not vary with the frequency it a constant that
will be added to all of the other factors when combined and
has the effect of moving the complete plot up or down
20 pLog10 i  20 pLog10
When this is plotted on a semilog graph ( the abscissa) this
is a straight line with a slope of 20p (p is negative if the sp term
is in the denominator of G(s)) … without out any other terms it
would pass through the point (w,MdB) = (1,0)
Computing Logarithms of G(s)
20 Log10 i  a  20 Log10  2  a 2

a  20 Log10  2  a 2  20 Log10a  constant

a  20 Log10  2  a 2  20 Log10
a is called the break frequency for this factor
For frequencies of less than a rad/sec, this is plotted as a
horizontal line at the level of 20Log10 a,
For frequencies greater than a rad/sec, this is plotted as a line
with a slope of ± 20 dB/decade, the sign determined by
position in G(s)
Computing Logarithms of G(s)
20 Log10  i   2n  i   n  20 Log10
2

n

n
2

n
2


2 2
 4 2n 2 2
20 Log10
n 2   2   4 2n 2 2  20Log10n 2  40Log10n  constant
20 Log10

2
n
2


2 2
 4 2n 2 2  20 Log10 2  40 Log10
wn is called the break frequency for this factor
For frequencies of less than wn rad/sec, this is plotted as a
horizontal line at the level of 40Log10 wn,
For frequencies greater than wn rad/sec, this is plotted as a
line with a slope of ± 40 dB/decade, the sign determined by
position in G(s)
Bode Plot Construction
 When constructing Bode plots, there is no need to draw the
curves for each factor: this was done to show you where the
parts came from.
 The best way to construct a Bode plot is to first make a table
of the critical frequencies and record that action to be taken
at that frequency.
 You want to start at least one decade below the smallest
break frequency and end at least one decade above the last
break frequency. This will determine how semilog cycles you
will need for the graph paper.
 This will be shown by the following example.
Example
 Plot the Bode magnitude plot of
( s  0.2)( s 2  2s  25)
G  s   10
s( s  3)( s 2  4s  16)
Break
Frequency
Factor
Effect
Cum
value
Cum
Slope dB/dec
0.01
K=10
20Log10(10)=20
20
…
0.01
s
Line -20db/dec
Thru (1,0)
20-slope for two
decades (40) =60
-20
0.2
s+0.2
+20Log10(.2)=
-13.98
60+6.02=46.02
0
3
s+3
-20Log10(3)=
-9.54
46.02-9.54=
36.48
-20
4
s2+4s+16
-40Log10(4)=
- 24.08
36.48- 24.08=
12.4
-60
5
s2+2s+25
+40Log10(5)=
27.96
32.4+27.96=
40.36
-20
Example
Constructed Bode
Actual Bode
Phase
s  i  u(t )  eit  cos  i   i sin  i 
y (t )  G (i )eit  Me
i  t
where M  G (i ) is the magnitude or gain of G  i  and
 Im  G  i   
  tan 
 is the phase angle or argument of G  i 
 Re  G  i   


1
 For a sinusoidal input, phase represents the lag of the system or,
alternatively, the processing time of the system to produce an output
from the input
 Phase is measured as an angle




A cycle of the input is consider to take 2p radians or 360 degrees
Phase is the angular distance it takes for the output to represent the input
Thus it is normal that as the frequency increases that the phase also increase
In the case where the phase exceeds 180 degrees, the output appears to
“lead” the input. This is particularly evident in the range of 270 to 360
degrees.
Phase
 Im  G  i   
G ( s)    tan 

 Re  G  i   


1
 As with magnitude there are 4 factors to consider which can be
added together for the total phase angle.
 We will consider, in turn,
K
s n
( s  a )
( s 2  2n s  n 2 )
 The sign will be positive if the factor is in the numerator and
negative if the factor is in the denominator
Phase Computations
K is the angle of a real number; therefore it is always 0 (or 180 if the number were negative)
sn  (in ) n
If n is an even number, then i n will either be +1 or -1, both real numbers
n
an odd number, i n will be -1: sn  180  the angle of complete turns (n 4 )
2
n
for an even number, i n will be +1: sn  0  the angle of complete turns (n 4 )
2
If n is an odd number, then i n will either be  i or - i, both pure imaginary numbers
for
n 1
an odd number, i n will be  i: s n  90  the angle of complete turns (n 4 )
2
n 1
for
an even number, i n will be - i: s n  270  the angle of complete turns (n 4 )
2
1
Examples: s  90 s 1    90 s 2  -180
s
s 3 = -270 s 5  90  360  450
for
Phase Computations

 
( s  a )  (i  a )  arctan  
a
This now involves the variable, , which is plotted on the semilog scale
Where 
 
a, arctan    0
a
Where 
 
a, arctan    900
a
Where
a
 
   10a, arctan    a line from 0 to 90
10
a
 2  
( s 2  2n s  n 2 )  (n 2   2   i 2n )  arctan  2 n 2 
 n   
This now also involves the variable, 
Where 
Where 
Where
 
n , arctan    0
a
 
n , arctan    1800
a
a
 
   10a, arctan    a line from 0 to 180 with a correction for 
10
a
Example
 Plot the Bode Phase Plot for
G( s ) 
14
s2  s2  s  9
s 2  180 determines the starting phase angle
  s2  s  9 ,
n  9  3
2n  1    0.1667
The slope will be -90 per decade as the term is in the denominator
for a total transition of -180 or starting at -180 and ending at -360
 3

Starting from  , 180  , draw a line to  30, 360 
 10

Then, because the damping ratio is small, sketch in a correction
using the chart as a guideline
Example
 Plot the Bode Phase Plot for
 Again a table is useful:
n

( s  0.2)( s 2  2s  25)
G  s   10
s( s  3)( s 2  4s  16)
Frequency
Factor
Effect
Cum Value Cum Slope
0
s-1
constant -90
0.02
s+0.2
break up at 45/dec
45/dec
0.3
s+3
Break down at -45/dec
0
0.4
( s 2  4s  16)
4
0.5 Break down at -90/dec
0.5
( s 2  2s  25)
5
0.2
-90
-
-90/dec
Break up at 90/dec
0
2
s+0.2
take out 45/dec
-45/dec
30
s+3
take out -45/dec
0
40
( s2  4s  16)
take out -90/dec
90/dec
take out 90/dec
0
50
( s 2  2s  25)
total exponent
3-4=-1 >> -90
-90
Example
Asymptotic Bode Phase
Actual Bode Phase
Full Bode Plot
Constructed Bode
Actual Bode
Asymptotic Bode Phase
Actual Bode Phase
Matlab Command bode(sys)
System Identification
 It is not unusual for a field engineer to be shown a piece of
equipment and then asked if he can put a control system on it or
replace the control system for which there are no parts.
 The task of determining how an unknown structure responds is
called “System Identification”.
 To identify a system, there are many tools are your disposal
 First and foremost, what should the system structure look like?
 Motors are often first order transfer functions (G( s)  K ) which you then
sa
attempt to identify the constants
 Perform step tests and see what the response looks like
 Perform tests with sinusoidal outputs and use the Bode plot to
identify the system
 Apply statistical/time series methods such as ARMAX and RELS
Using Bode Plots for
System Identification
 The overall order of the system will be the high frequency phase divided by 90 degrees
 The exponent of the “s” term will be the slope on the magnitude plot at the lowest frequency
divided by 20
 Alternatively, the exponent of “s” is the lowest frequency phase divided by 90 degrees.
 The system gain constant (Kt) in dB will be the height value at the extension of the “s” term
line on the magnitude plot to where it crosses1 rps
 Starting from the left (the lowest frequency) on the magnitude plot, determine the structural
components using the change in slopes in increments of 20 degrees either up or down
 Then by using the intersection of the lines at those places match to the test curve, determine
the break frequencies
 Write the transfer function in the form
  s2
s
 s
  s
2 z1s   s 2
2 s 

1

1
...

1

 1 
 2  1 ...

b
 b
 b
2
2
nz1
nz 2
  m
  nz1
  nz 2

G( s)  Kt  1   2
2
2
 s
2 s   s
2 s 
 s
 s
  s
s p   1   1 ...   1 
 p1  1  
 p 2  1 ...
2
2

  np1
  np 2

np1
np 2
 a1   a2
  aq



Example
-40 dB/dec
Kt=35 dB
-60 dB/dec
20 dB/dec
40 dB/dec
60 dB/dec
80 dB/dec
-40 dB/dec
-80 dB/dec
-40 dB/dec
0.52
2.2
4
20
exp of s = -2
overall order =3
Example
-40 dB/dec
-40 dB/dec
35
Log10  K t  
20
K t  10
G ( s )test
35
20
Kt=35 dB
-60 dB/dec
20 Log10  Kt   35
-80 dB/dec
 56.23
2
2 z s 
 s
 s

1
 1

 2 
20
 2.2   20

 56.23
2
2 s 
 s
 s
s2 
 1  2  p  1
4
 0.52   4

-40 dB/dec
0.52
Total order of G( s) test  (1  2)  (2  1  2)  2
2.2
4
20
Doesn't match phase!
There is a mistake.!
exp of s = -2
overall order =-3
Example
-40 dB/dec
Kt=35 dB
-60 dB/dec
20 Log10  K t   35
Log10  K t  
K t  10
35
20
-40 dB/dec
35
20
-80 dB/dec
-60 dB/dec
-40 dB/dec
 56.23
 s
 s

 1   1

 2.2   20 
G ( s )test  56.23
2
2 s 
s
 s
2
s 
 1  2  p  1
4
 0.52   4
 0.52
Total order of G ( s )test  (1  1)  (2  1  2)  3
 is fairly large  
4
20
0.6
exp of s = s -2
 2.2  s  20 
s 2  s  0.52   s 2  4.8s  16 
G ( s )test  10.63
2.2
overall order =-3
Example
 s  2.2  s  20
s  s  0.52   s 2  4.8s  16
 s  3 s  9 
G( s )actual  25 2
s  s  0.4   s 2  4s  16
G( s )test  10.63
2
Summary
 Phase
 For a sinusoidal input, phase represents the lag of the system or,
alternatively, the processing time of the system to produce an
output from the input
 Phase Computations
K
 Full Bode Plot
s n
( s  a )
 System Identification
( s 2  2n s  n 2 )
 The task of determining how an unknown structure responds is
called “System Identification”.
 Using Bode Plots for System Identification
Next Class: Laplace Transforms