#### Transcript Lecture 19: Bode Plots

```Constructed Bode
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
Actual Bode
University of Toledo
Bode Magnitude Plots
Outline of Today’s Lecture
 Review






Poles and Zeros
Plotting Functions with Complex Numbers
Root Locus
Plotting the Transfer Function
Effects of Pole Placement
Root Locus Factor Responses
 Frequency Response
 Computing Logarithms of |G(s)|
 Bode Magnitude Plot Construction
Root Locus
 The root locus plot for a system is based on solving the system
characteristic equation
 The transfer function of a linear, time invariant, system can be factored as a fraction
of two polynomials
 When the system is placed in a negative feedback loop the transfer function of the
closed loop system is of the form
NG ( s)
Output ( s )
KG ( s )
KN G ( s )
DG ( s )



Input ( s ) 1  KG ( s ) 1  K N G ( s ) DG ( s )  KN G ( s )
DG ( s )
K
 The characteristic equation is
1  KG( s)  0  DG ( s)  KNG ( s)
 The root locus is a plot of this solution for positive real values of K
 Because the solutions are the system modes, this is a powerful design
tool
Root Locus
 The root locus plot for a system is based on solving the system
characteristic equation
 The transfer function of a linear, time invariant, system can be factored as a fraction
of two polynomials
 When the system is placed in a negative feedback loop the transfer function of the
closed loop system is of the form
NG ( s)
Output ( s )
KG ( s )
KN G ( s )
DG ( s )



Input ( s ) 1  KG ( s ) 1  K N G ( s ) DG ( s )  KN G ( s )
DG ( s )
K
 The characteristic equation is
1  KG( s)  0  DG ( s)  KNG ( s)
 The root locus is a plot of this solution for positive real values of K
 Because the solutions are the system modes, this is a powerful design
tool
The effect of placement
on the root locus jw
Imaginary axis
jwd
Real Axis
s  -zwn
s
• The magnitude of the vector to
pole location is the natural frequency
of the response, wn
The vertical component (the imaginary
part) is the damped frequency, wd
The angle away from the vertical is the
inverse sine of the damping ratio, z
The effect of placement
on the root locus jw
Imaginary axis
jwd
Real Axis
s  -zwn
s
• The magnitude of the vector to
pole location is the natural frequency
of the response, wn
The vertical component (the imaginary
part) is the damped frequency, wd
The angle away from the vertical is the
inverse sine of the damping ratio, z
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  iw  u(t )  eiwt  cos  iw   i sin  iw 
y (t )  G (iw )eiwt  Me
iw  t
where M  G (iw ) is the magnitude or gain of G  iw  and
 Im  G  iw   
  tan 
 is the phase angle or argument of G  iw 
 Re  G  iw   


-1
Frequency Response
 The response to the input is amplified by M and time shifted by
the phase angle.
 To represent this response we need two curves:
 one for the magnitude at any frequency and
 one for phase shift
 These curves when plotted are called the Bode Plot of the system
Amplifies
Attenuates
Input
Response
The magnitude is in decibels
also, cycle
Note: The scale for w 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
Note
 The book does not plot the Magnitude of the Bode Plot in
decibels.
 Therefore, you will get different results than the book where
decibels are required.
 Matlab uses decibels.
Sketching Semilog Paper
 First, determine how many cycles you will need:
 Ideally, you will need at least one cycle below the smallest zero
or pole and
 At least one cycle above the largest pole or zero
 Example
G ( s) 
s  20
 you want to plot from 0.02 to 200
s  s  .2  s  5 
the cycles needed will start with 0.01, 0.1, 1, 10, 100 or 5 cycles
 Knowing how many cycles needed, divide the sketch area into
these regions
Sketching Semilog Paper
 For this example assume
0.1
G ( s) 
1
 3 cycles
( s  1)( s  2)
1.0
10
Frequency w rps
100
Sketching Semilog Paper
 For each cycle,
 Estimate the 1/3 point and label this 2 x the starting point for the cycle
 Estimate the midway point and label this 3 * the starting point for the cycle
 At the 2/3 point, label this 5 x * the starting point for the cycle
 Half way between 3 and 5 place a tic representing 4
 Halfway between 5 and 10 place a tic and label this 7
 Halfway between 5 and 7, place a tic for 6
 Divide the space between 7 and 10 into thirds and label these 8 and 9
 These are not exact but useful for sketching purposes
0.1
0.2 0.3 0.5 0.7
1.0
2
3
5
7
10
Frequency w rps
20 30 50 70
100
Computing Logarithms of |G(s)|
s  iw  u(t )  eiwt  cos  iw   i sin  iw 
y (t )  G (iw )eiwt  Me
iw  t
where M  G (iw ) is the magnitude or gain of G  iw 
G( s) 
 s  b1  s  b2  ...  s  bm  where p  q  n
b( s )
K p
a( s)
s  s  a1  s  a2  ...  s  aq 
Complex numbers will only occur as pair(s) of complex conjugates
If the roots are complex conjugates, we should recombine these in the form
 s  s  i  s  s - i   s 2  2s s  s 2   2  s 2  2zwn s  wn 2
Because we will compute the Log Magnitude of G (iw ) in dB, we must compute 20Log10 G (iw )
Then 20Log10 G (iw )  20 Log10 K  20 Log10 iw  b1  20 Log10 iw  b2  ...  20 Log10 iw  bm
- 20 pLog10 iw - 20 Log10 iw  a1 - 20Log10 iw  b2 - 20 Log10 iw  aq
 20 Log10  iw   2z 2wn 2  iw   wn 2 2  20 Log10  iw   2z 2wn 2  iw   wn 2 2  ...
2
2
Therefore, in plotting the magnitude portion of the Bode plot, we can
compute each term separately and then add them up for the result
Computing Logarithms of G(s)
20 Log10 G(iw )  20 Log10 K  20 Log10 iw  b1  20Log10 iw  b2  ...  20Log10 iw  bm
- 20 pLog10 iw - 20Log10 iw  a1 - 20Log10 iw  b2 - 20Log10 iw  aq
 20Log10 s 2  2z 1wn1s  wn12  20Log10  iw   2z 2wn 2  iw   wn 2 2  ...
2
There are four possible factors to compute:
20Log10 K from the gain K,
20 pLog10 iw from the s p factor of G ( s ),
20 Log10 iw  c from the ( s  a ) and ( s  b) factors of G( s), and
20 Log10  iw   2zwn  iw   wn 2 from the recombined quadratic factors of G ( s )
2
The sign will be + if the term is in the numerator and - if in the denominator of
 s  b1  s  b2  ...  s  bm   s 2  2z 1wnz1s  wnz12  s 2  2z z 2wnz 2 s  wnz 2 2  ...
G(s)=K p
s  s  a1  s  a2  ...  s  aq   s 2  2z 1wn1s  wn12  s 2  2z 2wn 2 s  wn 2 2  ...
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 iw  20 pLog10w
When this is plotted on a semilog graph (w 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)
p is often called the “type” of the system
Static Error Constants
 If the system is of type 0 at low frequencies will be level.
 A type 0 system, (that is, a system without a pole at the origin,)
will have a static position error, Kp, equal to
lim G ( jw )  K  K p
w 0
 If the system is of type 1 (a single pole at the origin) it will
have a slope of -20 dB/dec at low frequencies
 A type 1 system will have a static velocity error, Kv, equal to the
value of the -20 dB/dec line where it crosses 1 radian per second
 If the system is of type 2 ( a double pole at the origin) it will
have a slope of -40 dB/dec at low frequencies
 A type 2 system has a static acceleration error,Ka, equal to the
value of the -40 dB/dec line where it crosses 1 radian per second
Computing Logarithms of G(s)
20 Log10 iw  a  20 Log10 w 2  a 2
w
a  20 Log10 w 2  a 2  20 Log10a  constant
w
a  20 Log10 w 2  a 2  20 Log10w
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)
Example
 Assume we have the transfer function G  s   14
s4
 To compute the magnitude part of the Bode plot
K  14  20 Log10 (14)  22.92
 s  4  4
20 Log10(4)  -12.04
Plot the line with a negative slope since it is in the denominator
+
Example
 Bode plot of G  s  
14
s4
Constructed Plot
Actual Bode Magnitude Plot
Our plotting technique produces an “asymptotic Bode Plot”
Computing Logarithms of G(s)
20 Log10  iw   2zwn  iw   wn  20 Log10
2
w
wn
w
wn
2
w
n
2
-w

2 2
 4z 2wn 2w 2
20 Log10
wn 2 - w 2   4z 2wn 2w 2  20Log10wn 2  40Log10wn  constant
20 Log10
w
2
n
2
-w

2 2
 4z 2wn 2w 2  20 Log10w 2  40 Log10w
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)
Example
 Construct a Bode magnitude plot of G( s) 
14
s2  s2  s  9
K  14  20 Log10 (14)  22.92
s 2  indicates a line with a slope of -40 (in the denominator) passing through (1,0)
s
2
 s  9   break frequency is at wn  9  3 radians/second
the level part is at -40 Log10 3  - 19.08
then the line will break down (negative slope) at -40 dB/decade
+
Note: there are two
lines here!
Example
Actual Bode Magnitude
Asymptotic Bode Magnitude
Corrections
 You seen on the asymptotic Bode magnitude plots, there
were deviations at the break frequencies.
Note that these could be either positive of negative corrections depending
on whether or not the term is in the numerator or denominator
For (s+a) type terms
pertains to a phase correction which will
be discussed next class
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
-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
Note:
M  G (iw ) is the magnitude or gain of G  iw 
 s  b1  s  b2  ...  s  bm   s 2  2z z1wnz1s  wnz12  s 2  2z z 2wnz 2 s  wnz 2 2  ...
b( s )
G( s) 
K p
a( s)
s  s  a1  s  a2  ...  s  aq   s 2  2z p1wnp1s  wn 2  s 2  2z p 2wnp 2 s  wnp 2 2  ..
The transfer function is often put in the time response form of
  s2
s
 s
  s
2z z1s   s 2
2z 2 s 

1

1
...

1


1

 1 ...


 b
2
2
K  a1a2 ...aqwnp12wnp 2 2   b1   b2
w
w
w
w
nz1
nz 2
  m
  nz1
  nz 2

G( s) 
2
2
 b1b2 ...bmwnz1 wnz 2 ... s p  s  1  s  1 ...  s  1  s 2  2z p1s  1  s 2  2z p 2 s  1 ...
a

 a
  wnp12 wnp1
  wnp 2 2 wnp 2

 1   a2
  q



 s2
2z z1s   s 2
2z 2 s 
 1 

 1 ...
 z1s  1 z 2 s  1 ...  zm s  1  2 
2
w
w
w
w
nz1
nz 2
 nz1
  nz 2

=K
 s2
2z p1s   s 2
2z p 2 s 
p
s  p1s  1 p1s  1 ...  p1s  1 


1

 1 ...
 
2
 wnp12 wnp1

wnp 2

  wnp 2

1) the time constants of the 1st order terms can be directly read
2) When constructing Bode plots the part of the curves up to the
break frequencies are 0 (20Log101=0). The level parts have been
taken up in the constant gain, Kt
Summary
 Frequency Response
 s  b1  s  b2  ...  s  bm   s 2  2z 1wnz1s  wnz12  s 2  2z z 2wnz 2 s  wnz 22  ...
G(s)=K p
s  s  a1  s  a2  ...  s  aq   s 2  2z 1wn1s  wn12  s 2  2z 2wn 2 s  wn 2 2  ...
for s  iw
 Computing Logarithms of |G(s)|
20Log10 K  constant
20 Log10 iw  a  20 Log10
20 pLog10 iw  20 pLog10w
w2  a2
w
a  20 Log10 w 2  a 2  20 Log10a  constant
w
a  20 Log10 w 2  a 2  20 Log10w
20 Log10  iw   2zwn  iw   wn 2  20 Log10
2
wn 2 - w 2   4z 2wn 2w 2
2
w
wn
20 Log10
wn 2 - w 2   4z 2wn 2w 2  20Log10wn 2  40Log10wn  constant
w
wn
20 Log10
wn 2 - w 2   4z 2wn 2w 2  20Log10w 2  40Log10w
2
2
 Bode Magnitude Plot Construction
Next: Bode Phase Plots
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