1. Bode Magnitude Plots Constructed Bode Actual Bode
Professor Walter W. Olson
Department of Mechanical, Industrial and Manufacturing Engineering
University of Toledo
Bode Magnitude Plots

2. 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
Reading the Bode Plot
Computing Logarithms of |G(s)|
Bode Magnitude Plot Construction

3. 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
The characteristic equation is
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

4. 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
The characteristic equation is
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

5. The effect of placement on the root locus
Imaginary axis
Real Axis jw s
jwd wn
s = -zwn
sin-1(z)
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

6. The effect of placement on the root locus
Imaginary axis
Real Axis jw s
jwd wn
s = -zwn
sin-1(z)
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

7. Frequency Response
General form of linear time invariant (LTI) system was previously expressed as
We now want to examine the case where the input is sinusoidal. The
response of the system is termed its frequency response.

8. 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

9. Reading the Bode Plot Amplifies Attenuates Input Response
Note: The scale for w is logarithmic
The magnitude is in decibels
Amplifies
Attenuates
decade
also, cycle
Input
Response

10. 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 ; 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

11. 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.

12. 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
Knowing how many cycles needed, divide the sketch area into these regions

13. Sketching Semilog Paper
0.1
1.0 10
100
Frequency w rps
For this example assume

14. 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
1.0 10
100
Frequency w rps
0.3 3 30 2 5 7
0.5
0.2
0.7 20 50 70

15. Computing Logarithms of |G(s)|
Therefore, in plotting the magnitude portion of the Bode plot, we can
compute each term separately and then add them up for the result

16. Computing Logarithms of G(s)

17. Computing Logarithms of G(s)
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
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

18. 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
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

19. Computing Logarithms of G(s)
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)

20. + Example Assume we have the transfer function
To compute the magnitude part of the Bode plot +

21. Example Bode plot of Constructed Plot Actual Bode Magnitude Plot
Our plotting technique produces an “asymptotic Bode Plot”

22. Computing Logarithms of G(s)
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)

23. Example + Construct a Bode magnitude plot of Note: there are two
lines here!

24. Example
Actual Bode Magnitude
Asymptotic Bode Magnitude

25. 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
For quadratic terms
pertains to a phase correction which will
be discussed next class

26. 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.

27. 20-slope for two decades (40) =60
Example
Plot the Bode magnitude plot of
Break
Frequency
Factor
Effect
Gain Cum
value
Cum
Slope dB/dec
0.01
K=10
20Log10(10)=20 20 … s
Line -20db/dec
Thru (1,0)
20-slope for two decades (40) =60
-20
0.2
s+0.2
+20Log10(.2)=
-13.98
=46.02 3
s+3
-20Log10(3)=
-9.54 =
36.48 4
s2+4s+16
-40Log10(4)= =
12.4
-60 5
s2+2s+25
+40Log10(5)=
27.96 =
40.36

28. Example
Actual Bode
Constructed Bode

29. Note: This form has certain advantages:
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

30. Summary Frequency Response Reading the Bode Plot
Computing Logarithms of |G(s)|
Bode Magnitude Plot Construction
Next: Bode Phase Plots