Professor Walter W. Olson Department of Mechanical, Industrial and Manufacturing Engineering University of Toledo Bode Magnitude Plots Actual Bode Constructed.
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Slide 1
Professor Walter W. Olson Department of Mechanical, Industrial
and Manufacturing Engineering University of Toledo Bode Magnitude
Plots Actual Bode Constructed Bode
Slide 2
Outline of Todays 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
Slide 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
Slide 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
Slide 5
The effect of placement on the root locus Imaginary axis Real
Axis jj jdjd nn n sin -1 ( ) The magnitude of the vector to pole
location is the natural frequency of the response, n The vertical
component (the imaginary part) is the damped frequency, d The angle
away from the vertical is the inverse sine of the damping
ratio,
Slide 6
The effect of placement on the root locus Imaginary axis Real
Axis jj jdjd nn n sin -1 ( ) The magnitude of the vector to pole
location is the natural frequency of the response, n The vertical
component (the imaginary part) is the damped frequency, d The angle
away from the vertical is the inverse sine of the damping
ratio,
Slide 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.
Slide 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
Slide 9
Reading the Bode Plot Note: The scale for is logarithmic The
magnitude is in decibels Amplifies Attenuates decade also, cycle
Input Response
Slide 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.logarithmic unit (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)ISBN 0-7381-2601-2 Because decibels is traditionally used
measure of power, the decibel value of a magnitude, M, is expressed
as 20 Log 10 (M) 20 Log 10 (1)=0 implies there is neither
amplification or attenuation 20 Log 10 (100)= 40 decibels implies
that the signal has been amplified 100 times from its original
value 20 Log 10 (0.01)= -40 decibels implies that the signal has
been attenuated to 1/100 of its original value
Slide 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.
Slide 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
Slide 13
Sketching Semilog Paper For this example assume 0.11.010 100
Frequency rps
Slide 14
Sketching Semilog Paper 0.11.010 100 Frequency rps
0.33302570.50.20.7205070 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
Slide 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
Slide 16
Computing Logarithms of G(s)
Slide 17
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 ( the abscissa) this is a straight line with a
slope of 20p (p is negative if the s p 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
Slide 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, K
p, 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, K v, 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,K a, equal to the value of the -40 dB/dec
line where it crosses 1 radian per second
Slide 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 20Log 10 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)
Slide 20
Example Assume we have the transfer function To compute the
magnitude part of the Bode plot +
Slide 21
Example Bode plot of Our plotting technique produces an
asymptotic Bode Plot Actual Bode Magnitude Plot Constructed
Plot
Slide 22
Computing Logarithms of G(s) w n is called the break frequency
for this factor For frequencies of less than w n rad/sec, this is
plotted as a horizontal line at the level of 40Log 10 w n, For
frequencies greater than w n rad/sec, this is plotted as a line
with a slope of 40 dB/decade, the sign determined by position in
G(s)
Slide 23
Example Construct a Bode magnitude plot of Note: there are two
lines here! +
Slide 24
Example Asymptotic Bode Magnitude Actual Bode Magnitude
Slide 25
Corrections You seen on the asymptotic Bode magnitude plots,
there were deviations at the break frequencies. For (s+a) type
terms For quadratic terms pertains to a phase correction which will
be discussed next class Note that these could be either positive of
negative corrections depending on whether or not the term is in the
numerator or denominator
Slide 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.
Slide 27
Example Plot the Bode magnitude plot of Break Frequency
FactorEffectGain Cum value Cum Slope dB/dec 0.01K=1020Log 10
(10)=2020 0.01sLine -20db/dec Thru (1,0) 20-slope for two decades
(40) =60 -20 0.2s+0.2+20Log 10 (.2)= -13.98 60+6.02=46.020
3s+3-20Log 10 (3)= -9.54 46.02-9.54= 36.48 -20 4s 2 +4s+16-40Log 10
(4)= - 24.08 36.48- 24.08= 12.4 -60 5s 2 +2s+25+40Log 10 (5)= 27.96
32.4+27.96= 40.36 -20
Slide 28
Example Actual Bode Constructed Bode
Slide 29
Note: This form has certain advantages: 1) the time constants
of the 1 st order terms can be directly read 2) When constructing
Bode plots the part of the curves up to the break frequencies are 0
(20Log 10 1=0). The level parts have been taken up in the constant
gain, K t
Slide 30
Summary Frequency Response Reading the Bode Plot Computing
Logarithms of |G(s)| Bode Magnitude Plot Construction Next: Bode
Phase Plots