bode plots overview
TRANSCRIPT
Bode Plots Overview Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plots
Examples
BodePlotGui
Rules Table
Printable
Bode plots are a very useful way to represent the gain and phase of a
system as a function of frequency. This is referred to as the frequency
domain behavior of a system. This web page attempts to demystify the
process. The various parts are more-or-less stand alone, so if you want to
skip one or more, that should not be a problem. If you are only interested in a
quick lesson on how to make Bode diagrams, go to part 4, "Making Plots." A
MatLab program to make piecewise linear Bode plots is described in part 6.
The documents are:
1. What is the frequency domain response? In other words, "What does a Bode
Plot represent?"
2. Why is the Bode representation used? There are many ways to represent
frequency response. Why are magnitude and phase plots used? This includes
some animations.
3. How are the piecewise linear asymptotic approximations derived?
4. Rules for making Bode plots. This is a quick "How to" lesson for drawing Bode
plots.
5. Some examples (1, 2, 3, 4, 5, 6) - (combined into one file).
6. BodePlotGui: A software tool for generating asymptotic Bode plots.
7. A MatLab program for making semi-logarithmic paper for drawing your own
Bode plots.
8. A table summarizing Bode rules
9. The MATLAB files discussed in these documents.
Also available is a compilation of items 1 through 8, for easy printing.
What Bode Plots Represent: The Frequency Domain
Freq Domain
Why Sine Waves?One of the most commonly used test functions for a circuit or system is
the sine wave. This is not because sine waves are a particularly common
signal. They are in fact quite rare - the transmission of electricity (a 60 Hz
sine wave in the U.S., 50 Hz in much of the rest of the world) is one example.
The reason sine waves are important is complex and involve a branch of
Mathematics called Fourier Theory. Briefly put: any signal going into a circuit
can be represented by a sum of sine waves of varying frequency and
amplitude (often an infinite sum). As a simple example, consider the graphs
shown below. The top graph shows a triangle wave. The middle graph
shows a number of sine waves of varying frequency and amplitude. The
bottom graph shows the sum of the sine waves (red) and the original triangle
wave (dotted black).
Clearly even just a few sine waves are sufficient in this case to closely
approximate the original function. Fourier states that any function (with some
very minor restrictions that won't concern us), can be represented in this way.
This is why sine waves are important. Not because they are common,
but because we can represent arbitrarily complex functions using only these
very simple function.
Determining system output given input and transfer function
Given that sine waves are important, how can we analyze the response
of a circuit or system to sinusoidal inputs? There are many ways to do this,
depending on your mathematical sophistication. Let's use a fairly basic
explanation that uses phasors. If you are unfamiliar with phasors, you can
find a description in almost any circuits or systems textbook.
Using complex impedances it is possible to find the transfer function of a
circuit. For example, the circuit below is described by the transfer function,
H(s), where s=jω.Circuit Transfer Function
Consider the case where R=1 and C=0.1. In that case:
Generally we know the input Vi and want to find the output Vo. We can do
this by simple multiplication
If we have a phasor representation for the input and the transfer function, the
multiplication is simple (multiply magnitudes and add phases). Finding the
output becomes easy. Let's look at some examples:Example 1
and the transfer function evaluates to
The output is just the product of the input and the transfer function (evaluated as phasors)
Note that Vo has an amplitude of 0.95 and lags Vi by 72°. It is a phase "lag" because
the output lags, or follows, the input (the input goes up before the output so the output is
following the input).Example 2
Change input phase
Both input and output have shifted 40° from those in Example 1.Example 3
Change input frequency
Frequency has changed, magnitude of output has increased, but phase lag has
decreased (to 45°).Example 4
Cosine Input
Note: all angles are given in degrees. They should be changed to radians before evaluation
by calculator or computer.
Key Concept: Sine waves can be used to represent other functions
This document explained briefly why sine waves are important and how to find
the output of a system given a sinusoidal input (i.e., represent input and transfer
function as phasors and multiply to determine output as a phasor).
Why Use Magnitude and Phase Plots? Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Rules Table
Printable
The previous document made the case that the study of sinusoidal inputs
is important, and showed how phasor representations of the input and the
system transfer function can be used to easily determine system output. This
document will explain why the standard way of representing the transfer
function is with two plots: magnitude vs. frequency and phase vs. frequency.
At the bottom of this page there is an animation to help develop an intuitive
understanding of the concepts describe in this page.
The difficulty in representing the transfer function comes about because
we need to plot a complex number, H(s) or H(jω), as a function of frequency.
Consider the transfer function
To graph this, the most straightforward way (with a computer) might be to plot
the value of H(s) as the frequency changes. This yields the blue line in the
three-dimensional plot shown below.
It would obviously be hard to get accurate information about the real and
imaginary parts of H(s) from such a plot. It is easier if we plot the real and
imaginary parts as a function of frequency (the red and green projections of
the blue line). Clearly, in this case, two 2-dimensional graphs (one for real
and one for imaginary) are superior to a single 3-dimensional graph.
However, in the last document we showed that to easily determine the
output given the input, we would like to have the transfer function in phasor
notation. This means that we should make a plot of magnitude and phase.
Again, we could make a single 3-dimensional plot (the blue line), but it would
be easier to interpret the results if we make two 2-dimensional plots (the
magenta and cyan lines).
To clarify further, lets make separate plots of the magnitude and phase.
Note: Standard Bode plots are logarithmic on the frequency axis, and plot the magnitude in
dB's (deciBels). We'll explore that in the next installment.
Consider the examples from the previous document.
Example 1
and the transfer function evaluates to
The output is just the product of the input and the transfer function (evaluated as phasors)
Note that Vo has an amplitude of 0.95 and lags Vi by 72°.Example 2
Change input phase
Both input and output have shifted 40° from those in Example 1.Example 3
Change input frequency
Frequency has changed, magnitude of output has increased, but phase lag has
decreased (to 45°).Example 4
Cosine input (i.e, a change of phase).
The magnitude and phase plots determine the phasor representation of
the transfer function at any frequency. On the graphs below we can see that
at 10 rad/sec the phasor representation of the transfer function has a
magnitude of 0.707 and a phase of -45°. This means that at 10 rad/sec the
magnitude of the output will be 0.707 times the magnitude of the input and the
output will lag the input by 45°.
Key Concept: The frequency response is shown with two plots, one for magnitude and one for phase.
The frequency response of a system is presented as two graphs: one showing
magnitude and one showing phase. The phasor representation of the transfer
function can then be easily determined at any frequency. The magnitude of the
output is the magnitude of the phasor representation of the transfer function (at a
given frequency) multiplied by the magnitude of the input. The phase of the output is
the phase of the transfer function added to the phase of the input.
An animationTo get a more intuitive idea of what the frequency response represents,
consider the system below. (Hit start button to show animation)
For an animation of an analogous electrical system, go here.umbk
2.5-2.5uy
Animation by Ames Bielenberg
The transfer function of the system is given by (with m=1, b=0.5, k=1.6,
u=input to system, y=output (the position of the mass):
You can see by the animation that at low frequencies the input and output are
equal, and in phase. At intermediate frequencies the system is somewhat
resonant, and the output actually gets larger than the input (but there is a
growing phase lag). As frequency increases further, the output decreases.
The outline of the peaks of the output plot is similar to the magnitude plots
above (the phase plot is not obvious, but it obviously starts at 0° and then
decreases - if you type "ctrl +" you can zoom in to see the phase shift). In this
case, the magnitude plot would start at one (output=input) at low frequencies,
it would then increase, followed by a decrease.
Bode plots (introduced next) formalize a particular method for drawing
magnitude and phase plots (as a function of frequency) associated with a
given transfer function.
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College
The Asymptotic Bode Diagram: Derivation of Approximations
Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Rules Table
Printable
Contents: Asymptotic approximation for...
A Constant
Real Pole
Real Zero
Pole at Origin
Zero at Origin
Complex Pole
Complex Zero
IntroductionGiven a transfer function, such as
the question naturally arises: "How can we display this function?" In
the previous document the argument was made that the most useful way to
display this function is with two plots, the first showing the magnitude of the
transfer function and the second showing its phase. One way to do this is by
simply entering many values for the frequency, calculating the magnitude and
phase at each frequency and displaying them. This is what a computer would
naturally do. For example if you use MATLAB® and enter the commands
>> MySys=tf(100*[1 1],[1 110 1000])
Transfer function:
100 s + 100
------------------------------
s^2 + 110 s + 1000
>> bode(MySys)
you get a plot like the one shown below. The asymptotic solution is given
elsewhere.
However, there are reasons to develop a method for drawing Bode
diagrams manually. By drawing the plots by hand you develop an
understanding about how the locations of poles and zeros effect the shape of
the plots. With this knowledge you can predict how a system behaves in the
frequency domain by simply examining its transfer function. On the other
hand, if you know the shape of transfer function that you want, you can use
your knowledge of Bode diagrams to generate the transfer function.
The first task when drawing a Bode diagram by hand is to rewrite the
transfer function so that all the poles and zeros are written in the form
(1+s/ω0). The reasons for this will become apparent when deriving the rules
for a real pole. A derivation will be done using the transfer function from
above, but it is also possible to do a more generic derivation. Let's rewrite the
transfer function from above.
Now lets examine how we can easily draw the magnitude and phase of this
function when s=jω.
First note that this expression is made up of four terms, a constant (0.1),
a zero (at s=-1), and two poles (at s=-10 and s=-100). We can rewrite the
function (with s=jω) as four individual phasors.
We will show (below) that drawing the magnitude and phase of each
individual phasor is fairly straightforward. The difficulty lies in trying to draw
the magnitude and phase of H(jω). We can write H(jω) as a single phasor:
Drawing the phase is fairly simple. We can draw each phase term separately,
and then simply add them. The magnitude term is not so straightforward
because of the fact that the magnitude terms are multiplied, it would be much
easier if they were added - then we could draw each term on a graph and
just add them. A method for doing this is outlined below.
The Magnitude PlotOne way to transform multiplication into addition is by using the
logarithm. Instead of using a simple logarithm, we will use a deciBel (named
for Alexander Graham Bell). (Note: Why the deciBel?) The relationship between a
quantity, Q, and its deciBel representation, X, is given by:
So if Q=100 then X=40; Q=0.01 gives X=-40; X=3 gives Q=1.41; and so on.
If we represent the magnitude of H(s) in deciBels we get
The advantage of using deciBels (and of writing poles and zeros in the form
(1+s/ω0)) are now revealed. The fact that the deciBel is a logarithmic term
transforms the multiplication of the individual terms to additions. Another
benefit is apparent in the last line that reveals just two types of terms, a
constant term and terms of the form 20log10(|1+jω/ω0|). Plotting the constant
term is trivial, however the other terms are not so straightforward. These plots
will be discussed below. However, once these plots are drawn for the
individual terms, they can simply be added together to get a plot for H(s).
The Phase PlotIf we look at the phase of the transfer function, we see much the same
thing: The phase plot is easy to draw if we take our lead from the magnitude
plot. First note that the transfer function is made up of four terms. If we want
Again there are just two types of terms, a constant term and terms of the form
(1+jω/ω0). Plotting the constant term is trivial; the other terms are
discussed below.
A more generic derivationThe discussion above dealt with only a single transfer function. Another
derivation that is more general, but a little more complicated mathematically
is here.
Making a Bode DiagramFollowing the discussion above, the way to make a Bode Diagram is to
split the function up into its constituent parts, plot the magnitude and phase of
each part, and then add them up. The following gives a derivation of the plots
for each type of constituent part. Examples, including rules for making the
plots follow in the next document, which is more of a "How to" description of
Bode diagrams.
A Constant TermConsider a constant term,
MagnitudeClearly the magnitude is constant
PhaseThe phase is also constant. If K is positive, the phase is 0° (or any even
multiple of 180°). If K is negative the phase is -180°, or any odd multiple of
180°. We will use -180° because that is what MATLAB® uses. Expressed in
radians we can say that if K is positive the phase is 0 radians, if K is negative
the phase is -π radians.
Example: Bode Plot of Gain Term
Key Concept: Bode Plot of Gain Term
For a constant term, the magnitude plot is a straight line.
The phase plot is also a straight line, either at 0° (for a positive constant) or ±180°
(for a negative constant).
A Real PoleConsider a simple real pole
The frequency ω0 is called the break frequency, the corner frequency or the 3
dB frequency (more on this last name later).
MagnitudeThe magnitude is given by
Let's consider three cases for the value of the frequency:
Case 1) ω<<ω0. This is the low frequency case. We can write an
approximation for the magnitude of the transfer function
The low frequency approximation is shown in blue on the diagram below.
Case 2) ω>>ω0. This is the high frequency case. We can write an
approximation for the magnitude of the transfer function
The high frequency approximation is at shown in green on the diagram
below. It is a straight line with a slope of -20 dB/decade going through the
break frequency at 0 dB. That is, for every factor of 10 increase in frequency,
the magnitude drops by 20 dB.
Case 3) ω=ω0. The break frequency. At this frequency
This point is shown as a red circle on the diagram.
To draw a piecewise linear approximation, use the low frequency
asymptote up to the break frequency, and the high frequency asymptote
thereafter.
The resulting asymptotic approximation is shown highlighted in pink. The
maximum error between the asymptotic approximation and the exact
magnitude function occurs at the break frequency and is approximately 3 dB.
The rule for drawing the piecewise linear approximation for a real pole
can be stated thus:
For a simple real pole the piecewise linear asymptotic Bode plot for
magnitude is at 0 dB until the break frequency and then drops at 20 dB per
decade (i.e., the slope is -20 dB/decade).
PhaseThe phase of a single real pole is given by is given by
Let us again consider three cases for the value of the frequency:
Case 1) ω<<ω0. This is the low frequency case. At these frequencies
We can write an approximation for the phase of the transfer function
The low frequency approximation is shown in blue on the diagram below.
Case 2) ω>>ω0. This is the high frequency case. We can write an
approximation for the phase of the transfer function
The high frequency approximation is at shown in green on the diagram
below. It is a straight line with a slope at -90°.
Case 3) ω=ω0. The break frequency. At this frequency
This point is shown as a red circle on the diagram.
A piecewise linear approximation is not as easy in this case because the
high and low frequency asymptotes don't intersect. Instead we use a rule that
follows the exact function fairly closely, but is also arbitrary. Its main
advantage is that it is easy to remember. The rule can be stated as
Follow the low frequency asymptote until one tenth the break frequency
(0.1 ω0) then decrease linearly to meet the high frequency asymptote at ten
times the break frequency (10 ω0). This line is shown above. Note that there
is no error at the break frequency and about 5.7° of error at one tenth and ten
times the break frequency.Example 1: Real Pole
The first example is a simple pole at 10 radians per second. The low frequency asymptote is the dashed blue line, the exact function is the solid black line, the cyan line represents 0.
Example 2: Repeated Real Pole
The second example shows a double pole at 30 radians per second. Note that the
slope of the asymptote is -40 dB/decade and the phase goes from 0 to -180°.
Key Concept: Bode Plot for Real Pole
For a simple real pole the piecewise linear asymptotic Bode plot for magnitude is at 0
dB until the break frequency and then drops at 20 dB per decade (i.e., the slope is -
20 dB/decade). An nth order pole has a slope of -20·n dB/decade.
The phase plot is at 0° until one tenth the break frequency and then drops linearly to
-90° at ten times the break frequency. An nthorder pole drops to -90°·n.
A Real ZeroThe piecewise linear approximation for a zero is much like that for a pole
Consider a simple zero:
MagnitudeThe development of the magnitude plot for a zero follows that for a pole.
Refer to the previous section for details. The magnitude of the zero is given
by
Again there are three cases:
1. At low frequencies, ω<<ω0, the gain is approximately zero.2. At high frequencies, ω>>ω0, the gain increases at 20 dB/decade and goes through
the break frequency at 0 dB.3. At the break frequency, ω=ω0, the gain is about 3 dB.
The rule for drawing the piecewise linear approximation for a real zero
can be stated thus:
For a simple real zero the piecewise linear asymptotic Bode plot for
magnitude is at 0 dB until the break frequency and then increases at 20 dB
per decade (i.e., the slope is +20 dB/decade).
PhaseThe phase of a simple zero is given by:
The phase of a single real zero also has three cases:
1. At low frequencies, ω<<ω0, the phase is approximately zero.2. At high frequencies, ω>>ω0, the phase is 90°.3. At the break frequency, ω=ω0, the phase is 45°.
The rule for drawing the phase plot can be stated thus:
Follow the low frequency asymptote until one tenth the break frequency
(0.1 ω0) then increase linearly to meet the high frequency asymptote at ten
times the break frequency (10 ω0).Examples
This example shows a simple zero at 30 radians per second. The low frequency
asymptote is the dashed blue line, the exact function is the solid black line, the cyan line
represents 0.
Key Concept: Bode Plot of Real Zero:
For a simple real zero the piecewise linear asymptotic Bode plot for magnitude is at 0
dB until the break frequency and then rises at +20 dB per decade (i.e., the slope is
+20 dB/decade). An nth order zero has a slope of +20·n dB/decade.
The phase plot is at 0° until one tenth the break frequency and then rises linearly to
+90° at ten times the break frequency. An nthorder zero rises to +90°·n.
A Pole at the OriginA pole at the origin is easily drawn exactly. Consider
MagnitudeThe magnitude is given by
This function is represented by a straight line on a Bode plot with a slope of -
20 dB per decade and going through 0 dB at 1 rad/ sec. It also goes through
20 dB at 0.1 rad/sec, -20 dB at 10 rad/sec...
The rule for drawing the magnitude for a pole at the origin can be thus:
For a pole at the origin draw a line with a slope of -20 dB/decade that
goes through 0 dB at 1 rad/sec.
PhaseThe phase of a simple zero is given by:
The rule for drawing the phase plot for a pole at the origin an be stated thus:
The phase for a pole at the origin is -90°.Example: Real Pole at Origin
This example shows a simple pole at the origin. The black line is the Bode plot, the
cyan line indicates a zero reference (dB or °).
Key Concept: Bode Plot for Pole at Origin
For a simple pole at the origin draw a straight line with a slope of -20 dB per decade
and going through 0 dB at 1 rad/ sec. An nthorder pole has a slope of -20·n
dB/decade.
The phase plot is at -90°°. An nth order pole is at -90°·n.
A Zero at the OriginA zero at the origin is just like a pole at the origin but the magnitude
increases, and the phase is positive.
Key Concept: Bode Plot for Zero at Origin
For a simple zero at the origin draw a straight line with a slope of +20 dB per decade
and going through 0 dB at 1 rad/ sec. An nthorder zero has a slope of +20·n
dB/decade.
The phase plot is at +90°°. An nth order zero is at +90°·n.
A Complex Conjugate Pair of PolesThe magnitude and phase plots of a complex conjugate (underdamped)
pair of poles is more complicated than those for a simple pole. Consider the
transfer function:
MagnitudeThe magnitude is given by
Let's consider three cases for the value of the frequency:
Case 1) ω<<ω0. This is the low frequency case. We can write an
approximation for the magnitude of the transfer function
The low frequency approximation is shown in red on the diagram below.
Case 2) ω>>ω0. This is the high frequency case. We can write an
approximation for the magnitude of the transfer function
The high frequency approximation is at shown in green on the diagram
below. It is a straight line with a slope of -40 dB/decade going through the
break frequency at 0 dB. That is, for every factor of 10 increase in frequency,
the magnitude drops by 40 dB.
Case 3) ω≈ω0. It can be shown that a peak occurs in the magnitude plot
near the break frequency. The derivation of the approximate amplitude and
location of the peak are given here. We make the approximation that a peak
exists only when
0<ζ<0.5
and that the peak occurs at ω0 with height 1/(2·ζ).
To draw a piecewise linear approximation, use the low frequency
asymptote up to the break frequency, and the high frequency asymptote
thereafter. If ζ<0.5, then draw a peak of amplitude 1/(2·ζ) Draw a smooth
curve between the low and high frequency asymptote that goes through the
peak value.
As an example For the curve shown below,
The peak will have an amplitude of 5.00 or 14 dB.
The resulting asymptotic approximation is shown as a black dotted line, the
exact response is a black solid line.
The rule for drawing the piecewise linear approximation for a complex
conjugate pair of poles can be stated thus:
For the magnitude plot of complex conjugate poles draw a 0 dB at low
frequencies, go through a peak of height,
.
at the break frequency and then drop at 40 dB per decade (i.e., the slope is -
40 dB/decade). The high frequency asymptote goes through the break
frequency.
PhaseThe phase of a complex conjugate pole is given by is given by
Let us again consider three cases for the value of the frequency:
Case 1) ω<<ω0. This is the low frequency case. At these frequencies
We can write an approximation for the phase of the transfer function
The low frequency approximation is shown in red on the diagram below.
Case 2) ω>>ω0. This is the high frequency case. We can write an
approximation for the phase of the transfer function
The high frequency approximation is at shown in green on the diagram
below. It is a straight line at -180°.
Case 3) ω=ω0. The break frequency. At this frequency
The asymptotic approximation is shown below, followed by an
explanation
A piecewise linear approximation is not easy in this case, and there are
no hard and fast rules for drawing it. The most common way is to look up a
graph in a textbook with a chart that shows phase plots for many values of ζ.
Three asymptotic approximations are given here. We will use the
approximation that connects the the low frequency asymptote to the high
frequency asymptote starting at
and ending at
If ζ<0.02, the approximation can be simply a vertical line at the break
frequency.
The rule for drawing phase of an underdamped pair of poles can be
stated as
Follow the low frequency asymptote at 0° until
then decrease linearly to meet the high frequency asymptote at -180° at
Key Concept: Bode Plot for Complex Conjugate Poles
For the magnitude plot of complex conjugate poles draw a 0 dB at low frequencies,
go through a peak of height,
.
at the break frequency and then drop at 40 dB per decade (i.e., the slope is -40
dB/decade). The high frequency asymptote goes through the break frequency. Note
that the peak only exists for
0 < ζ < 0.5
To draw the phase plot simply follow low frequency asymptote at 0° until
then decrease linearly to meet the high frequency asymptote at -180° at
If ζ<0.02, the approximation can be simply a vertical line at the break frequency.
Other magnitude and phase approximations are given here.
A Complex Conjugate Pair of ZerosNot surprisingly a complex pair of zeros yields results similar to that for a
complex pair of poles. The differences are that the magnitude has a dip
instead of a peak, the magnitude increases above the break frequency and
the phase increases rather than decreasing.Example: Complex Conjugate Zero
The graph below corresponds to a complex conjugate zero with
The dip in the magnitude plot will have a magnitude of 0.2 or -14 dB.
Key Concept: Bode Plot of Complex Conjugate Zeros
For the magnitude plot of complex conjugate zeros draw a 0 dB at low frequencies,
go through a dip of magnitude:
.
at the break frequency and then rise at +40 dB per decade (i.e., the slope is +40
dB/decade). The high frequency asymptote goes through the break frequency. Note
that the peak only exists for
0 < ζ < 0.5
To draw the phase plot simply follow low frequency asymptote at 0° until
then increase linearly to meet the high frequency asymptote at 180° at
Other magnitude and phase approximations are given here.
Brief review of page: This document derived piecewise linear
approximations that can be used to draw different elements of a Bode
diagram. A synopsis of these rules can be found in a separate document.
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College
Rules for Constructing Bode Diagrams Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Rules Table
Printable
This document will discuss how to actually draw Bode diagrams. It
consists mostly of examples.
Key Concept - To draw Bode diagram there are four steps:
1. Rewrite the transfer function in proper form.2. Separate the transfer function into its constituent parts.3. Draw the Bode diagram for each part.4. Draw the overall Bode diagram by adding up the results from part 3.
1. Rewrite the transfer function in proper form.A transfer function is normally of the form:
As discussed in the previous document, we would like to rewrite this so
the lowest order term in the numerator and denominator are both unity.
Some examples will clarify:Example 1
Note that the final result has the lowest (zero) order power of numerator and
denominator polynomial equal to unity.Example 2
Note that in this example, the lowest power in the numerator was 1.Example 3
In this example the denominator was already factored. In cases like this, each
factored term needs to have unity as the lowest order power of s (zero in this case).
2. Separate the transfer function into its constituent parts.The next step is to split up the function into its constituent parts. There
are seven types of parts:
1. A constant2. Poles at the origin3. Zeros at the origin4. Real Poles5. Real Zeros6. Complex conjugate poles7. Complex conjugate zeros
We can use the examples above to demonstrate again.
Example 1
This function has
a constant of 6,
a zero at s=-10,
and complex conjugate poles at the roots of s2+3s+50.
The complex conjugate poles are at s=-1.5 ± j6.9 (where j=sqrt(-1)). A more
common (and useful for our purposes) way to express this is to use the standard notation
for a second order polynomial
In this case
Example 2
This function has
a constant of 3,
a zero at the origin,
and complex conjugate poles at the roots of s2+3s+50, in other words
Example 3
This function has
a constant of 2,
a zero at s=-10, and
poles at s=-3 and s=-50.
3. Draw the Bode diagram for each part.The rules for drawing the Bode diagram for each part are summarized
on a separate page. Examples of each are given later.
4. Draw the overall Bode diagram by adding up the results from step 3.
After the individual terms are drawn, it is a simple matter to add them
together. See examples, below.
Examples: Draw Bode Diagrams for the following transfer functions
These examples are compiled on the next page.Example 1
A simple pole
Full SolutionExample 2
Multiple poles and zeros
Full SolutionExample 3
A pole at the origin and poles and zeros
Full SolutionExample 4
Repeated poles, a zero at the origin, and a negative constant
Full SolutionExample 5
Complex conjugate poles
Full SolutionExample 6
A complicated function
Full Solution
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College
Bode Plot Examples
Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Rules Table
Printable
Several examples of the construction of Bode Plots are included in this
file. Click on the transfer function in the table below to jump to that example.Examples (Click on Transfer Function)
1
(a real pole)
2
(real poles and zeros)
3
(pole at origin)
4
(repeated real poles,negative constant)
5
(complex conj. poles)
6
(multiple poles atorigin,
complexconj zeros)
7
(time delay)
Bode Plot: Example 1Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 0 polynomial, the denominator is order 1.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 2 components:
A constant of 3.3
A pole at s=-30
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 3.3 is equal to 10.4 dB). The phase is
constant at 0 degrees.
The pole at 30 rad/sec is the blue line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency (3 rad/sec) then drops linearly down to -90 degrees at 10 times the break
frequency (300 rad/sec).
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The overall asymptotic plot is the translucent pink line, the exact
response is the black line.
Bode Plot: Example 2Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 1 polynomial, the denominator is order 2.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 4 components:
A constant of 0.1
A pole at s=-10
A pole at s=-100
A zero at s=-1
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 0.1 is equal to -20 dB). The phase is
constant at 0 degrees.
The pole at 10 rad/sec is the green line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency (1 rad/sec) then drops linearly down to -90 degrees at 10 times the break
frequency (100 rad/sec).
The pole at 100 rad/sec is the blue line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency (10 rad/sec) then drops linearly down to -90 degrees at 10 times the break
frequency (1000 rad/sec).
The zero at 1 rad/sec is the red line. It is 0 dB up to the break frequency, then rises
at 20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (0.1 rad/sec)
then rises linearly to 90 degrees at 10 times the break frequency (10 rad/sec).
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The overall asymptotic plot is the translucent pink line, the exact
response is the black line.
Bode Plot: Example 3Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 1 polynomial, the denominator is order 2.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 4 components:
A constant of 33.3
A pole at s=-3
A pole at s=0
A zero at s=-10
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 33.3 is equal to 30 dB). The phase is
constant at 0 degrees.
The pole at 3 rad/sec is the green line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency (0.3 rad/sec) then drops linearly down to -90 degrees at 10 times the break
frequency (30 rad/sec).
The pole at the origin. It is a straight line with a slope of -20 dB/dec. It goes through
0 dB at 1 rad/sec. The phase is -90 degrees.
The zero at 10 rad/sec is the red line. It is 0 dB up to the break frequency, then rises
at 20 dB/dec. The phase is 0 degrees up to 1/10 the break frequency (1 rad/sec)
then rises linearly to 90 degrees at 10 times the break frequency (100 rad/sec).
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The overall asymptotic plot is the translucent pink line, the exact
response is the black line.
Bode Plot: Example 4Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 1 polynomial, the denominator is order 3.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 4 components:
A constant of -10
A pole at s=-10
A doubly repeated pole at s=-1
A zero at the origin
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 10 is equal to 20 dB). The phase is
constant at -180 degrees (constant is negative).
The pole at 10 rad/sec is the blue line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency then drops linearly down to -90 degrees at 10 times the break frequency.
The repeated pole at 1 rad/sec is the green line. It is 0 dB up to the break frequency,
then drops off with a slope of -40 dB/dec. The phase is 0 degrees up to 1/10 the
break frequency then drops linearly down to -180 degrees at 10 times the break
frequency. The magnitude and phase drop twice as steeply as those for a single
pole.
The zero at the origin is the red line. It has a slope of +20 dB/dec and goes through
0 dB at 1 rad/sec. The phase is 90 degrees.
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The overall asymptotic plot is the translucent pink line, the exact
response is the black line.
Bode Plot: Example 5Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form.Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 1 polynomial, the denominator is order 2.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 4 components:
A constant of 6
A zero at s=-10
Complex conjugate poles at the roots of s2+3s+50,
with
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 6 is equal to 15.5 dB). The phase is
constant at 0 degrees.
The zero at 10 rad/sec is the green line. It is 0 dB up to the break frequency, then
rises with a slope of +20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency then rises linearly to +90 degrees at 10 times the break frequency.
The plots for the complex conjugate poles are shown in blue. They cause a peak of:
at a frequency of
This is shown by the blue circle. The phase goes from the low frequency asymptote
(0 degrees) at
to the high frequency asymptote at
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The exact response is the black line.
Bode Plot: Example 6Draw the Bode Diagram for the transfer function:
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 2 polynomial, the denominator is order 3.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 4 components:
A constant of 1
A pole at s=-100
A repeated pole at the origin (s=0)
Complex conjugate zeros at the roots of s2+s+25,
with
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 1 is equal to 0 dB). The phase is
constant at 0 degrees.
The pole at 100 rad/sec is the green line. It is 0 dB up to the break frequency, then
falls with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency then falls linearly to -90 degrees at 10 times the break frequency.
The repeated poles at the origin are shown with the blue line. The slope is -40
dB/decade (because pole is repeated), and goes through 0 dB at 1 rad/sec. The
slope is -180 degrees (again because of double pole).
The complex zero is shown by the red line. The zeros give a dip in the magnitude
plot of
at a frequency of 5 rad/sec (because ζ is small, ωr≈ω0). This is shown by the red
circle. The phase goes from the low frequency asymptote (0 degrees) at
to the high frequency asymptote at
Again, because ζ is so small, this line is close to vertical.
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The exact response is the black line.
Bode Plot: Example 7Draw the Bode Diagram for the transfer function:
This is the same as "Example 1," but has a 0.01 second time delay. We have
not seen a time delay before this, but we can easily handle it as we would any
other constituent part of the transfer function. The magnitude and phase of a
time delay are described here.
Step 1: Rewrite the transfer function in proper form. Make both the lowest order term in the numerator and denominator
unity. The numerator is an order 0 polynomial, the denominator is order 1.
Step 2: Separate the transfer function into its constituent parts.
The transfer function has 3 components:
A constant of 3.3
A pole at s=-30
A time delay of 0.01 seconds (magnitude and phase of time delay described here).
Step 3: Draw the Bode diagram for each part.This is done in the diagram below.
The constant is the cyan line (A quantity of 3.3 is equal to 10.4 dB). The phase is
constant at 0 degrees.
The pole at 30 rad/sec is the blue line. It is 0 dB up to the break frequency, then
drops off with a slope of -20 dB/dec. The phase is 0 degrees up to 1/10 the break
frequency (3 rad/sec) then drops linearly down to -90 degrees at 10 times the break
frequency (300 rad/sec).
The time delay is the red line. It is 0 dB at all frequencies. The phase of the time
delayis given by -0.01·ω rad, or -0.01·ω·180/π° (at ω=100 rad/sec, the phase is -
0.01·100·180/π≈-30°). There is no asymptotic approximation for the phase of a time
delay. Though the equation for the phase is linear with frequency, it looks
exponential on the graph because the horizontal axis is logarithmic.
Step 4: Draw the overall Bode diagram by adding up the results from step 3.
The exact response is the black line.
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College
BodePlotGui: A Tool for Generating Asymptotic Bode Diagrams
Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Rules Table
Printable
BodePlotGui is a graphical user interface written in the MATLAB®
programming language. It takes a transfer function and splits it into its
constituent elements, then draws the piecewise linear asymptotic
approximation for each element. It is hoped that the BodePlotGui program
will be a versatile program for teaching and learning the construction of Bode
diagrams from piecewise linear approximations.
Files for the program are found here
Use of program.A Simple Example.
Consider the transfer function:
This function has three terms to be considered when constructing a Bode diagram, a
constant (100), a pole at ω=10 rad/sec, and a zero at the origin. The following MATLAB®
commands begin execution of the GUI:
>>MySys=tf(1000*[1 0],[1 10]); %define Xfer function
>>BodePlotGui(MySys) %Invoke GUI
The GUI generates a window as shown below.
Starting in the upper left and going counterclockwise, the windows show:
1. The magnitude plot, both the piecewise linear approximation for all three terms as well as the asymptotic plot for the complete transfer function and the exact Bode diagram for magnitude. Also shown is a zero reference line.
2. The phase plot.3. A list of the systems in the user workspace.4. Several checkboxes that let the user format the image. In particular there is a check-
box that determines whether or not to display the asymptotic plot for the complete transfer function; sometimes it gets in the way of seeing the other plots, so you may want to hide it.
5. Buttons for help and a web resource (the web resource are the Bode Plot pages you are currently looking at).
6. The legend identifying individual terms on the plot.7. A box that shows elements excluded from the plot. This box is empty in this display
because the diagram displays all three elements of the transfer function.8. A 'Legend' box that shows elements displayed in the plot.9. Several check-boxes that allow the user to display how the plots are displayed
Also shown in the upper right hand corner is the transfer function, H(s).
Modifying what is displayedThe function displayed can be manipulated term by term to illustrate the
effect of each term. For example, the zero at the origin can be excluded
simply by clicking on it in the lower left hand box. The figure below shows the
result.
Note that the zero at the origin is no longer included in the plot. Each
term can be likewise included or excluded by simply clicking on it.
The next plot shows the plot modified to have thicker lines, a grid, phase
in radians and with the asymptotic plot of the complete transfer function. In
the previous graph, the phase of the asymptotic plot obscured that of the real
pole; this is an example when it might be convenient not to show the
asymptotic approximation.
Underdamped termsUnderdamped poles (and zeros) present a difficulty because they cause
a peak (dip) in the magnitude plot. The program show this with a simple circle
showing the peak height. For example the transfer function
yields the output shown below. The peak due to the underdamped pole
is clearly shown.
A more complicated example
The example below is more complicated. It shows underdamped terms, repeated
poles, and a pole at the origin.
Limitations of the software:The software has known limitations. It does not calculate the phase
associate with a time delay. It only decomposes transfer functions with up to
20 separate terms (a limit only of the number of colors used). Repeated poles
and zeros are treated as a single term (i.e., it is not possible to move only one pole of a repeated pair from the �Elements included in plot� window to the
�Elements excluded from plot� window of the GUI).
→Get the files →
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College
Rules for Drawing Bode Diagrams Overview
Freq Domain
Why Bode?
Asymptotic plots
Making Plot
Examples
BodePlotGui
Summary
Printable
The table below summarizes what to do for each type of term in a Bode
Plot. This is also available as a Word Document or PDF
Term Magnitude Phase
Constant: K 20log10(|K|)K>0: 0° K<0: ±180°
Pole at Origin
(Integrator)
-20 dB/decade passing through 0 dB at ω=1
-90°
Zero at Origin
(Differentiator)
+20 dB/decade passing through 0 dB at ω=1
(Mirror image of Integrator about 0 dB)
+90° (Mirror image of Integrator
about 0°)
Real Pole1. Draw low frequency
asymptote at 0 dB2. Draw high frequency
asymptote at -20 dB/decade
3. Connect lines at ω0.
1. Draw low frequency asymptote at 0°
2. Draw high frequency asymptote at -90°
3. Connect with a straight line from 0.1·ω0 to 10·ω0
Real Zero 1. Draw low frequency asymptote at 0 dB
2. Draw high frequency asymptote at +20
1. Draw low frequency asymptote at 0°
2. Draw high frequency asymptote at +90°
dB/decade3. Connect lines at ω0.
(Mirror image of Real Pole about 0 dB)
3. Connect with a straight line from 0.1·ω0 to 10·ω0
(Mirror image of Real Pole about 0°)
Underdamped Poles
(Complex
conjugate poles)
1. Draw low frequency asymptote at 0 dB
2. Draw high frequency asymptote at -40 dB/decade
3. If ζ<0.5, then draw peak at ω0 with amplitude |H(jω0)|=-20·log10(2ζ)
4. Connect lines
1. Draw low frequency asymptote at 0°
2. Draw high frequency asymptote at -180°
3. Connect with a straight line from
You can also look in a
textbook for examples
Underdamped Zeros
(Complex
conjugate zeros)
1. Draw low frequency asymptote at 0 dB
2. Draw high frequency asymptote at +40 dB/decade
3. If ζ<0.5, then draw peak at ω0 with amplitude |H(jω0)|=+20·log10(2ζ)
4. Connect lines
(Mirror image of Underdamped Pole about 0 dB)
1. Draw low frequency asymptote at 0°
2. Draw high frequency asymptote at +180°
3. Connect with a straight line from
You can also look in a
textbook for examples.
(Mirror image of Underdamped Pole about
0°)
For multiple order poles and zeros, simply multiply the slope of the
magnitude plot by the order of the pole (or zero) and multiply the high and low
frequency asymptotes of the phase by the order of the system.
For example:Second Order Real
Pole1. Draw low frequency
asymptote at 0 dB2. Draw high frequency
asymptote at -40 dB/decade3. Connect lines at break
frequency.
-40 db/dec is used because of
order of pole=2. For a third order
1. Draw low frequency asymptote at 0°
2. Draw high frequency asymptote at -180°
3. Connect with a straight line from 0.1·ω0 to 10·ω0
-180° is used because order of
pole=2. For a third order pole,
pole, asymptote is -60 db/dechigh frequency asymptote is at -
270°.
References
© Copyright 2005-2012 Erik Cheever This page may be freely used for educational purposes.
Comments? Questions? Suggestions? Corrections?Erik Cheever Department of Engineering Swarthmore College