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