Bode' Plots
Why Bode' Plots?
What are Bode' Plots?
First Order Systems
Bode' Plots for
Second Order Systems
Bode Plots for Larger Systems (Examples)
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Why Bode' Plots?

        Bode plots are the most widely used means of displaying and communicating frequency response information. There are many reasons for that.

Bode' plots are really log-log plots, so they collapse a wide range of frequencies (on the horizontal axis) and a wide range of gains (on thevertical axis) into a viewable whole.

In Bode' plots, commonly encountered frequency responses have a shape that is simple. That simple shape means that laboratory measurementscan easily be discerned to have the common factors that lead to those shapes. For example, first order systems have two straight line asymptotes and if you take data and plot a Bode' plot from the data, you can pick out first order factors in a transfer function from the straight line asymptotes.

        You may have used Bode' plots without knowing it. Stereo equipment - amplifiers, speakers, microphones, headsets, etc. - often have frequency response specifications, and when you buy that kind of equipment, you may have seen a Bode' plot used to communicate frequency response specifications.

        All in all, Bode' plots are widely used, not just to specify or show a frequency response, but they also give useful information for designing control systems. Stability criteria can be interpreted on Bode' plots and there are numerous design techniques based on Bode' plots.

        You need to know how to use Bode' plots when you encounter them in those situations, so this lesson will help you to understand the basics of Bode' plots.

        What do you need to learn about Bode' plots? Here is a short summary:

What Are Bode' Plots?

        Bode' plots are:

    Plots of frequency response. Gain and phase are displayed in separate plots.
    Logarithmic plots.
    The horizontal axis is frequency - plotted on a log scale. It can be either f or w.
    The vertical axis is gain, expressed in decibels - a logarithmic measure of gain.
    Sometimes, the vertical axis is simply a gain on a logarithmic scale.
        Given these characteristics, you still need to know what a Bode' plot looks like.  Our strategy in this lesson will be to examine some simple systems - first order and second order systems - to see what Bode' plots for the frequency response of those systems look like.  We'll start with the simplest system first, and work from there.  We will end by looking at how those simple systems can be combined to make more complex systems with more complex Bode' plots.  Remember one of our goals above.         That's what we will start with for first order systems.
Bode' Plots For First Order Systems

        In this section we will work on that general goal for first order systems.  Let's look at an example Bode' plot for a first order system. Here's a plot for a sample transfer function.

        G(jw) = 1/(jwt+ 1) with t = .001

Here's the Bode' plot. Examine the following points for this plot.

The low frequency asymptote,

The high frequency asymptote,

The "mid point" where

wt = 1
That's at f = 159 Hz.
Let's look at the low frequency asymptote first. Here's the transfer function.

G(jw) = 1/(jwt+ 1)

        If w is small, then the imaginary term in the denominator is small, and we have:

G(jw) ~= 1/(j0 + 1) = 1

The low frequency behavior of the plot shows that the plot is flat at a value of 1.

Now, let's look at the high frequency asymptote. Here's the transfer function.
G(jw) = 1/(j wt+ 1)

If w is large, then the imaginary term in the denominator dominates, and we have:

G(j w) ~= 1/j wt

The magnitude of the gain is:

|G(jw)| ~= 1/wt

        The gain drops off inversely with frequency, but the Bode' plot drops off as a straight line. Hmmmm?  That's very interesting - that it is a straight line.  The straight line high frequency asymptote shouldn't be cause for consternation. If we have:

|G(w)| ~= 1/wt

        Remember that the Bode' plot is log gain vs log frequency, so let's look at the logarithm of the magnitude of the gain.

log(|G(jw)|) = log(1/wt)
= -log(wt)

= -log(w) - log(t)

        So, log gain depends linearly upon the log of frequency (w) for higher frequencies.

        That's an important point to remember, and it is also a reason Bode' plots are used so much.  When the asymptotic behavior - both at high frequencies and low frequencies - is straight line behavior, it makes Bode' plots easier to sketch and easier to understand.

        Actually, we need to note that the slope of this plot - at high frequencies - is just -1. Look again at the asymptotic high frequency relationship between the gain and frequency.

log(|G(jw)|) = -log(w) - log(t)
When frequency increases by a factor of 10, log(w) increases by 1.
Therefore, when frequency increases by a
      factor of 10, log(|G(jw)|)decreases by 1.

Therefore, when frequency increases by a
      factor of 10, |G(jw)| decreases by a factor of 10.

        From this discussion, we need to draw a conclusion.
When frequency increases by a factor of 10, |G(jw)| decreases by a factor of 10.

        Check that conclusion on the plot to be sure you understand what it means.

        Here is a plot with the lower limit extended.

Check going from f = 300 to f = 3000.

        Does the gain decrease by a factor of 10 when the frequency increases by a factor of 10?

        The last point we need to examine is the behavior of the frequency response for frequencies between high frequency and low frequency - what we referred to as the mid-point earlier. If the frequency response function is given by:

G(jw) = 1/(jwt + 1)

If w = 1/tthen (taking that frequency as the mid-point), we have:

G(jw) = 1/(j + 1)
The magnitude of the gain is:
|G(jw)| = 1/|j + 1| = 1/sqrt(2)

~= 0.707

This point is at w = 1000, or f =159Hz.
        There are some interesting things to note about this frequency response. Consider the interactive graph below. On that graph you can see the low frequency asymptote, the high frequency asymptote and the point where the gain is .707 of the low frequency gain.

Check the intersection of the two lines.

The intersection of the two lines occurs where w = 1/t.

For obvious reasons, this intersection is called the corner frequency.

Problem 1  What is the corner frequency for a system with this transfer function?

G(s) = 22/( 17 s + 1)

Enter your answer in the box below, then click the button to submit your answer.  You will get a grade on a 0 (completely wrong) to 100 (perfectly accurate answer) scale.

Your grade is:

Problem 2

        Here is a Bode' plot like the one we have been examining. Determine the corner frequency, in Hz, for this system.

Enter your answer in the box below, then click the button to submit your answer.

Your grade is:

Problem 3

        Here is another Bode' plot like the one we have been examining. Determine the corner frequency, in Hz, for this system.

Enter your answer in the box below, then click the button to submit your answer.
Your grade is:

        There's one last point to observe regarding first order systems. The general first order system has a transfer function of this form.
G(jw) = Gdc/(jwt + 1)
        The point to note is that there is a DC gain term in the numerator.  This really is the DC gain. Let the frequency, w, be zero:
G(j0) = Gdc/(j0 + 1) = Gdc
        The effect of DC gain is to raise or lower the entire plot.  You need to understand the effect of a DC gain on a Bode' plot. Let's look at the entire transfer function.
G(jw) = Gdc/(jwt + 1)

log(|G(jw)|) = log(Gdc) - log(1/((wt)2 + 1))

This really says that log(Gdc) is added at every frequency.

            Here is a movie where you can set the gain and see how the gain changes the Bode' plot.

Adding log(Gdc) at every frequency shifts the entire plot up by log(Gdc).

Phase in 1st Order Bode' Plots

        We have looked exclusively at the magnitude portion of the Bode' plots we have examined. We need to look at the phase plot as well.

The transfer function is: G(jw) = Gdc/(jwt + 1)

The phase angle at an angular frequency w is: Angle(G(jw)) = - tan-1(jwt)

The phase plot - against frequency - is important in many systems.

We will plot the phase for this transfer function- the one used earlier in this section:
G(jw) = 1/(jwt + 1) with t = .001
Note the following:
The phase starts at 0o at low frequencies.

The phase goes to -90o at high frequencies.

The phase is -45o at a frequency of 159 Hz - the corner frequency.

There are several things to note at this point
          We'll start with second order factors in the denominator, i.e. second order poles.  We're not done with Bode' plots. Remember:

Decibels And More

        When we introduced Bode' plots, we noted that the vertical scale of a Bode' plot is often in terms of decibels. It's time you got acquainted with decibels if you haven't heard of them before. Here's a start.

Originally, decibels were used to measure power gains.

If a system had an output power, Po, and an input power, Pi, then the ratio of output power to input power - the power gain - is:


The decibel gain is proportional to the logarithm - to the base ten (10) - of the power gain

The gain can be expressed as the logarithm -  to the base ten (10) - of the power gain

Gain = log10(Po/Pi)

When expressed this way, the units are bels.

A decibel is one tenth of a bel, so the gain expressed in decibels is:

Gaindb = 10 log10(Po/Pi)

The unit bel is something of a story in itself.

Alexander Graham Bell did a lot of work with the deaf, and he was recognized for his work with an honorary doctorate in 1880 by Gaulladet College in Washington, D.C.(and he also delivered the commencement address) He is more famous for his founding of the National Geographic Society, and other work he did.
Alexander Graham Bell was also honored by having a unit named in his honor - the bel.

Today, the decibel is a commonly used unit to measure sound intensity and it is well known that high decibel levels contribute to deafness - a very ironic closing of the circle.

Today, power is not so much an issue. We're more interested in voltage gain of an amplifier.  There's an interesting transition from power to voltage that will help us understand how gain - expressed in decibels - is viewed today.

In an amplifier, if the amplifier has an input resistance R1, then the power input to the amplifier is given by:

Power In = V12/R1

Similarly, the output power into a resistor Ro is given by:

Power Out = Vo2/Ro

Now, look at the ratio of output power to input power:

Power Out/Power In = (Vo2/Ro)/(Vi2/Ri)

Now, compute the decibel gain:

               Gaindb = 10 log10(Po/Pi) = 10 log10(Po/Pi)

                        = 10 log10(Vo2/Ro)/(Vi2/Ri)

                         = 10 log10(Vo2/Vi2) + 10 log10(Ri/Ro)

                         = 20 log10(Vo/Vi) + 10 log10(Ri/Ro)

The final result has a term in it that depends upon the resistors.

Gaindb = 20 log10(Vo/Vi) + 10 log10(Ri/Ro)

        Today, engineers are often more concerned with things like voltage gain. The resistances and power involved are not a concern at all when analyzing control systems, so the resistance term is ignored, and we take the gain, in db, of a system to be:

Gaindb = 20 log10(Vo/Vi)

        We should realize that we can plot gain, in db, for a system as a function of frequency.  The ratio of output voltage to input voltage is simply the ratio of output amplitude to input amplitude at some frequency - our old friend, frequency response.

        OK! You know about decibels. But there's some other things you need to know about Bode' plots. The vertical axis on a true Bode' plot is scaled in db. The horizontal axis is scaled using a logarithmic frequency scale. Here's some not-so-obvious facts about the frequency scale.

An increase of frequency by a factor of 10 is referred to as a decade. That's a fairly obvious reference. Ten years is a decade when speaking of time. Our currency is based on a decimal system because it's based on factors of 10.

An increase of frequency by a factor of 2 is referred to as an octave. We're getting into the Latin and Greek roots here. Decade is based on a Latin root - referring to the number 10.  Octave is based on a classical root referring to the number two - or is it? Right or wrong?

       Wrong! Octave refers to eight, not two. The reason a doubling of frequency is called an octave is that the musical world defined the term far earlier than we ever thought of it. An octave is a doubling of frequency, but it's eight notes in the scale to go up an octave.

        Ok, now we're going to put this all together.  Here's a Bode' plot for a first order system. It has a DC gain of 20db, and a corner frequency near f = 80 Hz. Now, look at the slope of the high frequency portion of the plot.

        Check that this is the slope for any decade, from 1000 to 10,000 or from 3000 to 30,000 Hz.         If we go back to the transfer function for a first order system, we can re-examine the high frequency behavior. Here's the transfer function.

G(jw) =  1/(jwt+ 1)

G(jw) ~= 1/jwt

|G(jw)| ~= 1/wt

log(|G(jw)|) = log(1/wt)
= -log(wt)

= -log(w) - log(t)

Express things in terms of decibels.

log(|G(jw)|) = -log(w) - log(t)

Gaindb = 20 log10(|G(jw)|) = -20 log(w) - 20 log(t)

Now, if we start with some frequency, wo, we can calculate the gain at the frequency.
Gaindb(wo) = -20 log(wo) - 20 log(t)
Now, take a frequency one decade higher, at 10wo.
Gaindb(10wo) = -20 log(10wo) - 20 log(t)
We can calculate the difference in the db gain at these two frequencies.
Gaindb(10wo) - Gaindb(wo)
= [-20 log(10wo) - 20 log(t)] - [-20 log(10wo) - 20 log(t)]
The difference is:
Gaindb(10wo) - Gaindb(wo)

= -20 log(10wo) + 20 log(wo)

= -20 log(10) - 20 log(wo) + 20 log(wo)

= -20 log(10) = -20 db - in one decade!

        Reflecting on the derivation above, we realize that this derivation says that the slope is -20 db/decade for the high frequency asymptote in the Bode' plot.  It's also possible to express that another way. If we consider two frequencies that are an octave apart, we can see that the slope can also be said to be -6db/octave.

        The difference in the frequency response between the two frequencies is:

Gaindb(2wo) - Gaindb(wo)

= -20 log(2wo) + 20 log(wo)

= - 20 log(2) - 20 log(wo) + 20 log(wo)

= -20 log(2) = -6.0206 db - in one decade - and it's usually just rounded to -6db/octave.

         It's time to leave this topic. However consider this. We've only looked at one first order system. Higher order systems - even second order systems - are bound to have some differences in their Bode' plot behavior. High frequency asymptotes will drop off at different slopes, for example, although we'll find that they drop off at integral multiples of -20db/decade or -6 db/octave.

        There are lots of interesting things you need to know, and you can start looking at second order systems now.

Bode' Plots For 2nd Order Systems

      We've looked at first order systems. Remember our general goal:

      Second order systems exhibit behavior that you will never see in a first order system. We're going to work on that goal for second order systems - systems that have this general transfer function.

       If we have this transfer function:

        Here's a Bode' plot for a second order system. This system has the following parameters:

       This system also has at least one unexpected feature - the "hump" in the frequency response between f = 100 and f = 200 - a resonant peak. It's important to understand how that peak in the frequency response comes about.  Let's look at the transfer function of a second order system. Here's a general form for such a system. Examine how that system behaves for different frequencies.

        That means that the high frequency gain drops off at -40 db/decade.         We will start by looking at the interesting things that happen at the intermediate frequencies. Here's the transfer function again, with s replaced now by jw.

       Now we can find an explanation for the hump in the frequency response.

The formula for the gain of the frequency response at w = wn is interesting because:

        Now, recall the other important behavior at low frequencies and high frequencies. Here we assume that the natural frequency is fn = 20. For this example, we'll assume z = 0.1.  Remember:
Gdc= 10, and z = 0.1,
so this works out to be a gain of 50 at the resonant peak, the equivalent of 34 db.  Do we have a problem here?

        Let's examine the parameters here again to be sure that his all hangs together.  The system parameters were:

With these paramters, note the following in the plot.         All of these observations confirm the calculations, and they really point out that it can be important to understand how the resonant peak depends upon the damping ratio.

        To make that correspondence between resonant peak and damping ratio as clear as possible, we have here an example of a frequency response for another system. We'll let you control the damping ratio, but we're going to set the DC gain and the natural frequency. Hopefully, you'll see how this peak depends upon the system's damping ratio.  Use the right and left arrow controls to step the movie a single step forward or backward.

        What should we note about the second order system response in the movie?         Finally, we have to deal with the phase. A Bode' plot isn't complete until you have the phase plot. Here's a phase plot for a system with:

Notice the following for this plot.

        How the phase plot depends upon damping ratio is something you should know. Next, we have a movie of phase shift as a function of damping ratio.

        For the system in the plot, the parameters are:

        Now, at this point you've seen Bode' plots for second order system with complex poles.  Second order systems with real poles are really combinations of two first order systems, and they will be covered in the next section.

        At this point, one direction to continue would be to continue to the next section.  However, you might want to go in the direction of looking at Nyquist plots for the systems discussed above.  In that case, use this link to go to the lesson on Nyquist plots.

Nyquist Plots

Sketching Bode' Plots For Larger Systems - Examples

        There will be times when you will need to have some sense of what a Bode' plot looks like for a larger system. A useful skill is to be able to sketch what the plot should look like so that you can anticipate what you'll get. That's particularly helpful when you have a complex system and you enter a large transfer function.  It's not only helpful. You can often gain insight by playing "What if?" games with a notepad and pencil.

        In this section, we will look at some larger systems and examine some overall properties of Bode' plots for those systems.

        We will start with a system that is not all that large - a second order system with two real poles. Just for discussion, we'll use the system with the transfer function shown below.

At slightly higher frequencies, the (.002s + 1) term will start to have an effect.

        Now, let us make this slightly more complicated.  Here's another transfer function.

Start by looking at the DC gain - as before. And don't forget we still have one more corner frequency. so let's add the last corner frequency.

  • The straight line approximation is off at the corners, but gives a pretty good idea of where the actual Bode' plot lies.
        This is a good point for some reflection.

        You should be able to sketch even higher order systems. Real poles and zeroes give rise to straight forward corner frequencies - either pole or zero factors - and you can account for them.  Problems could arise if the corner frequencies were more "bunched" than they were in the examples. We deliberately chose two examples which had some separation between corner frequencies and that allowed the Bode' plots to straighten out between corners.

        We have yet to talk about two other topics - phase plots and second order factors with low damping - i.e. with resonant peaks.  We will first look at resonant peaks on Bode' plots.

  • We'll work with an example related to the one we've done with real poles. The transfer function is shown below.
  • We start by looking at the DC gain.
  • Letting w = 0 in G(jw), we get: G(0) = 10. (20db)
  • At frequencies below 100 Hz, the quadratic factor will stay essentially 1.
  • However, as we go up in frequency, the (.01s + 1) term will have an effect,
  • We get a corner frequency from the (.01s + 1) term.
  • The corner frequency is at:
    • f = 100/2p = 15.9Hz.
  • Now, we need to consider the quadratic factor.
  • The natural frequency, wn = 1000.
  • The damping ratio is z = .05, since:
    • 2zwn = 100
  • The damping ratio predicts a resonant peak of:
    • (1/2z) = 10 (or 20 db)
    • at f = 1000/2p = 159Hz.
  • We also know that the quadratic factor is going to add -40 db/decade above the resonant peak at the natural frequency.
        Finally, you can compare what we have generated with what a detailed plot shows.

        It's amazing how it all fits together, isn't it?

  • Again, we are off at the corner frequency, but overall, the straight line gives a good idea of where the Bode' plot lies.
        There are numerous other kinds of examples we should examine. One simple example is a case in which the resonant peak is below the corner frequency. You should be able to predict what happens in that case. But just in case you can't we'll examine that situation. Here's a transfer function of that type.

Here's what you need to do.

  • Calculate the DC gain.
  • Check which critical frequency is lowest - the corner frequency or the natural frequency in the quadratic factor.
  • Think that through now before moving on.
                Now, we will show you what we got.
  • The DC gain is 10 (20db).  Here's a plot with a DC gain of 20 db.  It shows 20 db at all frequencies, and we know that needs to be fixed.
  • The natural frequency is at:
    • f = 100/2p = 15.9Hz.
  • The corner frequency is at:
    • f = 1000/2p  = 159Hz.
  • The resonant peak is a decade below the corner frequency.
  • Here is our estimate of the frequency response.
  • We've shown the asymptotes,  We also can figure the damping ratio.  We need to do that because we know there is a resonant peak.  We have the following:
    • wn2 = 10,000, so wn = 100
    • 2zwn = 10, so z = 10/200 = .05
  • Knowing the damping ratio allows us to predict how high the resonant peak will be above the straight line approximation.
    • Gain above approximation = 1 /j2z
    • If the damping ratio is .05, then this works out to a gain of 10, which is equivalent to 20 db.
    • That allows us to add one more point to our approximate Bode' plot.  Here's the plot.
Now, compare our estimate with the actual response.

From the straight line approximation you could generate a fairly accurate Bode' plot.

Summarizing this section, you should note the following.

  • You actually get a lot of insight about the shape of the plot if you can sketch it. The visualization helps a great deal.
  • There are times when a plot might surprise you, and you now have a technique that can eliminate those surprises.
  • We haven't covered all of the ground. There's more yet.

  • Problems


    To steal ideas from one person is plagiarism;
            To steal from many is research.

    These Links will take you to related files

    Second Order Quiz
    A Bode' Plot Puzzle