Bode' Plots
Why Bode' Plots?
What are Bode' Plots?
First Order Systems
Decibels
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 is a Bode' plot?
• How is magnitude plotted? (dbs)
• How is phase plotted? (degrees)
• How is frequency plotted?  (on a logarithmic scale)
• Given a Transfer Function:
• Be able to plot the Bode' plot, manually or with a math analysis program.
• Know that the Bode' plot you generated "makes sense".
• Given a Bode Plot for a System:
• Determine the transfer function of thesystem represented by the Bode' plot.

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.
• Given a Transfer Function:
• Be able to plot the Bode' plot, manually or with a math analysis program.

• Know that the Bode' plot you generated "makes sense".
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)

Problem 2

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

Problem 3

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

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

• Any transfer function is a ratio of polynomials - and those polynomials have real coefficients.
• Polynomials with real coefficients have real roots - first order factors - and complex conjugate pairs of roots - second order factors.
• Our discussion of this first order system model is really only addressing systems with one pole - one real root - in the denominator.
• More interesting systems have second order factors.
We'll start with second order factors in the denominator, i.e. second order poles.  We're not done with Bode' plots. Remember:

• Our Bode' plots so far were all plotted with a log scale on the vertical (gain) axis. Decibels are more often used and you need to learn about them.
• Second order systems have interesting Bode' plots - and it is important to know about them. Click here to look at Bode' plots for second order systems.

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:

Po/Pi

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.

• Every decade increase causes the same decrease in dbs.
• Actually, every octave increase causes equal decreases in dbs.
• The slope appears to be -20 db/decade.
Check that this is the slope for any decade, from 1000 to 10,000 or from 3000 to 30,000 Hz.
• Not so obviously, the slope could be expressed as -6 db/octave.
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)

• If w is large (and only if it is large!), then the imaginary term in the denominator dominates, and we have:
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:

• Given a Transfer Function:
• Be able to plot the Bode' plot, manually or with a math analysis program.

• Know that the Bode' plot you generated "makes sense".
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:

• A little reflection will probably tell you some things.
• For example, this system could have two complex roots.
• It's not obvious, but to have two complex roots, the only thing necessary is that the damping ratio, z, be less than one.
Here's a Bode' plot for a second order system. This system has the following parameters:
• z- the damping ratio = 0.1
• wn- the undamped natural frequency = 1000.
• Gdc- the DC gain of the system = 1.0.

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.

• Substitute s = jw, to get the frequency response.
• For small w, the gain is just Gdc.
• For large w, the gain is Gdc/w2.
That means that the high frequency gain drops off at -40 db/decade.
• There are intermediate frequencies where interesting things happen!
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.
• We will examine what happens when w = wn.
• At the natural frequency, the (jw)2 term becomes -wn2, cancelling out the last term in the denominator, the wn2 term, since j2 = -1.
• Now, the really interesting things start to happen. When those terms cancel the denominator just has one term left, and we have:

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

• The only term that involves the damping ratio is the one left in the denominator when w = wn.
• The damping ratio is in the denominator, so the smaller the damping ratio, the larger the frequency response is going to be.
• At w = wn, the magnitude of the frequency response function is:
• or   G(jwn) =Gdc/j2z

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

• It depends only upon the DC gain and the damping ratio, and, the smaller the damping ratio, the higher the gain at the natural frequency.
Now, recall the other important behavior at low frequencies and high frequencies.
• For small w, the gain is just Gdc.
• For large w, the gain Gdc/w2.
• For small w, the gain is just Gdc, assuming Gdc = 10 (or 20 db) on the plot.
• For large w, the gain is Gdc/w2, - dropping off at -40 db/decade.
Here we assume that the natural frequency is fn = 20.
• And, we can insert the point at the resonant frequency, using our formula.
• G(jwn) =Gdc/j2z
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?
• The peak is well above either of the asymptotes at the natural frequency.
• We should believe all of the math we've done.
• Is there really a problem here? Should we look at the actual frequency response? Here it is. There's  the peak. It does exist.

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

• Gdc= 10,
• z = 0.1,
• wn = 2 p 20, (since that natural frequency was 20 Hz.)
With these paramters, note the following in the plot.
• The DC gain is 20 db which corresponds to a gain of 10.
• The resonant peak is pretty much right at 20 Hz as it should be.
• The resonant peak is about 13 or 14 db high.
• A gain of 50 would be 14 db, do that also checks!
• The high frequency slope looks to be around -12 db/octave or -20 db/decade.
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.

• Gdc = 1.0.
• Natural frequency = 159 Hz.
• Damping ratio - variable and controllable by user.
What should we note about the second order system response in the movie?
• There is a resonant peak in the second order system response.
• The size of the resonant peak depends upon the damping ratio.
• For damping ratios less than about 0.5 the peak is relatively insignificant.
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:
• A damping ratio of 0.1
• An undamped natural frequency of 159 Hz. (1000 rad/sec.)

Notice the following for this plot.

• The phase starts at zero degrees for low frequencies.
• The phase asymptotically approaches -180o for high frequencies.
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:

• Gdc = 1.0.
• Natural frequency = 159 Hz.
• Damping ratio - variable.
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.

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.

• If we wanted to sketch this Bode' plot we could start by looking at the DC gain.
• Remember that the DC gain is just G(jw) with w = 0.
• Letting w = 0 in G(jw), we get:
• G(0) = 10. (20db)
• At low frequencies, the (.002s + 1) term in the denominator will still look pretty much like 1.0.
• However, as we go up in frequency, the (.01s + 1) term will have an effect.
• The (.01s + 1) term introduces a corner frequency which we discussed earlier in the section on Bode' plots for first order systems.
• The corner frequency is at:
• f = 100/2p = 15.9Hz.

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

• The (.002s + 1) term will add another -20db/decade slope to the plot, for a total of -40 db.decade.
• We get -40 db/decade because we now have two poles contributing to the roll-off, and 2*(-20db/dec) = -40 db/dec.
• The second corner frequency is at f = 500/2p = 79.5Hz.

• The straight line approximation is high at the corners, but gives a pretty good idea of where the actual Bode' plot lies.
Now, let us make this slightly more complicated.  Here's another transfer function.

Start by looking at the DC gain - as before.
• Remember that the DC gain is just G(jw) with w = 0.
• Letting w = 0 in G(jw), we get:
• G(0) = 10. (20db)
• As we go up in frequency from DC, the (.01s + 1) term will have an effect.
• The (.01s + 1) term introduces a corner frequency - as before.
• The corner frequency is at f = 100/2p = 15.9Hz.
• Check the slope. It should be -20 db/decade.
• At slightly higher frequencies, the (.002s + 1) term will start to have an effect.
• The (.002s + 1) term will add another -20db/decade - or wait a minute - is that +20 db/decade?
• Because it is a zero, it is +20db/dec, and the corner frequency is at:
• f = 500/2p = 79.5Hz.
• For frequencies above 79.5 Hz, the gain would be 10*.002/.01 = 2 or 6db.
And don't forget we still have one more corner frequency. so let's add the last corner frequency.
• We have another corner frequency at:
• f = (1/.0001)/2p = 1590Hz. - Call that 1600 Hz.
• Above 400 Hz, we have another -20 db/decade added, but the total will now be -20 db/decade.