The basic implementation of the time-domain Schroedinger equation into the FDTD formulation

4.4 Bode PlotsDecibels

Before launching into this section, we will consider the unit of decibels, which is widely used in engineering. The amplitude of a frequency domain function can be written in decibels by

.

The phase of is expressed the same as before.

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Convert to decibels (db):

i.

ii.

iii.

iv.

Solution

i.

ii.

iii.

iv.

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Bode PlotsSuppose we have a system with the transfer function

(4.4.1)This is a one-pole, low pass filter. I know that for any frequency, we will be most interested in the magnitude and phase. However, instead of just magnitude, it will prove to be advantageous to look at the answer in decibels, which is

.What will that give us? When ,

and the logarithm of one zero. What happens when omega is much greater than?

.and

,e.g., at

,

,i.e., it drops at the rate of 20 dB per decade. What about right at ?

Now consider the phase. We calculate this by

When , . When , . The other times, we have to calculate it. A Bode plot can be generated by the MATLAB commands:

g = tf( [ 1000],[1 1000 ])

bode(g)

This is shown in Fig. 4.4.1a. The following MATLAB program offers better control of the figures:% My_bode.mw = (10^1:2:10^5);

WW = log10(w);

% A one-pole LPRnum = [1e3];

den = [1 1e3];

g = tf (num,den)

H = freqs(num,den,w);

% Bode plots

subplot(2,1,1)

semilogx(w,20*log10(abs(H)),'k')

axis ( [ 10^2 10^5 -40 5 ])

set(gca,'fontsize',14)

ylabel('|H| db')

grid on

title('My-bode')

subplot(2,1,2)

semilogx(w,(180/pi)*angle(H),'k');

axis ( [ 10^1 10^5 -100 10 ])

set(gca,'YTick',[ -180 -90 -45 0 45 90 ])

set(gca,'fontsize',14)

xlabel('\omega (rad/sec)');

ylabel('/_ H (degrees)')

grid onThe results of this program are shown in Fig. 4.4.1.b.

(a) Bode plot of Eq. (4.4.1) using the MATLAB command bode.

(a) Bode plot of Eq. (4.4.1) using my_bode.m

Figure 4.4.1. Bode plots of the one-pole low pass filter of Eq. (4.4.1).

If you are constructing a bode plot by hand, first draw the asymptotes, i.e, 0 between , and minus 20 dB per decade for . At , it must be -3 db.Then connect the lines smoothly. To draw the phase, it is zero for and 90 degrees for . At , it is 45 degrees. So draw the curve that shows this.

Now suppose the transfer function were

. (4.4.2)

This is clearly a two pole low pass filter. We could go through the same procedure, and find that above, the magnitude decreased at 40 db per decade, and at it is 6db. The phase ranges from 0 to -180 degrees, and right at it is -90 degrees. This is shown in Fig. 4.4.2.

Figure 4.4.2. Bode plot of the two pole low pass filter in Eq. (4.4.2).

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Example

This example demonstrates one of the primary reasons for the Bode plots. Suppose the input to the two-pole low pass filter of Eq. (4.4.2) is

.(4.4.3)What comes out? SolutionTo determine this, we need only go to the Bode plot and find the corresponding frequency of rad/sec, as shown in Fig. 4.4.3. The magnitude is at about -14 dB. This corresponds to

or

Similarly, we read a phase shift of -125 degrees. So we would say the output is

(4.4.4)Check

Figure 4.4.3. Using the Bode plot to determine attenuation and phase shift of Eq. (4.4.2) at rad/sec.

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Suppose I had

.(4.4.5)

The zero at means it will start rising at a rate of 20 db/decade. Note that above is looks like a differentiator.

Figure 4.4.4. Bode plot of Eq. (4.4.5).

Now let us look at the filter described by

.(4.4.6)Equation (4.4.6) has a zero, the s term in the numerator. What does this transfer function look like when:

.This looks like a differentiator for low frequencies. Remember that for both Laplace and Fourier transforms, or means the Fourier (Laplace) transform of the derivative of x(t). It looks like a differentiator until it gets near the first pole at . To find out what happens, lets see what it is at:

.In other words, it has little frequency dependence, at least until it starts approaching . Above , what happens?

.

Now I would say it has started to look like an integrator. The Bode plot is given in Fig. 4.4.5.

Figure 4.4.5. Bode plot of the band pass filter in Eq. (4.4.6).

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Example

Suppose the following function, T = 10 ms, is put through the filter described by Eq. (4.4.6). How many terms do I really need to describe the output?

SolutionWe found earlier that the Fourier series is

For

Figure 4.4.6. Bode plot of Eq. (4.4.6) with the frequencies of the example marked at the corresponding position.

Clearly the dc term will pass undisturbed. At n = 1, the filter attenuates about 2 db or

At n= 5, its about -11 db,

Note also, that the magnitudes of the input terms are attenuated as the frequency increases. By n=5, the ration of amplitudes between the n=1 term and the n= 5 term is

So you will probably not be making a significant error by only using the first five terms. {Actually,.}

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Example The following function goes through a filter with the following Bode plot. What comes out?

.

Figure 4.4.8. Bode plot of a Filter.Solution

At 2 kHz, the attenuation is about -13 db and -125 degrees.

So

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Drawing Bode Plots

In drawing the magnitude, the following rules are used in first drawing the asymptotes:

1. Each pole, i.e., a term like, will result in a negative slope of -20 db/decade once the frequency is above . A term means the plot starts with a -20 db/decade slope.

2. Each zero, i.e., a term like, will result in a positive slope of 20 db/decade once the frequency is above . A term means the plot starts with a 20 db/decade slope.

In drawing the phase plot, the following rules apply:

3. Each pole will result in a negative 90 degree phase shift one decade beyond. It has no effect a decade before , and right at it results in a -45 degree phase shift.4. Each zero will result in a positive 90 degree phase shift one decade beyond. It has no effect a decade before , and right at it results in a 45 degree phase shift.

----------------------------------------------------------------------------------------------------------Example

Draw a Bode plot for the following transfer function

Solution

I see that there are zeros at s = 0 and and a pole at . The zero at s = 0 means it is originally going up at 20 db/decade. Then at is flattens, and at it starts up at 20 db/decade again. To get a reference point, I evaluate at

Furthermore, I see the angle

From there I can draw the asymptotes.

Figure 4.4.9. Asymptotic approximation of the bode plot for .The MATLAB commands

g=tf[1e-3* [ 1 1e4 0]],[1 1e2] ]

bode(g)

produce the following:

Figure 4.4.10. Bode plot of

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Example

Draw the Bode plot of the transfer function of the following circuit assuming the following values: L = 10 mH, C = 1 , and R = 25 .

Figure 4.4.11. An RLC circuit.Solution

I see that the denominator is second order, so I will look for the position of the two poles

Form this I see that the corresponding corner frequency is . What is H(s) at this value? (For simplicity, I will use s=j1000}

Instead of the -6 bd attenuation that we expect at the corner frequency of a two pole low pass filter, this has a gain. This gain is known as overshoot. At the resonant frequency of the circuit, this value of R is not enough to completely dampen the resonance. The Bode plot is given below.

Figure 4.4.12. Bode plot of .------------------------------------------------------------------------------------------------------------

What value of R would be enough to get the kind of damping we would expect from a two-pole LPF?

So if R were 200 , then , we would the Bode plot below:

Figure 4.4.13. The Bode plot of .

Determining the Transfer Function from a Bode Plot.

The Bode plot can be thought of as a detailed, quantitative description of a function, usually a transfer function. As we have seen, if we have the Bode plot describing the transfer function we know how a signal at any frequency will be effected. In this section, we will describe how to construct the transfer function from a Bode plot. This is often useful in determining the qualitative features of a transfer function. For instance, in Fig. 4.4.14, we can see that this is the transfer function of a band pass filter. Among other things, dc components will not pass through, and high frequency components will be attenuated. However, if the problem at hand is to determine how a signal is effected by a transfer function and the Bode plot is available, there is no reason to fist construct the transfer function from the Bode plot.

Figure 4.4.14. In the above Bode plot, at the far left side, we see that there is a 20 db/decade rise. Among other things, this says that no dc term will pass. We also see that on the far right side, there is a minus 20 db/decade slope, so frequency components at rad/sec and above will be attenuated.

First, write down the poles and zeros by observing the slopes on the magnitude plot. Then from a flat point on the slope, evaluate the magnitude. This will give the multiplying constant.

Sometimes, plots created from a transfer function with multiple poles and zeros can be evaluated with the help of the phase plot. For instance, if the plot ends at -90 degrees, it means there were two more poles than zeros. If it starts at 90 degrees, it means that there is a term .

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Example. Find the transfer function corresponding to the Bode plot in Figure 4.4.15.Solution

It this case more can be learned by looking at the angle. For the very low frequencies, the angle is -90 degrees, so there must by a factor 1/s. Then at it passes through 45 degrees on its way up, so there is a zero at , giving a factor in the numerator. Then it passes through -45 degrees on its way back down to degrees at . So far, I suspect my transfer function looks something like

,

where K is an unknown constant. To get K I have to look at the magnitude. At it appears to be -40 db, so

Figure 4.4.15. Bode plot of a transfer function.

----------------------------------------------------------------------------------------------------------Exercises

4.4.1. Draw by hand the bode plot of

.4.4.2. Repeat problem 4.4.1 using a MATLAB program.4.4.3 The pattern illustrated below repeats at 0.5 msec intervals. It is run through a filter whose transfer function is described by the Bode plot. What comes out (to within 10% accuracy)?

PAGE 194.4 Bode Plots

9/30/2011

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