# frequency transfer

Discussion in 'General Electronics Chat' started by full, Nov 8, 2014.

1. ### full Thread Starter Member

May 3, 2014
225
2
hello everyone

I am study frequency transfer in AC circuit(bode plots) , I studied a lot in book and YouTube but I not understand a lot .
there is example I am understand from images 1 to 4 and in images from 5 to 8 I am not understand how I am can do the magnitude & phase table and bode plots.

please help my how I can do magnitude & phase table and bode plots in images 5,6,7,8?

there are any tutorials and notices can help my same pdf , youtube , ppt or web site ?

thanks

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2. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hello,

Which of those images do you not understand?

3. ### full Thread Starter Member

May 3, 2014
225
2
images 5,6,7,8

please if you can explain how I can do magnitude & phase table and bode plots?

thanks sir

4. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Since you say you understand how the equations in Figure 3 come about, let's start there. Figure 6 is merely a plot of the equation in Figure 3. Since the equation in Figure 3 is a sum of four terms, you can plot four separate graphs, one for each term, and then add the plots together to get Figure 6. All Figure 5 is doing is giving you a guide for how to do that, in an approximate way, for each of the terms in Figure 3. This is called the "Asymptotic Bode Plot" -- it is not exact, particularly at the points where the chart changes, but it is close enough for most design work.

So let's see where this comes from.

Consider the response for a first order low pass filter with a DC gain of 100 and a corner frequency of 10 kHz. The transfer function for this would be:

$
H(j\omega) \. = \. \frac{100}{1-j\frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)}}
$

If we want this in dB, we take 20log_10 of the magnitude to get:

$
\left| H \right| _{dB}(j\omega) \. = \. 20log_{10} \left( \left| \frac{100}{1-j\frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)}} \right| \right)
$

$
\left| H \right| _{dB}(j\omega) \. = \. 20log_{10} \left( \frac{100}{ \sqrt{ 1^2 +\left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)^2 }} \right)
$

$
\left| H \right| _{dB}(j\omega) \. = \. 20log_{10} \left( 100 \right) - 20log_{10} \left( \sqrt{ 1 +\left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)^2 } \right)
$

to get

$
\left| H \right| _{dB}(j\omega) \. = \. 40dB - 10log_{10} \left( 1 +\left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)^2 \right)
$

This is exact, but that last term is a bit hard to get our minds around. So let's consider JUST that last term (let's call it H_10kHz) and what it looks like at three different spots: at frequencies well below the cutoff frequencies, at the cutoff frequency, and at frequencies well above the cutoff frequency. This is where the notion of "asymptotic behavior" comes from.

$
\left| H_{10kHz} \right| _{dB}(j\omega) \. = \. - 10log_{10} \left( 1 +\left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)^2 \right)
$

At frequencies far below 10 kHz, the fraction is very small compared to 1 and this reduces to:

$
\left| H_{10kHz} \right| _{dB}(j\omega) \. = \. - 10log_{10} \left( 1 \right) \. = \. 0 dB
$

Next let's look at the behavior AT the corner frequency:

$
\left| H_{10kHz} \right|_{dB}(j\omega=j(2\pi \frac{rad}{cyc}) (10 kHz)) \. = \. - 10log_{10} \left( 1+1 \right) \. = \. -10log_{10}(2) \. = \. -3.01dB
$

Finally, let's look at the behavior at frequencies well above the corner frequency where the fraction is very large compared to 1 and the expression reduces to:

$
\left| H_{10kHz} \right| _{dB}(j\omega) \. = \. - 10log_{10} \left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)^2 \. = \. - 20log_{10} \left( \frac{\omega}{(2\pi \frac{rad}{cyc}) (10 kHz)} \right)
$

Now let's introduce the notion of "decades". All this means is the number of factors of ten that the numerator is greater than the denominator. So 1300 is two decades greater than 13 and one decade greater than 130 and one decade less than 13000. Hopefully you can see that this is directly related to the base-ten log of the ratio, namely log_10(1300/13) = 2 decades while log_10(1300/13000) = -1 decade. With this in mind, we see that the slop of this function is

Let's also look at the slope in this region:
$
\frac{d\left| H_{10kHz} \right| _{dB}(j\omega)}{d \omega} \. = \. -20 dB/decade
$

We can also see that the slope at frequencies well below the corner frequency is 0 dB/decade (i.e., it's flat).

With this you should be able to now see (or better yet, derive) that, for zeros of H(jw) the asymptotic behavior will be +20 dB/decade below the zero and flat above the zero while for poles it will be flat below the pole and -20 db/decade above the pole. See if that helps you understand what Figure 5 is telling you. Then see if you can apply this same thinking to the phase plots.

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5. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522

Hello again,

Take a look at image 3 and image 6 and forget about the table for now.
Note that in image 3 whenever we have a "jw" we have a number underneath it, like 2 or 10, except for the term 20 log10|jw|. However, we can look at that last term as if that 'jw' had a '1' under it, so we would have: 20 log10|jw/1|. Now every term has a number under the jw, either 1, 2, or 10.
Now look at image 6 and note that for the terms that have a 'jw' in them they all have dotted lines that cross (or leave) the 0db line (the solid horizontal line is 0db) at the same angular frequency as the number in the denominator of the terms in image 3, namely 1, 2, and 10.

So for the single zero we have, its dotted line crosses the 0db line at w=1, and it slants upward at +20db/decade.
The next term after that in image 3 has a jw/2 in it, and that line starts at w=2, and it slants down at -20db/decade.
The next term has a jw/10 in it, and that line starts at w=10, and also slants down at -20db/decade.
So we see that zeros slant up at +20db/decade and cross odb at their break frequencies, and poles slant down at -20db/decade and begin at 0db at their break frequencies.
The constant term in image 3 is 20 log10 10, which comes out to +20db, so that is shown as a horizontal line all the way across with no slant. So we see that constant gains show up as a horizontal line across the whole graph, and that is because they are not dependent on frequency like the other terms are.

Next we simply add up all the lines working from left to right, and that's how we get the red line which is the approximation to the response.
To add the lines, start at the left and add all the values together, and that gives us 0db at the very left, because the 20db constant gain adds to the single zero line at the left as +20-20=0 (the zero has value -20 at the far left side).
Since adding a slanted line to a constant line just increases the offset of the slanted line, we get the entire first part of the red line. That's the part that slants upward.
What happens next is we get to the first pole at w=2, and that is where the next part of the response kicks in. Note it is assumed that it has no effect up to that point so it had no influence on the first part of the red line. Once it hits it's break frequency however which is w=2, then it starts to affect the red line. Since it's slope is -20db/decade and the red line is +20db/decade the two add together at w=2 and come out to 0db/decade, thus the red line now turns into a 0db/decade slope which of course is a horizontal line. This would not have happened yet if the pole was at w=3 so the line would have continued up higher, but it did have a break frequency of w=2 so it happened at w=2.
So we continue on with the horizontal line until we reach the next break frequency, which is w=10. At w=10 we have another -20db/decade kick in so it adds to the red line of 0db/decade and so we get -20db/decade now, and that means the line slants down at -20db/decade starting at w=10.

So you see it is a matter of adding up the constant and the slopes, but only after those frequencies take effect, and if they are poles they only take effect when we reach their break frequencies which occur in the denominators of the terms in image 3, and if they are zeros they cross 0db at their break frequencies.. If you take it step by step like that it makes more sense and then you are only adding the next part of the response to the current response calculated up to that point.
In reality the parts of the response that dont affect the total response yet dont affect it yet because they are still at 0db. It's only when they leave 0db that they have an effect.

For a quick note that may not apply to your studies yet, complex poles have to be handled a little differently.

Last edited: Nov 8, 2014
6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I quite like teaching videos by Prof Gopal of NPTEL. You might something on the Bode plot in one of his lectures.

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7. ### full Thread Starter Member

May 3, 2014
225
2
thanks sir a lot for explain
if I do magnitude table I should do all the time in question same this:

Last edited: Nov 9, 2014
8. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hello,

Yes if you absolutely must make a table then yes you call the poles 0db any place where the frequency is less than their break frequencies.

You do the same for the phase, except the rules involve a certain phase shift for a zero and another for a pole. Can you figure out what phase shifts to use for a zero, a pole, and a constant ? It would probably be good for you to try to figure this out. Take a little while to think about it and if you still dont know we can cover that next.

9. ### full Thread Starter Member

May 3, 2014
225
2
is mean :
zero=0 is all 90 degree
and in plots=2 there are two 45 degree and there are two 90 degree
and in plots =10 there are two 45 degree and there is one 90 degree
Taking into account the signals (positive or negative)
true same this? or there are another method?

thanks sir

Last edited: Nov 9, 2014
10. ### full Thread Starter Member

May 3, 2014
225
2
there is any standard for angle?
I can do same angle in green ?

Last edited: Nov 9, 2014
11. ### MrAl Distinguished Member

Jun 17, 2014
2,573
522
Hello again,

For the constant, we have no phase shift so it has no influence over the phase response plot.
For the zero, we have a 90 degree phase shift over all frequencies.
For a pole, we have two different associated frequencies wA and wB for a pole at w:
wA=w/10, and wB=w*10.
So for example a pole at w=3 would give us wA=3/10 and wB=3*10, or simply wA=0.3 and wB=30.
Here the poles influence on the frequency kicks in sooner than the actual pole frequency, and ends at a frequency at 10 times the pole frequency. This is the approximation to the real response because of the inverse tangent function behavior. So the response would be -45 degrees/decade at wA and +45 degrees/decade at wB, where the phase shift at wB cancels out the original influence of the phase shift at wA.
The way they do it in the table is they associate a constant -90 degrees at wB.

So when you add up the phases, you have to consider that the particular pole only influences the phase response between wA and wB, which is unlike the magnitude response. It is an approximation used so that you can get an idea what the circuit is doing without actually plotting it. You should realize that this technique was invented before the age of the computer as we know it, so a more modern method would be to simply plot the response. Sometimes however knowing this method will allow you to get an idea about what is happening without plotting anything by just thinking about it.

The exact phase response for the function given (in degrees) is:
PhaseShift=(180/pi)*atan2((200*w*(20-w^2))/((w^2+4)*(w^2+100)),(2400*w^2)/((w^2+4)*(w^2+100)))