Bode plot correct?

ronsimpson said:

If you are making this kind of plots then look at this graph paper.
I made several kinds of these but this one is best overall. Print the PDF.

A 1K resistor is a line from left to right through the center of the paper.
A 0.1uf cap is a line from top left to a point at the bottom in the center. "\"
Inductors are lines "/".

Draw a line for every L, C and R in your circuit.

For a 1k resistor and 0.1uf cap low pass filter; draw the lines for the two parts.
At low frequencies (at the left side) the capacitor has no effect so start with the 1k ohm line. Follow along that line until it hits the 0.1uF line then follow along the capacitor line.

Click to expand...

ericgibbs said:

Hi foudalnoor,
For reference this image is the LTSpice plot for the 3 circuits, for comparison.

Click to expand...

MrAl said:

Hi,

Oh wow a passive RC filter with a gain of 60db! I love it
You might want to included what we sometimes call a 'sanity check' just to make sure you have something reasonable. I mention this because a 60db gain in a passive low pass filter is just not ever going to happen. I cant stress this enough. Once you have the solution, go over it briefly to see if you spot anything that looks like it wont work or is off by a large amount. Maybe it takes time to learn this kind of thing but it's worth thinking about.

You were going pretty good there, with your low pass example. You came up with:
Hs=1/(sRC+1)

and then you divided top and bottom by RC:
Hs=(1/RC)/(s+1/RC)

and you reasoned that you have a pole at 1/RC (of course you know this is really -1/RC). You did very well there.
But now you KEPT that form to figure out the gain, so you of course got the gain 1/RC.
BUT, in the previous discussion, we had to get the equation into the form with the factors (s*tau+1) in order to find the gain K. For some reason you skipped this step and that's how you got the gain of 1/RC, which of course isnt right.
Getting back into the s*tau+1 factors form, we have again:
Hs=1/(sRC+1)
with tau=RC. Now the only thing left in the numerator is the '1', so the gain K is equal to 1.
So try to remember which form is which, and that you need the transfer function in the s*tau+1 factors form to get the gain to stand alone in the numerator.

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MrAl said:

Hi,

It appears that you were trying to do the first one (C=1nf, R1=100k, R2=10k) so we'll do that one first. You'll want to note that the zero and the pole cause opposite sign slopes, one +20db/decade and the other -20db/decade.

The transfer function Hs is:
Hs=(s*C*R1*R2+R2)/(s*C*R1*R2+R2+R1)

At zero frequency the gain K is equal to:
K=R2/(R1+R2)

and dividing numerator and denominator by C*R1*R2 we get:
Numerator: N=1/(C*R1)+s
Denominator: D=1/(C*R2)+1/(C*R1)+s

So we have constant gain:
K=1/11

and the numerator is:
N=s+10000

and the denominator is:
D=s+110000

Thus we have a zero at w=10k and a pole at w=110k, and constant gain -21db.

The two frequencies are 10k and 110k, so we start out at w=0 with the constant gain of -21db which is a straight horizontal line (like you have drawn) and at the frequency of 10k we have a zero so we get a slope of +20db/decade starting at 10k. This continues until we reach 110k and there we have another frequency of 110k which is a pole so we get a slope of -20db/decade starting at 110k, and this added to the previous slope means we again get a line with a slope of 0 which is a horizontal line.

So the plot looks like a horizontal line at -21db that goes from w=0 to w=10k, then slopes upward at +20db/decade, then at 110k it becomes a horizontal line again. Since the slope of the upward line is +20db/decade and it starts at -21db and goes from 10k to 110k, that means it goes upward about +22db and this means we end up at +1db, but the horizontal portion from there is around 0db because since the cap conducts much more than the 10k resistor at infinite frequency, the response is clearly 0db above w=110k. Note that since the distance from the 10k to 110k is really 1.1 decades, we would multiply 20db times 1.1 and get 22db as the rise so this would lead us to +1db rather than 0db. But the quick check with the cap and 10k resistor shows this to really be 0db. So we've corrected this a little by doing that.

Also note that when you get a pole after a zero it cancels out the slope of the zero because the pole has the exact opposite slope as the zero.



So the plot looks like this:

Code:

   0                 ------------------
                    /
                   /
                  /
                 /
 -21  -----------
      0        10k  110k              +inf

Click to expand...

fouadalnoor said:

Also, is the transfer function for the third (bottom circuit) correct: H(s) = RC1s/((RC2s+1)(RC1s+1))?

Click to expand...

MrAl said:

Hi fouad,

You seem to be having a problem with the gain so lets start with that.

To find the gain you factor the Hs into this form:

[LATEX]
\ \ \ \ H(s)=\frac{K(s\tau_{1}+1)(s\tau_{2}+1)...}{(s\tau_{3}+1)(s\tau_{4}+1)...}
[/LATEX]

If you look at that you'll see that the numerator is factored into terms with s times tau plus 1, and so is the denominator, and when factored like this the constant gain K stands out in the numerator as a single number.

To illustrate this with your next circuit (the one on the right) we start with the transfer function:
Hs=(s*C1*R2)/(s^2*C1*C2*R1*R2+s*C2*R2+s*C1*R2+s*C1*R1+1)

and yes because all the R's are the same and all the C's are the same this simplifies to:
Hs=(s*C*R)/(s^2*C^2*R^2+3*s*C*R+1)

but now we substitute the values for the components R and C and we get:
Hs=(1e-4*s)/(1e-8*s^2+3e-4*s+1)

Now we have numerator:
N=1e-4*s

and denominator:
1e-8*s^2+3e-4*s+1

First thing we'll do is multiply N and D by the reciprocal of the factor of s^2 which is 1e-8, so we get:

N=10000*s
D=s^2+30000*s+1e8

Now we factor D and we get two roots:
r1=-3819.660112501051
r2=-26180.33988749895

What this means is our transfer function can now be written as:
Hs=10000*s/((s-r1)*(s-r2))

(note we subtract the roots not add them to reconstruct the denominator, and since they were negative they now will come out positive)

which numerically is:
Hs=10000*s/((s+3819.660112501051)*(s+26180.33988749895))

From this we can see that we have two poles one at w=3819.660112501051 and one at w=26180.33988749895.

Now we want to get this new Hs into the form shown above with all the tau's in it, so we take the first pole in the denominator and divide that by the negative of the root:
(s+3819.660112501051)/(3819.660112501051)

and this of course gives us:
s/3819.660112501051+1

and you see now that part is in the "s times tau plus 1" form where tau=1/(-r1)=1/3819.660112501051.

But since we divided that by 3819.660112501051 to keep that part of the equation the same we have to also multiply by that, so we get this form for that one part of the denominator:
3819.660112501051*(s/3819.660112501051+1)

Now we have this in the form 1/tau times the "s times tau plus 1" form.

We have another pole, so we do the same thing with that one and we end up with:
26180.33988749895*(s/26180.33988749895+1)

Again note that this is also in the form (1/tau)*(s*tau+1).

Now we put the Hs back together and we have:
10000*s/((3819.660112501051*(s/3819.660112501051+1))*(26180.33988749895*(s/26180.33988749895+1)))

and since we have two constants in the denominator we can lump them together:
10000*s/(3819.660112501051*26180.33988749895*(s/3819.660112501051+1))*(s/26180.33988749895+1))

This is now in the form N/(Ka*Kb*(s*tau1+1)*(s*tau2+1))

so we simply multiply Ka times Kb which is 3819.660112501051*26180.33988749895 and we get a single number:
Ka*Kb=1e8

So now we have:
10000*s/(1e8*(s/3819.660112501051+1)*(s/26180.33988749895+1))

Now all we do is divide top and bottom by that 1e8 and we end up with:
1e-4*s/((s/3819.660112501051+1)*(s/26180.33988749895+1))

This is now in the form: Kn*s/((s*tau1+1)*(s*tau2+1)).

Since there is no additive term in the numerator we're just about done.
Now all we do is:
Kn(db)=20*log10(Kn)=20*log10(1e-4)=20*(-4)=-80db

so the constant gain is -80db.

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MrAl said:

Hi,

There are a number of ways to do these problems, i just happened to like starting by dividing out the leading coefficient sometimes. There's no rule you have to simplify a quadratic with no coeff of s^2 because the quadratic formula takes that into account too. If you do divide out the leading coeff, you end up with a slightly simpler version of the quadratic formula.

Did you take into account that the graph Eric's plot shows has the running variable of regular frequency f and not angular frequency w? Most of these calculations are done in terms of w not f, so we end up having to convert at some point for some real problems.

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MrAl said:

Hello,


Sooner or later you have to start thinking for yourself
You have to start learning to put two and two together and come up with the right answer. You have plenty of resources in this thread that you can turn to, so you should be able to figure out how to do this now. You also have to learn to check the validity of your results at least to some degree. If you dont do that you wont be able to handle this when there is no one around to ask what to do for every little tiny thing

So what you should do is review all the data in this thread, take a minute or two (or more) to think it over, then figure out what to do. I cant keep telling you how to do little things like algebraic manipulation because that varies from problem to problem and you really have to learn that yourself. If you dont know how to factor you probably should have covered that before getting this far, or you should review that. This is nothing more than algebraic and if you follow the guidelines i presented with the illustration (especially the numerical one with all the factoring) then you should be able to solve this now. The idea is to think once in a while, and review some basic algebra.

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