No, you must select all the variables and parameters so that the parts values are positiv.

The calculation of R2 the denominator would always be negative if you use alpha as the square root of 2 and a gain greater than 1. I must be missing something here. Can you help clear this up for me please?

 

The answer is simple: The expressions as given in the table are valid for the assumption R3=R4 only. Therefore, we cannot assume that we are free to select all filter parameters without any restriction. In this case, the midband gain will be always <1 or equal to unity.
Note that for α=SQRT(2) and g=1 the resistor R2 can be omitted (R2 infinite). This one of the well-known alternatives of the MFB bandpass.

 

Hello,

Yes various topologies for same purposes.

There is a very general technique which starts from drawing several Bode plots (hand drawn actual Bode plots NOT frequency response plots) and a few simple rules to follow that allow a quick design to unfold. I havent used this in years though so i'd have to try to remember the rules but it's very informative if i get inclined i'll post the simple rules and procedure.

Basically each (true) Bode plot is associated with one resistor and one capacitor (or one or two resistors), so you just add one of each for each Bode plot you draw and the Bode plots graphically add up to create the entire response. Very informative even if you dont use the technique.

There is also a very general analysis technique which holds for all analog filters and some switched capacitor/resistor filters. I'll try to post this as it is very informative when you wish to verify a design procedure.

 

Hello, my co-worker gave me the a quick guide for filter design.

View attachment 193450
Has anyone seen something like this in filter design? I've working the numbers to see if it is valid.

Thanks
Joe

Hello,

Is that filter topology correctly shown?

I ask because i get a very different value for w0.

To test, set all resistors equal to 1 and and all capacitors equal to 1.
See what you get for w0.
The formula gives w0=1, but that is not what i got at all.
The formula for the gain g=R4/R1 is the same as i got though.

Anyone else try this?

 

Hello,
Is that filter topology correctly shown?
I ask because i get a very different value for w0.

To test, set all resistors equal to 1 and and all capacitors equal to 1.
See what you get for w0.
The formula gives w0=1, but that is not what i got at all.
The formula for the gain g=R4/R1 is the same as i got though.

Anyone else try this?

I think, the topology as well as the expression for wo (pole frequency identical to 3dB cut-off for Butterworth) is correct.

 

I think, the topology as well as the expression for wo (pole frequency identical to 3dB cut-off for Butterworth) is correct.

Hello,

But did you test that?
Or do a simulation?

I did both, found the 3db cutoff frequency to be different than stated.
See what you can come up with we can compare notes.
I could have made a mistake but then i cant see how the simulation would have come up the same as my results rather than the stated text results.

 

Hello,

But did you test that?
Or do a simulation?
I did both, found the 3db cutoff frequency to be different than stated.
See what you can come up with we can compare notes.
I could have made a mistake but then i cant see how the simulation would have come up the same as my results rather than the stated text results.

No, I did not test that by simulation.
Moreover, I did not check all the other (to me: unknown in this form) expressions for designing the filter (parts values).
Perhaps is there an error? In any case, I am very unhappy with the "coupling factor α".
I do not know any filter documentation (book or paper) in which this expression is used instead of the pole quality factor (Qp=1/α).

However, I have calculated the expression from the circuit diagram and - more than that - the result corresponds with the written documentation.
Here is the transfer function:
H(s)=N/D(s) with N=R4/R1 and D(s)=[1+sC5(R3+R4+R3R4/R1)+s²R3R4C2C5]

From D(s) it is clear that the given expression for the pole frequency is correct because for α=SQRT(2) we have Qp=0.7071 (Butterworth) and the pole frequency is identical to the 3dB cutoff wo.

 

No, I did not test that by simulation.
Moreover, I did not check all the other (to me: unknown in this form) expressions for designing the filter (parts values).
Perhaps is there an error? In any case, I am very unhappy with the "coupling factor α".
I do not know any filter documentation (book or paper) in which this expression is used instead of the pole quality factor (Qp=1/α).

However, I have calculated the expression from the circuit diagram and - more than that - the result corresponds with the written documentation.
Here is the transfer function:
H(s)=N/D(s) with N=R4/R1 and D(s)=[1+sC5(R3+R4+R3R4/R1)+s²R3R4C2C5]

From D(s) it is clear that the given expression for the pole frequency is correct because for α=SQRT(2) we have Qp=0.7071 (Butterworth) and the pole frequency is identical to the 3dB cutoff wo.

Well something is wrong because if we use the formula in the first post:
w0=1/sqrt(C2*R3*R4*C5)

and make all those values equal to 1, we get of course:
w0=1/1=1

and that's not what the simulation shows. The simulation shows w0 close to 0.374 but feel free to try it yourself. If not i'll post the graph of amplitude vs w0.

Here's the graph. Note the frequency where the amplitude reaches 0.7071

 

Independent on the chosen values - for 3 equal resistors and two equal capacitors the quality factor is 1/3 and not 0.7071 (as required for the Butterworth response).
In any case (as I have mentioned already), the given expression for wo in the table gives the POLE FREQUENCY which is identical to the 3dB frequency for the Butterworth reponse only. However, for the parts values as given in the simulation, we have no Butterworth response....in contrary, we have two real poles which could be realized even with a passive second order RC combination.
Example: Use three equal resistors and make the capacitor ratio C2/C5=4.5.
This will give a Butterworth response - and the expression under discussion (wo) is the 3dB frequency.

 

Independent on the chosen values - for 3 equal resistors and two equal capacitors the quality factor is 1/3 and not 0.7071 (as required for the Butterworth response).
In any case (as I have mentioned already), the given expression for wo in the table gives the POLE FREQUENCY which is identical to the 3dB frequency for the Butterworth reponse only. However, for the parts values as given in the simulation, we have no Butterworth response....in contrary, we have two real poles which could be realized even with a passive second order RC combination.
Example: Use three equal resistors and make the capacitor ratio C2/C5=4.5.
This will give a Butterworth response - and the expression under discussion (wo) is the 3dB frequency.

So the expression given is wrong in general, and a general expression is what we expect from a filter design procedure.

Make all resistors equal to R1, make all caps equal to 1/R1, then:
w0=sqrt(sqrt(53)-7)/sqrt(2)=0.374

and that includes all values equal to 1.
The formula gives w0=1, therefore it is WRONG.

I think what you are saying is that we MUST use a Butterworth filter by forcing R1 to be some certain value.
That might be true, but i dont think it is a good idea to specify something in that way.

"The value of w0 is f(a,b,c,d) BUT ONLY IF YOU CHOOSE the constant e to be equal to x Ohms.".
Or alternately force C2 to be some value like 3/sqrt(2) Farads if we want w0=1.
Apparently they are trying to force a Butterworth response and try to make the calculations easier.
We then get in the analysis:
(g*w0^2)/sqrt((w0^2-w^2)^2+a^2*w^2*w0^2)=g/sqrt(2)

and when w=w0 we get:
g/abs(a)=g/sqrt(2)

or
a=sqrt(2)

I guess that has some merit to it but a more general formula is better i think.

For the true LP type filter only:
|Vout(w)|=Vout(0)/sqrt(2)

then solve for w.

And to force Butterworth:
w0=1/sqrt(C2*R3*R4*C5)

subject to the constraint:
C2=((R2+R1)*R3+R1*R2)/(sqrt(2)*w0*R1*R2*R3)

or any variation of that expression.

 

MrAl - I must admit that I do not understand the content of your last contribution.
Question 1: Did you simulate the numeric example I have proposed in my former post?
Question 2: Was the simulation result in accordace with the mentioned formula for wo?

I can assure you and without any doubt - the formula is correct, but it must be interpreted correctly!

I repeat: The formula wo=1/SQRT(R3R4C2C5) gieves the POLE FREQUENCY for the MFB lowpass circuit.
And there is one single case only where this pole frequency is identical with the the 3dB CUT-OFF FREQUENCY - and this is the Butterworth response with Qp=0.7071. And only this special case is treated in the corresponding table (as given by the questioner).
For all other lowpass responses (Bessel, Chebyshev...) with Qp values other than Qp=0.7071 the pole frequency wo (as given with the mentioned formula) will NOT be identical to the 3dB cutoff !!
I have mentioned this fact already in my first contribution (post#6)

Finally - this formula for the pole frequency wo is mentioned in all relevant filter books and it is rather easy to derive it.

 

MrAl - I must admit that I do not understand the content of your last contribution.
Question 1: Did you simulate the numeric example I have proposed in my former post?
Question 2: Was the simulation result in accordace with the mentioned formula for wo?

I can assure you and without any doubt - the formula is correct, but it must be interpreted correctly!

I repeat: The formula wo=1/SQRT(R3R4C2C5) gieves the POLE FREQUENCY for the MFB lowpass circuit.
And there is one single case only where this pole frequency is identical with the the 3dB CUT-OFF FREQUENCY - and this is the Butterworth response with Qp=0.7071. And only this special case is treated in the corresponding table (as given by the questioner).
For all other lowpass responses (Bessel, Chebyshev...) with Qp values other than Qp=0.7071 the pole frequency wo (as given with the mentioned formula) will NOT be identical to the 3dB cutoff !!
I have mentioned this fact already in my first contribution (post#6)

Finally - this formula for the pole frequency wo is mentioned in all relevant filter books and it is rather easy to derive it.

Hi,

Yes no problem with that. My comments were about a more straightforward way to describe what they wanted to show. As you say the interpretation is important, and the key is at the top not within the formula itself. So i decided to rewrite it so that it was in a more straightforward form.

I did not simulate anything else because i already calculated the response the way they wanted to do it and it came out OK. I just dont like the way they had shown that technique that's all. It makes it look wrong, and it is wrong for all values except those that fit the constraint i gave in my previous post. So that formula is really subject to a constraint.