# Second order passive low pass filter design

#### pinkyponky

Joined Nov 28, 2019
319
Hello again,

Hey that's pretty good. You got the transfer function right although you went through a lot of trouble to do that you could have used just Algebra, but you did get it right and that's great.

To design a filter first you have to know the application, and then you have to know how sharp you want the response to be. When you change the ratio of L/C (or C/L) you get either sharper or less sharp of a response and that has a large effect on your application. Because you also have R in series with L that means the response will be less sharp because of that also.

What you could do is look at some known filter types and their responses, or just plot some responses yourself with different L and C. You will start to get an idea how this works. You may be able to work from the ratios of L/C and L/C^2 and with the rest of the coefficients being just gains you can get an idea of the shape of the response from that and decide what ratio of L/C you want.

I guess we can look at a few different ways to do this if you like.
You can also probably use a curve fitting technique where you define two or three points on the response curve you want to use as reference points.
Hi MrAl!

I already said the application. This filter will be used to filter the input signal of the ADC and output of the DAC.

You asked me to derive the transfer function. Eventually, Done!. Ok. How can I plot the response from this transfer function?.

How to design the R, L, and C values?. These values will be used to design a filter that attenuate the signal above 100kHz.

S=jw, w=2*pi*f, and j2=-1. After substitution of these values, the final Transfer Function is derived and attached below. Please have a look and tell me how to plot the response from the below transfer function.

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#### Ian0

Joined Aug 7, 2020
7,441
I have to commend you for working through the maths. I think that having done it once, you can now give yourself permission just to look it up in the tables!

#### MrAl

Joined Jun 17, 2014
10,070
Hi MrAl!

I already said the application. This filter will be used to filter the input signal of the ADC and output of the DAC.

You asked me to derive the transfer function. Eventually, Done!. Ok. How can I plot the response from this transfer function?.

How to design the R, L, and C values?. These values will be used to design a filter that attenuate the signal above 100kHz.

S=jw, w=2*pi*f, and j2=-1. After substitution of these values, the final Transfer Function is derived and attached below. Please have a look and tell me how to plot the response from the below transfer function.
Hi,

In the purest way, you convert the denominator to one real part and one imaginary part, then multiply top and bottom by the conjugate of that, simplify, then take the norm, then plot that amplitude vs w (or if you prefer f).

The conjugate of a number is the same number but with the sign of the imaginary part reversed, so if you have:
a+b*j
then the conjugate is:
a-b*j
and of course if the number is:
a-b*j
then the conjugate is:
a+b*j

To find the norm which is the amplitude, you square the two parts and take the square root:
sqrt(a^2+b^2)

and since this will be a function of w (or f if you prefer) then you can plot that amplitude vs w or f. That shows you the way the curve changes with frequency.

See the attachment for an example, and note how the response changes with increasing L. If you increase R1 you will also see the response die down a bit and not be so peaked. The response you are probably looking for is the one where L is closer to 100 Henries but that is with different values of C not the one you used.

This is the same circuit with the output taken across the cap and low value for R1. R2 is the load.

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#### pinkyponky

Joined Nov 28, 2019
319
I have to commend you for working through the maths. I think that having done it once, you can now give yourself permission just to look it up in the tables!
Which table?

#### Ian0

Joined Aug 7, 2020
7,441
Which table?
Post #22.
There are a whole series of tables, for Bessel, Butterworth and all varieties of Chebyshev, as well as a few others.

#### pinkyponky

Joined Nov 28, 2019
319
Hi,

In the purest way, you convert the denominator to one real part and one imaginary part, then multiply top and bottom by the conjugate of that, simplify, then take the norm, then plot that amplitude vs w (or if you prefer f).

The conjugate of a number is the same number but with the sign of the imaginary part reversed, so if you have:
a+b*j
then the conjugate is:
a-b*j
and of course if the number is:
a-b*j
then the conjugate is:
a+b*j

To find the norm which is the amplitude, you square the two parts and take the square root:
sqrt(a^2+b^2)

and since this will be a function of w (or f if you prefer) then you can plot that amplitude vs w or f. That shows you the way the curve changes with frequency.

See the attachment for an example, and note how the response changes with increasing L. If you increase R1 you will also see the response die down a bit and not be so peaked. The response you are probably looking for is the one where L is closer to 100 Henries but that is with different values of C not the one you used.

This is the same circuit with the output taken across the cap and low value for R1. R2 is the load.
Hi,

Not clearly understand. Please could you show me the example one so that I will do this.

#### MrAl

Joined Jun 17, 2014
10,070
Hi,

Not clearly understand. Please could you show me the example one so that I will do this.
Ok sure, but if you like i can get back here with a more all encompassing example and if i beat you back here i may be able to show that next anyway.

Say we have this transfer function (ill keep it very simple for now):
1/(s+1)
which with s=jw we have:
1/(j*w+1)
and written in a slightly more conventional way:
1/(1+j*w)
and to keep this really simple say w=1, so we end up with:
1/(1+j)
The conjugate of the denominator is 1-j so we multiply top and bottom by that::
[1*(1-j)]/[(1+j)*(1-j)]
and with a little algebra we end up with:
(1-j)/(1^2-j^2)
which reduces to:
(1-j)/(2)
and note that now the denominator is real with no imaginary part.
Now we just expand the expression:
1/2-j/2
and now we can identify the real and imaginary parts:
real=1/2
imag=-1/2
The amplitude is the norm of this:
sqrt(real^2+imag^2)=sqrt(1/4+1/4)
which equals
sqrt(1/2) or 1/sqrt(2)
So the amplitude is 1/sqrt(2).
We would plot this vs w (in the actual calculation we would keep w a variable so we can plot this later) so in the actual calculation with w still a variable we would end up with:
ampl=1/sqrt(w^2+1)
and that is what we would plot, with w on the x axis and ampl on the y axis.
With a 2nd order filter you do the same thing just a little more math, and you come out with a single complex number a+b*j and find the norm and then plot that vs w.

Does that help ?

#### pinkyponky

Joined Nov 28, 2019
319
The conjugate of the denominator is 1-j so we multiply top and bottom by that::
[1*(1-j)]/[(1+j)*(1-j)]
Why we need to do this step?.

#### pinkyponky

Joined Nov 28, 2019
319
Post #22.
There are a whole series of tables, for Bessel, Butterworth and all varieties of Chebyshev, as well as a few others.
I have done that in the earlier posts, the problem is from the table is that I come up with the large inductor value after calculations done.

#### Ian0

Joined Aug 7, 2020
7,441
I have done that in the earlier posts, the problem is from the table is that I come up with the large inductor value after calculations done.
And as the tables are derived from the same equations as you are working on, what makes you think the outcome will be any different?
You can do a whole exercise book of algebra, but you will end up with L≈R/(2πf)

#### LvW

Joined Jun 13, 2013
1,644
Hi,
Not clearly understand. Please could you show me the example one so that I will do this.
Hi pinkyponky,
as I have told you already before, it is not necessary (and very involved and time consuming) to do all the math with complex figures.
Nobody is doing this (because you would start at "zero".)
This was already done long time ago - and the results are summarized in tables which allow to design a second order filter (or even higher order) for the desired approximation.

In post#177 I gave you the transfer function for your second-order RLC filter.
For your convenience, I repeat it again: The part names are according to your hand written figure in your post#174.
The parallel connection is Rp=R||RL.

H(s)=Ao / {1 + s[(RpC + L/(R+RL)] + s²AoLC}

Writing the transfer function in this form (so called "standard form") has the advantage that we immediately can identify the three main parameters which define the lowpass response:
* DC gain is Ao
* Pole frequency is wp=SQRT(1/AoLC)
* Quality factor (pole-Q) can be found by 1/Qp=wp[(RpC + L/(R+RL)]

* For DC the gain Ao can immediately found to be RL/(R+RL). You are free to select R undf RL according to your requirements.
* The pole frequency wp is in rather vicinity to the wanted 3dB-cutoff frequency wc (for a maximum flat magnitude, Butterworth approximation, it is identical wp=wc). Because Ao now is known you can compute the values for L and C together with the required pole-Q (next point). This results in a set of two equations for the two unknown values L and C.
* The required Q-values can be found in corresponding tables: Q=1/SQRT(2)=0.7071 for max. flat (Butterworth) response. Larger Q-values give a Chebyshev response with a slight "peak" in the vicinity of the pole frequency (tabulated for a peak ("ripple") of 0.05dB, 0.1dB, 0.3dB.........3dB).

Based on these information you immediately can design your filter.
I hope this helps.

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#### pinkyponky

Joined Nov 28, 2019
319
And as the tables are derived from the same equations as you are working on, what makes you think the outcome will be any different?
You can do a whole exercise book of algebra, but you will end up with L≈R/(2πf)
Ohh... Really!.

I don't know why you people are guiding me in different directions, since I'm struggling to design a simple RLC circuit.

I was already done the calculation according to post #22. But, I got the higher Inductance value. Later when I ask you guys that how to minimize the inductance values. Then, MrAL rise that in order to design filter circuit (like good one), he propose me that, first I need to derive the Transfer Function of the circuit. I though that If I follow this way may be I would have design a filter circuit with less value of Inductor. Later, he asked me to plot the graph of response. So, I'm working on this, but I feeling that It's bit hard to plot the response from my transfer function.

Now Ian0 asking me that why can't you use the table which is posted in #22.

I don't know how to end this task.

#### LvW

Joined Jun 13, 2013
1,644
I don't know how to end this task.
pinkyponky - because I am continuously trying to improve my teaching methods:
I have described the classical steps in modern filter design - based on tables with pole data.
I have designed active/passive filters for about 30 years (and I have published a book on filters - theory and practice).

#### MrAl

Joined Jun 17, 2014
10,070
Why we need to do this step?.
That is the most direct way to isolate the real part from the imaginary part so you can compute the amplitude using the norm. If you can compute the norm another way that's fine too.

In order to calculate the response we have to calculate the amplitude as it changes with frequency w or f.

#### Ian0

Joined Aug 7, 2020
7,441
because I am continuously trying to improve my teaching methods:
I have described the classical steps in modern filter design - based on tables with pole data.
I have designed active/passive filters for about 30 years (and I have published a book on filters - theory and practice).
forgive me for encouraging laziness in your pupils bay suggesting they should look things up in tables when they should work it out from first principles!
Why don’t you give your book a plug?

#### LvW

Joined Jun 13, 2013
1,644
forgive me for encouraging laziness in your pupils bay suggesting they should look things up in tables when they should work it out from first principles!
Why don’t you give your book a plug?
Yes - this is my approach. Using the standard form of the transfer function and corresponding expressions - together with tabulated figures - for wp and Qp.
The book is in German.

#### MrAl

Joined Jun 17, 2014
10,070
Did somebody remove a post here or did i forget to post it?
It was a complete worked example of a DIFFERENT system not the one at hand.

#### ericgibbs

Joined Jan 29, 2010
17,419
hi Al,
No record of post removal, when did you post it?

Mod

#### MrAl

Joined Jun 17, 2014
10,070
hi Al,
No record of post removal, when did you post it?

Mod
A few hours ago but ok i must have made a mistake. I'll post it here in a few.

#### MrAl

Joined Jun 17, 2014
10,070
Hello again,

Ok here is a complete worked example showing how you can get the amplitude which you would want to plot, and you can get the phase shift also from the final real+j*imag expression. This is NOT your transfer function i did not want to post that work because you should do this yourself and you will find it interesting im sure.
Also note that there are other ways to get the amplitude but this one is probably the most basic so i wanted you to see that. If you use math software you can get these results in minutes without writing anything down on paper.
You will see the expression followed by the procedure that was used to get that expression, in curly brackets. Each result follows from the previous.

(Ro*s)/(Ro*s^2*C*L+s*L+Ro), {this is the transfer function}
s/(s^2+s+1), {Ro=1, C=1, L=1}
(j*w)/(j^2*w^2+j*w+1), {s=j*w}
[j*w,j^2*w^2+j*w+1], {separate numerator and denominator)
[j*w,1-w^2+j*w], {simplify the denominator}
1-w^2-j*w, {calculate the conjugate of the denominator}
D2=w^4-w^2+1, {multiply denominator by the conjugate, becomes new denominator D2}
N2=w^2+j*(w-w^3), {multiply numerator by the conjugate, becomes the new numerator N2}
N2=[w^2,j*(w-w^3)], {separate the real part and imag part of the numerator}
w^2/(w^4-w^2+1)+j*(w-w^3)/(w^4-w^2+1)
in this last step we divide the real part by the new denominator D2 and also the
imag part by the new denominator D2, and keep the parts separate but added together.
That makes the form of the expression into a+j*b which is what we were after.
Next, we separate the two parts and just square the two separage parts separately:
[w^4/(w^4-w^2+1)^2,((w-w^3)^2)/(w^4-w^2+1)^2]
note the second part above is the imaginary part squared without the 'j'.
Now we add the two back together without any 'j':
w^4/(w^4-w^2+1)^2+((w-w^3)^2)/(w^4-w^2+1)^2
and if we factor that we get:
w^2/(w^4-w^2+1)
then take the square root:
sqrt(w^2/(w^4-w^2+1))
and we end up with:
w*sqrt(1/(w^4-w^2+1))
which is the amplitude. We can then plot this amplitude vs w.

If you want the phase shift, go back to the a+j*b form and do:
ph=atan2(b,a)
and that is the principal phase shift.

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