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Is that much difficult to design a RLC filter with my specifications?.
Provide me other type of filter (not RC filter) to design a RLC filter to reduce the inductance value, from the copy of Zverev.
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Second order passive low pass filter design
- Thread starter pinkyponky
- Start date
Provide me other type of filter (not RC filter) to design a RLC filter to reduce the inductance value, from the copy of Zverev.
Hi Ian,
I have tried with RCRC second order filter. But I'm not achieve the results even with RC filter, the results are fc=31 kHz@-3dB and only 3.7dB/Octave measure instead 12dB/Octave since it is second order filter.
I'm not achieving the result even through I followed your instruction. See the below image, the filter is designed for fc=100 kHz.
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In my experience, when given a low-pass filter attenuation slope of N dB/octave, it refers to the straight-line slope of the attenuation curve that is well past the cutoff frequency.
Do you need an RLC circuit or can you use an op amp circuit with just R's and C's ?
To design an RLC filter usually you figure out what kind of response you want then go from there.
Besides being a good idea to begin with, your choice of R, L, and C can shape the response based on more than just the cutoff frequency. The cutoff frequency is just one aspect of a filter like this. Since you already decided on the topology you can go forward with an analysis from there. You start with the transfer function.
So with that in mind, you can write your transfer function and then solve for L, C, and R in terms of the other two. You can then choose one or two values and then solve for the others. Do you know how to write the transfer function for this circuit in the frequency domain (only AC circuit analysis is required) ?
I don't know how to write the transfer function of the circuit. Please could you help me with that?.
hi pp,
Read this linked Article.
E
https://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SDE.html
Read this linked Article.
E
https://lpsa.swarthmore.edu/Representations/SysRepTransformations/TF2SDE.html
I understand the derivation. But, not understanding how they start with the differential equation. Left side contains 4 terms and Right side contains 3 terms as I shown below. If it is second order filter, then, do I need to chose 3 terms on the right side and 2 terms on the left side. Am I correct?.
Please could you explain to me how to write differential equation for the below circuit. After that I will start to derive the transfer function.
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Why - for heaven's sake - do you want to set up the differential equation in the time domain?
Filter response is always defined in the frequency domain - and the transfer function is quite simple to find with the calculation in the frequency domain.
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I don't have an idea either do I need to setup differential equation or idea equation. I have just followed the post #167. Then, I got the question and posted at #168.
Ofcourse, It's very simple for you. But, It's difficult for the beginners like me.
If I ask a question, most of the people curious to know about my work, that's good and I believe that I will get the answer when I provided my circuits. Later, you guys guiding me to do like this (or) follow this method (or) something else. That's also nice and I'm very glad that you guys are helping me. If I struck somewhere in that execution, and If I ask why it is not working, Then, I will not get proper answer for that issue, like why it is not working, even if I submitted all information. In this case, for example post #150 and post #162. I don't get the answer to my questions why it is not working even I followed your guidelines.
Again, the ericgibbs asked me to follow the document (post #167) to know how to derive the transfer function. I have read the article and I have some questions, When I ask them, you guys are saying that is easy. But, here, the people asking the questions are not related to either the question is easy or tough. That relates to the answer and we are expecting the answers and guidelines from you guys (like you as an experts).
Still, I'm interesting to know how to draw the derivation (Transfer Function) of my circuit and design the RLC filter.
I know RLC filter circuit design is very simple for you. I'm also provided all my circuit even though I'm not getting the answer to my issue.
Please help me guys (I'm not saying that you guys are not helping). You guys are helping but I'm not getting the answer to my question.
Please don't take it negatively.
Thank you, Guys!
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OK - let me say the following:
* Do you know the impedance expressions in the frequency domain for L (jwL) and C (1/jwC) ?
* I suppose you know Ohms law which also applies to those impedances.
* With regard to your last drawing (with R2+R3=R23) - can you apply the formula for voltage division and calculate the signal voltage at node A?
* Start: V_A = V1 [(R23||1/jwC1)]/[R1+jwL1+(R23||1/jwC1)].
* Thats all because after some rearranging you have the transfer function V_A/V1=(constant)/(2nd degree polynominal)
Do you understand how to use complex algebra? That's just the idea of using complex numbers to calculate things with rather than just regular numbers.
For example, one value might be:
2.54+j*1.2887
where the 'j' indicates the 'imaginary' part.
Using this kind of math you can quickly calculate the transfer functions for circuits like this.
An even simple method is to just declare a variable which we call 's' which is just a lower case English letter "S". You can then redefine your reactive elements as such:
L goes to "s*L"
C goes to "1/(s*C)"
R stays at "R"
and then you use your knowledge of combining resistances.
For example, two resistors in series would be R1+R2.
A resistor in series with an inductor would be R1+s*L1.
A resistor in series with a capacitor would be R1+1/(s*C1).
A resistor in parallel with an inductor would be (R1*s*L1)/(R1+s*L1).
All we do is remember to use the proper 's' form (called an impedance) rather than the part component number alone.
Once you do that, you can then substitute s=j*w where w=2*pi*frequency and then you can use complex numbers to compute the response at that frequency, and when you compute the response at several frequencies you get the shape of the response and that tells you how well the filter behaves.
You can also use a graphical method historically known as a "Bode Plot" and that gives you an approximation to the response, but gets a little tricky when you get into strongly reactive 2nd order systems.
The prerequisite for this is you have to know how to combine resistors in series and parallel combinations and understand the 'voltage divider' concept. If you understand complex numbers you will be able to do this quickly or else you'd have to learn how to do that, although it would not take long to learn the basics and get going on this.
Hi MrAl,
In the above post you guys asked me to derive the transfer function of above circuit.
I have attached the home work of transfer function of second order filter with load. Please can you have a look and let me know, is that correct or not?. If so, later, what I have to do to design RLC filter?.
In the above post you guys asked me to derive the transfer function of above circuit.
I have attached the home work of transfer function of second order filter with load. Please can you have a look and let me know, is that correct or not?. If so, later, what I have to do to design RLC filter?.
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.
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 pinkyponky - MrAl told you that you could have used simple algebra to find the transfer function.
In the following I like to show you (a) how simple this is and (b) how you should rewrite the transfer function .
Using the voltage divider rule and the form Ra||Rb=1/(1/Ra + 1/Rb) we immediately can write:
H(s)=[1/(sC + 1/RL)] / {[1/(sC + 1/RL)] + R +sL]}=[1 + (R+sL)(sC + 1/RL]
After evaluation of the denominator and multiplying numerator and denominator with Ao=RL/(R+RL) we get
H(s)=Ao / {1 + s[(RpC + L/(R+RL)] + s²AoLC}
Writing the transfer function in this form (so called "nominal 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)]
Special case: For Qp=SQRT(0.5)=0.7071 we have a Butterworth response with wp=wo (3dB- cutoff)
In the following I like to show you (a) how simple this is and (b) how you should rewrite the transfer function .
Using the voltage divider rule and the form Ra||Rb=1/(1/Ra + 1/Rb) we immediately can write:
H(s)=[1/(sC + 1/RL)] / {[1/(sC + 1/RL)] + R +sL]}=[1 + (R+sL)(sC + 1/RL]
After evaluation of the denominator and multiplying numerator and denominator with Ao=RL/(R+RL) we get
H(s)=Ao / {1 + s[(RpC + L/(R+RL)] + s²AoLC}
Writing the transfer function in this form (so called "nominal 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)]
Special case: For Qp=SQRT(0.5)=0.7071 we have a Butterworth response with wp=wo (3dB- cutoff)
I did it myself. whats wrong with that?.
No - I think, it is correct. However, a very lengthy calculation. I have tried to show a shorter way (my post'177),
