Hello All,

Please can my work be checked. I started this section on Filters and need guidance.

The below question was taken from a past year examination paper.

Thank you,



My Attempted answer:

 

It is correct that the order is 2nd, but how did you find that the order of the filter should be second? Show us the calculation.
The prototype filter should have an impedance of 1 Ohm and a load of 1 Ohm. Now you can use the coefficients from the table directly on the schematic. Then you scale in frequency and finally you scale to match the impedance.

 

Hi Sir, I used the formula to work out the gain is around 13dB, so from there the value approximated to be 13 and it was the second order filter.

I can take a pic from my book.

Also since this is a passive low butterworth filter, will this circuit contain just the Inductor and capacitor?

Will is contain a op-amp?

thank you

 

Hello,

If you are doing a passive filter why would you need an op amp?

You should check your results by calculating the Zin at least, also the response at the cutoff frequency.

Without checking we can always make a mistake and wont know it. If it was a test we'd get that answer wrong and that would cost us a grade or more.

Checking the results should be taken as more important than actually calculate the results. We make many mistakes but if we catch them then we get everything right, all
the time.

So regardless what way you choose to calculate, do the checking afterwards. If you dont know how to check then you should think about how you might go about it for each problem.

 

Hi Sir,

Thanks mind my stupidity, you are right no need for op amp.

Let me re-do this and show u what I have done.

thank you very much.

 

Hi Sir,

Please find the attached worked out solution.

 

The units for an inductor are not μF, they are in Henrys. I get 11.255 mH (milliHenrys) using your numbers. IIRC you need the factor of 2π when doing frequency scaling since the coefficients are normalized for a radian frequency of 1 radian per second.

 

The units for an inductor are not μF, the are in Henrys. I get 11.255 mH (milliHenrys) using your numbers. IIRC you need the factor of 2π when doing frequency scaling since the coefficients are normalized for a radian frequency of 1 radian per second.

Hi Sir. Oops yes its Henrys for the Inductor. My mistake. Is the other calculations correct?
Is the circuit partially correct then?
What do you mean factor of by 2pi?
Thank you.

 

Hi Sir,

Thanks mind my stupidity, you are right no need for op amp.

Let me re-do this and show u what I have done.

thank you very much.

Hi,

Well i didnt think it was a stupid question really i just wondered what made you think that there should be an op amp.

But one question, the resistor you have there is 50 ohms, and that is the source resistance right? So that means the impedance looking into the LC alone should be 50 ohms too right?
I just want to make sure i understand what you are looking for.
And also i guess that means the network will be loaded with a 50 ohm resistor right?
So the source resistance is 50 and the load is 50 ohms pure resistive, right?

 

The circuit is correct. As you know frequency can be measured in Hz. (aka cycles per second) and also in radians per second. Cycles and radians are both dimensionless so the units of frequency are (1/sec). The constant of proportionality between Hz. and radians per second is 2pi. Thus:

\(\omega\;=\;2\pi f\)

I just asked if you need to include the factor of 2pi when you scale the Butterworth coefficients which are calculated for a radian frequency of 1 radian per second, when using frequencies in Hz.

 

Hi Sir, is the load and source has to be 50ohms.
Yes. I finally figured the difference between passive and active filters
Thank you.

 

The circuit is correct. As you know frequency can be measured in Hz. (aka cycles per second) and also in radians per second. Cycles and radians are both dimensionless so the units of frequency are (1/sec). The constant of proportionality between Hz. and radians per second is 2pi. Thus:

\(\omega\;=\;2\pi f\)

I just asked if you need to include the factor of 2pi when you scale the Butterworth coefficients which are calculated for a radian frequency of 1 radian per second, when using frequencies in Hz.

Hi Sir, im not too sure if I have to add the factor radian per second. Do you suggest I should do that?
Thank you Sir

 

The circuit is correct. As you know frequency can be measured in Hz. (aka cycles per second) and also in radians per second. Cycles and radians are both dimensionless so the units of frequency are (1/sec). The constant of proportionality between Hz. and radians per second is 2pi. Thus:

\(\omega\;=\;2\pi f\)

I just asked if you need to include the factor of 2pi when you scale the Butterworth coefficients which are calculated for a radian frequency of 1 radian per second, when using frequencies in Hz.

Hi Sir, im not too sure if I have to add the factor radian per second. Do you suggest I should do that?
Thank you Sir

 

Hello again,


This is interesting because i am running into a problem also, depending on how we interpret the question.

The source/load impedance is 50 ohms, so you would think that they want to match the input and output impedances with the network also. But that doesnt seem to be the case because i dont think there is any way to get 50 ohms input impedance with a 50 ohm load and still get the Butterworth response.

I get the same value for L as Papa, but when i calculate the input impedance Zin i get around 57.7 Ohms. A little mismatch, so i guess that's just too bad
I do get about -12db at 2 times the -3db frequency though.

 

Hi Sir, im not too sure if I have to add the factor radian per second. Do you suggest I should do that?
Thank you Sir

When you calculate the response you use radian frequency, normalized so that the corner frequency
\(\omega_0\;=\;1\)
and the response is
\(\frac{1}{\sqrt{2}}\)
at the corner frequency. I think I recall using that factor when scaling component values, but it was a long time ago.

Also the table value of 0.2251 appears to be incorrect. I read the table value as 1.414 or

\(\sqrt{2}\)

Ah....so

\(\frac{\large 1.414}{\large 2\pi}\;=\;0.2251\)

My simulation, using your values, is not producing good results so there may still be something wrong.

OK got it. When you simulate a passive filter with equal source and load impedances there is an insertion loss of 6 dB that must be accounted for. If you remove the filter you see the output will always be 6 dB blow the input. Now everything makes sense.

 

Hi Sir, Yoh. this is beginning to stress me now. I pray I don't make a mistake in the exams for this type of questions.

So I think this is ideal time to iron out all the problems and questions I have.

What should I do? Must I request the memo from the lecturer?

Because I noticed at our institute they always have ambiguous questions and then some of us get part marks due to the fact we don't fully understand the question.

Papabravo/MrAI do you know a good link on the internet that can provide me with good questions and answers on this topic?

thank you.

 

When you calculate the response you use radian frequency, normalized so that the corner frequency
\(\omega_0\;=\;1\)
and the response is
\(\frac{1}{\sqrt{2}}\)
at the corner frequency. I think I recall using that factor when scaling component values, but it was a long time ago.

Also the table value of 0.2251 appears to be incorrect. I read the table value as 1.414 or

\(\sqrt{2}\)

Ah....so

\(\frac{\large 1.414}{\large 2\pi}\;=\;0.2251\)

My simulation, using your values, is not producing good results so there may still be something wrong.

OK got it. When you simulate a passive filter with equal source and load impedances there is an insertion loss of 6 dB that must be accounted for. If you remove the filter you see the output will always be 6 dB blow the input. Now everything makes sense.

Hi Sir, so this means my working is not correct when using 0.2251 table value?
Thank you

 

Hello again,


This is interesting because i am running into a problem also, depending on how we interpret the question.

The source/load impedance is 50 ohms, so you would think that they want to match the input and output impedances with the network also. But that doesnt seem to be the case because i dont think there is any way to get 50 ohms input impedance with a 50 ohm load and still get the Butterworth response.

I get the same value for L as Papa, but when i calculate the input impedance Zin i get around 57.7 Ohms. A little mismatch, so i guess that's just too bad
I do get about -12db at 2 times the -3db frequency though.

It is not true that you want to match source and load impedances in a filter. You want the impedance to be such that you get the desired characteristic; in this case maximally flat in the passband and less than or equal to some value in the transition band. Impedance matching is primarily done to transfer the maximum amount of power from the source to the load. In most filtering applications power transfer is irrelevant.

 

Hi Sir, so this means my working is not correct when using 0.2251 table value?
Thank you

Either you took the value from a "standard" table (1.414) and divided it by the value of 2pi to it to get 0.2251 or you used a non-standard table and pulled the value 0.2251 from it without realizing what you were doing. You would do well to do all of your filter work with the "standard" tables and radian measure, converting back to Hz. as the last step. Others will be able to follow your work without so much confusion. That's just my opinion -- you do whatever you feel comfortable doing.

Here is what I consider to be the "standard" table

http://www.crbond.com/papers/btf2.pdf

 

Either you took the value from a "standard" table (1.414) and divided it by the value of 2pi to it to get 0.2251 or you used a non-standard table and pulled the value 0.2251 from it without realizing what you were doing. You would do well to do all of your filter work with the "standard" tables and radian measure, converting back to Hz. as the last step. Others will be able to follow your work without so much confusion. That's just my opinion -- you do whatever you feel comfortable doing.

Here is what I consider to be the "standard" table

http://www.crbond.com/papers/btf2.pdf

Hi Sir. So this means my working is still correct?
You are correct I just pulled that value from the table.
Thank you.