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I've been trying to understand this exercise proposed by my teacher.

He asked us to find a combination of 2 ideal filters to generate a filter within the conditions of the image below. (That means, band-stop,band-pass,low band and high band)

That means, I need to reject 100-200 Hz and (here comes what I don't understand) reject with 50% of attenuation 300-400Hz.

Afterwards he asked to generate them with a Butterworth first order, which is not a problem, except that I don't know what kind of combination I should use.

Thank you all for your patience.

Cahier des Charges

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  • \$\begingroup\$ Does your problem restrict you to only using first order Butterworth filters, or are other types allowed? \$\endgroup\$ – alphasierra Mar 7 at 23:21
  • \$\begingroup\$ Any Ideal filter. After I have to find a Butterworth equivalent. \$\endgroup\$ – Lucas Tonon Mar 8 at 6:53
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You can't have a 1st order bandpass or bandstop, so he must've meant to combine the two 1st orders to give a resultant 2nd order. Then, if you account that a Butterworth lowpass prototype is

$$H(s)=\frac{1}{s^2+\sqrt2s+1}$$

then the resultant bandstop would be

$$H_1(s)=\frac{s^2+1}{s^2+\sqrt2s+1}$$

For the second one, you have to ensure that the transfer function doesn't attenuate ad infinitum, and that also has to happen for a specific bandwidth, in a bandstop configuration. So introduce the \$s\$ term in the numerator so that the ratios of attenuations are ½ at the desired frequency, and 1 in rest:

$$H_2(s)=\frac{s^2+\frac{\sqrt2}2+1}{s^2+\sqrt2s+1}$$

If you plot the frequency response of \$H_1(s)H_2(s)\$, with \$f_1=0.1\text{Hz}\$ and \$f_2=10\text{Hz}\$, this is what comes up:

freq

Notice that due to the relative proximity of the center frequencies and the low orders, the filters interact with each other to give a less than unity response at \$1\text{Hz}\$, but this is to be expected. Also, if your teacher actually wants to fulfill the requirements as seen in the picture, an order (much) greater than 2 would be needed. How much? How to build them? That's for you to figure out. Note that I also didn't include explicit bandwidth definitions for them, I'll also leave that detail to you, it should come out easy from frequency transformations.


[edit for the comment]

To get to this 2nd order bandpass, or bandstop, you start from the 1st order lowpass prototype:

$$H(s)=\frac1{s+1}$$

and carry on with frequency transformations, which generates a bandpass/bandstop having a central frequency and a bandwidth. For example, for a bandstop you'd have:

$$H(\frac{BW s}{s^2+\omega_0^2})=\frac{s^2+\omega_0^2}{s^2+BW s+\omega_0^2}$$

which, represented as a generic 2nd order transfer function, is:

$$H(s)=\frac{a_2s^2+a_1s+a0}{b_2s^2+b_1s+b_0}$$

but with \$a_1=0\$. Some use \$b\$ for numerator, it doesn't matter, really, adjust accordingly. In simple terms, the \$a_1\$ and \$b_1\$ terms define the quality factor, so for the case of interest, bandstop, the quality factor is infinite in the numerator (\$a_0\rightarrow 0\$), and the ratio \$\frac{a_1}{b_1}=0\$, which shows the attenuation at the central frequency. If you could force a specific attenuation, you would have to impose a value for \$a_1\$, and for this case, in particular, you need minimum ½, so then the ratio becomes:

$$\frac{a_1}{b_1}=\frac{a_1}{\sqrt2}=\frac12 \Rightarrow a_1=\frac{\sqrt2}2$$

Don't forget that this is valid for a 2nd order transfer function, for higher orders it gets messy: you no longer have a lowpass prototype from which you can derive the transfer function, you have a shelf lowpass, which is a different thing. But the frequency transformations go just for any other case, it's just that the result will be more "exotic".

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  • \$\begingroup\$ Thank you very much, the only thing I didn't quite grasp was how to regulate the attenuation in the second filter. Would you explain it please? I mean, how can I set methodically the numerator willing to ajust the attenuation? \$\endgroup\$ – Lucas Tonon Mar 8 at 12:09
  • \$\begingroup\$ Thank you once again, that was really helpful =) \$\endgroup\$ – Lucas Tonon Mar 8 at 19:45
  • \$\begingroup\$ I'm quite impressed how well it worked, is it there any bibliography I can find this content with more exploited mathematical explanations? \$\endgroup\$ – Lucas Tonon Mar 8 at 19:56
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    \$\begingroup\$ @LucasTonon No need to worry about details, for now, you'll learn them as you go. If you think this answered your problem, mark it down (the check mark) so that others, in the future, searching for similar matters, can find the answer easier. \$\endgroup\$ – a concerned citizen Mar 8 at 20:11
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    \$\begingroup\$ @LucasTonon There's really not much to add. The numerator has infinite quality factor for default 2nd order bandstop (\$s^2+1\$), which means the zeroes are purely imaginary, thus straight on the \$j\omega\$ axis => infinite attenuation. The moment you impose a value for \$a_1\$, the roots become a complex conjugate pair => the quality factor is reduced => the attenuation is less. How much less? Numerator divided by denominator less. Also, you should try, at its simplest, num=[1,(f2-f1)*alfa,4*pi^2*f1*f2], den=[1,f2-f1,4*pi^2*f1*f2], alfa variable, can be greater than 1. \$\endgroup\$ – a concerned citizen Mar 9 at 16:47

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