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I have heard that said, and it sounds vaguely plausible, but my maths is not up to proving it. Here is why: The Ideal n-th order Butterworth response, in terms of S-parameters, is $$ |s21|^2 = \frac{1}{1 + \omega^{2n}} $$ On the other hand, the Chebychev ideal resonse is $$ |s21|^2 = \frac{1}{1 + \epsilon^2 T_n^{2}(\omega)} $$ As I see it, as \$\epsilon\$ approaches zero, \$|s21|^2\$ approaches one identically. Is my initial assumption wrong, or is there some mathematical magic that can come up with a different form of the limit for Chebychev with \$\epsilon\$ approaching zero?

For example, I can imagine the (semi-)circular pole-zero plot for Butterworth "morphing" from a circle into an ellipse, but I can't relate that to the transfer function.

[EDIT] On thinking about this more, perhaps the comparison can only be made when the cutoff specifications for the two types are identical (3db frequency, 1db etc). In this case, the "cutoff frequency" for zero ripple Chebychev would be \$\infty\$, making a direct comparison impossible.

I think this just needs a restatement of the Butterworth power transfer function as follows $$ |s21|^2 = \frac{1}{1 + k\omega^{2n}} $$ to allow a cutoff value other than 3db and enable the direct comparison where appropriate.

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  • \$\begingroup\$ I don't understand your question title. \$\endgroup\$
    – Andy aka
    Commented Oct 17 at 14:49
  • \$\begingroup\$ @Andyaka I have read somewhere that the Butterworth filter is the limiting case of Chebychev, when \$\epsilon\$ approaches zero. I would like to know whether this is true, and if so how to show it. \$\endgroup\$
    – m4r35n357
    Commented Oct 17 at 15:25
  • \$\begingroup\$ My advice is that you find that assertion on line and link it into your question. \$\endgroup\$
    – Andy aka
    Commented Oct 17 at 17:30
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    \$\begingroup\$ FWIW, I understood the title and question perfectly (I think); but I don't have the analytical theory necessary to address it. \$\endgroup\$ Commented Oct 17 at 19:57
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    \$\begingroup\$ Perhaps these snaps from Section 8.5 of Van Valkenburg may assist you in some way? 1 2 3 4 5 \$\endgroup\$
    – Ste Kulov
    Commented Oct 18 at 19:16

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Compare a low-pass Cheby filter with a low-pass Butterworth filter. Both have a -3dB point near 100 MHz:

  • Chebychev:
  • passband ripple: 0.000005 (a fairly small value above zero)
  • cutoff frequency: 16.3MHz (-3dB will be at a much higher frequency...100 MHz)
  • RL=RS=50 ohms (arbitrary choice)

The Butterworth filter has similar RL,RS, and a simple 100 MHz cutoff frequency of 100 MHz. That is, the -3dB point is at 100 MHz.:

schematic

simulate this circuit – Schematic created using CircuitLab

These examples seem to converge.

The problem is that convergence fails when Chebychev passband ripple input specification goes to 0dB, as it should to match flat Butterworth passband. In this example, Chebychev passband ripple has been made very, very small, and cutoff frequency adjusted to 16.3 MHz (which results in a -3dB point near 100 MHz).

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    \$\begingroup\$ The empirical result seems motivating -- but the question asks for an answer in terms of poles; can this be proven analytically? \$\endgroup\$ Commented Oct 17 at 19:56

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