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I am having trouble understanding the exact derivation of the butterworth filter and how it results in the output of the poles. I have researched multiple lecture series and textbooks and this is my understanding so far:

  1. An idealised low-pass filter, i.e. a brick wall, can be realised with the equation below as as $n$ tends to infinity? $$ \lvert H(j\omega)\rvert = \frac{1}{\sqrt{1+\omega^{2n}}} $$

    This is the amplitude response - i.e. the magnitude of the frequency response?

  2. Now by squaring this amplitude response and setting $s = j\omega$ you get the following: $$ \lvert H(s)\rvert^2 = \frac{1}{1+\left(\frac sj\right)^{2n}} $$ What does squaring the magnitude of the frequency response get you? I understand that $\lvert X(f)\rvert^2 = X(f)\cdot X^*(f)$?

  3. The result of these two terms(/transfer functions?) $H(s)$ and $H(-s)$. Subsequently you are only interested in the $H(s)$, why? I understand it's poles lie on the left of $s$-plane and therefore the system is stable? but what about $H(-s)$?

  4. You then obtain the poles from the resulting derivations and get a resulting transfer function. This transfer function and its related co-efficients can be implemented physically with the appropriate R and C values?

Any help would be greatly appreciated!

Regards

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  • $\begingroup$ can't spend time on an answer now. you get the poles for the value of $s$ that sets the denominator to zero. $\endgroup$ – robert bristow-johnson Aug 13 '16 at 4:51
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Note that the Butterworth filter is only one of many possible approximations of an ideal low pass filter. It is a Taylor series approximation (at $\omega=0$) of the desired response, and, consequently, the magnitude response of the Butterworth filter is maximally flat at $\omega=0$.

Approximating the squared magnitude response (instead of the complex frequency response, or the magnitude response) has the advantage that the function to be approximated is an analytic real-valued function of a real-valued variable, which simplifies the approximation problem.

We have

$$|H(j\omega)|^2=H(j\omega)H^*(j\omega)=H(j\omega)H(-j\omega)=H(s)H(-s)\Big |_{s=j\omega}\tag{1}$$

because $H^*(j\omega)=H(-j\omega)$ for real-valued systems, i.e., for systems with a real-valued impulse response, or, equivalently, with real-valued transfer function coefficients.

Since for a Butterworth low pass filter

$$|H(j\omega)|^2=\frac{1}{1+\omega^{2N}}\tag{2}$$

we have

$$H(s)H(-s)=\frac{1}{1+(s/j)^{2N}}=\frac{1}{1+(-s)^{2N}}\tag{3}$$

From the squared magnitude function we need to derive the transfer function $H(s)$, which is the function that we actually want to implement. From $(3)$ we know that the poles of $H(s)H(-s)$ (i.e., the zeros of the denominator of $(3)$) lie one a circle in the complex plane. Since we can only implement a causal and stable transfer function, we choose the $N$ poles in the left half-plane for $H(s)$, and consequently, the mirrored poles in the right half-plane are the poles of $H(-s)$.

Now we know $H(s)$ (i.e., its numerator and denominator coefficients), and, after having chosen a cut-off frequency and a filter structure, we can compute the values of the passive elements necessary to implement the given transfer function.

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  • $\begingroup$ Hi Matt, thanks for answering my question. Think i've got my head round it now, Cheers! $\endgroup$ – ConfusedCheese Aug 14 '16 at 11:01

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