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I noticed that for a low pass Butterworth filter, as the cut off frequency reduces, the poles get closer to the unit circle. Is it applicable to all filters and why is this so?

As the normalized cut off frequency reduces from $\pi/2$ to $0$ (close to $0$- really low cut off frequency), the poles get closer to unit circle (the distance between the poles and unit circle reduces monotonically). Similar behavior is observed when we increase the cut off frequency from $\pi/2$ to $\pi$ (close to $\pi$- really high cut off frequency), the poles get closer to unit circle again- at $\pi/2$ they seem to be farthest away from the unit circle.

I would really appreciate if anybody could help me understand why this is so intuitively and also the math behind this (if there is any). Thanks in advance.

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The Butterworth filter is in fact a continuous-time filter, and what you are seeing is very probably a bilinear transformation applied to this filter in order to get a discrete-time filter.

Software such as Octave or Matlab do this when you ask for a digital Butterworth filter.

The original Butterworth filter, in continuous time, $H(s)$ has all poles equally distributed around a semicircle in the left of the $s$ plane. The radius of this circle is the cutoff frequency (in $rad/s$).

The relationship between frequencies in discrete-time ($\theta$) and continuous-time ($\omega$) given by the bilinear transformation is: $\omega = \frac{2}{T} \tan (\theta/2)$ (where $T$ is an arbitrary sampling period used during design).

You can imagine the left $s$ semiplane as mapped into the unit circle in $z$, or the imaginary axis mapped to the unit circumference.

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  • $\begingroup$ Makes sense. As the cut off in digital domain moves towards pi or 0, in analog domain it will tend to 0 (inverse tan curve), which mean poles in analog domain tend to the origin which translates to unit circle in digital domain. Did I understand it correctly? Thank you so much. Truly appreciate it. $\endgroup$ – Curious91 Oct 1 '15 at 18:59

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