A Butterworth lowpass filter of order n, with a cutoff frequency at distance D0 from the origin, has the transfer function H(u, v) given by
\(\frac{1}{1+\left[\frac{D(u,v)}{D_0}\right]^{2n}}\)
A lowpass filter must pass low frequencies (\(H \to 1\) near the origin) and attenuate high frequencies (\(H \to 0\) far from it). D(u, v) is the distance of the point (u, v) from the centre of the frequency rectangle.
The Butterworth lowpass filter of order n is \(H(u,v) = \frac{1}{1+\left[\frac{D(u,v)}{D_0}\right]^{2n}}\). At \(D = 0\), \(H = 1\); as \(D \to \infty\), \(H \to 0\); and at the cutoff \(D = D_0\), \(H = \frac{1}{2}\).
With the ratio inverted, \(\frac{D_0}{D(u,v)}\) becomes huge near the origin, so H is close to 0 at low frequencies and close to 1 at high ones: that is the Butterworth highpass filter, not a lowpass one.
The exponent of the Butterworth response is 2n (the squared magnitude response of an n-th order Butterworth filter), so the form with power n does not describe a filter of order n.
Hence, the answer is \(\frac{1}{1+\left[\frac{D(u,v)}{D_0}\right]^{2n}}\).
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