A Butterworth lowpass filter of order n, with a cutoff frequency at distance…
2014
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
- A.
\(\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{2n}}\) - B.
\(\frac{1}{1+\left[\frac{D(u, v)}{D_{0}}\right]^{n}}\) - C.
\(\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{2n}}\) - D.
\(\frac{1}{1+\left[\frac{D_{0}}{D(u, v)}\right]^{n}}\)
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Correct answer: A
Answer: The Butterworth lowpass filter of order n with cutoff at D0 is H(u, v) = 1 / (1 + [D(u, v)/D0]^{2n}).
Definition: The one-dimensional Butterworth magnitude response is given by 1 / (1 + (f/fc)^{2n}); in two dimensions replace the frequency variable f by the distance D(u, v) from the origin to get the image-domain form.
Distance meaning: D(u, v) denotes the distance from the origin in the frequency plane (commonly D(u, v) = sqrt(u^2 + v^2) when the center is at the origin).
Role of n: The filter order n controls the steepness of the transition between passband and stopband. Larger n yields a sharper cutoff.
Why the other formulas are incorrect:
Using exponent n instead of 2n (for example 1 / (1 + [D(u, v)/D0]^n)) does not match the Butterworth definition and gives a different frequency response.
Inverting the ratio (for example 1 / (1 + [D0/D(u, v)]^{2n})) changes the filter to a highpass-like behavior rather than lowpass; combining inversion with the wrong exponent compounds the errors.