Bessel filter

3.4

The Bessel filter is given by the normalized transfer function

Normalized transfer function for the Bessel filter

where n is the order of the filter and qn are the Bessel polynomials

Normalized Bessel polynomials

with coefficients

Coefficients of the Bessel polynomials

The transfer function above is normalized (i.e., it is presented for the Bessel low pass filter with cutoff frequency 1). The Bessel low pass filter can be obtained from the transfer function above with the substitution S = s / ωc, where s = jω, ωc is the cutoff frequency of the filter, and ω is the angular frequency spanning the frequency spectrum between 0 and π. The substitution S = ωc / s produces the Bessel high pass filter. The substitution S = (s2 + ωc2) / (B s) produces the Bessel band pass filter, where ωc is the midpoint of the pass band and B is the width of the band. The substitution S = B s / (s2 + ωc2) produces the Bessel band stop filter.

The Bessel filter is said to have an almost flat group delay (delay of the amplitude envelope for various frequencies). In other words, the Bessel filter has close to the same delay for all frequencies.

Example: High pass Bessel filter of the third order

Set n = 3 and use the substitution S = ωc / s. The transfer function of the third order Bessel high pass filter is

Transfer function of the high pass Bessel filter of order three

The bilinear transformation s = 2 (z – 1) / (z + 1) allows us to rewrite the transfer function using the Z transform as follows.

Transfer function for the high pass Bessel filter of order three after the Z transform

Say that the cutoff frequency of the filter is ωc = 0.6 (technically, ωc = 2 arctan(0.6/2) ≈ 0.583, because of the warping of the frequency domain by the bilinear transformation). The transfer function of this example Bessel high pass filter is

Transfer function of an example high pass Bessel filter of order three

and the filter itself is

y(k) = 0.747496 x(k) – 2.242488 x(k – 1) + 2.242488 x(k – 2) - 0.747496 x(k – 3)
+ 2.435790 y(k – 1) - 1.995366 y(k – 2) + 0.548811 y(k – 3)

Suppose that the sampling frequency is 2000 Hz. The cutoff frequency then is ωc = (0.6 * 2000) / (2 π) = 191 Hz. The magnitude response of the filter is shown in the graph below.

Magnitude response of the example high pass Bessel filter of order three

Example: Band stop Bessel filter of the second order

Set n = 2 and S = B s / (s2 + ωc2). The transfer function of the second order band stop Bessel filter is

Transfer function for the band stop Bessel filter of order two

After the bilinear transformation s = 2 (z – 1) / (z + 1), the transfer function becomes

Transfer function for the band stop Bessel filter of order two after the Z transform

If, for example, we use the midpoint frequency ωc = 0.6 and the stop band width B = 1, and we scale the coefficients to obtain b0 = 1, then a0 = a4 = 0.654084, a1 = a3 = -2.184281, a2 = 3.131742, b0 = 1, b1 = -2.685262, b2 = 3.039987, b3 = -1.683299, b4 = 0.399923. If we suppose that the sampling frequency is 2000 Hz, then ωc = (0.6 * 2000) / (2 π) = 191 Hz, B = (1 * 2000) / (2 π) = 318 Hz and the magnitude response of the filter is as follows.

Magnitude response of an example band stop Bessel filter of order two

Phase response of the Bessel filter

The Bessel filter is said to have an almost flat group delay (delay of the amplitude envelope for various frequencies). In other words, the Bessel filter has close to a linear phase response. A comparison of the phase delay for the second order low pass Bessel filter and the second order low pass Butterworth filter is shown below.

Phase response of an example low pass Bessel filter and Butterworth filter of order two



  Rating
Rate This Page: Poor Great   |  Rate Content |
Average rating:  3.4   
10202
12345
Number of Ratings : 5
  Comments
Add Comment
No Comments Yet


Copyright 2006 by Kaliopa Publishing, LLC