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Bessel-Thomson characteristics and 3rd order TF

Short answer

The Bessel-Thomson filter features maximally flat group delay, minimal overshoot, and gradual magnitude roll-off. Its third-order transfer function, normalized for 1 s delay, is T(s) = 15/(s^3 + 6s^2 + 15s + 15).

What are the characteristics of Bessel-Thomson filter? Find the transfer function for 3rd order Bessel Thomson filter.

Characteristics of the Bessel-Thomson filter

The Bessel-Thomson filter is an all-pole low-pass filter designed for maximally flat group delay at ω=0\omega = 0 rather than a flat magnitude.

  • Maximally flat delay: τ(ω)\tau(\omega) is constant over the passband (as many derivatives as possible are zero at ω=0\omega=0), so the phase is almost linear.
  • Very little overshoot in the step response (under about 1%) and minimal ringing, so the pulse shape is preserved.
  • Gradual magnitude roll-off: the transition band is wider than for Butterworth or Chebyshev of the same order; the −20n-20n dB/decade slope is reached only far from the cutoff.
  • The denominator is a Bessel polynomial Bn(s)B_n(s): Bn=(2n−1)Bn−1+s2Bn−2B_n = (2n-1)B_{n-1} + s^2B_{n-2}.
  • It is normalized for a delay of 1 s; frequency scaling (s→sτ0s\to s\tau_0) sets a delay τ0\tau_0, and the 3 dB frequency then depends on nn.
  • Frequency transformation to high-pass or band-pass does not keep the constant-delay property.

Transfer function of the third-order Bessel-Thomson filter

Approach (Storey/Thomson): the ideal delay of 1 s is T(s)=e−s=1sinh⁡s+cosh⁡sT(s) = e^{-s} = \dfrac{1}{\sinh s + \cosh s}. Write

coth⁡s=cosh⁡ssinh⁡s=1s+13s+15s+17s+⋯\coth s = \frac{\cosh s}{\sinh s} = \frac{1}{s} + \cfrac{1}{\cfrac{3}{s} + \cfrac{1}{\cfrac{5}{s} + \cfrac{1}{\cfrac{7}{s}+\cdots}}}

Truncating after nn terms gives coth⁡s≈m(s)/n(s)\coth s \approx m(s)/n(s), where mm (even) approximates cosh⁡s\cosh s and nn (odd) approximates sinh⁡s\sinh s. Then T(s)≈K/[m(s)+n(s)]T(s) \approx K/[m(s)+n(s)].

For order 3 keep the first three terms (1/s1/s, 3/s3/s, 5/s5/s). First simplify the inner part:

3s+15/s=3s+s5=s2+155scoth⁡s≈1s+5ss2+15=s2+15+5s2s(s2+15)=6s2+15s3+15s\begin{aligned} \frac{3}{s} + \frac{1}{5/s} &= \frac{3}{s} + \frac{s}{5} = \frac{s^2+15}{5s} \\ \coth s &\approx \frac{1}{s} + \frac{5s}{s^2+15} = \frac{s^2 + 15 + 5s^2}{s(s^2+15)} = \frac{6s^2+15}{s^3+15s} \end{aligned}

So m(s)=6s2+15m(s) = 6s^2 + 15 and n(s)=s3+15sn(s) = s^3 + 15s, and

B3(s)=m(s)+n(s)=s3+6s2+15s+15B_3(s) = m(s)+n(s) = s^3 + 6s^2 + 15s + 15

Choosing K=15K = 15 for unity DC gain:

T(s)=15s3+6s2+15s+15T(s) = \frac{15}{s^3 + 6s^2 + 15s + 15}

Check: the delay at DC is b1/b0=15/15=1b_1/b_0 = 15/15 = 1 s. The poles are s=−2.3222s = -2.3222 and s=−1.8389±j1.7544s = -1.8389 \pm j1.7544 (normalized to 1 s delay). For a delay of τ0\tau_0, replace ss by sτ0s\tau_0.

✓ Answer checked by Claude Sonnet 5.5

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