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4th order Butterworth TF; phase and gain equalization

Short answer

Normalized 4th-order T(s)=1/[(s^2+0.7654s+1)(s^2+1.8478s+1)] = 1/(s^4+2.6131s^3+3.4142s^2+2.6131s+1). Gain equalization adds a network with inverse magnitude response to flatten overall gain. Phase equalization uses all-pass networks to linearize phase (constant group delay) without altering magnitude.

Determine the transfer function of a normalized 4th order Butterworth low pass approximation. What do you mean by phase and gain equalization?

Normalized 4th-order Butterworth transfer function

With ε\varepsilon = 1 and ωc\omega_c = 1 rad/s, ∣T(jω)∣2=1/(1+ω8)|T(j\omega)|^2 = 1/(1 + \omega^8). The poles are the left-half-plane roots of 1+(−s2)4=01 + (-s^2)^4 = 0, on the unit circle:

pk=−sin⁡θk±jcos⁡θk,θk=(2k−1)π8=22.5°, 67.5°p_k = -\sin\theta_k \pm j\cos\theta_k, \qquad \theta_k = \frac{(2k-1)\pi}{8} = 22.5°,\ 67.5°
θk\theta_kPolesFactor s2+2sin⁡θks+1s^2 + 2\sin\theta_k s + 1
22.5°−0.3827±j0.9239-0.3827 \pm j0.9239s2+0.7654s+1s^2 + 0.7654s + 1
67.5°−0.9239±j0.3827-0.9239 \pm j0.3827s2+1.8478s+1s^2 + 1.8478s + 1
T(s)=1(s2+0.7654s+1)(s2+1.8478s+1)=1s4+2.6131s3+3.4142s2+2.6131s+1\begin{aligned} T(s) &= \frac{1}{(s^2 + 0.7654s + 1)(s^2 + 1.8478s + 1)} \\ &= \frac{1}{s^4 + 2.6131s^3 + 3.4142s^2 + 2.6131s + 1} \end{aligned}

Phase and gain equalization

  • Gain (amplitude) equalization: adding a network whose magnitude response is the inverse of the unwanted variation of a system (e.g. a cable or telephone line whose loss rises with frequency), so the overall gain becomes flat over the band. Example: line equalizers, audio graphic equalizers.
  • Phase (delay) equalization: adding all-pass networks, ∣T∣=1|T| = 1, whose phase/delay complements that of a filter or channel, so the total phase becomes nearly linear (constant group delay) without changing the magnitude response. Example: equalizing the delay peak of a Chebyshev filter for data transmission.

✓ Answer checked by Claude Sonnet 5.5

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