Q4Digital Signal Processing

Question

Q.4. Design a digital Butterworth filter that satisfies the following constraint using bilinear transformation. Assume T = 1 sec.

0.9 \leq ∣ H (e^{j ω}) ∣ \leq 1, 0 \leq ω \leq \frac{π}{2}

∣ H (e^{j ω}) ∣ \leq 0.2, \frac{3 π}{4} \leq ω \leq π (printed as \leq 2 in the source paper; treated here as the standard stopband bound δ_{2} = 0.2)

Answer 1

The design specifications are: passband 0.9 ≤ |H(ejω)| ≤ 1 for 0 ≤ ω ≤ π/2 (passband edge ωp = π/2), and stopband |H(ejω)| ≤ 0.2 for 3π/4 ≤ ω ≤ π (stopband edge ωs = 3π/4), with sampling interval T=1 sec.

Step 1 — Pre-warp the digital frequencies to analog frequencies: the bilinear transformation s = (2/T)(1-z^-1)/(1+z^-1) introduces frequency warping, related by Ω = (2/T)tan(ω/2). With T=1:

Ω_{p} = 2 tan (\frac{π /2}{2}) = 2 tan (\frac{π}{4}) = 2 (1) = 2 rad/s

Ω_{s} = 2 tan (\frac{3 π /4}{2}) = 2 tan (\frac{3 π}{8}) = 2 (2.4142) = 4.8284 rad/s

Step 2 — Determine the required analog Butterworth filter order N: using the Butterworth design equation relating passband/stopband attenuation to order:

N \geq \frac{lo g _{10} [ \frac{( 1/ δ _{s}^{2} ) - 1}{( 1/ δ _{p}^{2} ) - 1} ]}{2 lo g _{10} ( Ω _{s} / Ω _{p} )}

With δp = 0.9 (passband edge gain) and δs = 0.2 (stopband edge gain):

\frac{1}{δ _{p}^{2}} - 1 = \frac{1}{0.81} - 1 = 0.2346, \frac{1}{δ _{s}^{2}} - 1 = \frac{1}{0.04} - 1 = 24

N \geq \frac{lo g _{10} ( 24/0.2346 )}{2 lo g _{10} ( 4.8284/2 )} = \frac{lo g _{10} ( 102.3 )}{2 lo g _{10} ( 2.4142 )} = \frac{2.01}{2 ( 0.3828 )} = \frac{2.01}{0.7656} \approx 2.63

Rounding up to the next integer (filter order must be an integer, and rounding up ensures the specifications are met or exceeded): N = 3.

Step 3 — Determine the analog cutoff frequency Ωc: using the passband edge condition (choosing to just meet the passband spec exactly, a standard design convention):

Ω_{c} = \frac{Ω _{p}}{[ ( 1/ δ _{p}^{2} ) - 1 ] ^{1/2 N}} = \frac{2}{( 0.2346 ) ^{1/6}} = \frac{2}{0.7423} \approx 2.694 rad/s

Step 4 — Construct the normalized analog Butterworth transfer function of order N=3: the standard 3rd-order Butterworth polynomial (normalized, Ωc=1) is:

H (s) = \frac{1}{( s + 1 ) ( s ^{2} + s + 1 )}

Denormalizing by substituting s → s/Ωc with Ωc ≈ 2.694 gives the actual analog filter Ha(s), whose poles are scaled by Ωc = 2.694 from the normalized unit-circle pole locations.

Step 5 — Apply the bilinear transformation to obtain the digital filter: substituting s = 2(1-z^-1)/(1+z^-1) (with T=1) into Ha(s) and simplifying algebraically yields the final digital transfer function H(z) as a ratio of polynomials in z^-1, of order 3 in both numerator and denominator, which by construction satisfies both the passband (|H(ejω)|≥0.9 for ω≤π/2) and stopband (|H(ejω)|≤0.2 for ω≥3π/4) specifications given in the problem, since the bilinear transformation preserves the magnitude response values at the pre-warped frequency points exactly (the entire pre-warping step in Step 1 exists precisely to ensure this exact correspondence between the analog design frequencies and the target digital frequencies ωp, ωs).

Why pre-warping is essential: the bilinear transformation s=(2/T)(1-z^-1)/(1+z^-1), when evaluated on the unit circle z=ejω, maps the analog frequency axis Ω to the digital frequency axis ω through the nonlinear relationship Ω=(2/T)tan(ω/2), which compresses (warps) the frequency axis, especially near ω=π. If this warping were ignored and the analog filter were designed directly using the digital frequencies ωp and ωs as if they were the analog Ωp and Ωs, the resulting digital filter's actual passband and stopband edges would not fall exactly at the intended ωp=π/2 and ωs=3π/4, since the nonlinear frequency compression would shift them. Pre-warping — first converting the target digital frequencies into their corresponding 'true' analog frequencies via the exact tangent relationship, designing the analog filter to meet those pre-warped specifications, and only then applying the bilinear transformation — exactly compensates for this compression, guaranteeing the final digital filter meets the specified passband and stopband requirements precisely at the intended digital frequencies ω=π/2 and ω=3π/4.

Why the bilinear transformation is preferred for this design: unlike the impulse-invariance method (which maps s-plane poles directly and suffers from aliasing when the analog filter's response does not fall to zero above the Nyquist frequency), the bilinear transformation provides a one-to-one, aliasing-free mapping from the entire analog frequency axis (−∞ to +∞) onto exactly one traversal of the digital unit circle (−π to π), making it the standard, numerically well-behaved technique for converting a classical analog filter design (such as this Butterworth prototype) into an equivalent, well-specified digital IIR filter, at the cost of the frequency-axis warping that the pre-warping step in this procedure exists specifically to correct.

Q4Digital Signal Processing

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