Question 6 of 6: Reading a Kaiser-window highpass design backwards

Question 6: Reading a Kaiser-window highpass design backwards (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A completed Kaiser-window highpass design whose outputs are quoted, together with the two Kaiser design formulas.

Find. (a) the tolerance $\delta$ and the two band-edge frequencies $\omega_s$ and $\omega_p$; (b) what must still be changed so the design meets its specification at every frequency including $\omega = \pi$.

Approach. Both Kaiser formulas are invertible: $\beta$ gives $A$, $A$ gives $\delta$, and $A$ together with $M$ gives the transition width $\Delta\omega$. The windowed prototype places its cut-off at the midpoint of the transition band, so $\omega_c$ and $\Delta\omega$ fix the two band edges. Part (b) is then a question about the parity of $M$.

1. Invert the $\beta$ formula to recover $A$. The quoted $\beta = 5.653$ is large, so try the top branch, which is valid for $A \gt 50$: $$\beta = 0.1102(A - 8.7) \;\Longrightarrow\; A = \frac{\beta}{0.1102} + 8.7 = \frac{5.653}{0.1102} + 8.7 = 60.0 \ \text{dB}.$$ The result satisfies $A \gt 50$, so the branch choice was consistent. Substituting back, $0.1102(60 - 8.7) = 5.653$, which confirms the round trip.
2. Convert the attenuation into a tolerance. By definition $A = -20\log_{10}\delta$, so $$\boxed{\,\delta = 10^{-A/20} = 10^{-60/20} = 10^{-3} = 0.001\,}$$ The Kaiser method produces essentially equal ripple in both bands, so this one number is both the passband tolerance and the stopband tolerance: the passband must stay within $1 \pm 0.001$ and the stopband must not exceed $0.001$, i.e. 60 dB of attenuation.
3. Invert the order formula to recover the transition width. Using the unrounded value $M = 96.58$ (the rounded 97 would introduce a small error), $$\Delta\omega = \frac{A - 8}{2.285\,M} = \frac{60 - 8}{2.285 \times 96.58} = \frac{52}{220.69} = 0.2356\ \text{rad/sample}.$$ Expressed in units of $\pi$ this is a round number, $$\boxed{\,\Delta\omega = 0.075\pi\ \text{rad/sample}\,}$$ and substituting it back reproduces $M = 96.58$ exactly.
4. Place the band edges. A windowed design puts the prototype cut-off in the middle of the transition band, not at its edge, so $\omega_c = \tfrac{1}{2}(\omega_s + \omega_p)$ and $\Delta\omega = |\omega_p - \omega_s|$. For a highpass the stopband is the lower band, so $\omega_s \lt \omega_c \lt \omega_p$: $$\omega_s = \omega_c - \frac{\Delta\omega}{2} = 0.525\pi - 0.0375\pi, \qquad \omega_p = \omega_c + \frac{\Delta\omega}{2} = 0.525\pi + 0.0375\pi.$$
5. Report the specification. Carrying out the arithmetic, $$\boxed{\,\omega_s = 0.4875\pi = 1.5315\ \text{rad/sample}, \qquad \omega_p = 0.5625\pi = 1.7671\ \text{rad/sample}\,}$$ so the complete specification reads: $|H(e^{j\omega})| \le 0.001$ for $0 \le \omega \le 0.4875\pi$, and $0.999 \le |H(e^{j\omega})| \le 1.001$ for $0.5625\pi \le \omega \le \pi$, with the transition band left unconstrained.
6. Part (b): diagnose the failure at $\omega = \pi$. The design returned $M = 97$, an odd value, so the impulse response has length $M + 1 = 98$, an even number of samples. A symmetric response of even length is a Type II FIR filter, and its axis of symmetry $M/2 = 48.5$ falls between two samples. That is not a cosmetic detail: for a Type II filter $$H(e^{j\pi}) = \sum_{n=0}^{M} h[n](-1)^{n} = 0 \quad\text{identically},$$ because $n$ and $M - n$ have opposite parity whenever $M$ is odd, so the symmetric pairs $h[n]$ and $h[M-n]$ enter the sum with opposite signs and cancel term by term.
7. State the remedy. A highpass filter must have gain close to unity at $\omega = \pi$, so a Type II response can never meet the specification there no matter how large $\beta$ or $M$ becomes. What is required is a change of parity: $$\boxed{\text{take } M \text{ EVEN} \;-\; \text{here } M = 98, \text{ length } 99, \text{ a Type I filter}}$$ Type I has an integer symmetry point $M/2 = 49$, imposes no structural zero at $z = -1$, and keeps the same $\beta = 5.653$.
8. Confirm the rest of the specification still holds. Increasing $M$ from 96.58 to 98 can only help, because the transition width the design achieves is inversely proportional to $M$: $\Delta\omega = 52/(2.285 \times 98) = 0.2322$ rad $= 0.0739\pi$, which is narrower than the required $0.075\pi$. The stopband attenuation is controlled by $\beta$ alone and is unchanged at 60 dB. So the single change $M = 97 \to 98$ fixes the neighbourhood of $\pi$ while leaving every other requirement satisfied with a small margin to spare.