Question 6 of 6: Kaiser-window highpass design — recovering the specifications and repairing the error at $\omega=\pi$

Question 6: Kaiser-window highpass design — recovering the specifications and repairing the error at $\omega=\pi$ (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 Kaiser-window highpass design with $\omega_{c} = 0.6\pi$ rad/sample, $\beta = 3.86$ and $M = 51$, the two Kaiser design formulas above, and the observation that the design meets its specifications everywhere except near $\omega = \pi$.

Find. (a) $\delta$, $\omega_{s}$ and $\omega_{p}$ — i.e. the specification set that produced these two numbers; and (b) what must change so the final design meets the specification at every frequency, with the reason.

Approach. Run both Kaiser formulas backwards: invert the $\beta$ formula for $A$, hence $\delta$; invert the $M$ formula for $\Delta\omega$; then place the two corner frequencies symmetrically about $\omega_{c}$. For (b), test the parity of $M$ against the four types of generalized linear-phase FIR filter.

1. Recover $A$ from $\beta$. Since $0 \lt \beta = 3.86$, the middle branch applies. Solving $$0.5842\,(A-21)^{0.4} + 0.07886\,(A-21) = 3.86$$ numerically gives $A - 21 = 22.99$, i.e. $$\boxed{A \approx 44\ \text{dB}}$$ (check: $A = 44$ returns $0.5842(23)^{0.4} + 0.07886(23) = 3.861$, matching the quoted $\beta = 3.86$ to three figures). Because $A = 44 \le 50$, the branch choice is self-consistent.
2. Convert $A$ to the tolerance $\delta$. From $A = -20\log_{10}\delta$, $$\boxed{\delta = 10^{-A/20} = 10^{-2.2} = 6.31\times 10^{-3}}$$ Note that the Kaiser method uses one number for both bands: the passband ripple and the stopband ripple are equal, $\delta_{1} = \delta_{2} = \delta$.
3. Recover the transition width $\Delta\omega$ from $M$. Rearranging the order formula, $$\Delta\omega = \frac{A - 8}{2.285\,M} = \frac{44 - 8}{2.285 \times 51} = \frac{36}{116.5} = 0.3088\ \text{rad/sample} = 0.0983\pi .$$
4. Place the corner frequencies. In the Kaiser (windowed ideal-filter) method the cutoff of the ideal prototype sits at the centre of the transition band, so $\omega_{s} = \omega_{c} - \Delta\omega/2$ and $\omega_{p} = \omega_{c} + \Delta\omega/2$ — and for a highpass filter the stopband is the lower band. Hence $$\boxed{\omega_{s} = 0.6\pi - 0.1544 = 1.7305\ \text{rad} = 0.551\pi}$$ $$\boxed{\omega_{p} = 0.6\pi + 0.1544 = 2.0394\ \text{rad} = 0.649\pi}$$ and $\omega_{p} - \omega_{s} = \Delta\omega$ as required.
5. (a) State the full specification. The design targets $$\left|H(e^{j\omega})\right| \le 6.31\times 10^{-3} \qquad 0 \le |\omega| \le 0.551\pi \quad\text{(stopband)},$$ $$1 - 6.31\times10^{-3} \le \left|H(e^{j\omega})\right| \le 1 + 6.31\times10^{-3} \qquad 0.649\pi \le |\omega| \le \pi \quad\text{(passband)},$$ with no constraint in the transition band $0.551\pi \lt |\omega| \lt 0.649\pi$; equivalently 44 dB stopband attenuation and $\pm 0.055$ dB passband ripple.

6. (b) Diagnose the failure at $\omega = \pi$. With $M = 51$ the impulse response occupies $0 \le n \le M$, so its length is $M+1 = 52$, an even number, and the symmetry point sits at $M/2 = 25.5$ — halfway between two samples. A window-designed filter inherits the symmetry $h[n] = h[M-n]$ of the ideal prototype, so an even-length symmetric response is a Type II generalized linear-phase FIR filter. Every Type II filter has a structural zero at $z = -1$: $$H\!\left(e^{j\pi}\right) = \sum_{n=0}^{M}h[n](-1)^{n} = 0 \qquad\text{identically, for any } h[n]=h[M-n] \text{ with } M \text{ odd},$$ because the samples pair up as $h[n]$ and $h[M-n]$ with opposite signs of $(-1)^{n}$ (the exponents $n$ and $M-n$ have opposite parity when $M$ is odd) and cancel exactly. A highpass filter, however, is required to have $\left|H(e^{j\pi})\right| \approx 1$. The response is therefore forced to zero at the very frequency where it must be unity, which is precisely the reported symptom: the error grows rapidly in the neighbourhood of $\pi$ and violates the tolerance no matter how large $\beta$ or $M$ is made. It is a structural defect, not a windowing accuracy problem.
7. (b) The repair: make $M$ even. What else is required is therefore a change of parity, not a change of window: $$\boxed{\text{take } M \text{ even — the next value is } M = 52 \text{ (length } 53) \text{, keeping } \beta = 3.86 }$$ Then $M/2 = 26$ is an integer, the symmetry point falls on a sample, and the design is a Type I filter, whose amplitude response is unconstrained at both $\omega = 0$ and $\omega = \pi$. Nothing else needs to move: the order formula with $M = 52$ gives $\Delta\omega = 36/(2.285\times 52) = 0.3029$ rad, slightly narrower than required, so the transition-band and stopband specifications remain satisfied with a little margin, and the realised $\left|H(e^{j\pi})\right|$ comes out within the $\pm\delta$ passband tolerance. Equivalently, one may design a lowpass prototype and use the spectral-inversion identity $h_{hp}[n] = \delta[n - M/2] - h_{lp}[n]$, which is only defined for even $M$ — the same requirement seen from the other side.
8. Confirm the fix. Evaluating both designs on the unit circle: the $M = 51$ (Type II) response falls monotonically to exactly zero at $\omega = \pi$, whereas the $M = 52$ (Type I) response holds $\left|H(e^{j\pi})\right| = 1.0001$, inside the specification. Across the stopband $0 \le \omega \le 0.551\pi$ the $M = 52$ design stays below $\delta$, and its passband ripple over $0.649\pi \le \omega \le \pi$ stays within $\pm\delta$, so all specifications are met at every $\omega$.