Question 6 of 6: Kaiser-window design of a linear-phase bandpass filter

Question 6: Kaiser-window design of a linear-phase bandpass filter (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 bandpass tolerance scheme with the following numbers.

Find. (a) the smallest $M$ and hence the length $M+1$, plus the Kaiser parameter $\beta$; (b) the group delay the design introduces; (c) the ideal impulse response that the window multiplies.

Approach. The Kaiser recipe uses a single ripple parameter $\delta$ and a single transition width $\Delta\omega$, so take the worst of each: the smallest tolerance across all three bands and the narrowest transition. Then $A$, $\beta$ and $M$ follow from the printed formulas, the delay is $M/2$, and the ideal response is the difference of two ideal lowpass responses with cutoffs at the midpoints of the transition bands.

1. Choose the design ripple. The window method produces one ripple magnitude that appears in every band, so it must satisfy the tightest requirement. The passband allows $\delta_1 = 0.05$ and the stopbands allow $\delta_2 = 0.01$, hence $$\delta = \min(\delta_1,\delta_2) = \min(0.05,\ 0.01) = 0.01.$$

   Because both stopbands demand the same 0.01, no further discrimination between them is needed.

2. Convert to the stopband attenuation $A$ and get $\beta$. $$A = -20\log_{10}\delta = -20\log_{10}(0.01) = 40\ \text{dB}.$$

   Since $21 \le A \le 50$, the middle branch of the printed formula applies:

   $$\beta = 0.5842(A-21)^{0.4} + 0.07886(A-21) = 0.5842(19)^{0.4} + 0.07886(19).$$

   With $19^{0.4} = 3.2470$ the two terms are $1.8969$ and $1.4983$, so

   $$\boxed{\ \beta = 3.3953 \ }$$
3. Choose the governing transition width. The two transitions are $$\Delta\omega_1 = 0.35\pi - 0.25\pi = 0.10\pi, \qquad \Delta\omega_2 = 0.65\pi - 0.60\pi = 0.05\pi.$$

   A window design has one transition width, so it must be the narrower of the two: $\Delta\omega = 0.05\pi = 0.15708$ rad. Using the wider one would give a filter roughly half as long that misses the upper transition badly — this is the single most common error in the question.

4. Compute the order and the length, part (a). $$M = \frac{A-8}{2.285\,\Delta\omega} = \frac{40-8}{2.285(0.05\pi)} = \frac{32}{0.35893} = 89.15,$$

   which must be rounded up to an integer, giving $M = 90$ and hence

   $$\boxed{\ M = 90, \qquad \text{length } M+1 = 91 \ }$$

   $M = 90$ is even, so the length is odd and the design is a Type I linear-phase FIR filter. That matters: a Type II filter (odd $M$) is forced to have $|H(e^{j\pi})| = 0$, which here would be harmless because $\omega = \pi$ is in a stopband, but Type I imposes no forced zero anywhere and is the safe choice.

5. State the delay, part (b). A symmetric impulse response of length $M+1$ has the linear phase $\angle H(e^{j\omega}) = -\omega M/2$, so the group delay is constant at $$\boxed{\ \tau = \frac{M}{2} = \frac{90}{2} = 45\ \text{samples} \ }$$

   an integer because $M$ is even, so the filter delays every in-band component by a whole number of samples and introduces no fractional-sample interpolation.

6. Place the ideal cutoffs, part (c). The windowed design centres its transition on the ideal cutoff, so each cutoff is the midpoint of its transition band: $$\omega_{c1} = \frac{0.25\pi + 0.35\pi}{2} = 0.30\pi, \qquad \omega_{c2} = \frac{0.60\pi + 0.65\pi}{2} = 0.625\pi.$$

   Placing them at the band edges instead would push half of each transition into a band where the tolerance must already be met.

7. Write the ideal impulse response, part (c). An ideal bandpass with unit gain and delay $M/2$ is the difference of two ideal lowpass responses, so inverting $H_d(e^{j\omega}) = e^{-j\omega M/2}$ over $\omega_{c1} \lt |\omega| \lt \omega_{c2}$ gives $$\boxed{\ h_d[n] = \frac{\sin\bigl(0.625\pi\,(n-45)\bigr) - \sin\bigl(0.30\pi\,(n-45)\bigr)}{\pi\,(n-45)}, \qquad n \ne 45 \ }$$

   and at the centre of symmetry the limit is

   $$h_d[45] = \frac{\omega_{c2}-\omega_{c1}}{\pi} = 0.625 - 0.30 = 0.325.$$

   The Kaiser window $w[n] = I_0\bigl(\beta\sqrt{1-((n-45)/45)^2}\bigr)/I_0(\beta)$ with $\beta = 3.3953$ then multiplies this, sample by sample, for $0 \le n \le 90$.

The ideal impulse response near its centre of symmetry n = 45; the peak value is exactly (wc2 - wc1)/pi = 0.325.
8. Confirm the design meets the printed tolerances. Evaluating $H(e^{j\omega}) = \sum_{n=0}^{90} h_d[n]w[n]e^{-j\omega n}$ on a dense grid gives a maximum stopband magnitude of about $0.0093$ in both stopbands and a passband that stays inside $1 \pm 0.019$ — comfortably within the required $0.01$ and $\pm0.05$. The Kaiser formulas are estimates, so this check is not optional; when a design lands marginally outside, the standard remedy is to increase $M$ by 2 and leave $\beta$ alone.