Question 2 of 7: FIR and IIR Notch Filter Design from a DFT Spectrum

Question 2: FIR and IIR Notch Filter Design from a DFT Spectrum (20 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. $F_s = 48$ kHz; a 64-point DFT of the sampled sequence; the two interfering components appear at bins $k = 8$ and $k = 16$, with their conjugate images at $k = 56$ and $k = 48$ as expected for a real sequence; a d.c. bin is also present.

Find. (a) the two analogue interferer frequencies; (b) the causal real-coefficient FIR notch $H_{notch}(z)$; (c) its pole-zero plot, a minimum-multiplier realisation and its operation count; (d) the frequencies blocked if the same filter is fed at 10 kHz; (e) a narrower-notch IIR replacement with its structure.

[Figure not reproduced: Figure 2 redrawn: magnitude of the 64-point DFT. The four heavy stems at k = 8, 16, 48 and 56 are the two interferers and their conjugate images; everything else is broadband signal content. See the official exam paper.]

Approach. Convert bin index to analogue frequency, convert that to digital radian frequency, place a conjugate pair of zeros on the unit circle at each unwanted frequency, then factor the result into second-order sections to minimise multipliers. For part (e), pull a matching pole in behind each zero along the same radial line.

1. (a) Map DFT bin index to analogue frequency. The $k$-th bin of an $N$-point DFT of a sequence sampled at $F_s$ sits at
   $$F_k = \frac{k F_s}{N} = \frac{k \times 48\,000}{64} = 750k \;\text{Hz}.$$

   Hence $F_1 = 750 \times 8 = 6\,000$ Hz and $F_2 = 750 \times 16 = 12\,000$ Hz. The stems at $k = 56$ and $k = 48$ are not extra tones: for real $x[n]$ the DFT is conjugate-symmetric, $X[N-k] = X^{*}[k]$, so each real sinusoid always shows up as a mirrored pair. Both frequencies satisfy the stated $0 \lt F_1 \lt F_2 \lt 24$ kHz, so neither is aliased.

   $$\boxed{F_1 = 6\;\text{kHz}, \qquad F_2 = 12\;\text{kHz}.}$$
2. (b) Place zeros on the unit circle at the offending digital frequencies. The digital radian frequencies are
   $$\omega_1 = \frac{2\pi F_1}{F_s} = \frac{2\pi (6000)}{48000} = \frac{\pi}{4}, \qquad \omega_2 = \frac{2\pi F_2}{F_s} = \frac{\pi}{2}.$$

   A zero exactly at $z = e^{j\omega_0}$ forces $H(e^{j\omega_0}) = 0$, which annihilates that tone completely. Real coefficients require zeros in conjugate pairs, and each pair contributes a real second-order factor:

   $$H_{notch}(z) = \prod_{i=1}^{2}\left(1 - 2\cos\omega_i\, z^{-1} + z^{-2}\right) = \left(1 - \sqrt{2}\,z^{-1} + z^{-2}\right)\left(1 + z^{-2}\right),$$

   using $2\cos(\pi/4) = \sqrt{2}$ and $2\cos(\pi/2) = 0$. Expanding,

   $$\boxed{H_{notch}(z) = 1 - \sqrt{2}\,z^{-1} + 2 z^{-2} - \sqrt{2}\,z^{-3} + z^{-4}.}$$

   All four taps are real and the filter is causal (only non-negative powers of $z^{-1}$), as required. The tap set $\{1, -\sqrt{2}, 2, -\sqrt{2}, 1\}$ is symmetric, so the filter also has exactly linear phase with a constant group delay of 2 samples.

3. (c) Pole-zero plot. The four zeros sit on the unit circle at $e^{\pm j\pi/4}$ and $e^{\pm j\pi/2}$; because $H_{notch}(z)$ is a fourth-order polynomial in $z^{-1}$, all four poles are stacked at the origin.
   Pole-zero diagram of H_notch(z): four unit-circle zeros at the notch frequencies, a fourth-order pole at the origin, ROC the whole plane except z = 0.

   Rather than implementing the five-tap transversal filter directly, keep the cascade of the two second-order sections: the first has the single non-trivial coefficient $-\sqrt{2}$, and the second has coefficients $\{1, 0, 1\}$, i.e. no multiplier at all. That is the minimum-multiplication structure.

   Minimum-multiplier realisation: two cascaded second-order transversal sections. Only one true multiplier appears, in section 1.

   Counting per output sample: section 1 forms $x[n] + x[n-2]$ (one addition) and then subtracts $\sqrt{2}\,x[n-1]$ (one multiplication and one subtraction); section 2 forms $w[n] + w[n-2]$ (one addition). Storage is two delay registers per section plus the one coefficient.

   $$\boxed{1 \text{ multiplication}, \quad 3 \text{ additions/subtractions}, \quad 4 \text{ delay registers} + 1 \text{ coefficient word.}}$$

   For comparison, the direct five-tap form needs four multiplications and four additions; exploiting the tap symmetry brings that to two multiplications and four additions. The cascade wins because the $\omega_2 = \pi/2$ section is coefficient-free, and even the remaining multiplier could be traded for a shift-and-add pair since $\sqrt{2} \approx 1 + \tfrac{1}{4} + \tfrac{1}{8} + \tfrac{1}{32}$ if a small coefficient error were acceptable.

4. (d) The filter blocks fixed digital frequencies. The zeros are pinned to $\omega = \pi/4$ and $\pi/2$ in the discrete-time domain; they do not know what sampling rate produced the samples. Re-mapping to analogue frequency at the new rate $F_s^{\prime} = 10$ kHz,
   $$F = \frac{\omega F_s^{\prime}}{2\pi} \;\Rightarrow\; F_1^{\prime} = \frac{(\pi/4)(10\,000)}{2\pi} = 1250 \text{ Hz}, \qquad F_2^{\prime} = \frac{(\pi/2)(10\,000)}{2\pi} = 2500 \text{ Hz}.$$

   So the same code now removes 1.25 kHz and 2.5 kHz — the notches scale with the sampling rate, which is exactly why a rate change silently de-tunes a fixed-coefficient filter.

   $$\boxed{F_1^{\prime} = 1.25\;\text{kHz}, \qquad F_2^{\prime} = 2.5\;\text{kHz.}}$$
5. (e) Add poles just behind the zeros to narrow the notches. The FIR notch is wide because there is nothing to restore the gain away from the null. Placing a pole at the same angle but at radius $r$ slightly less than 1 cancels the zero except very close to the notch frequency. Part (e) also asks for the d.c. component, so a real zero at $z = 1$ and a real pole at $z = r$ are added. With $r = 0.95$:
   $$H_{IIR}(z) = \frac{(1 - z^{-1})\left(1 - \sqrt{2} z^{-1} + z^{-2}\right)\left(1 + z^{-2}\right)}{(1 - 0.95 z^{-1})\left(1 - 1.34350 z^{-1} + 0.9025 z^{-2}\right)\left(1 + 0.9025 z^{-2}\right)},$$

   where $2r\cos(\pi/4) = 1.34350$ and $r^{2} = 0.9025$. The structure is elegant: every denominator coefficient is the corresponding numerator coefficient scaled by $r^{k}$, i.e.

   $$\boxed{H_{IIR}(z) = \frac{N(z)}{N(z/r)}, \qquad N(z) = 1 - 2.41421 z^{-1} + 3.41421 z^{-2} - 3.41421 z^{-3} + 2.41421 z^{-4} - z^{-5}.}$$

   All poles have radius $r = 0.95 \lt 1$, so the filter is stable, and the ROC $|z| \gt 0.95$ makes it causal.

   Cascade realisation of the IIR notch: a first-order d.c. section followed by two second-order notch sections.

   Each second-order section is implemented canonically in Direct Form II, which needs only two delay elements because the feedback and feedforward arms share them:

   Section 2 of the IIR notch in canonic Direct-Form II. The same skeleton, with its own coefficients, realises each section.

   The improvement is quantitative, not merely qualitative. Measuring the −3 dB width of the $\pi/4$ notch against the local passband level between the two nulls gives $0.4145$ rad/sample for the FIR ($\approx 3167$ Hz at 48 kHz) but only $0.0999$ rad/sample for the IIR ($\approx 763$ Hz) — a factor of $4.15$ narrower, so far less of the wanted signal is destroyed alongside the interference.

   Magnitude responses on a dB scale, each normalised to its own peak. The FIR notch (grey, dashed) removes a wide band around each tone; the IIR notch (blue) is nearly flat everywhere except in two very narrow nulls.