Design of FIR filters

Why it matters

FIR filters are the default choice for cleaning up sampled sensor data — removing mains hum, band-limiting before decimation, smoothing ECG or vibration records — because they are always stable and can have exactly linear phase, so pulse shapes are not distorted. Designing one means choosing a length and coefficients that meet a passband/stopband specification at an acceptable computing cost.

Key ideas

Structure. An FIR filter of length N (order N − 1) computes y[n] = Σ from k=0 to N−1 of h[k] x[n − k]. Its impulse response is the coefficient list, it has no feedback, all its poles are at z = 0, and it is therefore always stable.

Linear phase. If the coefficients are symmetric, h[n] = h[N − 1 − n], or antisymmetric, h[n] = −h[N − 1 − n], the phase is linear: every frequency is delayed by the same (N − 1)/2 samples, so the waveform shape in the passband is preserved.

• Type I: symmetric, N odd — any filter type (low-, high-, band-pass, band-stop).
• Type II: symmetric, N even — H(e^(jπ)) = 0, so no high-pass or band-stop.
• Type III: antisymmetric, N odd — zeros at ω = 0 and π; differentiators and Hilbert transformers only (band-pass).
• Type IV: antisymmetric, N even — zero at ω = 0; high-pass, differentiators.
• Zeros of a linear-phase filter occur in reciprocal pairs (z0 and 1/z0).

Window method

1. Specify the ideal response, e.g. a low-pass with cut-off ωc (rad/sample).
2. Its impulse response hd[n] = sin(ωc n)/(πn) is infinitely long and non-causal.
3. Truncate to N samples centred on n = 0 by multiplying by a window w[n], then delay by (N − 1)/2 to make it causal: h[n] = hd[n − (N − 1)/2]·w[n].

• Truncation causes ripples near the band edge (Gibbs phenomenon). With a rectangular window, the peak ripple stays fixed (about 9 %, giving only ~21 dB stopband attenuation) however large N is; increasing N only narrows the transition band.
• Tapered windows lower the sidelobes (more attenuation) at the cost of a wider transition band for the same N. The window chooses the attenuation; N chooses the transition width.

Typical window figures (approximate, as tabulated in most DSP textbooks — use your course's table):

• Rectangular: main-lobe width 4π/N, minimum stopband attenuation ≈ 21 dB, transition ≈ 1.8π/N.
• Hann (Hanning): 8π/N, ≈ 44 dB, ≈ 6.2π/N.
• Hamming: 8π/N, ≈ 53 dB, ≈ 6.6π/N.
• Blackman: 12π/N, ≈ 74 dB, ≈ 11π/N.
• Kaiser: adjustable through its parameter β.

Other methods. Frequency sampling (specify H at N equally spaced frequencies and take the IDFT) and optimal equiripple design (Parks–McClellan), which gives the shortest filter for a given specification.

FIR versus IIR. FIR: always stable, exact linear phase possible, robust to coefficient quantisation, but needs many more coefficients (and delay) for a sharp transition. IIR: few coefficients, sharp transitions, but non-linear phase and stability must be checked.

Formulas

• y[n] = Σ h[k] x[n − k] — FIR output.
• hd[n] = sin(ωc n)/(πn), hd[0] = ωc/π — ideal low-pass impulse response.
• h[n] = hd[n − (N − 1)/2] · w[n], n = 0 … N − 1 — windowed, causal design.
• w[n] = 0.54 − 0.46 cos(2πn/(N − 1)) — Hamming window; 0.5 − 0.5 cos(2πn/(N − 1)) — Hann.
• τg = (N − 1)/2 samples — group delay of a linear-phase FIR.
• ωc = (ωp + ωs)/2 — cut-off used for the ideal response.
• N ≈ (constant × π)/Δω, e.g. N ≈ 6.6π/Δω for Hamming — length from transition width.
• ω = 2πf/fs — digital frequency.
• H(e^(j0)) = Σ h[n] (DC gain); H(e^(jπ)) = Σ (−1)ⁿ h[n].

Symbols: N filter length (taps); h[n] coefficients (dimensionless); ωc, ωp, ωs cut-off, passband and stopband edges (rad/sample); Δω = ωs − ωp (rad/sample); f, fs (Hz); τg group delay (samples).

Worked examples

Example 1 (standard). Design a 7-tap low-pass FIR with ωc = π/2 using (a) a rectangular and (b) a Hamming window. Compare the DC gains.

1. Ideal response: hd[n] = sin(πn/2)/(πn), hd[0] = 0.5. For n = ±1, ±2, ±3: 0.3183, 0, −0.1061.
2. (a) Rectangular, delayed by 3: h = {−0.1061, 0, 0.3183, 0.5, 0.3183, 0, −0.1061} (symmetric ⇒ Type I, linear phase, delay 3 samples).
3. DC gain = Σ h = 0.924; at ω = π: |Σ(−1)ⁿ h[n]| = 0.076.
4. (b) Hamming for N = 7: w = {0.08, 0.31, 0.77, 1, 0.77, 0.31, 0.08}.
5. h = {−0.0085, 0, 0.2451, 0.5, 0.2451, 0, −0.0085}; DC gain = 0.973, gain at ω = π = 0.027.
6. Rectangular: DC gain 0.924, |H(e^(jπ))| = 0.076; Hamming: 0.973 and 0.027 — the window reduces ripple and improves attenuation, at the cost of a wider transition.

Example 2 (GATE level). A DAQ system samples at 8 kHz. Required: passband to 1 kHz, stopband from 1.5 kHz with at least 50 dB attenuation. Choose a window, the length, the cut-off and the delay.

1. 50 dB needs a window with at least that attenuation: Hamming (≈ 53 dB) suffices; Hann (≈ 44 dB) does not.
2. Transition width Δf = 1.5 − 1 = 0.5 kHz ⇒ Δω = 2π × 0.5/8 = 0.125π rad/sample.
3. N ≈ 6.6π/Δω = 6.6/0.125 = 52.8 ⇒ take the next odd value, N = 53 (Type I).
4. Cut-off ωc = 2π × 1.25/8 = 0.3125π rad/sample.
5. Delay = (53 − 1)/2 = 26 samples = 26/8000 s = 3.25 ms.
6. Hamming window, N = 53 taps, ωc = 0.3125π, delay 3.25 ms.

Common mistakes

• Forgetting to shift hd[n] by (N − 1)/2 — the filter is then non-causal.
• Expecting a longer rectangular window to remove the Gibbs overshoot. Only a different window lowers the ripple.
• Using a Type II (even-length symmetric) filter for a high-pass design — it has a forced zero at ω = π.
• Confusing "normalised frequency" conventions: f/fs, f/(fs/2) and ω = 2πf/fs all appear. State which you use.
• Taking the cut-off at the passband edge instead of mid-transition.
• Ignoring the delay: a 53-tap filter adds 26 samples of latency, which matters inside a control loop.

For GATE IN

Expect: identifying linear-phase types from coefficient symmetry and length, computing group delay, writing windowed coefficients for a small N, estimating N from a transition width and window table, choosing a window for a given attenuation, and comparing FIR with IIR. Practise hd[n] values for ωc = π/2, π/4 and DC/π gains as quick checks.

Quick check

1. Group delay of a 31-tap linear-phase FIR?
2. Which linear-phase type cannot realise a high-pass filter because H(e^(jπ)) = 0?
3. hd[0] for an ideal low-pass with ωc = 0.3π?
4. Does increasing N reduce the peak ripple with a rectangular window?
5. Where are the poles of an FIR filter? Answers: 1. 15 samples. 2. Type II (symmetric, even length). 3. 0.3. 4. No — it only narrows the transition. 5. All at z = 0.