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Describe FIR design by window method
Short answer
The window method truncates the ideal infinite impulse response h_d(n) (from inverse DTFT of the desired frequency response) with a finite-length window w(n). After specifying cutoff ωc=(ωp+ωs)/2 and shift α=(M−1)/2, choose a window and length M from attenuation and transition width, then form h(n)=h_d(n)w(n) for a causal linear-phase FIR filter.
Describe how digital FIR filter can be designed by window method.
The window method designs an FIR filter by truncating the infinite impulse response of an ideal filter with a finite-length window.
Steps
- Specify the ideal response H_d(e^jω), the pass-band edge ωp, the stop-band edge ωs and the stop-band attenuation αs.
- Choose the cut-off ωc = (ωp + ωs)/2 for a low-pass filter.
- Find the ideal impulse response by the inverse DTFT: h_d(n) = (1/2π) ∫ H_d(e^jω) e^{jωn} dω = sin(ωc(n − α)) / (π(n − α)), α = (M−1)/2 (for high-pass: δ(n−α) − sin(ωc(n−α))/(π(n−α)); for band-pass: [sin(ω₂(n−α)) − sin(ω₁(n−α))]/(π(n−α)).)
- Choose a window from αs: rectangular (about 21 dB), Bartlett (25 dB), Hanning (44 dB), Hamming (53 dB), Blackman (74 dB), or Kaiser for any attenuation.
- Choose the length M from the transition width Δω = ωs − ωp and the window's main-lobe width (for example, for Hamming and Hanning: M ≈ 8π/Δω). Take M odd for a Type-I filter, so that α is an integer.
- Form the filter: h(n) = h_d(n) · w(n), 0 ≤ n ≤ M−1. The shift by α makes it causal.
- Check the frequency response H(e^jω) against the specification; if it fails, increase M or use a better window.
Remarks
- Because h(n) = h(M−1−n), the phase is exactly linear: θ(ω) = −αω.
- Windowing multiplies in time and convolves in frequency, so the sharp edge of H_d is smoothed (a transition band appears) and ripples appear (Gibbs effect).
- Trade-off: a narrow main lobe gives a sharp transition; low side lobes give high attenuation. For a fixed length, a window cannot optimise both at once.
The method is simple and gives linear-phase, always-stable filters, but the pass-band and stop-band ripples cannot be controlled independently.
✓ Answer checked by Claude Opus 5.5 (high effort)
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