--- title: "Describe FIR design by window method" subject: "Digital Signal Analysis and Processing" chapter: 5 key: "semester_7/digital_signal_analysis_and_processing:window-method-describe" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/window-method-describe site: BCT Engineering --- # Describe FIR design by window method Digital Signal Analysis and Processing · Chapter 5: FIR Filter Design - Asked in: 2072 Chaitra, 5 marks; 2070 Chaitra, 4 marks (asked 2 times) - Video: [Design FIR Filter using Window Method Step by Step - Complete Basics & Formulae](https://www.youtube.com/watch?v=5JpqbuRCjEE) — Shrenik Jain - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/window-method-describe - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-5 ## Question Describe how digital FIR filter can be designed 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. ## Answer The **window method** designs an FIR filter by truncating the infinite impulse response of an ideal filter with a finite-length window. #### Steps 1. **Specify** the ideal response H_d(e^jω), the pass-band edge ωp, the stop-band edge ωs and the stop-band attenuation αs. 2. **Choose the cut-off** ωc = (ωp + ωs)/2 for a low-pass filter. 3. **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)).) 4. **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. 5. **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. 6. **Form the filter:** h(n) = h_d(n) · w(n), 0 ≤ n ≤ M−1. The shift by α makes it causal. 7. **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the low-pass cut-off chosen in the window method? A. wc = wp B. wc = ws C. wc = (wp + ws)/2 D. wc = ws - wp Answer: C. The cut-off is the midpoint of the transition band. (Key point: Cut-off choice) 2. What decides which window to use? A. The required stop-band attenuation B. The sampling rate C. Only the length M D. The pass-band phase Answer: A. Each window has a fixed minimum stop-band attenuation; pick the simplest meeting alpha_s. (Key point: Choosing window) 3. How is M chosen? A. From transition width and the window's main-lobe width, e.g. M >= 8pi/delta-w for Hamming/Hanning B. From the pass-band ripple only C. M = 21 always D. M = wc/pi Answer: A. Main-lobe width sets the transition width, so M >= 8pi/(ws - wp) for Hanning/Hamming. (Key point: Choosing length) 4. How is the final filter formed? A. h(n) = h_d(n) w(n), 0 <= n <= M-1, with shift alpha for causality B. h(n) = h_d(n) + w(n) C. h(n) = w(n) only D. h(n) = h_d(n)/w(n) Answer: A. Windowing multiplies the shifted ideal response by the window, giving symmetric h(n) and linear phase. 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