A signal contains two tones: one at 100 Hz with amplitude 1.0, and another at 102 Hz with amplitude 0.001 (60 dB weaker). With a rectangular window, the weak tone is buried under sidelobes from the strong one. What window property do you need to detect the weaker tone?
AA wider mainlobe to better resolve the 2 Hz separation
BLower sidelobes to suppress contamination from the 100 Hz tone, even at the cost of broader mainlobe
CA higher sampling rate to increase the DFT's frequency resolution
DA longer record length to reduce the mainlobe width below 2 Hz
When the goal is detecting a weak signal near a strong one, sidelobe attenuation is the critical property. The rectangular window has sidelobes at only −13 dB — the strong 100 Hz tone contaminates neighboring bins at 1/5 of its amplitude, easily burying a tone 60 dB weaker. A window with low sidelobes (e.g., Blackman at −58 dB, or a Kaiser window tuned to ≥60 dB) suppresses this contamination. The cost is a wider mainlobe (poorer frequency resolution), but that tradeoff is acceptable here since you need dynamic range, not close-frequency separation.
Question 2 Multiple Choice
Why does a pure sinusoid at a non-integer bin frequency produce spectral leakage in the DFT, even though it is a single-frequency signal?
AThe DFT cannot represent sinusoids; it can only represent sums of cosines
BThe DFT implicitly tiles the finite record, creating a discontinuity at the boundary when the sinusoid does not complete an integer number of cycles in the window
CNon-integer frequencies fall between DFT bins, causing the DFT to interpolate inaccurately
DLeakage occurs only for complex exponentials, not real sinusoids
The DFT treats the N samples as one period of a periodic signal — it implicitly tiles the record. If a sinusoid completes 5.3 cycles in the window, the tiled signal has a phase jump at every period boundary: the end of one period does not match the beginning of the next. This discontinuity contains energy at all frequencies, so it contaminates every spectral bin — spectral leakage. A sinusoid at exactly bin 5 (an integer number of cycles) has no jump and produces energy at only one bin. Leakage is entirely about the assumed periodicity of the DFT, not about the sinusoid being non-sinusoidal.
Question 3 True / False
Applying a Hann window to a signal eliminates spectral leakage by preventing signal energy from spreading to other frequency bins.
TTrue
FFalse
Answer: False
Window functions reduce leakage but do not eliminate it. Any finite-length analysis produces some spectral smearing because the window's frequency-domain representation always has nonzero values at frequencies other than DC. What window functions do is shape this smearing: a good window concentrates energy near the mainlobe and suppresses sidelobes, reducing contamination at distant bins without eliminating it entirely. The Hann window drops sidelobes to −31 dB versus the rectangular window's −13 dB — a major improvement. Complete elimination of leakage would require an infinitely long analysis window.
Question 4 True / False
Using a window with lower sidelobes always reduces your ability to resolve two closely spaced frequency components.
TTrue
FFalse
Answer: True
This is the fundamental tradeoff in window design: sidelobe attenuation and mainlobe width are inversely related. A rectangular window has a narrow mainlobe (1/N resolution) but high sidelobes (−13 dB). A Blackman window suppresses sidelobes to −58 dB but widens the mainlobe to about 3/N — signals must be separated by 3/N rather than 1/N to be resolved. The Kaiser window parameterizes this tradeoff continuously via β, letting you choose where to operate, but cannot escape the tradeoff itself. Resolving close frequencies and detecting weak signals near strong ones are competing requirements.
Question 5 Short Answer
Explain why computing the DFT of a finite signal segment is mathematically equivalent to multiplying by a rectangular window, and why this causes spectral leakage.
Think about your answer, then reveal below.
Model answer: A finite signal segment of length N is equivalent to multiplying the infinite signal by a rectangular pulse: 1 for samples 0 to N−1, 0 elsewhere. By the convolution theorem, multiplication in time equals convolution in frequency. So the DFT output is the convolution of the true spectrum with the Fourier transform of the rectangular window (a sinc-like function with a narrow mainlobe and slowly decaying sidelobes at −13 dB). This convolution spreads energy from each true frequency component across neighboring bins — spectral leakage. A tone at a non-integer bin has its energy convolved with this sinc, smearing it across all bins.
This perspective — finite duration equals rectangular window equals frequency-domain convolution — shows why leakage is fundamental, not incidental. All other window functions replace the rectangular window with one whose frequency-domain representation has better sidelobe properties, accepting a wider mainlobe in exchange. The goal is always to shape the convolution kernel, not to eliminate it.