Exercise · ZW-SHORT-TIME-FOURIER-TRANSFORM-010 · ★★★ Advanced

A signal \( x[n] \) has an STFT with a rectangular window of length \( L=32 \). Suppose \( x[n]=\cos(2\pi\,0.2\,n) \) for \( n<64 \) and \( x[n]=\cos(2\pi\,0.2\,n)+\cos(2\pi\,0.4\,n) \) for \( n\ge 64 \). Which statement about the spectrogram on \( n\in[80,112) \) is correct?

Choose one option.

Show answer
  • D. Peaks appear at \( \omega=0.4\pi \) and \( \omega=0.8\pi \)

Explanation

For \( n\ge 64 \) the signal contains both components, at \( \omega=0.4\pi \) and \( \omega=0.8\pi \). The 32-sample window over \( [80,112) \) lies entirely in that region, so both peaks appear. Reading only the initial component ignores the change at \( n=64 \).

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Review the concept: Short-time Fourier transform

The short-time Fourier transform (STFT, \( X(\tau,\omega) \)) describes how the frequency content of a signal varies over time using a sliding window.

\[X(\tau,\omega)=\sum_{n} x[n]\,w[n-\tau]\,e^{-j\omega n}\]

More practice problems: short-time fourier transform

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