Exercise · ZW-SHORT-TIME-FOURIER-TRANSFORM-006 · ★★☆ Intermediate
A signal equals \( x[n]=\cos(2\pi\,0.25\,n) \) for all time. At what normalized frequency does the peak of its STFT appear in any frame?
Choose one option.
A \( \omega=0.125\pi \) B \( \omega=\pi \) C \( \omega=0.25\pi \) D \( \omega=0.5\pi \) Show answer ← Back to the question
Explanation Since \( 2\pi\cdot 0.25 = 0.5\pi \), the peak is at \( \omega=0.5\pi \). The option with \( 0.25\pi \) confuses cyclic frequency with angular frequency.
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