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Showing analytically that sampling exactly 1 period of a sinusoid yields a spectrum with no lobes in the DFT

To solve the roots, set: $$e^{-j(\Omega-\omega_o)M/2}\frac{\sin \bigg(\frac{M+1}{2} (\Omega - \omega_o))\bigg)}{\sin((\Omega-\omega_0)/2)} = -e^{-j(\Omega+\omega_o)M/2}\frac{\sin \bigg(\frac{M+1}{2} (\...
Dan Boschen's user avatar
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Frequency resolution in STFT

The $N$ defining the full signal length is not the same as the $N$ in the STFT equation, which defines the DFT length. The STFT is a sliding DFT so the only value that matters is the latter. Without ...
Jdip's user avatar
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What is the Wrap-Around Phenomenon in FFT?

Just like the reciprocal case of padding a time domain signal with zeros, but padding in frequency domain, you effectively interpolate. You can think of it as if you’ve at a higher rate, and had no ...
Yair M's user avatar
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32-point DFT computation of a sampled sinusoidal signal

Hint: The DFT of a complex sine wave that has an integer number of periods in the DFT window is simply a Kronecker Delta at that frequency. We can write the time domain signal as $$x[n] = e^{j2\pi\...
Hilmar's user avatar
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