Why multiplying spectral density by 2/N bandwidth results in sinusoidal amplitude?

Question

The image is taken from DSP Guide, where plot [a] is the time domain to synthesize [b] is the frequency domain where the amplitude denotes spectral density and [c] shows the actual magnitude of the sinusoidal amplitude.

To convert spectral density to sinusoidal amplitude (i.e. from [b] to [a]), we multiply the spectral density by 2/N bandwidth. For instance, frequency sample number 4 has spectral density of 32, multiplying 32 by 2/N (or 2/32) gives us 2 - the actual sinusoidal amplitude. But why?

The spectral density counts the number of times the signal falls within a unit bandwidth (i.e. the value 32 for frequency number 4 means there are 32 samples that have frequency between 3.5 and 4.5 ). What the spectral density tells us is the count or how probable a specific range of frequency embedded within a signal. I don't see how it has any correlation to the sinusoidal amplitude, or why multiplying by the bandwidth can magically converts a probability density or a count into a physical sinusoidal amplitude such as voltage.

Answer 1

That's not spectral density, it's just real DFT.

If you have a $60 Hz$ sine that lasts for 1 sec, its DFT will be half as large as for the same sine that lasts for 2 secs, since there's half the number of samples and thus half the (unnormalized) correlation: sum(product). To get the same value for both, we divide out the sample count.

Why multiplying by 2/N results in sinusoidal amplitude?

Because any (non-Nyqist/DC; see "special cases" in the article) bin will correlate with integer periods as $N/2$:

$$ \sum_{n=0}^{N-1} (A \cos(2\pi k n)) \cdot \cos(2\pi k n) = A (N/2) \tag{1} $$

This holds for real and complete DFT. The units are carried over with $A$.

The article doesn't discuss finding amplitude of a sinusoid in a general signal from DFT (for which DFT is ill-suited), however - it's rather about normalizing DFT: real has half the number of bins of complete DFT, so we double the normalization.

What the spectral density tells us is ... how probable a specific range of frequency embedded within a signal.

There's no need for probabilities in interpreting results for deterministic signals - but if we insist, the normalizations remain accurate.

If this still doesn't make sense, then it's about understanding how DFT works fundamentally. In addition to a closer look at DSP guide, I recommend this clip.