Let's consider a real function of time $x(t)$ being a Fourier-transformable signal. The synthesis equation states that:

$$x(t)=∫_{-∞}^{+∞}X(f)⋅e^{2πjft} df $$

Meaning: $x(t)$ can be seen as the sum of infinite sine waves at different frequencies (spectral components) with amplitude |X(f)| and initial phase (delay) equal to ∠X(f).

Now, let's suppose to evaluate the Power Spectral Density. If we define it as power per frequency bandwidth, I'd evaluate it in the following way.The Fourier synthesis relation tells us that the elementary sinusoid is:

$$X(f)⋅e^{2πjft} df$$

Its average elementary power dP() is ½ peak amplitude (i.e. |X(f)df|) for the real part and the same for the imaginary part. So

$$dP= |X(f)df|^2$$


$$ PSD(f)=dP/df=|X(f)|^2 df$$

Ok, there is the frequency bin df inside the equation, which may look strange. However, I do not find any mistake in my derivation. Is it wrong?

This definition appears quite similar to the way a Spectrum Analyzer evaluates the power per frequency bandwidth. It applies a narrow Resolution Bandwidth bandpass filter (moving frequency filter), lets out only one sine wave and measures the Watts transported by that sine wave. These Watts Absorbed divided by the Resolution Bandwidth is the PSD (Watts/Hz) at that specific frequency.

However, the only definition I can find of PSD is the following one:

$$PSD(f) = lim_{T→+∞}⁡\frac{1}{T}|X_T (f)|^2$$

I really can't understand it and associate it with the definition "Power per bandwidth". If we want the power per bandwidth, why shouldn't we consider the power of the single spectral component (which is 1/2 times peak amplitude: no need for windowing the signal on period T).

  • 1
    $\begingroup$ Keep in mind most (all?) real-world signals have zero PSD (because of the limit). When the PSD of an actual signal is calculated, there is an underlying assumption: that the signal is a portion of a random (or unknown) signal of infinite duration. Then you can estimate the PSD, not of the actual signal, but of the larger infinite-duration signal. $\endgroup$
    – MBaz
    Oct 16, 2023 at 13:45

1 Answer 1


That's a bit of a philosophical question but here is my take.

The main purpose of windowing is practical: Sine waves do NOT exist in the real world. In order for a wave to be a sine wave in the sense of the Fourier Transform it needs to be infinitely long. For any real world signal, that's a non-starter, so in practice you always have a finite observation time and hence a finite observation bandwidth and frequency resolution.

A finite time/frequency resolution also has the nice advantage that the spectrum can change with time. That's what the Short Time Fourier Transform is all about.

If we want the power per bandwidth, why shouldn't we consider the power of the single spectral component

Because a "single spectral component" doesn't exist in the physical world. Because you have to to know the history back to he Big Bang and need to know what the signal will be doing until the heat death of the universe. Because there would be infinitely many components with infinitely small bandwidth.

So this idea is great for text books and pencil-and-paper analysis but not fully applicable to doing work with real world signals.


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