The Fourier transform of a damped cosine and the units of the result

Question

If I take a simple transient voltage signal of the form $$f(t) = V_p e^{-t/\tau} \cos(\omega_0 t)$$ and take the Fourier transform in the normal way $$F(\omega) = \frac{V_p}{\sqrt{2 \pi}} \int^{+\infty}_{0} e^{-t/\tau} \cos(\omega_0 t) e^{-i \omega t} \ \ dt $$ giving the result $$F(\omega) = \frac{V_p}{2\sqrt{w\pi}} \left( \frac{1}{1/\tau - i (\omega_{0} - \omega)} + \frac{1}{1/\tau + i (\omega_{0} + \omega)} \right)$$ Seeing as I am interested in a Lorentzian like function, I ignore the second term in the bracket -- and then take the absolute-squared value of what is left giving $$\left|F(\omega)\right|^{2} = \frac{V^{2}_{p}}{2 \pi}\frac{1}{(2/\tau)^{2} + 4(\omega - \omega_{0})^{2}}$$ which when making the approximation of $2/\tau = \Delta\omega$ the linewidth gives a quantity in units of $[V^{2}/Hz^{2}]$. Consistent with a power spectrum density.

This is the crux of my question as if I consider what really happens with a measurement say on a signal/spectrum analyser, the time transient voltage is dissipated over the input impedance of the signal/spectrum analyser, which will have a thermal Johnson-Nyqist noise of $[V/\sqrt{Hz}]$ -- and define my noise floor.

I feel that my result should somehow have the same units, that a spectral lineshape, originating from the FT of a time transient voltage signal should have either units of $[V/\sqrt{Hz}]$ or $[V^{2}/Hz]$. How can I rectify my inconsistency with units as in the end I wish to write something like $$PSD = \sqrt{\left|F(\omega)\right|^{2} + e_{n}^{2}}$$ Where $e_{n}$ (units of $[V/\sqrt{Hz}]$) is the noise floor of my spectrum.

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Some additional thought. I believe this problem can be rectified by the definition of the Power Spectrum density $$S(\omega) = \lim_{T \rightarrow \infty} \frac{1}{T} |F(\omega)|^{2}$$ this factor of $1/T$ takes care of the dimensionality problem -- however I am unsure how to apply this equation as it seems to be in the limit just sends the function to zero because of the $1/T$ factor.

Answer 1

You define PSD proportional to $|X(\omega)|^2$, whose unit is $V^2 \cdot s^2$, where $X(\omega)$ is the Fourier transform of the input signal $x(t)$ in units of Volts. But this quantity is not a PSD but ESD (Energy Spectral Density) used for energy signals instead.

You shall define PSD for periodic signals with a normalization by time:

$$ \text{PSD_x} = \frac{1}{T} |X(\omega)|^2 $$

Where $T$ is the observation wime with units of seconds. Without this division, longer signals would seem to have more power, which is not what you want to see; they have more energy but the same power.

Units of this PSD will be $V^2 \cdot s = V^2 / Hz$ and its square root $V / \sqrt{Hz}$.

Answer 2

if $v(t)$ is an electromotive force (a voltage expressed in units of volt) that is a function of time (let's say expressed in units of second), then the Fourier Transform (or the Laplace Transform):

$$ \mathscr{F} \Big \{ v(t) \Big \} \triangleq V(i \omega) = \int\limits_{-\infty}^{\infty} v(t) e^{-i \omega t} \, \mathrm{d}t $$

will have units of volt-second. And $\omega$ will be in radian/second.