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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.


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.

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    $\begingroup$ can we change your voltage symbol from "$f(t)$" to "$v(t)$" and reserve "$f$" for frequency, if we come to it? i am refraining from editing your question for a little while to let you respond. ... And does your damped sinusoid have a Heaviside unit step (EEs usually call it "$u(t)$") attached to it? $\endgroup$ – robert bristow-johnson Nov 7 at 2:03
  • $\begingroup$ No problem, I usually use $\nu$ or $\omega$ for frequency, but let's go with $f$. And I don't include a Heaviside as by changing my lower integration limit from $-\infty$ to $0$ I thought this takes care of it? $\endgroup$ – Q.P. Nov 7 at 11:24
  • $\begingroup$ ok you added the answer while I was preparing it :)) $\endgroup$ – Fat32 Nov 10 at 22:57
  • $\begingroup$ @Fat32 sort of you have the final piece I think! $\endgroup$ – Q.P. Nov 10 at 23:08
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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 $V / \sqrt{Hz}$.

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    $\begingroup$ THANK YOU! You're the first person to give me a straight answer! I have always been using the definition where $T \rightarrow \infty$ which obviously just tends to zero! Is there a proof that says that a finite time Fourier transform PSD can be normalised like this? Maybe related to my other question dsp.stackexchange.com/questions/61825/… $\endgroup$ – Q.P. Nov 10 at 23:08
  • $\begingroup$ The proof (derivation) is laid out in many standard signal processing and communication theory books. Essentially, the infinite limit will yield zero average power for energy signals but nonzero for power signals. The finite length division will always yield nonzero average power, however it will stay the same for power signals, while it will decrease to zero as $T$ increases for energy signals. $\endgroup$ – Fat32 Nov 10 at 23:18
  • $\begingroup$ The green tick is yours, incidently, could you recommend such a text? My field is not electronic engineering or signal analysis -- I work in experimental trapped ion physics. $\endgroup$ – Q.P. Nov 10 at 23:20
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    $\begingroup$ You can find the definition of power spectral density and energy spectral density from the following book: Communication Systems_Haykin_4E, page 44-50. However they use Random Process terminology and not practical measurements. More generally you may read the following texts: Signals and Systems_Oppenheim, Communication System Engineering_Proakis. They are heavily signal processing oriented however. $\endgroup$ – Fat32 Nov 10 at 23:31
  • $\begingroup$ Much obliged to you! $\endgroup$ – Q.P. Nov 10 at 23:37
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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.

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    $\begingroup$ Thanks for replying. Yes, or equivalently, $V/Hz$. I completely agree. But then there is an inconsistency in a number of places in my understanding. One in how I calculate my Fourier transform to be gin with -- because taking the absolute value squared of the Fourier transform should also square all units right? And then seeing as the signal is dissipated over a resistor (e.g. input impedance of a spectrum analyser) which has a thermal noise in units of $V/\sqrt{Hz}$, which I take to be the noise floor of my signal in frequency space. How do I reconcile the unit inconsistencies? $\endgroup$ – Q.P. Nov 7 at 11:31
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    $\begingroup$ i think i understand your concern. lemme ruminate and get back to this. $\endgroup$ – robert bristow-johnson Nov 7 at 22:49

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