Assume a function $f(t)$ which represents the voltage over time and its Fourier transform $\mathcal{F}(\omega)$. The unit of f(t) is [V] and the unit of $\mathcal{F}(\omega)$ will be [V/Hz].

I am interested in the voltage amplitude A (unit [V]) of a wave at a specific frequency $\omega_0$ ($Ae^{i\omega_0 t}$). The amplitude of the Fourier transform at this frequency is $\mathcal{F}(\omega_0)$. Is it correct to state that A=$\mathcal{F}(\omega_0)$? I think not because A has unit [V] and $\mathcal{F}(\omega_0)$ has unit [V/Hz]. How can the Fourier transform then help finding the amplitude A at a specific frequency?

Do I have to take the limit? $$\lim_{\Delta\omega \to 0}\int_{\omega_0-\Delta\omega}^{\omega_0+\Delta\omega} \mathcal{F}(\omega) \Delta\omega d\omega = \mathcal{F}(\omega_0)d\omega$$

But this result does not help me any further...

So, concrete my question: how can you find the voltage [V] (and not [V/Hz]) at a specific frequency using the Fourier transform? And what is the relation between the amplitude A (unit [V]) of a wave at a specific frequency $\omega_0$ and $\mathcal{F}(\omega_0)$ (unit [V/Hz])?

  • $\begingroup$ First of all, if $f(t)$ has units of $[V]$ then $\mathcal{F}(\omega)$ has units of $[V * s]$ as you integrated over time to get it. Second - you must define what you mean by 'voltage at specific frequency'? I would say that what $\mathcal{F}(\omega)d\omega$ is indeed the closest answer to that question ('how much voltage is at the frequency interval $[\omega,\omega+d\omega]$'). $\endgroup$ – yu-v Oct 30 at 12:30
  • $\begingroup$ Check your math. The units of the F.T. will be V$\cdot$s, or V/Hz. $\endgroup$ – garyp Oct 30 at 12:30
  • $\begingroup$ Thanks for the unit correction. By 'voltage at a specific frequency' I mean the amplitude A of the wave at this frequency $A e^{i\omega_0 t}$. The amplitude has unit [V] but the Fourier transform gives unit [V/Hz]? Is $\mathcal{F}(\omega_0)$ related to $A$? $\endgroup$ – Frederic Oct 30 at 12:52
  • $\begingroup$ $V/Hz = V/(1/s) = V\cdot s$. $\endgroup$ – JEB Oct 30 at 13:33
  • $\begingroup$ How is the amplitude of $\mathcal{F}(\omega_0)$ related to the amplitude A? Is there a way to trace back the value of A by calculating the Fourier transform $\mathcal{F}(\omega_0)$? $\endgroup$ – Frederic Oct 30 at 13:38

This is a great example of how checking units and dimensionality help us understand the physics.

If you apply a discrete set of voltages with different frequencies, each with amplitude $a_i$, that is $$f(t) = \sum_i a_i e^{i\omega_i t}$$ then the dimensions of $a_i$ are voltage and you will get it from the discrete Fourier transform, which has the appropriate units. If you apply a continuous set of frequencies $$f(t) = \int d\omega a(\omega) e^{i\omega t}$$ then $a(\omega)$ has dimensions of voltage-per-frequency (equivalently voltage-times-time), because it reflects a distribution of voltage amplitudes. So one can ask what is the total voltage not at frequency $\omega_0$ but at a small interval $[\omega_0, \omega_0+d\omega]$, and it will be $a(\omega_0)d\omega$. Asking what is the voltage at a specific frequency assumes a discrete set of frequencies. You get $a(\omega)$ by the continuous Fourier transform.

The two pictures can come together if we employ delta functions, and then $$f(t) = \sum_i a_i e^{i\omega_i t} = \int d\omega e^{i\omega t}\sum_i \delta(\omega-\omega_i)a_i$$ Note, however, that delta-functions are not dimensionless! indeed, $a_i \delta(\omega-\omega_i)$ has units of voltage-per-frequency, just as $a(\omega)$ -- which in fact it is! $a(\omega) = \sum_i a_i \delta(\omega-\omega_i)$ in this case.


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