0
$\begingroup$

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])?

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.