# Units of Fourier transform

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])?

• 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 $[\omega,\omega+d\omega]$'). – yu-v Oct 30 at 12:30
• Check your math. The units of the F.T. will be V$\cdot$s, or V/Hz. – garyp Oct 30 at 12:30
• 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 $A$? – Frederic Oct 30 at 12:52
• $V/Hz = V/(1/s) = V\cdot s$. – JEB Oct 30 at 13:33
• 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 $\mathcal{F}(\omega_0)$? – Frederic Oct 30 at 13:38

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.