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])?
 A: 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.
