Define the Fourier Transform of a certain signal in the time domain FT[$x(t)$]=$X(j\omega)$
$X(j\omega)$ = $\int$ $x(t)$ $e$^($j\omega$$t$)$ $dt
I'd like to ask what is the meaning of the value obtained from $X(j\omega)$ with certain frequency $\omega$
for example if we have a voltage signal of 1 Volt and found that
$$X(j100) = 100$$
– the unit is weird for me which will be voltage*sec- what does that 100 mean here?
Also: Why there is a factor of $\frac{1}{T}$ difference between the units of Fourier series and Fourier transform ?
I've asked at Signals processing/Math stack exchange but no answer
I've read this answer but it says:
The multiplication by $T$ in the limit is to account for the differences in definition between the Fourier series and Fourier transform: the series representation typically has a factor of $\frac{1}{T}$, while the transform does not. I don't know that there is a lot of insight to be gained via this analysis, but it shows that the series and transform representations are intimately related.
which didn't satisfy me.
Best Answer
Consider for a moment, the synthesis equation where we 'construct' $x(t)$ out of a weighted 'sum' (integral) of the orthonormal basis functions of time: $e^{j\omega t}$
$$x(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}\mathrm{d}\omega\:X(\omega)\:e^{j\omega t}$$
Here, it is clear that $X(\omega)$ is the amount or weight of $e^{j\omega t}$ that goes into 'constructing' $x(t)$ from these basis functions. We haven't specified what $X(\omega)$ is but we assert that, for a large class of $x(t)$, there is an appropriate $X(\omega)$ such that the above holds.
As you may have already concluded, there is a way to find $X(\omega)$ given an $x(t)$ which is
$$X(\omega) = \int_{-\infty}^{\infty}\mathrm{d}t\:x(t)\:e^{-j\omega t}$$
which (left as a fun exercise for the reader) can be verified by substituting the later expression into the former expression.
Since, in the first expression, we're integrating with respect to $\omega$ which has units of $\mathrm{s}^{-1}$, it must be that $X(\omega)$ has the units of $x(t)$ multiplied by $\mathrm{s}$.
If we were to make a discrete approximation of the first expression, it would be something like
$$\tilde{x}(t) = \frac{1}{2\pi}\sum_{n = -\infty}^\infty \frac{2\pi}{T} \:X(n\frac{2\pi}{T})e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:\frac{X(n\frac{2\pi}{T})}{T}e^{jn\frac{2\pi}{T}t} = \sum_{n = -\infty}^\infty \:a_n e^{jn\frac{2\pi}{T}t}$$
which is periodic with period $T$ (thus the tilde over). Using the usual method of integrating the product of both sides with $e^{-jm\frac{2\pi}{T}t}$ over a period $T$, we arrive at
$$\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jm\frac{2\pi}{T}t} = Ta_m$$
since the integral on the right hand side vanishes for $n \ne m$. And so,
$$ a_n= \frac{X(n\frac{2\pi}{T})}{T} = \frac{1}{T}\int_0^T\mathrm{d}t\:\tilde{x}(t)\:e^{-jn\frac{2\pi}{T}t}$$
So, the $\frac{1}{T}$ comes from the fact that the $a_n$ are the 'sampled' Fourier transform divided by the period $T$ as shown above.