Say I have a table of functions in time, and their corresponding Fourier transforms in the radian frequency domain:
How can I translate these Fourier transform expressions to the standard frequency domain, f, in Hz?
Is it as simple as replacing every instance of $\omega$ with $2 \pi f$?
I don't think it is, because say I have a function below that I want to take a Fourier transform of: $$ x(t) e^{\displaystyle{j \frac{2\pi n}{T}t}} = x(t) e^{\displaystyle{j 2\pi n ft}} $$
We can write this as: $$ x(t) e^{\displaystyle{j n \omega_0 t}} $$
The Fourier transform of this is a standard result: $X(\omega - n\omega_0)$.
But if I try to convert this to be in terms of $f$: $$X(2\pi f - n2\pi f_0) = X(2\pi(f-nf_0))$$
However, if I do the same question now by using the fact that: $$X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} \mathrm{d}t$$ I get that the Fourier transform is $X(f-nf_0)$
So, the two results are not the same, as far as I can see.
Have I made a mistake somewhere? Or is that not the way to convert between $f$ and $\omega$?
In general,
$$F(\omega)\neq G(f)$$ For every $f_i$ there is an $\omega_i=2\pi f_i$ such that $F(\omega_i)=G(f_i) \rightarrow F(2\pi f_i)=G(f_i).$
I think your confusion comes from using the same letters for $F$and $G$.
This is where $X(2\pi(f-nf_0))=X(f-nf_0)$ comes from..
To "get the one from the other", $$G(f)=F(2\pi f)$$ $$F(\omega)=G(\frac{\omega}{2\pi})$$
In you example, $$G(f-nf_o)=F(2\pi(f-n f_o))$$