Jtdylktuy

Suppose that a function$,$ $f(t)$ $,$has the Fourier transform
$,$$hat f(omega)=e^{-omega^4}$.
What general rules are there for computing the Fourier transforms of things like $,$$g(t)=f(2t)$, $,$ $h(t)=f(2t+1)$, $k(t)=f(2t+1)e^{it}$ and similar?

I know that the Fourier transforms are the once below and I know how to find them (by finding f(t) by the inverse transform and then transforming f(2t+1) etc.)
My question is if there is a quicker way of finding a Fourier transform of a function with a modified argument if the transform of the function is known.(?)

$$hat g (omega)=frac{1}{2} hat f left(frac{omega}{2}right)=frac{1}{2}e^{-frac{omega^4}{16}}$$

$$hat h (omega)=e^{i frac{omega}{2}} hat g (omega) =
frac{1}{2}e^{i frac{omega}{2}-frac{omega^4}{16}}$$

$$hat k (omega)= hat h (omega-1)=frac{1}{2}e^{i frac{(omega-1)}{2}-frac{(omega-1)^4}{16}}$$

There are indeed such laws!

Throughout, we will consider ourselves with a function $f(t)$ and its Fourier transform, and we will let $gamma$ be a real constant.

We will also suppose $f$ has a defined transform and inverse transform where needed, denoted $mathscr{F} left. [f(t)] right|_omega$ and $mathscr{F}^{-1} left. [f(omega)] right|_t$ respectively.

(Generally speaking I'm borrowing naming and notation conventions from my own textbook/lectures on the topic from this semester if the sudden change in such confuses you. I also modify these slightly so that they make slightly more sense or are at least easier to remember. I find them a lot nicer to work with than what you presented, no offense intended.)

Scalar Laws: (we further assume $gamma neq 0$ here)

$$mathscr{F} left. [f(gamma t)] right|_omega = frac{1}{|gamma |} mathscr{F} left. [f(t)] right|_omega$$

$$mathscr{F}^{-1} left. [f(gamma omega)] right|_t = frac{1}{|gamma |} mathscr{F}^{-1} left. [f(omega)] right|_t$$

Translation Laws:

$$mathscr{F} left. [f(t - gamma )] right|_omega = e^{-i2pi gamma omega} cdot mathscr{F} left. [f(t)] right|_omega$$

$$mathscr{F}^{-1} left. [f(omega - gamma )] right|_t = e^{i2pi gamma omega} cdot mathscr{F}^{-1} left. [f(omega)] right|_t$$

Equivalently, following directly from these, one can also state,

$$mathscr{F} left. [e^{i2pi gamma omega} cdot f(t)] right|_omega = mathscr{F} left. [f(t)] right|_{omega - gamma}$$

$$mathscr{F}^{-1} left. [e^{-i2pi gamma omega} cdot f(omega)] right|_t = mathscr{F}^{-1} left. [f(omega)] right|_{t - gamma }$$

You can utilize these as necessary to manipulate your function into a "nicer" place. You might need to utilize multiple identities in sequence

For example, in finding the transform of $f(2t+1)$, you'd first want to note that $f(2t+1)=f(2(t+frac{1}{2}))$. Then you can apply a translation identity with $gamma = -frac{1}{2}$, which gives you an $e^{(stuff)}$ factor times the transform of $f(2t)$. Then you apply the scaling identity to get a constant times the transform of $f(t)$.

Or, to just show the calculation, first we use the translation identity:

$$mathscr{F} left. [f(2t+1)] right|_omega = mathscr{F} left. left[ f left( 2 left( t+frac{1}{2} right) right) right] right|_omega = e^{-i2pi left( - frac{1}{2} right) omega} cdot mathscr{F} left. [f(2t)] right|_omega = e^{ipi omega} cdot mathscr{F} left. [f(2t)] right|_omega$$

Then we use the scaling identity on the transform we got:

$$mathscr{F} left. [f(2t)] right|_omega = frac{1}{2} cdot mathscr{F} left. [f(t)] right|_omega$$

Thus, overall,

$$mathscr{F} left. [f(2t+1)] right|_omega = frac{1}{2} cdot e^{ipi omega} cdot mathscr{F} left. [f(t)] right|_omega$$

I'll be the first to admit that it's entirely likely I made minor mistakes in the preceding statements of the identities/calculations. This was a mess to look at the LaTeX for and while I think everything checks out I could be completely wrong. Still, worst case, you could Google up "Fourier transform scalar translation identities" and pull up some results - or maybe even look in the index of your own text, OP, if you have one. I'd be surprised if a text didn't cover these identities. But yes, long story short, there are identities to cover cases such as those in the original post.