One common definition of a fourier transform for function f(x) is $$F(v)=\int_{-\infty}^\infty f(\tau)e^{2\pi i\tau v}d\tau$$ I know some definitions have an extra sqrt(2*pi). I shall ignore that.
However, there are 3 tau's in the formula and it does not specify how each one relates to x. For example it does not explain for the fourier transform of f(x+1), whether tau+1 replaces all the taus or only for some.
From looking at derivations of fourier transform properties I believe that you must state state both the function and the variable the fourier transform is of. However, if your function is say f((y-1)) convention assumes that the variable is y instead of say (y-1).
- Is that right?
- If so, is this the correct fourier transform for f(g(x)) with respect to x? $$F(v)=\int_{-\infty}^\infty f(g(\tau))e^{2\pi i \tau v}d\tau$$
The Fourier transform of $f(x)$ is $$ \mathcal F\{f(x)\}=F(\nu)=\int_{-\infty}^\infty f(x)\,\mathrm e^{-2\pi i x \nu}\,\mathrm d x $$ With this convention, the inverse Fourier transform is $$ \mathcal F^{-1}\{F(\nu)\}=f(f)=\int_{-\infty}^\infty F(\nu)\,\mathrm e^{+2\pi i x \nu}\,\mathrm d \nu $$ For $f(x+1)$ we have (time shift property) $$ \mathcal F\{f(x+1)\}=\int_{-\infty}^\infty f(x+1)\,\mathrm e^{-2\pi i (x+1) \nu}\,\mathrm d x=F(\nu)\,\mathrm e^{+2\pi i \nu} $$ For $f(g(x))=h(x)$ we have to evaluate $$ \mathcal F\{f(g(x))\})=\mathcal F\{h(x)\}=H(\nu)=\int_{-\infty}^\infty f(g(x))\,\mathrm e^{-2\pi i x\nu}\,\mathrm d x $$ There is no genral rule in this case. If $g$ is a bijection and smooth enough then, if integral exists $$ \mathcal F\{f(g(x))\}=\int_{-\infty}^\infty f(g(x))\,\mathrm e^{-2\pi i x\nu}\,\mathrm d x=\int_{-\infty}^\infty f(y)\,\mathrm e^{-2\pi i g^{-1}(y)\nu}\frac{1}{|g'(y)|}\,\mathrm d y $$