Inverse Fourier transform of scaled and shifted tanh function

Question

I have a function in frequency where I want to perform an inverse Fourier transform, $$F(\omega)=\tanh(a(\omega-b))$$

where I define the FT relations as $F(\omega)=\int_{-\infty}^{\infty}dte^{i\omega t}f(t)$ and $f(t)=\int_{-\infty}^{\infty} \tfrac{d\omega}{2\pi}e^{-i\omega t}F(\omega)$.

From this post, we can say that there is a Fourier transform pair $$\tanh(\omega) \leftrightarrow \dfrac{t}{2}\mathrm{csch}\dfrac{\pi t}{2}$$

Then using the scaling and shifting properties $F(\omega-\alpha) \to f(t)e^{-i\alpha t}$ and $F(\omega/\alpha) \to |\alpha| f(\alpha t)$ (are they correct?) we deduce that $$\mathcal{F}^{-1}\left[ \tanh(a(\omega-b)) \right] = |\tfrac{2}{a}|\tfrac{t}{2} e^{-ibt}\mathrm{csch}\dfrac{\pi at}{2}$$

Is this the correct way to deal with scaling and shifting properties? If not, is there a particular order in which they are considered?

Edit: Are there any good online references where I can find tables of these kinds of non-trivial Fourier transform pairs? e.g. Abramowitz and Stegun is one famous example but they don't cover FTs.

Answer 1

Well, let's do a quick derivation

$$\mathcal{F}^{-1}\left\{F\left(a[\omega-b]\right)\right\} = \dfrac{1}{2\pi}\int_{-\infty}^{\infty}F\left(a[\omega-b]\right)e^{-j\omega t}d\omega$$

and make the following substitutions (assuming $a>0$ to avoid the hassles with the absolute values):

$$\begin{align*}\omega'&=a(\omega -b) & \implies \omega = \dfrac{\omega'}{a}+b\\ \\d\omega' &= a d\omega &\implies d\omega = \dfrac{d\omega'}{a} \end{align*}$$

so

$$\begin{align*}\mathcal{F}^{-1}\left\{F\left(a[\omega-b]\right)\right\} &= \dfrac{1}{2\pi}\int_{-\infty}^{\infty}F\left(a[\omega-b]\right)e^{-j\omega t}d\omega\\ \\ &= \dfrac{1}{a}\dfrac{1}{2\pi}\int_{-\infty}^{\infty}F\left(\omega'\right)e^{-j\left(\frac{\omega'}{a}+b\right) t}d\omega'\\ \\ &= \dfrac{e^{-jbt}}{a}\dfrac{1}{2\pi}\int_{-\infty}^{\infty}F\left(\omega'\right)e^{-j\omega'\frac{t}{a}}d\omega'\\ \\ &= \dfrac{e^{-jbt}}{a}f\left(\dfrac{t}{a}\right)\\ \\ \end{align*}$$

Assuming this transform pair is correct

$$\tanh(\omega) \leftrightarrow \dfrac{t}{2}\mathrm{csch}\dfrac{\pi t}{2}$$

then your desired transform pair is

$$\tanh(a[\omega-b]) \leftrightarrow \dfrac{e^{-jbt}}{|a|}\dfrac{t}{2a}\mathrm{csch}\dfrac{\pi t}{2a}$$

Answer 2

It should be $F(\omega-\alpha) \to f(t)e^{\LARGE{\color{blue}{i\alpha t}}}$.

So $$\mathcal{F}^{-1}\left[ \tanh(a(\omega-b)) \right] =\mathcal{F}^{-1}\left[ \tanh(a\omega-ab) \right] \\ = \large |\tfrac{1}{a}|\tfrac{t}{2a} e^{it\frac{ab}{a}}\mathrm{csch}\dfrac{\pi t}{2a} = |\tfrac{1}{a}|\tfrac{t}{2a} e^{ibt}\mathrm{csch}\dfrac{\pi t}{2a}.$$

My tip: if you insist on using properties, do the $x$-translation first, followed by the scaling. For instance, let $g(\omega)=\tanh(\omega-ab)$ and $h(\omega)=g(a\omega)=\tanh(a\omega-ab)=\tanh(a(\omega-b))$.