How to find the Fourier Transform of a function that is periodic in an interval only?

Question

I know how to find the Fourier Series for a periodic function (Periodic for all inputs, from the definition). I also know how to find the Fourier Transform for non periodic functions. But, Which formula to use to calculate the Fourier transform of a function that is periodic in an interval only (Bounded), like shown here.. In this function, the period is 'r' and the function exists from 0 to M.

Answer 1

Let be $$ \mathcal F\{f(t)\}=F(\omega)=\int_{-\infty}^\infty f(t)\mathrm e^{-i\omega t}\mathrm d t=\int_{0}^Mf(t)\mathrm e^{-i\omega t}\mathrm d t $$ Let be $f_0(t)$ the basis function in the interval $[0,\, r]$ and $M=nr$ for some $n\ge 1$. We have $$ f(t)=\sum_{k=0}^{n-1}f_0(t-kr) $$ and then $$ \begin{align} F(\omega)&=\int_{0}^Mf(t)\mathrm e^{-i\omega t}\mathrm d t=\int_{0}^{nr} \sum_{k=0}^{n-1}f_0(t-kr)\mathrm e^{-i\omega t}\mathrm d t\\ &=\sum_{k=0}^{n-1}\int_{0}^{nr}f_0(t-kr)\mathrm e^{-i\omega t}\mathrm d t=\sum_{k=0}^{n-1}\int_{0}^{r}f_0(t)\mathrm e^{-i\omega t}\mathrm e^{-i\omega kr}\mathrm d t\\ &=F_0(\omega)\sum_{k=0}^{n-1}\mathrm e^{-i\omega kr}=F_0(\omega)\frac{1-\mathrm e^{-i\omega nr}}{1-\mathrm e^{-i\omega r}}\\ &=F_0(\omega)\frac{\sin \left(\frac{n \omega r}{2}\right)}{\sin \left(\frac{\omega r}{2}\right)}\,\mathrm e^{i\frac{\omega r}{2} (n-1)} \end{align} $$ where $F_0(\omega)=\mathcal{F}\{f_0(t)\}$