Electronic – Find fundamental period of a discrete-time signal

Apply the continuous to discrete time equation you've already given. The discrete time domain is itself a sampled version of the continuous time, sampled by the function \$x_d[n] = x(nT_s)\$

We need to play a bit with the layout of the expression. We have:

$$ F=\frac{(1+a)(1-a)}{(1-ae^{\ jw})(1-ae^{\ -jw})} =\frac{(1+a)}{(1-ae^{\ -jw})}\frac{(1-a)}{(1-ae^{\ jw})} $$

We can rewrite it as:

$$ =\left( \frac{1}{(1-ae^{\ -jw})}+\frac{a}{(1-ae^{\ -jw})}\right) \left( \frac{1}{(1-ae^{\ jw})}-\frac{a}{(1-ae^{\ jw})}\right) \\ $$

We factor out a \$-ae^{\ jw}\$ from the right-most terms and do the inverse transform:

$$ = \left( \frac{1}{(1-ae^{\ -jw})}+a\frac{1}{(1-ae^{\ -jw})}\right) \left( -\frac{1}{a}\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}+\frac{e^{\ -jw}}{(1-\frac{1}{a}e^{\ -jw})}\right) \\ \implies \left( a^nu[n]+a(a^nu[n]) \right) * \left( -\tfrac{1}{a}({\tfrac{1}{a}}^nu[n-1])+({\tfrac{1}{a}}^nu[n-1]) \right) $$

Finally, cleaning up:

$$ = \left( a^nu[n](1+a) \right) * \left (\tfrac{1}{a}^{n-1}u[n-1](1-\tfrac{1}{a}) \right) $$

Sorry if it is too messy. Fourier tends to be a lot of writing. Tell me where you think I may have made an error or isn't clear! If someone finds a mistake, please let me know.