This is the solution to a fourier series problem, of the function $sin(\omega_0t)$:
I understand how the author has used Euler's formula to split this function into two exponential terms. However, I don't understand how the coefficients were obtained.
They have used a method called inspection. However, they have also stated we can use the formula:
$$c_n=\frac{1}{T_0}\int^{T_0/2}_{-T_0/2} f(t)e^{-jn\omega_0t}\,dt$$
But what do we put in $T_0$ in this formula?
And what is the method of inspection?
Best Answer
without any other specification than that cited we assume that we are working over a period with $-\pi \le \omega_0t \le \pi$ so $T_0 = \frac{2\pi}{\omega_0}$
so we are looking for an expansion: $$ f(t) = \sum_{n=-\infty}^\infty c_n e^{ni\omega_0 t} $$
inspection generally means "just by looking at" - usually when we can directly guess the correct solution, perhaps with a quick bit of mental arithmetic. in applied situations this can save time by reducing formality, though there is always the danger of making an error through haste, or a false assumption etc.
however the fact that, knowing the expression given for the $\sin$ function, we immediately may write: $$ f(t) = \frac1{2j}\left( e^{j\omega_0 t} - e^{-j\omega_0 t}\right) \tag{1} $$ means inspection is quite reliable in the present case. the expression (1) is already the required expansion!
if you plug it into the integral as given you will find, from the orthogonality of the functions $e^{i\omega_0t}$ over the given range, that all coefficients except $c_{-1}$ and $c_1$ evaluate to zero.
for example: $$ c_1 = \frac{\omega_0}{2\pi}\int_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \frac1{2j}\left( e^{j\omega_0 t} - e^{-j\omega_0 t}\right) e^{-j\omega_0 t}dt \\ =\frac{\omega_0}{4\pi j}\int_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \left( 1- e^{-2j\omega_0 t}\right)dt \\ =\frac{\omega_0}{4\pi j} \left[ t +\frac1{2 j \omega_0} e^{-2j\omega_0 t}\right]_{-\frac{\pi}{\omega_0}}^{\frac{\pi}{\omega_0}} \\ = \frac1{2j} $$