Proof for determining Fourier coefficients

Question

While determining Fourier coefficients we have this equation $$\int^{T}_{0} x(t) e^{-jn\omega_0t} dt = \sum^{+\infty}_{k\ =\ -\infty} a_k [\int^{T}_{0} e^{j(k-n)\omega_0t}dt]$$

I want to ask that how can we get final equation of $a_n$ as: $$a_n= 1/T \int^{T}_{0} x(t)e^{-jn\omega_0t}dt$$ from the above equation. please any one help to explain this

Answer 1

For the periodic signal $x(t)$ with fundamental period $T=1/F_0$, proceeding from where you are:

$$ \Rightarrow \sum^{+\infty}_{k\ =\ -\infty} a_k \left[\int^{T}_{0} e^{j(k-n)\omega_0t}dt\right] = \sum^{+\infty}_{k\ =\ -\infty} a_k \left[\frac{e^{j(k-n)\omega_0t}}{j(k-n)\omega_0}\right]^{T}_{0} $$

1. For $k \neq n$ you got $0$ (you can prove that)

2. And for $k=n$ you see that $\int^{T}_{0}dt = T$

You therefore have:

$$\int^{T}_{0} x(t)e^{-jn\omega_0t} dt= a_kT = a_nT$$

You can proceed from there.

EDIT:

• We do not assume the existence of the two possible solutions. The right-hand side of the equation evaluates to that. It can be written in compact form using the unit impulse which has two distinct values depending on the index. (see the Kronecker delta in digital signal processing). Think $\delta_{kn}$.
• We still need the summation, but we can start with the integral. It's only that the integral is zero for all values of $k$ except at $k=n$ where it is $T$. So you remain with one term in your summation.

All in all, your initial equation equals: $\displaystyle \sum^{+\infty}_{k\ =\ -\infty} a_kT\delta_{kn}$