I was asked to do the following thing.
Consider the fourier series of a signal given by
$$x(t)=\sum_{k=-\infty}^{\infty} a_ke^{jk\omega_0t}$$
Let's consider an approaches to this series given by the truncated series.
$$x_N(t)=\sum_{k=-N}^{N} a_ke^{jk\omega_0t}$$
a- Show that if $x(t)$ is real then the series can be rewritten as
$$x_N(t)=\sum_{k=-N}^{N} A_k\cos(k\omega_0t +\phi_k)$$
b- How can we calculate $A_k$ and $\phi_k$ knowing $a_k$?
Here goes my attempt:
So I first thought of putting together the k term and the (-k term). I have a doubt about how to get the expression of the series by doing this but I will clear it up after my proof.
We will have then:
$$a_ke^{jk\omega_0t} + a_{-k}e^{-jk\omega_0t}$$
Because x is real then $a_{-k}=a_k*$ where $a_k*$ is the conjugate of $a_k$.
We will have then
$$a_ke^{jk\omega_0t} + a_k*e^{-jk\omega_0t}$$ $$a_ke^{jk\omega_0t} + (a_k e^{jk\omega_0t})*$$ $$2 \operatorname{Re} (a_ke^{jk\omega_0t})$$
If we make $a_k = A_k e^{\phi_k}$ we will have:
$$2 \operatorname{Re} (A_ke^{j(k\omega_0t + \phi_k)})$$ $$2 A_k \cos(k\omega_0t + \phi_k)$$
Now my main doubt here is that factor of 2 and how it will affect the way I will express my calculations.
My first thought was: I reached this expression because I summed up two terms of the series. That means one term will be half what I calculates. Therefore, the series will be:
$$x_N(t)=\sum_{k=-N}^{N} A_k\cos(k\omega_0t +\phi_k)$$
Therefore, to answer question b I will just have to remember that $$a_k = A_k e^{\phi_k}$$ and so I will get to:
$$A_k = |a_k|$$ $$\phi_k = \arg(a_k)$$
But I'm not sure if this last few steps are correct, if my elimination of the 2 factor is correct. Or should I do:
$$x_N(t)=\sum_{k=-N}^{N} 2 A_k\cos(k\omega_0t +\phi_k)$$ $$x_N(t)=\sum_{k=-N}^{N} A'_k\cos(k\omega_0t +\phi_k)$$
and then $$A'_k = |a_k|/2$$ $$\phi_k = \arg(a_k)$$
Can someone clarify this to me please? Which process is correct and why? Thanks!