Phase angle of a Fourier series

Question

I have read in my textbook that if a Fourier series consists of only sine terms(that is, the function is odd), its phase angle is 0. If the Fourier series consists of only cosine terms(that is, the function is even), then the phase angle is 90 degrees. Can anyone provide an explanation?

Answer 1

Well it has been a while since the question was submitted but anyway let's start...

So in Fourier series, for coefficients, we have by definition:

$c_k := \int_{<T>} x(\tau)\cdot e^{-j \frac{2\pi}{T} k \tau} d\tau$

where $c_k$ is the k-th coefficient of the Fourier series expansion, $T$ is the period of signal $x(t)$, and $<T>$ means that the integral is over one arbitrary period of signal $x(t)$. And by definition we have:

$x(t) = \sum_{k=-\infty}^{\infty} c_k \cdot e^{j \frac{2\pi}{T} kt}$

First we should show that if the signal is real-valued, $c_k = c^*_{-k}$, where $*$ indicates complex conjugate of operand below itself.

For proof, using the definition of $c_k$ written above we can continue:

$c_{-k}^* = (\int_{<T>} x(\tau)\cdot e^{-j \frac{2\pi}{T} (-k) \tau} d\tau)^*$

then we apply the conjugate operator into parentheses, since $x(\tau)$ is real-valued, it doesn't change and the operand only affects $j$ in the exponent, then we can write:

$c_{-k}^* = \int_{<T>} {x(\tau)\cdot e^{j \frac{2\pi}{T} (-k) \tau} d\tau} = c_k \, \blacksquare \hspace{5mm} (1)$

Next we should show that the Fourier series coefficients for $x(-t)$, are $c_{-k}$, if $c_k$ are coefficients for $x(t)$.

Using the synthesis equation, we can write:

$x(-t) = \sum_{k'=-\infty}^{\infty} c_{k'} \cdot e^{j \frac{2\pi}{T} k'(-t)}$

Now if we substitute $k'$ by $-k$, we'll have:

$x(-t) = \sum_{k=-\infty}^{\infty} c_{-k} \cdot e^{j \frac{2\pi}{T} kt}\hspace{5mm}(2)$

so we can conclude that the coefficients for $x(-t)$ are $c_{-k}$ given that coefficients for $x(t)$ are $c_{k}$.

For odd functions we know that $x(t) = -x(-t)$ and we can now, equate the k-th coefficients of both sides of the equation. Using relation (2) we have:

$c_k = -c_{-k}$

From (1) we have:

$c_k = c_{-k}^*$

Using above equations we can write:

$-c_{-k} = c_{-k}^*$

And now, considering complex numbers characteristics, we can conclude that $c_{-k}$ are imaginary numbers and so $c_k$, thus the phase angle for odd functions is $\pm90$.

For even functions similary we will have:

$c_{-k} = c_{-k}^*$

and we can conclude that $c_k$ are real so the phase angle should be 0 or 180. (180 is for negative coefficients)