# How can $a_n$ for the trigonometric Fourier series be $0$ for a non-odd square signal?

Can $a_n$ for the trigonometric Fourier series be $0$ for a non-odd wave? Since at the moment the square wave is not even nor odd. Square wave is given by: $$s_2(t)=\begin{cases} -4, & 0\leq t\leq 2.5\\ 6,& 2.5\leq t<5 \end{cases}$$

The signal looks like this:

• When I find $a_n$ it gives me $0$ but when you get a value of zero, doesn't this mean that the function is odd?
• OR DOES THE ZERO ALSO IMPLY THAT IT IS NOT AN EVEN and clearly the wave isnt even?

EDIT:

If this wave is odd, how is it odd because I thought it had to be $s(t)=-s(-t)$ so for e.g. $6$ doesn't = $-4$?

• When finding the Fourier series, it is assumed that you are working on one period of a periodic signal. Is the periodic extension of your signal odd or even?
– MBaz
Apr 7, 2017 at 16:10
• Wouldnt it be neither when repetitions occur? as its going from 6 to -4? 0-5s is one period Apr 7, 2017 at 16:12
• Well, it's not odd since $-f(1)=4 \neq f(-1)$. It's not even, either. What can you say aobut $a_n$ then? Hint: the relationship is "if the function is odd, then the coefficients have certain form". Does this imply that, if the coefficients have certain form, the function is odd?
– MBaz
Apr 7, 2017 at 16:26
• what are you implying when you reference certain form? Apr 7, 2017 at 16:30
• "certain form" = "they have a particular property" = "they are zero".
– MBaz
Apr 7, 2017 at 16:31

Indeed, if the cosine coefficients ($a_n$) of the Fourier series of a function $s(t)$ are all zero, then the function is the sum of sine terms, which are all odd, implying that the function $s(t)$ is odd.
In your case, however, not all $a_n$ are zero; in particular, $a_0=2$. This means that the wave has a "DC" or mean value of $a_0/2=1$. Since $a_n=0$ for $n>0$, then the function $$u(t)=s(t)-a_0/2$$ is indeed odd, and now $a_n=0$ for all $n$.
• I don't know if the answer is clear: the function isn't odd, and not all $a_n$ are zero, so there's no contradiction anywhere.