Given is the signal $y(t) = sgn(\sin{\frac{2πt}{T_0}})$ where $sgn(t)=1 $ for $ t \geq 0 $ and $sgn(t)=-1 $ for $ t < 0$. And $T_0$ is the period of the signal.
I need to find the complex fourier series coefficients $c_k$ such that $y(t)=\sum_{k=-\infty}^{\infty}c_k e^{\frac{2 \pi i k t}{T_0}}$. The solution says that $c_0=0$ and $c_k=\frac{1}{\pi i k}(1-e^{-\pi i k})$ which gives $0$ for even $k$ and $\frac{2}{\pi i k}$ for odd $k$.
Now, I have no clue about how they got that result. Also, I've read here https://www.physicsforums.com/threads/fouier-series-from-the-sign-function.342670/ that the signum function has no fourier series representation.
Do you have any idea/ hint about how to proceed ?
Best Answer
$ y(t)= sgn(\sin{\frac{2 \pi t}{T_0}})$ equals $1$ between $t=0$ and $t= $$T_0 \over 2$, and equals $-1$ between $t=$ $T_0 \over 2$ and $t = T_0$.
The fourier coefficients are: $c_k=\frac{1}{T_0}\int_{0}^{T_0}y(t)e^{-2 \pi i k t \over T_0 }dt=\frac{1}{T_0}\int_{0}^{T_0 \over 2}y(t)e^{-2 \pi i k t \over T_0 }+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0 }y(t)e^{-2 \pi i k t \over T_0 }= $
$\frac{1}{T_0}\int_{0}^{T_0 \over 2}1\cdot e^{-2 \pi i k t \over T_0 }+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0}(-1)\cdot e^{-2 \pi i k t \over T_0 }dt = $ $\frac{1}{-2\pi i k}(-1)^k + \frac{1}{2\pi i k} + \frac{1}{2\pi i k}+ \frac{1}{-2\pi i k}(-1)^k = $
$\frac{1}{\pi i k}+\frac{1}{\pi i k}(-1)^{k+1} = 0$ for $k$ even and $\frac{2}{\pi i k}$ for $k$ odd.
And $c_0= \frac{1}{T_0}\int_{0}^{T_0}y(t)dt = \frac{1}{T_0}\int_{0}^{T_0 \over 2}1 dt+\frac{1}{T_0}\int_{T_0 \over 2}^{T_0 }(-1)dt = \frac{1}{2} + 0 - 1 + \frac{1}{2} = 0$