0

I know that the signals Fourier serie looks like this:

$$x(t)=\frac{1}{2}-\frac{1}{\pi}\sum^{\infty}_{n=1}\frac{1}{n}\sin(\frac{n\pi t}{L})$$

And that the graph should look something like this (pretend the lines are straight, did my best):

Graph of the signal

To solve it I thought that the sum $\frac{1}{\pi}\sum^{\infty}_{n=1}\frac{1}{n}\sin(\frac{n\pi t}{L})$ have to be $\frac{1}{2}$ when $t=0$ and $- \frac{1}{2}$ when $t=T_0$. Because of this, $x(\frac{T_0}{2})$ should be equal to $\frac{1}{2}$ since the graph is linear between $0 \le t \le T_0$. Because of this we know that the sum should be equal to zero:

$$\sum^{\infty}_{n=1}\frac{1}{n}\sin(\frac{n\pi \frac{T_0}{2}}{L}) = 0$$

Now, the only way I can think of this being the case is if $\frac{n\pi \frac{T_0}{2}}{L} = 2\pi, \forall n, n \in N$ since if sinusoid is 0, $\frac{1}{n}$ is zero. So then I solve:

$$\frac{\pi \frac{T_0}{2}}{L} = 2\pi \implies T_0 = 4L$$

This was my answer, but the solution say that $T_0 = 2L$ and I can not for the life of me figure out why. Can anyone help me?

Salviati
  • 357

1 Answers1

1

When a function has a (finite) discontinuity, the Fourier series will converge to the mid value.
So, in the case of the triangular wave that you have depicted, the Fourier series is converging to $1/2$ either at $t=0$ and $t=T_0$.

And actually, the period of the series is the minimum $T$ for which you have $x(t+T) =x(t)$, so: $$ \eqalign{ & x(t + T) = x(t) = \cr & = \frac{1} {2} - \frac{1} {\pi }\sum\limits_{n = 1}^\infty {\frac{1} {n}\sin \left( {\frac{{n\pi t}} {L} + \frac{{n\pi T}} {L}} \right)} = \cr & = \frac{1} {2} - \frac{1} {\pi }\sum\limits_{n = 1}^\infty {\frac{1} {n}\sin \left( {\frac{{n\pi t}} {L}} \right)} \quad \Rightarrow \cr & \Rightarrow \quad \frac{{\pi T}} {L} = 2\pi \quad \Rightarrow \quad T = 2L \cr} $$

Following your reasoning, then we have $$ \hat x(t) = F^{\, - \,1} \left( {F\left( {x(t)} \right)} \right):\quad \hat x(0) = \hat x(T_0 ) = 1/2 $$ and also $$ \hat x(T_0 /2) = x(T_0 /2) = 1/2 $$

Therefore $$ \sin \left( {{{n\pi T_0 /2} \over L}} \right) = 0\quad \Rightarrow \quad {{\pi T_0 /2} \over L} = \pi $$ not $2 \pi$, because the zeros of $\sin x$ are at distance $\pi$, while the period is $2 \pi$.

G Cab
  • 35,272