Period and Fourier series

Question

For signal x(t) I have to show it is periodic by finding its period. After that, I need to find Fourier coefficient $$ a_n $$ and use them to reconstruct signal $$ x(t) $$ taking a finite number of samples. $$x(t) = \sum_{n=-\infty}^{\infty} x_0\left(t + 2n\right)$$ $$\quad x_0(t) = u\left(t + \frac{1}{4}\right) - u\left(t - \frac{1}{4}\right)$$ $$\quad u(t) = \begin{cases} 1 & \text{for } t \geq 0, \\ 0 & \text{for } t < 0. \end{cases}$$

Answer 1

I have tried something like this, but I have no idea am I doing right. Any help would be great.

To show that the signal $( x(t) )$ is periodic, we first need to find its period $( T )$. By the definition of the signal, we can observe that the period is $( 2 )$ because the signal repeats every $( 2 )$ seconds, as given by $( t \rightarrow t + 2n )$.

To find the Fourier coefficients $( a_n )$, we first have to find the Fourier series of the signal $( x_0(t) )$. The Fourier series for a periodic signal with period $( T )$ is defined as:

$x(t) = \sum_{n=-\infty}^{\infty} a_n e^{i n \omega_0 t},$

where $(\omega_0 = \frac{2\pi}{T})$, and $(a_n)$ are the Fourier coefficients given by the formula:

$a_n = \frac{1}{T} \int_{-T/2}^{T/2} x(t) e^{-i n \omega_0 t} dt.$

In this case, $( T = 2 )$ and $( \omega_0 = \pi )$. Thus, $( a_n )$ can be given by the formula:

$a_n = \frac{1}{2} \int_{-1}^{1} x(t) e^{-i n \pi t} dt.$

To find $( a_n )$, we need to find an expression for $( x(t) )$ in terms of $( x_0(t) )$. We can then use the formula for $( x_0(t) )$ to express $( x(t) )$ as:$x(t) = \sum_{n=-\infty}^{\infty} x_0(t+2n).]$

Next, we find an expression for $( x_0(t) )$ in terms of $( u(t) )$: [x_0(t) = u(t+1/4) - u(t-1/4).$

Now we can express $( x(t) ) = x(t) = \sum_{n=-\infty}^{\infty} \big[ u(t+1/4+2n) - u(t-1/4+2n) \big].$