Fundamental period of periodic delta function

I have the following signal:

$$x[n] = X\sum_{k=-\infty}^{\infty} (\delta[n − 4k] + \delta[n − 4k − 1] − \delta[n − 4k − 2] − \delta[n − 4k − 3])$$

The fundamental period of this signal is $4$. Is this purely because of the factor in front of $k$? Secondly, if I were to sketch the signal for 1 period, would the value for $x = \delta + \delta[-1] - \delta[-2] - \delta[-3]$? Or would I have to take into consideration the value of k?

• I mean, draw one element of the sum – for example, the one you get when setting $k=0$. Then change $k$ by one. You get periodicity without overlap, so yes, of course, the factor of 4 makes the period. – Marcus Müller Mar 21 '17 at 7:22

Okay let's go with the definition of periodicity:

\begin{equation*} x[n]=x[n+T] \end{equation*}

which in turn yields:

\begin{eqnarray*} x[n+T] = \sum\limits_{k=-\infty}^{\infty} (\delta[n+T-4k]+\delta[n+T-4k-1]-\delta[n+T-4k-2]-\delta[n+T-4k-3]) \\ = \sum\limits_{k=-\infty}^{\infty} (\delta[n-(4k-T)]+\delta[n-(4k+1-T]-\delta[n-(4k+2-T)]-\delta[n-(4k+3-T)]) \end{eqnarray*}

On inspection, if we allow T=4,

\begin{eqnarray*} x[n+4] = \sum\limits_{k=-\infty}^{\infty} (\delta[n-(4k-4)]+\delta[n-(4k+1-4]-\delta[n-(4k+2-4)]-\delta[n-(4k+3-4)]) \\ = \sum\limits_{k=-\infty}^{\infty} (\delta[n-4(k-1)]+\delta[n-(4k-3)]-\delta[n-(4k-2)]-\delta[n-(4k-1)]) \end{eqnarray*}

Now let k-1=m or m=k+1. Note that limits on m remain the same:

\begin{eqnarray*} x[n+4] = \sum\limits_{m=-\infty}^{\infty} (\delta[n-4m]+\delta[n-4m-1)]-\delta[n-4m-2]-\delta[n-4m-3)]) \\ = x[n] \end{eqnarray*}

This explains why the period of x[n] is 4. Naturally, the period is directly related to the coefficient of k, as can be seen in the above equations.

As for x, n takes the value of 0, but k continues to range from -infinity to infinity.

\begin{equation*} x = \sum\limits_{k=-\infty}^{\infty} (\delta[-4k]+\delta[-4k-1]-\delta[-4k-2]-\delta[-4k-3]) \end{equation*}