How will the signal $\sum_{n=-\infty}^\infty \Delta (t-n)u_{-1}(t-n)$ look like? How will the signal $\sum_{n=-\infty}^\infty \Delta (t-n)u_{-1}(t-n)$ look like?


I understand that $\sum_{n=-\infty}^\infty \Delta (t-n)$ is a periodic signal with period $1$, where the triangular pulse $\Delta(t)$ is shifted by $1$ units for each value of $n$.

where,
$$\Delta(t) =
\begin{cases}
1-2\vert t\rvert,&amp; \lvert t\rvert&lt;\frac{1}{2} \\
0,&amp;\lvert t\rvert&gt;\frac{1}{2} 
\end{cases}$$
From this I infer that $\sum_{n=-\infty}^\infty \Delta (t-n)u(t-n)$ would probably be a signal in which the RIGHT HALF of $\Delta (t)$ will be shifted by $1$ unit for every value of $n$. But here I have a doubt that $u(t-n)$ will be added with the previous values of $u(t-n)$ i.e $u(t-(n-1))$ and will have a staircase effect. Thus the signal will no longer be periodic.(Which is a cause of concern as we have been asked to find the Fourier series of the signal, which means the signal should be periodic)
I also don't understand what the subscript $-1$ in $u_{-1}(t-n)$ means.


Please help me visualize this signal(It would be highly helpful if you would  draw the graph as well to help me visualize the signal)
Initially, don't worry about the sum. Just try to figure out how the signal $\Delta(t)u_{-1}(t)$ looks like. As soon as you know that, you just need to shift it by integer values of $t$, and add them all up. If the shifted versions do not overlap, you simply get copies of the basic signal shifted to integer values of $t$. From the solution in the solution manual (as given in a comment), the basic signal is the right half of a triangle, and the period equals $1$. The symbol $u_{-1}(t)$ refers to the unit step function, which is more frequently referred to as $u(t)$ (see also this question and its answer(s)).