1. **The Signals**: You have a signal $x[l]$ and a system's impulse response $h[n-l]$. 2. **For Each Position \( n \)**: You're determining how ($x$) and ($h$) interact. You do this by "sliding" \( h \) across \( x \) (or vice versa) to each position \( n \). For each position \( n \), you calculate the overlap of \( x \) and \( h \) and sum up the products of their overlapping values. 3. **The Summation**: The formula ( $\sum_{l=-\infty}^{\infty} x[l] \cdot h[n-l]$) calculates this overlap. For each value of ($l$), you're multiplying the corresponding value of \( x \) with a value of \( h \). ( l ) ranges over all possible values so that you account for the entire overlap of the two signals. This sum gives the value of the convolved signal at the position \( n \). 4. **Convolution Operator**: This whole process is represented by \( x[n] * h[n] \), which gives the output signal as a function of \( n \). So, for each \( n \), you get a different value of the output signal, which is a result of the convolution operation. In simpler terms: - The convolution operation (represented by the asterisk) is like asking, "For each position \( n \) in the output, how much do \( x \) and \( h \) overlap, and what is the sum of the products of their overlapping values?" So, to directly answer your question: Yes, taking the sum from \( -\infty \) to \( \infty \) for the products of the overlapping values of \( x \) and \( h \) for each \( n \) gives us the convolved signal \( x[n] * h[n] \). ### Helpful facts For a system's impulse response $h[n]$ if $h[n] = 0$ for all $n < 0$, the system is considered causal. $y[n] = x[n] \ast h[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n-k]$ Notice that if $h[n]$ is not 0 for all n < 0, the system will not be causal. You will have to 'peek into the future' in order to determine $y[n]$. So, if you graph a system's impulse response and see stuff in the negative time axis, you can safely conclude that the system is non-causal.