For an accumulator, defined as shown in the image below, why would I define $B_x=1$? $u[n]$ is defined at zero so my (possibly misguided intuition) is telling me that I'd choose $B_x = 0$ to not eliminate that point.
1 Answer
$\begingroup$
$\endgroup$
$B_x$ is a bound on the magnitude of the input signal. So if $x[n]=u[n]$ you have $|x[n]|\le1$ and, consequently, $B_x=1$ is a valid bound on $|x[n]|$. $B_x=0$ would mean that the input signal is zero.