0
$\begingroup$

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.

description of a dsp accumulator's stability

$\endgroup$

1 Answer 1

0
$\begingroup$

$B_x$ 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.