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:<span style="color:blue"> TA's comments: If <math>x[n]</math> is a bounded signal then <math>x[n-1]</math> is also bounded. This is a direct result since all the values <math>|x[n]|</math> are bounded for all <math>n</math>, and thus time shifting the signal will not affect the values themselves but rather their place with respect to the time axis. Regarding convolution, I believe that flipping the signal that has longer duration should make the convolution easier.</span> | :<span style="color:blue"> TA's comments: If <math>x[n]</math> is a bounded signal then <math>x[n-1]</math> is also bounded. This is a direct result since all the values <math>|x[n]|</math> are bounded for all <math>n</math>, and thus time shifting the signal will not affect the values themselves but rather their place with respect to the time axis. Regarding convolution, I believe that flipping the signal that has longer duration should make the convolution easier.</span> | ||
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| + | [[2011 Spring ECE 301 Boutin|Back to 2011 Spring ECE 301 Boutin]] <br> | ||
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Revision as of 15:00, 2 February 2011
I'm having a hard time proving 1a is stable or unstable. If (|x[n]|<m) then is it also true that (|x[n-1]|<m)? I'm assuming the product of two bounded signals also gives a bounded signal.
I understand that convolution is commutative, but I was wondering if there are any good general rules as to picking the order. In other words, is there a good way of determining if computing the integral (wrt tau) of x(tau)h(t-tau) is easier than computing the same of h(tau)x(t-tau), or is this something that we will pick up on after some practice?
- TA's comments: If $ x[n] $ is a bounded signal then $ x[n-1] $ is also bounded. This is a direct result since all the values $ |x[n]| $ are bounded for all $ n $, and thus time shifting the signal will not affect the values themselves but rather their place with respect to the time axis. Regarding convolution, I believe that flipping the signal that has longer duration should make the convolution easier.
