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Hope that helps. -pm </span> | Hope that helps. -pm </span> | ||
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| + | ==About Invertibility in Question 1== | ||
| + | <span style="color:blue"> TA's comments: I received a couple of questions about invertibility in Question 1. In order to prove that a system is not invertible, it is sufficient to find just one pair of two different inputs that yield the same output. Otherwise, if you want to say that the system is invertible you have to find its inverse system. To do that, just try to write the input of the system at hand in terms of the output. The inverse system can be any general system that can recover the original input from the output. Hint: A system can depend not only on its input but on shifted versions of its output as well.</span> --[[User:Ahmadi|Ahmadi]] 18:40, 3 February 2011 (UTC) | ||
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[[HW3_ECE301_Spring2011_Prof_Boutin|Back to HW3]], | [[HW3_ECE301_Spring2011_Prof_Boutin|Back to HW3]], | ||
[[2011 Spring ECE 301 Boutin|Back to 2011 Spring ECE 301 Boutin]] | [[2011 Spring ECE 301 Boutin|Back to 2011 Spring ECE 301 Boutin]] | ||
Revision as of 13:40, 3 February 2011
Contents
Discussion for HW3, ECE301, Spring 2011, Prof. Boutin
About Stability in Question 1
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.
- 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 $, and thus time shifting the signal will not affect the values themselves but rather their place with respect to the time axis.
About convolution
I understand that convolution is commutative, but I was wondering if there are any good general rules as to picking the order. In other words, is there a good way of determining if computing the integral (wrt tau) of x(tau)h(t-tau) is easier than computing the same of h(tau)x(t-tau), or is this something that we will pick up on after some practice?
- TA's comments: Regarding convolution, I believe that generally flipping the signal that has longer duration should make the convolution easier.
About Question 2
Instructor's comment: I was just asked in an email whether one can answer Question 2 if the system given is not LTI. The answer is yes, you can find the unit impulse response of any system, not just LTI systems. Finding the unit impulse response is easy: just plug a unit impulse ($ \delta $) in place of the input signal!
Example: if $ y(t)=x(t+1)-3 $ then $ h(t)=\delta (t+1)-3 $.
Hope that helps. -pm
About Invertibility in Question 1
TA's comments: I received a couple of questions about invertibility in Question 1. In order to prove that a system is not invertible, it is sufficient to find just one pair of two different inputs that yield the same output. Otherwise, if you want to say that the system is invertible you have to find its inverse system. To do that, just try to write the input of the system at hand in terms of the output. The inverse system can be any general system that can recover the original input from the output. Hint: A system can depend not only on its input but on shifted versions of its output as well. --Ahmadi 18:40, 3 February 2011 (UTC)
