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| + | ==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. | 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. | ||
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| + | ==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? | 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? | ||
:<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 generally 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 generally flipping the signal that has longer duration should make the convolution easier.</span> | ||
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| + | ==About Question 2== | ||
| + | From your instructor: 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 system. Finding the unit impulse response is easy: just plug a unit impulse (<math>\delta</math>) in place of the input signal! | ||
| + | Example: if <math>y(t)=x(t+1)-3 </math> then <math>h(t)=\delta (t+1)-3</math>. | ||
| − | + | Hope that helps. -pm | |
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Revision as of 14:32, 2 February 2011
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.
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: 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 $, and thus time shifting the signal will not affect the values themselves but rather their place with respect to the time axis. Regarding convolution, I believe that generally flipping the signal that has longer duration should make the convolution easier.
About Question 2
From your instructor: I was just asked in an email whether one can answer Question 2 if the system given is not LTI. The answer is yes, you can find the unit impulse response of any system, not just LTI system. Finding the unit impulse response is easy: just plug a unit impulse ($ \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
