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| + | =Discussion for [[HW3_ECE301_Spring2011_Prof_Boutin|HW3], [[ECE301]], Spring 2011, Prof. [[user:mboutin|Boutin]]= | ||
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==About Stability in Question 1== | ==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. | ||
| − | + | :<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. </span> | |
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==About convolution== | ==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: | + | :<span style="color:blue"> TA's comments: 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== | ==About Question 2== | ||
| − | + | <span style="color:green">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 (<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>. | Example: if <math>y(t)=x(t+1)-3 </math> then <math>h(t)=\delta (t+1)-3</math>. | ||
| − | Hope that helps. -pm | + | Hope that helps. -pm </span> |
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| + | [[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]] |
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Revision as of 14:37, 2 February 2011
Contents
Discussion for [[HW3_ECE301_Spring2011_Prof_Boutin|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
