| Line 2: | Line 2: | ||
<br> This problem could be solved using euler's rule that e^jwt=cos(wt)+jsin(wt) | <br> This problem could be solved using euler's rule that e^jwt=cos(wt)+jsin(wt) | ||
<br> However, for this case I will simply observe that the system is: | <br> However, for this case I will simply observe that the system is: | ||
| − | <br> | + | <br> y(t)=tx(-t) |
| − | <br> | + | <br> So: x(t)=cos(2t) ;we make t=-t and multiply by t |
| − | <br> | + | <br> y(t)=t*cos(-2*t);Note that cos(-t)=cos(t) |
<br>Therefore: y(t)=t*cos(-2*t) | <br>Therefore: y(t)=t*cos(-2*t) | ||
Latest revision as of 08:17, 18 September 2008
Hw3.B basics of linearity
This problem could be solved using euler's rule that e^jwt=cos(wt)+jsin(wt)
However, for this case I will simply observe that the system is:
y(t)=tx(-t)
So: x(t)=cos(2t) ;we make t=-t and multiply by t
y(t)=t*cos(-2*t);Note that cos(-t)=cos(t)
Therefore: y(t)=t*cos(-2*t)
