Difference between revisions of "HW3.B Ben Horst ECE301Fall2008mboutin" - Rhea

Revision as of 10:57, 19 September 2008

Homework 3 Ben Horst: A :: B :: C

If the system is linear (it is) then the following should be true:

Thus, if we say

x1 = exp(2jt) and x2 = exp(-2jt)

then

y1 = t exp(-2jt) and y2 = t exp(2jt)

That means that:

x1 + x2 ->|system|-> y1 + y2

Since $\cos(2t) = {e^{j2t} + e^{-j2t} \over 2}$

Then we can suggest that ${1 \over 2t}x1 + {1 \over 2t}x2 -> |system| -> {1 \over 2t}y1 + {1 \over 2t}y2$

This leads to the conclusion that the input of cos(2t) would be ${1 \over 2t}x1 + {1 \over 2t}x2$

Which, simplified, is: $\cos(2t) \over t$ .

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 Homework 3 Ben Horst:  [[HW3.A Ben Horst _ECE301Fall2008mboutin| A]]  ::  [[HW3.B Ben Horst _ECE301Fall2008mboutin| B]]  ::  [[HW3.C Ben Horst _ECE301Fall2008mboutin| C]]
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+==Answer==
+If the system is linear (it is) then the following should be true:
+Thus, if we say
+x1 =  exp(2jt)  and  x2 = exp(-2jt)
+then
+y1 = t exp(-2jt)  and  y2 = t exp(2jt)
+That means that:
+x1 + x2 ->|system|-> y1 + y2
+Since    <math>\cos(2t) =  {e^{j2t} + e^{-j2t}  \over 2}</math>
+Then we can suggest that <math>{1 \over 2t}x1 + {1 \over 2t}x2 -> |system| -> {1 \over 2t}y1 + {1 \over 2t}y2 </math>
+This leads to the conclusion that the input of cos(2t) would be <math>{1 \over 2t}x1 + {1 \over 2t}x2</math>
+Which, simplified, is: <math>\cos(2t) \over t</math>.