Difference between revisions of "HW3.B Zachary Curosh ECE301Fall2008mboutin" - Rhea

Latest revision as of 06:47, 17 September 2008

According to the definition of linearity given in class,

x1(t) -> system -> y1(t) -> *a -> ay1(t) }

                               } -> + -> ay1(t) + by2(t)

x2(t) -> system -> y2(t) -> *b -> by2(t) }

Now, according to the problem statement, and let a=b=1 since it is not specified in the problem,

exp(2jt) -> system -> texp(-2jt) -> *1 -> texp(-2jt) }

                               } -> + -> texp(-2jt) + texp(2jt)

exp(-2jt) -> system -> texp(2jt) -> *1 -> texp(2jt) }

I've decided to use Euler's formula to change the exponential into a sum of sines and cosines.

exp(2jt)=cos(2t)+jsin(2t) exp(-2jt)=cos(2t)-jsin(2t)

After putting each of these into the system,

cos(2t)+jsin(2t) -> system -> t(cos(2t)-jsin(2t)) -> *1 -> t(cos(2t)-jsin(2t)) }

                                                  } -> + -> t(cos(2t)-jsin(2t)) + t(cos(2t)-jsin(2t))

cos(2t)-jsin(2t) -> system -> t(cos(2t)+jsin(2t)) -> *1 -> t(cos(2t)-jsin(2t)) }

Next, t(cos(2t)-jsin(2t)) + t(cos(2t)-jsin(2t)) = 2tcos(2t), which is the response to 2cos(t). From here, it is a simple matter of dividing by 2.

The system's response to cos(2t) is tcos(2t).

@@ Line 7: / Line 7: @@
-Now, according to the problem statement,
-exp(2jt) -> system -> exp(-2jt) -> *1 -> exp(-2jt)  }
-                                 } -> + -> exp(-2jt) + exp(2jt)
+Now, according to the problem statement, and let a=b=1 since it is not specified in the problem,
-exp(-2jt) -> system -> exp(2jt) -> *1 ->  exp(2jt)  }
+exp(2jt) -> system -> texp(-2jt) -> *1 -> texp(-2jt)  }
+                                 } -> + -> texp(-2jt) + texp(2jt)
+exp(-2jt) -> system -> texp(2jt) -> *1 ->  texp(2jt)  }
+I've decided to use Euler's formula to change the exponential into a sum of sines and cosines.
+exp(2jt)=cos(2t)+jsin(2t)
+exp(-2jt)=cos(2t)-jsin(2t)
+After putting each of these into the system,
+cos(2t)+jsin(2t) -> system -> t(cos(2t)-jsin(2t)) -> *1 -> t(cos(2t)-jsin(2t))  }
+                                                   } -> + -> t(cos(2t)-jsin(2t)) + t(cos(2t)-jsin(2t))
+cos(2t)-jsin(2t) -> system -> t(cos(2t)+jsin(2t)) -> *1 ->  t(cos(2t)-jsin(2t))  }
+Next, t(cos(2t)-jsin(2t)) + t(cos(2t)-jsin(2t)) = 2tcos(2t), which is the response to 2cos(t).  From here, it is a simple matter of dividing by 2.
+The system's response to cos(2t) is tcos(2t).