Line 12: Line 12:
 
                                 } -> + -> texp(-2jt) + texp(2jt)
 
                                 } -> + -> texp(-2jt) + texp(2jt)
 
exp(-2jt) -> system -> texp(2jt) -> *1 ->  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))  }

Revision as of 11:23, 16 September 2008

The Basics of Linearity

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)) }

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman