Revision as of 11:28, 16 September 2008 by Zcurosh (Talk)

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)) }

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).

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett