Revision as of 06:47, 17 September 2008 by Zcurosh (Talk)

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang