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Now, according to the problem statement, and let a=b=1 since it is not specified in the problem, | 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) } | exp(2jt) -> system -> texp(-2jt) -> *1 -> texp(-2jt) } | ||
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After putting each of these into the system, | After putting each of these into the system, | ||
| + | |||
| + | |||
cos(2t)+jsin(2t) -> system -> t(cos(2t)-jsin(2t)) -> *1 -> t(cos(2t)-jsin(2t)) } | cos(2t)+jsin(2t) -> system -> t(cos(2t)-jsin(2t)) -> *1 -> t(cos(2t)-jsin(2t)) } | ||
Latest revision as of 06:47, 17 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)) }
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).
