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<math>x(t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2}</math> | <math>x(t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2}</math> | ||
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| + | I contend that the <math>\omega_0=2</math> since both functions are periodic based on it. | ||
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| + | <math>a_7=\frac{1}{2j}</math> | ||
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| + | <math> a_{-7} = \frac{-1}{2j}</math> | ||
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| + | <math> a_1 = \frac{1+3j}{2} </math> | ||
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| + | <math> a_{-1} = \frac{1+3j}{2} </math> | ||
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| + | We can write this as a sum: | ||
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| + | <math> x(t)=\sum^{\infty}_{k = -\infty} a_k e^{j2k}\,</math> | ||
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| + | Where | ||
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| + | <math> a_1=a_{-1}=\frac{1+3j}{2} </math> | ||
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| + | <math> a_7 = -a_{-7} = \frac{1}{2j} </math> | ||
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| + | <math> a_k = 0, k \= 1, -1, 7, -7 </math> | ||
Revision as of 15:19, 25 September 2008
Periodic CT Signal
The first signal that comes to mind when i think of a periodic CT signal is one involving sines and cosines, so let's work with one of those.
Let $ x(t) = sin(14t)+(1+3j)cos(2t) $
This can also be expressed as
$ x(t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2} $
I contend that the $ \omega_0=2 $ since both functions are periodic based on it.
$ a_7=\frac{1}{2j} $
$ a_{-7} = \frac{-1}{2j} $
$ a_1 = \frac{1+3j}{2} $
$ a_{-1} = \frac{1+3j}{2} $
We can write this as a sum:
$ x(t)=\sum^{\infty}_{k = -\infty} a_k e^{j2k}\, $
Where
$ a_1=a_{-1}=\frac{1+3j}{2} $
$ a_7 = -a_{-7} = \frac{1}{2j} $
$ a_k = 0, k \= 1, -1, 7, -7 $
