- #1
Jncik
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Homework Statement
suppose that we have the continuous time signal
x(t) = cos(4πt) with fundamental period of T=1/2
Homework Equations
[tex]a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt[/tex]
where [tex]\omega_{0}[/tex] is obviously [tex]\frac{2\pi}{1/2} = 4\pi[/tex]
well the problem is that this integration becomes
[tex]
a_{k} = \frac{1}{T} \int_{T}{}x(t)e^{-j\omega_{0}kt}dt =
2\int_{0}^{\frac{1}{2}}cos(4 \pi t)e^{-j4 \pi kt}dt=
...=0
[/tex]
I find as a result 0, the calculation is complicated but using wolfram alpha's calculator I have the same result
here is the picture
[PLAIN]http://img52.imageshack.us/img52/6729/36554157.gif
as you can see
we have
[tex] 1 - e^{-2\pi j k} = 1 - ( cos(2\pi k) - jsin(2\pi k)) = 1 - 1 = 0[/tex]
hence it must be 0
now my book says the following
The nonzero FS coefficients of x(t) are [tex]a_{1} = a_{-1} = 1/2 [/tex]
but I have 0 as a result, and if I plug 1 or -1 I will get 0 again, what's wrong here?
thanks in advance
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