Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT Signal
- $ x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4}) $
This is a signal with period $ T = {2\pi \over w_0} $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j 2w_0 t + {\pi \over 4})}+e^{-(j 2w_0 t + {\pi \over 4})}] $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}] $
- $ e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $
- $ e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}] $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) [e^{j 2w_0 t}]+ {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) [e^{-j 2w_0 t}] $
Thus, the Fourier series coefficients for this are:
$ a_0 = 1 $
$ a_1 = {1 \over 2j} $
$ a_{-1} = {-1 \over 2j} $
$ a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) $
$ a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) $
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