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<math>\,x(t)=\frac{3\pi}{2}(\frac{e^{j(\frac{3\pi}{2}t+\pi)}+e^{-j(\frac{3\pi}{2}t+\pi)}}{2})(\frac{e^{j(\frac{3\pi}{4}t+\frac{\pi}{2})}-e^{-j(\frac{3\pi}{4}t+\frac{\pi}{2})}}{2j})\,</math> | <math>\,x(t)=\frac{3\pi}{2}(\frac{e^{j(\frac{3\pi}{2}t+\pi)}+e^{-j(\frac{3\pi}{2}t+\pi)}}{2})(\frac{e^{j(\frac{3\pi}{4}t+\frac{\pi}{2})}-e^{-j(\frac{3\pi}{4}t+\frac{\pi}{2})}}{2j})\,</math> | ||
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| + | <math>\,x(t)=\frac{3\pi}{8j}(e^{j(\frac{3\pi}{2}t)}e^{j\pi}+e^{-j(\frac{3\pi}{2}t)}e^{j\pi})(e^{j(\frac{3\pi}{4}t)}e^{j\frac{\pi}{2}}-e^{-j(\frac{3\pi}{4}t)}e^{j\frac{\pi}{2}})\,</math> | ||
Revision as of 09:32, 25 September 2008
Given the periodic CT signal
$ \,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\, $
compute its Fourier series coefficients.
Answer
We can rewrite the signal $ x(t) $ as
$ \,x(t)=\frac{3\pi}{2}(\frac{e^{j(\frac{3\pi}{2}t+\pi)}+e^{-j(\frac{3\pi}{2}t+\pi)}}{2})(\frac{e^{j(\frac{3\pi}{4}t+\frac{\pi}{2})}-e^{-j(\frac{3\pi}{4}t+\frac{\pi}{2})}}{2j})\, $
$ \,x(t)=\frac{3\pi}{8j}(e^{j(\frac{3\pi}{2}t)}e^{j\pi}+e^{-j(\frac{3\pi}{2}t)}e^{j\pi})(e^{j(\frac{3\pi}{4}t)}e^{j\frac{\pi}{2}}-e^{-j(\frac{3\pi}{4}t)}e^{j\frac{\pi}{2}})\, $
