| Line 12: | Line 12: | ||
<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> | <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> | ||
| + | |||
| + | <math>\,x(t)=\frac{3\pi}{8j}e^{j\pi}e^{j\frac{\pi}{2}}(e^{j(\frac{3\pi}{2}t)}+e^{-j(\frac{3\pi}{2}t)})(e^{j(\frac{3\pi}{4}t)}-e^{-j(\frac{3\pi}{4}t)})\,</math> | ||
Revision as of 09:34, 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}})\, $
$ \,x(t)=\frac{3\pi}{8j}e^{j\pi}e^{j\frac{\pi}{2}}(e^{j(\frac{3\pi}{2}t)}+e^{-j(\frac{3\pi}{2}t)})(e^{j(\frac{3\pi}{4}t)}-e^{-j(\frac{3\pi}{4}t)})\, $
