| Line 7: | Line 7: | ||
== Answer == | == Answer == | ||
| − | We can rewrite the signal <math>x(t)</math> as | + | We can rewrite the signal <math>\,x(t)\,</math> as |
<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> | ||
| Line 19: | Line 19: | ||
<math>\,x(t)=-\frac{3\pi}{8}(e^{j\frac{9\pi}{4}t}-e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t}-e^{-j\frac{9\pi}{4}t})\,</math> | <math>\,x(t)=-\frac{3\pi}{8}(e^{j\frac{9\pi}{4}t}-e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t}-e^{-j\frac{9\pi}{4}t})\,</math> | ||
| − | <math>\,x(t)= | + | <math>\,x(t)=-\frac{3\pi}{8}e^{j\frac{\pi}{4}t}+\frac{3\pi}{8}e^{j\frac{3\pi}{4}t}-\frac{3\pi}{8}e^{-j\frac{3\pi}{4}t}+\frac{3\pi}{8}e^{-j\frac{\pi}{4}t}\,</math> |
| − | -\frac{3\pi}{8}e^{j\frac{\pi}{4}t}+ | + | |
| − | \frac{3\pi}{8}e^{j\frac{3\pi}{4}t}- | + | |
| − | \frac{3\pi}{8}e^{-j\frac{3\pi}{4}t}+ | + | This form of <math>\,x(t)\,</math> is in the format of a Fourier series, so we can directly get the fundamental frequency <math>\,\omega_o\,</math> and the coefficients <math>\,a_k\,</math>. Therefore, the answer is: |
| − | \frac{3\pi}{8}e^{-j\frac{\pi}{4}t} | + | |
| − | \,</math> | + | <math>\,\omega_o=\frac{\pi}{4}\,</math> |
| + | |||
| + | and the coefficients are | ||
| + | |||
| + | <math>\,a_{-3}=-\frac{3\pi}{8}\,</math> | ||
| + | |||
| + | <math>\,a_{-1}=\frac{3\pi}{8}\,</math> | ||
| + | |||
| + | <math>\,a_1=-\frac{3\pi}{8}\,</math> | ||
| + | |||
| + | <math>\,a_3=\frac{3\pi}{8}\,</math> | ||
| + | |||
| + | <math>\,a_k=0\,</math> for all other <math>\,k\in\mathbb{Z}\,</math> | ||
Revision as of 08:57, 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})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j(\frac{3\pi}{2}+\frac{3\pi}{4})t}-e^{j(\frac{3\pi}{2}-\frac{3\pi}{4})t}+e^{j(-\frac{3\pi}{2}+\frac{3\pi}{4})t}-e^{j(-\frac{3\pi}{2}-\frac{3\pi}{4})t})\, $
$ \,x(t)=-\frac{3\pi}{8}(e^{j\frac{9\pi}{4}t}-e^{j\frac{3\pi}{4}t}+e^{-j\frac{3\pi}{4}t}-e^{-j\frac{9\pi}{4}t})\, $
$ \,x(t)=-\frac{3\pi}{8}e^{j\frac{\pi}{4}t}+\frac{3\pi}{8}e^{j\frac{3\pi}{4}t}-\frac{3\pi}{8}e^{-j\frac{3\pi}{4}t}+\frac{3\pi}{8}e^{-j\frac{\pi}{4}t}\, $
This form of $ \,x(t)\, $ is in the format of a Fourier series, so we can directly get the fundamental frequency $ \,\omega_o\, $ and the coefficients $ \,a_k\, $. Therefore, the answer is:
$ \,\omega_o=\frac{\pi}{4}\, $
and the coefficients are
$ \,a_{-3}=-\frac{3\pi}{8}\, $
$ \,a_{-1}=\frac{3\pi}{8}\, $
$ \,a_1=-\frac{3\pi}{8}\, $
$ \,a_3=\frac{3\pi}{8}\, $
$ \,a_k=0\, $ for all other $ \,k\in\mathbb{Z}\, $
