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<math>x(t) = 2 + \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{2j5\pi t}{6}} + 2je^{\frac{-2j5\pi t}{6}}</math> | <math>x(t) = 2 + \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{2j5\pi t}{6}} + 2je^{\frac{-2j5\pi t}{6}}</math> | ||
<br> | <br> | ||
| − | Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br> | + | Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br><br> |
Also, we can get coefficients <math>a_0</math>,<math>a_2</math>,<math>a_{-2}</math>,<math>a_5</math>, | Also, we can get coefficients <math>a_0</math>,<math>a_2</math>,<math>a_{-2}</math>,<math>a_5</math>, | ||
| − | <math>a_{-5}</math>.<br> | + | <math>a_{-5}</math>.<br><br> |
<math>a_0 = 2, a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j</math> | <math>a_0 = 2, a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j</math> | ||
Revision as of 16:27, 21 September 2008
CT signal
$ x(t) = 2 + cos({\frac{2\pi t}{3}})+ 4sin({\frac{5\pi t}{3}})\, $
Coefficients
$ cos({\frac{2\pi t}{3}}) = \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} $
$ 4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = 2 + \frac{1}{2}e^{\frac{j2\pi t}{3}} + \frac{1}{2}e^{\frac{-j2\pi t}{3}} -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}} $
$ x(t) = 2 + \frac{1}{2}e^{\frac{2j2\pi t}{6}} + \frac{1}{2}e^{\frac{-2j2\pi t}{6}} -2je^{\frac{2j5\pi t}{6}} + 2je^{\frac{-2j5\pi t}{6}} $
Then we can know the fundamental frequency is $ \frac{\pi}{3} $.
Also, we can get coefficients $ a_0 $,$ a_2 $,$ a_{-2} $,$ a_5 $,
$ a_{-5} $.
$ a_0 = 2, a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j $
