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<math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br> | <math>4sin({\frac{5\pi t}{3}}) = -2je^{\frac{j5\pi t}{3}} + 2je^{\frac{-j5\pi t}{3}}</math><br> | ||
| − | <math>x(t) = | + | <math>x(t) = \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}}</math> |
<br> | <br> | ||
| − | <math>x(t) = | + | <math>x(t) = \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><br> | Then we can know the fundamental frequency is <math>\frac{\pi}{3}</math>. <br><br> | ||
| − | Also, we can get coefficients | + | Also, we can get coefficients <math>a_2</math>,<math>a_{-2}</math>,<math>a_5</math>, |
<math>a_{-5}</math>.<br><br> | <math>a_{-5}</math>.<br><br> | ||
| − | <math> | + | <math>a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j, <math>a_k = 0, k != 2,-2,5,-5</math> </math> |
Revision as of 17:34, 25 September 2008
CT signal
$ x(t) = 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) = \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) = \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_2 $,$ a_{-2} $,$ a_5 $,
$ a_{-5} $.
$ a_2 = a_-2 = \frac{1}{2}, a_5 = -2j, a_-5 = 2j, <math>a_k = 0, k != 2,-2,5,-5 $ </math>
