| (7 intermediate revisions by the same user not shown) | |||
| Line 15: | Line 15: | ||
<big><math>x[n] = 1e^{j\pi n}</math></big> | <big><math>x[n] = 1e^{j\pi n}</math></big> | ||
| − | + | Comparing the simplified x[n] with the fourier series, we can get <math>k\frac{2\pi}{N} = \pi n</math> | |
| − | + | N = 2k, where k is the smallest integer for N to be an integer. Therefore k = 1 and N =2. | |
| − | <math> | + | <math>a_0 = \frac{1}{2}\displaystyle\sum_{n=0}^{1}e^{j\pi n} = \frac{1}{2}e^{j\pi 0} + \frac{1}{2}e^{j\pi} = 0 </math> |
| − | + | <math>a_1 = \frac{1}{2}\displaystyle\sum_{n=0}^{1}e^{j\pi n}e^{-j\pi n} = 1 </math> | |
| + | |||
| + | Therefore, <Big><math>x[n] = a_0e^{j\pi n} + a_1e^{j\pi n} = e^{j\pi n}</math></Big> | ||
Latest revision as of 13:22, 26 September 2008
For periodic DT signal, x[n] with fundamental period N:
$ x[n]=\displaystyle\sum_{k=0}^{n-1}a_ke^{jk\frac{2\pi}{N}n} $
The Fourier series coefficients can be calculated with:
$ a_k = \frac{1}{N}\displaystyle\sum_{n=0}^{N-1}x[n]e^{-jk\frac{2\pi}{N}n} $
Let us look for the Fourier series coefficients for the DT signal $ x[n] = cos(3\pi n) $
$ x[n] = cos(5\pi n) = \frac{e^{j5\pi n}+e^{-j5\pi n}}{2} = \frac{1}{2}e^{j4\pi n}e^{j\pi n} + \frac{1}{2}e^{-j4\pi n}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{-j\pi n} = \frac{1}{2}e^{j\pi n} + \frac{1}{2}e^{j\pi n} $
Finally,
$ x[n] = 1e^{j\pi n} $
Comparing the simplified x[n] with the fourier series, we can get $ k\frac{2\pi}{N} = \pi n $
N = 2k, where k is the smallest integer for N to be an integer. Therefore k = 1 and N =2.
$ a_0 = \frac{1}{2}\displaystyle\sum_{n=0}^{1}e^{j\pi n} = \frac{1}{2}e^{j\pi 0} + \frac{1}{2}e^{j\pi} = 0 $
$ a_1 = \frac{1}{2}\displaystyle\sum_{n=0}^{1}e^{j\pi n}e^{-j\pi n} = 1 $
Therefore, $ x[n] = a_0e^{j\pi n} + a_1e^{j\pi n} = e^{j\pi n} $
