| Line 21: | Line 21: | ||
All other <math>a_k=0</math>. | All other <math>a_k=0</math>. | ||
| − | The signal is then <math>x[n]=\frac{1}{2}(e^{\frac{3\pi}{4}jn}+e^{\frac{5\pi}{4}jn})</math> | + | The signal is then |
| + | |||
| + | <math>x[n]=\frac{1}{2}(e^{\frac{3\pi}{4}jn}+e^{\frac{5\pi}{4}jn})</math> | ||
| + | |||
| + | <math>x[n]=e^{j\pi n}\frac{1}{2}(e^{\frac{-\pi}{4}jn}+e^{\frac{\pi}{4}jn})</math> | ||
| + | |||
| + | <math>x[n]=(-1)^{n}\cos(\frac{\pi}{4}n)</math> | ||
Latest revision as of 17:19, 25 September 2008
Guess the Signal
1. $ x[n] $ is periodic with N=8.
2. $ \sum_{n=2}^{9}(-1)^{\frac{3}{4}n}x[n]=4 $.
3. $ a_3=a_5 $.
4. $ \sum_{k=0}^{7}a_k=1 $.
So What Is x[n]?
Given the first property of the signal, we can rewrite the bounds of summation in the second property since the zeroth term is the same as the eighth term and the first term is the same as the ninth term. Furthermore, we can change the $ (-1)^{\frac{3}{4}n} $ in the summation to something a little easier to manipulate. This will leave us with
$ \sum_{n=0}^{7}e^{\frac{3}{4}j\pi n}x[n]=4 $.
The astute observer will notice that, when divided by the period, this is the formula for computing $ a_3 $. This result along with hints 3 and 4 give the rest of the Fourier coefficients.
$ a_3=\frac{1}{2} $
$ a_5=\frac{1}{2} $
All other $ a_k=0 $.
The signal is then
$ x[n]=\frac{1}{2}(e^{\frac{3\pi}{4}jn}+e^{\frac{5\pi}{4}jn}) $
$ x[n]=e^{j\pi n}\frac{1}{2}(e^{\frac{-\pi}{4}jn}+e^{\frac{\pi}{4}jn}) $
$ x[n]=(-1)^{n}\cos(\frac{\pi}{4}n) $
