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==Answer== | ==Answer== | ||
Because the signal is even, a2=a(N-2) and a3=a(N-3. Using Parseval's relation, we find that the x[n]in part 3 is equal to ak. Since the sum of all 4 given ak values is 144, we know that all other values of ak =0. From here we just plug these 4 ak values into the x[n] equation. | Because the signal is even, a2=a(N-2) and a3=a(N-3. Using Parseval's relation, we find that the x[n]in part 3 is equal to ak. Since the sum of all 4 given ak values is 144, we know that all other values of ak =0. From here we just plug these 4 ak values into the x[n] equation. | ||
| + | <br> | ||
| + | <math>x[n] = \sum_{n=0}^{N-1}a_k e^{jk\frac{2\pi}{N}n}\,</math> | ||
| + | <br> | ||
| + | <math>x[n]=6e^{j\pi/3*n}+6e^{j\pi/2*n}+6e^{j3\pi/2*n}+6e^{j5\pi/4*n}</math> | ||
Latest revision as of 10:13, 26 September 2008
Question
1. DT signal is real and even with period N=12
2. a2=a3=6
3. $ 1/12\sum_{n=0}^{11}|x[n]|^2=144 $.
Answer
Because the signal is even, a2=a(N-2) and a3=a(N-3. Using Parseval's relation, we find that the x[n]in part 3 is equal to ak. Since the sum of all 4 given ak values is 144, we know that all other values of ak =0. From here we just plug these 4 ak values into the x[n] equation.
$ x[n] = \sum_{n=0}^{N-1}a_k e^{jk\frac{2\pi}{N}n}\, $
$ x[n]=6e^{j\pi/3*n}+6e^{j\pi/2*n}+6e^{j3\pi/2*n}+6e^{j5\pi/4*n} $
