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| − | 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. | + | 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. |
Revision as of 10:08, 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.
