Problem: Find the periodic DT signal x[n], given:
1.) The fundamental period N = 2.
2.) $ \sum_{n=1}^2 x[n] = 6 $
3.) $ 2 \sum_{k=0}^1 {|x[k]|}^2 = 100 $
4.) $ a_k \geq 0 \ \ \forall a_k $
Solution:
- Statement 1 tells us that we can represent x[n] as:
$ x[n]= \sum_{k=0}^{(2)-1} a_k e^{jk \pi n} = a_0 + a_1 e^{j \pi n} $
- Statement 2 looks similar to how one would find $ a_0 $:
$ a_0 = \frac{1}{2} \sum_{k=0}^1 x[k] $
$ = \frac{1}{2} (x[0] + x[1]) $
$ = \frac{1}{2} (x[0+2] + x[1]) $ (since x is periodic with period 2)
$ = \frac{1}{2} \sum_{k=1}^2 x[k] $
$ = \frac{1}{2} (6) $ (by statement 2)
$ \Rightarrow a_0 = 3 $
- Statement 3 looks kind of cryptic, until one realizes that Pareseval's relation can be applied:
$ 2 \sum_{k=0}^1 {|x[k]|}^2 $
$ = 4 (\frac{1}{2} \sum_{k=0}^1 {|x[k]|}^2) $
$ = 4 (\sum_{k=0}^1 {|a_k|}^2) $
$ = 4(|{a_0}|^2 + |{a_1}|^2) = 100 $
Since $ a_0 = 3 $, then $ a_1 = 4 \text{or} -4 $.
- Statement 4 tells us that $ a_1 = 4 $ and not -4.
So, our function is: $ x[n] = 3 + 4 e^{j \pi n} $.
