Guess Signal
The signal is DT periodic with period of 4
- $ x[1]=0 \, $
- $ \sum_{n=2}^{10} x[n]= 8 $
- $ \sum_{n=3}^7 x[n]e^{-j\pi n}=2 $
x[n] has min power among all signals that satisfy the above.
Solution
- $ a0=\dfrac{1}{2T}\sum_{n=2}^{10} x[n] $
- $ a0=1/8*8=1 \, $
- $ \sum_{n=5}^7 x[n]e^{-j\pi n}=2 $ looks like $ ak=\dfrac{1}{N}\sum_{n=0}^{N-1} x[n]e^{-jk2\pi n /N} $
N=4,so $ ak=\dfrac{1}{4}\sum_{n=1}^{4-1} x[n]e^{-jk2\pi n /4} $ for the exponent to be -j\pi n k has to equal 2,
- $ a2=\dfrac{1}{4}\sum_{n=1}^{3} x[n]e^{-j\pi n } $
