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now we know <math>x[n]= 1 + a1e^{-j\pi/2}+\dfrac{1}{2}e^{-j\pi/2}+a3e^{-j\pi/2}</math> | now we know <math>x[n]= 1 + a1e^{-j\pi/2}+\dfrac{1}{2}e^{-j\pi/2}+a3e^{-j\pi/2}</math> | ||
Since the power is minimum all the other ak values are zero. | Since the power is minimum all the other ak values are zero. | ||
| − | so x[n]= 1 + \dfrac{1}{2}e^{-j\pi/2} | + | so, <math>x[n]= 1 + \dfrac{1}{2}e^{-j\pi/2} |
Revision as of 06:05, 26 September 2008
Guess Signal
The signal is DT periodic with period of 4
- $ \sum_{n=2}^{10} x[n]= 8 $
- $ \sum_{n=4}^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=4}^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=0}^{4-1} x[n]e^{-jk2\pi n /4} $ for the exponent to be -j2\pi n k/4 has to equal 2,
- $ a2=\dfrac{1}{4}\sum_{n=0}^{3} x[n]e^{-j\pi n } $
- $ \sum_{n=0}^{3} x[n]e^{-j\pi n }=\sum_{n=4}^7 x[n]e^{-j\pi n}=2 $
- $ a2=1/4*2=1/2 \, $
now we know $ x[n]= 1 + a1e^{-j\pi/2}+\dfrac{1}{2}e^{-j\pi/2}+a3e^{-j\pi/2} $ Since the power is minimum all the other ak values are zero. so, $ x[n]= 1 + \dfrac{1}{2}e^{-j\pi/2} $
