(→Finding x(t) by using given information) |
(→Finding x(t) by using given information) |
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<math> \sum^{2}_{-1} |a_k|^2 = 200 \,</math> | <math> \sum^{2}_{-1} |a_k|^2 = 200 \,</math> | ||
| − | <math> |a_ | + | <math> |a_ -1|^2 + |a_1|^2 + |a_0|^2 + |a_2|^2 = 200 \,</math> |
Then <math> a_0 = a_2 = 0. \,</math> | Then <math> a_0 = a_2 = 0. \,</math> | ||
<math> x[n] = \sum^{2}_{-1} a_k e^{j\frac{2\pi}{4}kn}\,</math> | <math> x[n] = \sum^{2}_{-1} a_k e^{j\frac{2\pi}{4}kn}\,</math> | ||
Revision as of 18:09, 25 September 2008
Information of x(t)
$ N = 4 $
$ a_5 = 10 $
x(t) is a real and even signal.
$ \frac{1}{4}\sum^{3}_{0} |x[n]|^2 = 200\, $
Finding x(t) by using given information
$ a_1 = a_5 = 10\, $
x(t) is a even siganl,so $ a_-1 = 10\, $
Using parseval's relation
$ \sum^{2}_{-1} |a_k|^2 = 200 \, $
$ |a_ -1|^2 + |a_1|^2 + |a_0|^2 + |a_2|^2 = 200 \, $
Then $ a_0 = a_2 = 0. \, $
$ x[n] = \sum^{2}_{-1} a_k e^{j\frac{2\pi}{4}kn}\, $
