(New page: == Information of x(t) == <math> N = 4 </math> <math> a_5 = 10 </math> x(t) is a real and even signal. <math> \frac{1}{4}\sum^{3}_{0} |x[n]|^2 = 200\,</math> == Finding x(t) by usin...) |
(→Finding x(t) by using given information) |
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== Finding x(t) by using given information == | == Finding x(t) by using given information == | ||
| − | <math> a_1 = a_5 = 10</math> | + | <math> a_1 = a_5 = 10\,</math> |
| − | + | x(t) is a even siganl,so <math> a_{-1} = 10\,</math> | |
| − | + | Using parseval's relation | |
| − | <math> \sum^{2}_{-1} |a_k|^2 = 200 </math> | + | <math> \sum^{2}_{-1} |a_k|^2 = 200 \,</math> |
| − | <math> |a_-1|^2 + |a_1|^2 + |a_0|^2 + |a_2|^2 = 200 </math> | + | <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> |
| + | |||
| + | <math> x[n] = 10e^{j\frac{2\pi}{4}n} + 10e^{-j\frac{2\pi}{4}n}\,</math> | ||
| + | |||
| + | <math> x[n] = 10e^{j\frac{\pi}{2}n} + 10e^{-j\frac{\pi}{2}n}\,</math> | ||
Latest revision as of 18:13, 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}\, $
$ x[n] = 10e^{j\frac{2\pi}{4}n} + 10e^{-j\frac{2\pi}{4}n}\, $
$ x[n] = 10e^{j\frac{\pi}{2}n} + 10e^{-j\frac{\pi}{2}n}\, $
