(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) |
||
| Line 13: | Line 13: | ||
== 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, <math> so 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> |
Revision as of 18:08, 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}\, $
