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}\, $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva