CT Periodic Signal : $ x[n] = 5\cos(6\pi n + \pi) + 7\cos(3\pi n)\, $
$ N_1 = \frac{2\pi}{6\pi} k = \frac{1}{3} k $
$ N_2 = \frac{2\pi}{3\pi} k = \frac{2}{3} k $
$ k\, $ that makes both $ N_1 , N_2 \, $ integer is $ k = 3\, $
As $ N_2 > N_1\, $ thus $ N = 2\, $
Through Matlab, the following values can be calculated:
$ x[0] = 2\, $
$ x[1] = -12\, $
$ x[2] = 2\, $
$ x[3] = -12\, $
$ a_k = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^{-jk\pi n}\, $
$ a_0 = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^0\, $
$ = \frac{1}{2}\sum^{1}_{n = 0} x[n]\, $
$ = \frac{1}{2}(2 - 12)\, $
$ = -5\, $
$ a_1 = \frac{1}{2}\sum^{1}_{n = 0} x[n] e^{-j\pi n}\, $
$ = \frac{1}{2}(x[0] + x[1]e^{-j\pi})\, $
$ = \frac{1}{2}(2 - 12(-1))\, $
$ = \frac{1}{2}(14)\, $
$ = 7\, $
We can write the function as follows:
$ x(t) = \sum^{1}_{k = 0} a_k e^{jk\pi n}\, $ where
$ a_0 = -5\, $
$ a_1 = 7\, $
$ a_k = 0 , k \neq 0,1\, $
