Fourier Series Coefficients for a DT signal

Lets look at the signal $ \,x[n] = 7cos[5\pi n] $


$ \,N = \frac{2\pi}{5\pi}K $ where K is the smallest integer that makes N an integer.
$ \,N = 2 $ after setting K = 5.


$ \,x[0] = 7 $
$ \,x[1] = -7 $
$ \,x[2] = 7 $
$ \,x[3] = -7 $


$ \,a_0 = \frac{7+(-7)}{2} = 0 $ because $ \,a_0 $ is the average value of the signal over 1 period.
$ \, a_1 = \frac{1}{2}\sum^{1}_{n=0}x[n]e^{-j\pi n} = \frac{1}{2}(x[0]e^{0} + x[1]e^{-j\pi}) = \frac{1}{2}(7 + 7) = 7 $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman