DT signal:

$ X[n] = 6\cos(3 \pi n + \pi)\, $

$ N = \frac{2\pi}{3\pi} K \, $

Where K is the smallest integer that makes N an integer. K would be 3.

$ N = \frac{2\pi}{3\pi} 3 \, $

$ N = 2 \, $


$ X[0] = -6 \, $
$ X[1] = 6 \, $
$ X[2] = -6 \, $
$ X[-1] = 6 \, $


$ a_k = \frac{1}{N} \sum^{N-1}_{n = 0} X[n] e^{-jk\frac{2\pi}{N} n} $
$ 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} $
$ a_0 = \frac{1}{2} (-6 + 6) = 0 $

$ a_1 = \frac{1}{2} \sum^{1}_{n = 0} X[n] e^{-j\pi n} $
$ a_1 = \frac{1}{2} (-6 * e^0 + 6 * e^{-j\pi}) $
$ a_1 = \frac{1}{2} (-6 * 1 + 6 * -1) = -6 $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010