Revision as of 17:33, 25 September 2008 by Kschrems (Talk)

Define a DT LTI System

$ \,\ x[n] = 5*u[n-5] + 6*u[n+6] $

a) h[n] and H(z)



We obtain $ h[n] $ by finding the response of $ x[n] $ to the unit impulse response ($ \delta[n] $).

$ \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] $

$ \,\ H[z] = \sum_{m=-\infty}^\infty h[m] * Z $($ -m $)

$ \,\ H[z] = \sum_{m=-\infty}^{\infty} (5*\delta[n-5] + 6*\delta[n+6]) * Z $($ -m $)

By the sifting property, this sum equals:
$ \,\ H[z] = 5*Z $-5$ \,\ + 6*Z $6


b) Response of Signal in Question 1


From Question 1: Unfortunately, I did not not read ahead and make a DT signal for Parts 1/2. Instead of going back and redoing my work, I am going to "steal" the work of a Mr. Collin Phillips (my apologies and gratitude to Mr. Collin Phillips).

According to his work:

  • $ \,\ X[n] = 3cos(3\pi n + \pi) $
  • $ \,\ a_0 = 0 $
  • $ \,\ a_1 = -3 $
  • $ \,\ a_k = 0 $ elsewhere
  • $ \,\ N = 2 $

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