Revision as of 17:19, 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:

  • $ \,\ x(t) = 5cos(2t) - 4sin(5t) $

  • $ \,\ T = 2\pi $

  • $ \,\ x(t) = \frac{5}{2} * e $(j2t) $ \,\ + \frac{5}{2} * e $(-j2t)$ - \frac{4}{2j} * e $(j5t) $ + \frac{4}{2j} * e $(-j5t)
  • $ \,\ w_0 = 1 $

  • $ \mathbf{a_2} = \mathbf{\frac{5}{2}} $

  • $ \mathbf{a} $-2$ = \mathbf{\frac{5}{2j}} $

  • $ \mathbf{a_5} = \mathbf{-2} $

  • $ \mathbf{a} $-5$ = \mathbf{\frac{2}{j}} $

  • $ \,\ a_x = 0 $ elsewhere

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