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DT LTI System Part a



$ h[n] = e^{-n}u[n] $

and the input signal,

$ x[n] = 1 + e^{j({2\pi \over N})n}[{1 \over 2j} + {5 \over 2}] - e^{-j({2\pi \over N})n}[{1 \over 2j} - {5 \over 2}] - ({7 \over 2j}e^{-j{\pi \over 2}})e^{-j2({2\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j2({2\pi \over N}n)} $


$ H(e^{jw}) = \sum_{k=0}^{\infty} e^{-n}e^{-jwn} = \sum_{k=0}^{\infty} e^{(-jw-1)n} $


$ = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n $


$ = {1 \over 1 - e^{-jw-1}} $


Applying this to y[n],

$ y[n] = 1 + H(e^{{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] $

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva