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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}} $
- $ = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n $
Applying this to y[n],
$ y[n] = 1 + H(e^{j{2\pi \over N}})e^{j({2\pi \over N})n}[{1 \over 2j}+{5 \over 2}] + H(e^{-j{2\pi \over N}})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)} $
