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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 2}e^{-j2({2\pi \over N}n)} + {7 \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}] - H(e^{-j{4\pi \over N}}){7 \over 2}e^{-j({4\pi \over N}n)} + H(e^{j{4\pi \over N}}) {7 \over 2}e^{j({4\pi \over N}n)} $
- $ = 1 + {1 \over 1 - e^{-j{2\pi \over N} - 1}} e^{j({2\pi \over N})n} [{1 \over 2j}+{5 \over 2}] + {1 \over 1 - e^{j{2\pi \over N} - 1}} e^{-j({2\pi \over N})n} [{1 \over 2j} - {5 \over 2}] - {1 \over 1 - e^{j{4\pi \over N} - 1}}{7 \over 2}e^{-j2({2\pi \over N}n)} + {1 \over 1 - e^{j{4\pi \over N} - 1}}{7 \over 2}e^{-j({4\pi \over N}n)} $
