(DT LTI System Part a)
(DT LTI System Part a)
 
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== DT LTI System Part a ==
 
== DT LTI System Part a ==
 
<br><br>
 
<br><br>
<math> h[n] = cos[{\pi \over 3} n]u[n]</math><br><br>
+
<math> h[n] = e^{-n}u[n]</math><br><br>
 
and the input signal, <br><br>
 
and the input signal, <br><br>
<math>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)}</math><br><br><br>
+
<math>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)}</math><br><br><br>
<math> H(e^{jw}) = \sum_{k=0}^{\infty} cos[{\pi \over 3} n]e^{-jwn} = \sum_{k=0}^{\infty} {1 \over 2}(e^{j{\pi \over 3}n} + e^{-j{\pi \over 3}n})e^{-jwn}</math><br><br><br>
+
<math> H(e^{jw}) = \sum_{k=0}^{\infty} e^{-n}e^{-jwn} = \sum_{k=0}^{\infty} e^{(-jw-1)n}</math><br><br><br>
:::<math> = \sum_{k=0}^{\infty} {1 \over 2} (e^{j({\pi \over 3} - w)n} + e^{-j({\pi \over 3} + w)n})
+
:::<math> = \sum_{k=0}^{\infty} [e^{(-jw-1)}]^n </math><br><br><br>
 +
:::<math> = {1 \over 1 - e^{-jw-1}}</math><br><br><br>
 +
Applying this to y[n],<br><br>
 +
<math> 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)} </math><br><br><br>
 +
::<math> = 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)}</math><br><br><br>

Latest revision as of 17:24, 26 September 2008

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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}} $


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)} $


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