(A Periodic DT Signal)
(A Periodic DT Signal)
 
(8 intermediate revisions by the same user not shown)
Line 1: Line 1:
 +
== Preview ==
 +
    This is only a preview; changes have not yet been saved! (????)
 +
 
== A Periodic DT Signal ==
 
== A Periodic DT Signal ==
 
<br>
 
<br>
Line 6: Line 9:
 
<br><br>
 
<br><br>
  
<math> x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}+{\pi \over 2}n)}-e^{-j({4\pi \over N})n}]
+
<math> x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}+e^{-j({2\pi \over N})n}] + {7 \over 2j}[e^{j({4\pi \over N}n-{\pi \over 2})}-e^{-j({4\pi \over N}n - {\pi \over 2})}]</math>
 +
<br><br>
 +
 
 +
<math> x[n] = 1 + {1 \over 2j}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] + {5 \over 2}[e^{j({2\pi \over N})n}-e^{-j({2\pi \over N})n}] </math>
 +
<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^{-j({4\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j({4\pi \over N}n)}</math>
 +
<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><br><br>
 +
 
 +
<math> a_0 = 1</math><br><math> a_1 = {1 \over 2j} + {5 \over 2} </math><br><math> a_{-1} = -{1 \over 2j} + {5 \over 2}</math><br><math> a_2 = {7 \over 2}</math><br><math> a_{-2} = {7 \over 2}</math><br><br>

Latest revision as of 15:31, 26 September 2008

Preview

   This is only a preview; changes have not yet been saved! (????)

A Periodic DT Signal


$ x[n] = 1 + sin({2\pi \over N})n + 5cos({2\pi \over N})n + 7sin({4\pi \over N}n - {\pi \over 2}) $

The signal above x[n] is periodic with period N.

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

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

$ 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^{-j({4\pi \over N}n)} + ({7 \over 2j}e^{j{\pi \over 2}})e^{j({4\pi \over N}n)} $

$ 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)} $




$ a_0 = 1 $
$ a_1 = {1 \over 2j} + {5 \over 2} $
$ a_{-1} = -{1 \over 2j} + {5 \over 2} $
$ a_2 = {7 \over 2} $
$ a_{-2} = {7 \over 2} $

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn