(20 intermediate revisions by the same user not shown)
Line 2: Line 2:
 
A certain periodic signal has the following properties:
 
A certain periodic signal has the following properties:
  
1. N = 6
+
1. N = 4
  
2. <math>\sum_{n=0}^{5}x[n] = 4</math>
+
2. <math>\sum_{n=0}^{3}x[n] = 4</math>
  
3. <math>\sum_{n=1}^{6}(-1)^nx[n] = 2</math>
+
3. <math>\sum_{n=1}^{4}(-1)^nx[n] = 2</math>
  
4. <math>a_k = a_{k+3}\,</math>
+
4. For even <math>k\,</math>'s, <math>a_k = a_{k+1}\,</math>
  
  
 
== Answer ==
 
== Answer ==
  
From 1. we know that <math>x[n] = \sum_{n=0}^{5}a_k e^{jk\frac{\pi}{3}n}\,</math>
+
From 1. we know that <math>x[n] = \sum_{n=0}^{3}a_k e^{jk\frac{\pi}{2}n}\,</math>
  
 
Using 2., it is apparent that this is the formula for <math>a_k\,</math>. Specifically, for <math>a_0\,</math>, since the only thing under the sum is <math>x[n]\,</math>. So,
 
Using 2., it is apparent that this is the formula for <math>a_k\,</math>. Specifically, for <math>a_0\,</math>, since the only thing under the sum is <math>x[n]\,</math>. So,
  
<center><math>\frac{1}{6}\sum_{n=0}^{5}x[n] = 4\,</math>, and</center>
+
<center><math>\frac{1}{4}\sum_{n=0}^{3}x[n] = \frac{1}{4}*4\,</math>, and</center>
  
  
<center><math>= \frac{4}{6} = \frac{2}{3} = a_0\,</math></center>
+
<center><math>= 1 = a_0\,</math></center>
  
Now that we know <math>a_0\,</math>, we know that <math>x[n] = \frac{2}{3} + \sum_{n=1}^{5}a_k e^{jk\frac{\pi}{3}n}\,</math>
+
Now that we know <math>a_0\,</math>, we know that <math>x[n] = 1 + \sum_{n=1}^{4}a_k e^{jk\frac{\pi}{2}n}\,</math>
  
  
Since <math>\omega_0 = \frac{\pi}{3}\,</math>, let's try and find <math>a_3\,</math>,
+
Since <math>\omega_0 = \frac{\pi}{2}\,</math>, let's try and find <math>a_2\,</math>,
  
<math>a_3 = \frac{1}{6}\sum_{n=0}^{5}x[n] e^{-3j\frac{\pi}{3}n}\,</math>
+
<math>a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-2j\frac{\pi}{2}n} = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-j\pi n}\,</math>
 +
 
 +
 
 +
Using the property that <math>e^{-j\pi n} = (e^{-j\pi})^{n} = (-1)^{n} \,</math>, we can change the above equation to
 +
 
 +
<math>a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n](-1)^{n}\,</math>
 +
 
 +
 
 +
Since the function is periodic and the <math>a_k\,</math>'s repeat every 4 integers, we are able to shift the bounds of summation by one.
 +
 
 +
According to 3. <math>\sum_{n=1}^{4}(-1)^nx[n] = 2</math>, and
 +
 
 +
<math>a_2 = \frac{1}{4}\sum_{n=1}^{4}x[n](-1)^{n}\,</math>
 +
 
 +
<math>a_2 = \frac{1}{4} * 2\,</math>, and
 +
 
 +
<math>a_2 = \frac{1}{2}\,</math>
 +
 
 +
 
 +
Since we know the 2 even <math>a_k\,</math>'s in the fundamental period, by property 4, we can find the 2 odd <math>a_k\,</math>'s.
 +
 
 +
<math>a_0 = a_1 = 1\,</math>
 +
 
 +
<math>a_2 = a_3 = \frac{1}{2}\,</math>
 +
 
 +
 
 +
Now, the periodic signal has been found to be
 +
 
 +
<math>x[n] = \sum_{n=0}^{3}a_k e^{jk\frac{\pi}{2}n}\,</math>
 +
 
 +
<math>x[n] = 1 + \frac{1}{2}e^{j\frac{\pi}{2} n} + e^{j\pi n} + \frac{1}{2}e^{j\frac{3\pi}{2}n}\,</math>

Latest revision as of 14:01, 25 September 2008

Guess the Periodic Signal

A certain periodic signal has the following properties:

1. N = 4

2. $ \sum_{n=0}^{3}x[n] = 4 $

3. $ \sum_{n=1}^{4}(-1)^nx[n] = 2 $

4. For even $ k\, $'s, $ a_k = a_{k+1}\, $


Answer

From 1. we know that $ x[n] = \sum_{n=0}^{3}a_k e^{jk\frac{\pi}{2}n}\, $

Using 2., it is apparent that this is the formula for $ a_k\, $. Specifically, for $ a_0\, $, since the only thing under the sum is $ x[n]\, $. So,

$ \frac{1}{4}\sum_{n=0}^{3}x[n] = \frac{1}{4}*4\, $, and


$ = 1 = a_0\, $

Now that we know $ a_0\, $, we know that $ x[n] = 1 + \sum_{n=1}^{4}a_k e^{jk\frac{\pi}{2}n}\, $


Since $ \omega_0 = \frac{\pi}{2}\, $, let's try and find $ a_2\, $,

$ a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-2j\frac{\pi}{2}n} = \frac{1}{4}\sum_{n=0}^{3}x[n] e^{-j\pi n}\, $


Using the property that $ e^{-j\pi n} = (e^{-j\pi})^{n} = (-1)^{n} \, $, we can change the above equation to

$ a_2 = \frac{1}{4}\sum_{n=0}^{3}x[n](-1)^{n}\, $


Since the function is periodic and the $ a_k\, $'s repeat every 4 integers, we are able to shift the bounds of summation by one.

According to 3. $ \sum_{n=1}^{4}(-1)^nx[n] = 2 $, and

$ a_2 = \frac{1}{4}\sum_{n=1}^{4}x[n](-1)^{n}\, $

$ a_2 = \frac{1}{4} * 2\, $, and

$ a_2 = \frac{1}{2}\, $


Since we know the 2 even $ a_k\, $'s in the fundamental period, by property 4, we can find the 2 odd $ a_k\, $'s.

$ a_0 = a_1 = 1\, $

$ a_2 = a_3 = \frac{1}{2}\, $


Now, the periodic signal has been found to be

$ x[n] = \sum_{n=0}^{3}a_k e^{jk\frac{\pi}{2}n}\, $

$ x[n] = 1 + \frac{1}{2}e^{j\frac{\pi}{2} n} + e^{j\pi n} + \frac{1}{2}e^{j\frac{3\pi}{2}n}\, $

Alumni Liaison

EISL lab graduate

Mu Qiao