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A certain periodic signal has the following properties: | A certain periodic signal has the following properties: | ||
| − | 1. N = | + | 1. N = 4 |
| − | 2. <math>\sum_{n=0}^{ | + | 2. <math>\sum_{n=0}^{3}x[n] = 4</math> |
| − | 3. <math>\sum_{n=1}^{ | + | 3. <math>\sum_{n=1}^{4}(-1)^nx[n] = 2</math> |
| − | 4. <math>a_k = a_{k+ | + | 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}^{ | + | 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}{ | + | <center><math>\frac{1}{4}\sum_{n=0}^{3}x[n] = \frac{1}{4}*4\,</math>, and</center> |
| − | <center><math>= | + | <center><math>= 1 = a_0\,</math></center> |
| − | Now that we know <math>a_0\,</math>, we know that <math>x[n] = \frac{2}{3} | + | 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}{2}\,</math>, let's try and find <math>a_2\,</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 13:01, 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,
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}\, $
