(→b) Response of Signal in Question 1) |
(→Define a DT LTI System) |
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== Define a DT LTI System == | == Define a DT LTI System == | ||
| − | <math> \,\ | + | <math> \,\ y[n] = 5 * x[n-5] + 6 * x[n-6] </math> |
| − | === a) h[n] and H | + | === a) h[n] and H[z] === |
---- | ---- | ||
| − | |||
| − | + | In order to find h[n], we input <math> x[n] = \delta [n] </math> to y[n]. h[n] is then the unit impulse response.<br><br> | |
| − | <math> \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] </math> | + | <math> \,\ x[n] = \delta [n] </math><br> |
| + | <math> \,\ h[n] = 5 * \delta [n-5] + 6 * \delta [n+6] </math><br><br> | ||
| + | |||
| + | H[z] is the system's function, and is defined by:<br><br> | ||
| + | <math> \sum_{m = -\infty}^{\infty} h[m] * Z</math><sup>-m</sup> | ||
| + | <br> | ||
| + | <math> \,\ H[z] = \sum_{m = -\infty}^{\infty} ( 5 * \delta [n-5] + 6 * \delta [n+6] ) * Z</math><sup>-m</sup> | ||
<br><br> | <br><br> | ||
| − | + | This function will only be valid when n = 5 or n = -6 due to the <math> \delta </math>s. | |
| − | + | Therefore, the end result is:<br> | |
| − | <math> | + | <math> \,\ H[z] = 5 * Z</math><sup>5</sup><math> \,\ + 6 * Z</math><sup>-6</sup> |
| − | + | ||
| − | + | ||
| − | <math> \,\ H[z] = 5*Z</math><sup> | + | |
| − | + | ||
=== b) Response of Signal in Question 1 === | === b) Response of Signal in Question 1 === | ||
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From Question 1: | From Question 1: | ||
| − | Unfortunately, I did not | + | Unfortunately, I did not make a DT signal for Parts 1/2. Instead of going back and redoing my work, I am going to "steal" the work of a Mr. Collin Phillips (my apologies and gratitude to Mr. Collin Phillips). |
According to his work: | According to his work: | ||
| Line 31: | Line 32: | ||
* <math> \,\ a_k = 0 </math> elsewhere | * <math> \,\ a_k = 0 </math> elsewhere | ||
* <math> \,\ N = 2 </math> | * <math> \,\ N = 2 </math> | ||
| + | * <math> \,\ k = 3 </math> | ||
| + | |||
| + | |||
| + | Response of the System <math> \,\ y[n] = \sum_{k = <N>}^{\infty} a_k H[e</math><sup><math> j2\pi k/N</math></sup><math> \,\ ]e</math><sup><math> jk(2\pi /N) n</math></sup> | ||
| + | |||
| + | Because <math> a_k </math> only has one value, this shouldn't be that hard to calculate. | ||
| + | |||
| + | <math> \,\ a_k </math> is only valid at <math> a_1 = -3 </math>. Therefore... | ||
| + | |||
| + | <math> \,\ Z = e</math><sup><math> jk(2\pi /N) </math></sup> | ||
| + | |||
| + | <math> \,\ y[n] = -3 * [5 * e</math><sup><math>(j2\pi5*3/2)</math></sup><math> \,\ + 6 * e</math><sup><math>(j2\pi*-6*3/2)</math></sup><math> \,\ ] * e</math><sup><math>(j3*2\pi n/2)</math></sup> | ||
| + | <br><br> | ||
| + | <math> \,\ y[n] = -3 * [5 * e</math><sup><math>(j\pi15)</math></sup><math> \,\ + 6 * e</math><sup><math>(-j\pi*18)</math></sup><math> \,\ ] * e</math><sup><math>(j3*\pi* n)</math></sup> | ||
| + | <br><br> | ||
| + | <math> \,\ y[n] = -15 * e </math><sup><math>j3\pi(5 + n)</math></sup><math> \,\ - 18 * e </math><sup><math>j3\pi(n - 6)</math></sup> | ||
Latest revision as of 06:56, 26 September 2008
Define a DT LTI System
$ \,\ y[n] = 5 * x[n-5] + 6 * x[n-6] $
a) h[n] and H[z]
In order to find h[n], we input $ x[n] = \delta [n] $ to y[n]. h[n] is then the unit impulse response.
$ \,\ x[n] = \delta [n] $
$ \,\ h[n] = 5 * \delta [n-5] + 6 * \delta [n+6] $
H[z] is the system's function, and is defined by:
$ \sum_{m = -\infty}^{\infty} h[m] * Z $-m
$ \,\ H[z] = \sum_{m = -\infty}^{\infty} ( 5 * \delta [n-5] + 6 * \delta [n+6] ) * Z $-m
This function will only be valid when n = 5 or n = -6 due to the $ \delta $s.
Therefore, the end result is:
$ \,\ H[z] = 5 * Z $5$ \,\ + 6 * Z $-6
b) Response of Signal in Question 1
From Question 1: Unfortunately, I did not make a DT signal for Parts 1/2. Instead of going back and redoing my work, I am going to "steal" the work of a Mr. Collin Phillips (my apologies and gratitude to Mr. Collin Phillips).
According to his work:
- $ \,\ X[n] = 3cos(3\pi n + \pi) $
- $ \,\ a_0 = 0 $
- $ \,\ a_1 = -3 $
- $ \,\ a_k = 0 $ elsewhere
- $ \,\ N = 2 $
- $ \,\ k = 3 $
Response of the System $ \,\ y[n] = \sum_{k = <N>}^{\infty} a_k H[e $$ j2\pi k/N $$ \,\ ]e $$ jk(2\pi /N) n $
Because $ a_k $ only has one value, this shouldn't be that hard to calculate.
$ \,\ a_k $ is only valid at $ a_1 = -3 $. Therefore...
$ \,\ Z = e $$ jk(2\pi /N) $
$ \,\ y[n] = -3 * [5 * e $$ (j2\pi5*3/2) $$ \,\ + 6 * e $$ (j2\pi*-6*3/2) $$ \,\ ] * e $$ (j3*2\pi n/2) $
$ \,\ y[n] = -3 * [5 * e $$ (j\pi15) $$ \,\ + 6 * e $$ (-j\pi*18) $$ \,\ ] * e $$ (j3*\pi* n) $
$ \,\ y[n] = -15 * e $$ j3\pi(5 + n) $$ \,\ - 18 * e $$ j3\pi(n - 6) $
