Difference between revisions of "HW4.4 Kathleen Schremser ECE301Fall2008mboutin" - Rhea

Revision as of 17:57, 25 September 2008

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
$\,\ N = 2$

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...

$\,\ y[n] = -3 *$

@@ Line 1: / Line 1: @@
-== Define a DT LTI System ==
-<math> \,\ x[n] = 5*u[n-5] + 6*u[n+6] </math>
-=== a)  h[n] and H(z) ===
-----
-<br>
-We obtain <math> h[n] </math> by finding the response of <math> x[n] </math> to the unit impulse response (<math> \delta[n] </math>).
-<math> \,\ h[n] = 5*\delta[n-5] + 6*\delta[n+6] </math>
-<br><br>
-<math> \,\ H[z] = \sum_{m=-\infty}^\infty h[m] * Z</math><sup>(<math>-m</math>)</sup><br><br>
-<math> \,\ H[z] = \sum_{m=-\infty}^{\infty} (5*\delta[n-5] + 6*\delta[n+6]) * Z</math><sup>(<math>-m</math>)</sup>
-By the sifting property, this sum equals:<br>
-<math> \,\ H[z] = 5*Z</math><sup>-5</sup><math> \,\ + 6*Z</math><sup>6</sup>
 === b)  Response of Signal in Question 1 ===
 ----
 From Question 1:
-Unfortunately, I did not not read ahead and 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).
+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:
@@ Line 31: / Line 11: @@
 * <math> \,\ a_k = 0 </math> elsewhere
 * <math> \,\ N = 2 </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> \,\ y[n] = -3 *

Difference between revisions of "HW4.4 Kathleen Schremser ECE301Fall2008mboutin" - Rhea

Revision as of 17:57, 25 September 2008

b) Response of Signal in Question 1

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