Difference between revisions of "HW4.2 Spencer Erekson ECE301Fall2008mboutin" - Rhea

Revision as of 11:57, 26 September 2008

$x(t)=10cos(4\pi n + 2\pi)+5sin(2\pi n + 4\pi)\!$

In order to find the period of the signal below, we need to find a value of K that will make N an integer.

$N_1 = \frac{2\pi}{\omega_0} K \!$

$N_1 = \frac{2\pi}{4\pi} K \!$

$N_1 = 2K \!$

$N_2 = \frac{2\pi}{\omega_0} K \!$

$N_2 = \frac{2\pi}{2\pi} K \!$

$N_2 = K \!$

Since both numbers are integers before multiplying by K, we can just let K = 1.

@@ Line 1: / Line 1: @@
-<math>x(t)=10cos(4\pi n + 2\pi)\!</math>
+<math>x(t)=10cos(4\pi n + 2\pi)+5sin(2\pi n + 4\pi)\!</math>
-In order to find the period of the signal below, we need to find the smallest value of K that will make N an integer.
+In order to find the period of the signal below, we need to find a value of K that will make N an integer.
-<math>N = \frac{2\pi}{\omega_0} K \!</math>
+<math>N_1 = \frac{2\pi}{\omega_0} K \!</math>
-<math>N = \frac{2\pi}{2\pi} K \!</math>
+<math>N_1 = \frac{2\pi}{4\pi} K \!</math>
-<math>N = \frac{2\pi}{2\pi} K \!</math>
+<math>N_1 = 2K \!</math>
-<math>N = K\!</math>
+<math>N_2 = \frac{2\pi}{\omega_0} K \!</math>
-So the smallest value that K can have is 1.
+<math>N_2 = \frac{2\pi}{2\pi} K \!</math>
-<math>K = 1\!</math>
+<math>N_2 = K \!</math>
+Since both numbers are integers before multiplying by K, we can just let K = 1.