Difference between revisions of "HW3.B Scott Erdbruegger ECE301Fall2008mboutin" - Rhea

Revision as of 07:34, 18 September 2008

Given

e^{2 x i}=t e^{-2 x i}\,

e^{-2 x i}=t e^{2 x i}\,

The Signal is Linear

\cos x = \dfrac{e^{i x}+e^{-i x}}{2}

\cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2}

The Systems response to $\cos 2x$ is $\ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2}$ but

e^{2 x i}=\cos 2x + i \sin 2x \,

and

e^{-2 x i}=\cos 2x - i \sin 2x \,

so the response is equal to

\cos 2t

@@ Line 14: / Line 14: @@
 <math>\ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} </math>
 but
-:<math>e^{2 x i}=\cos 2x + i \sin 2x \, </math> and <math>e^{-2 x i}=\cos 2x - i \sin 2x \, </math> so the response is
+:<math>e^{2 x i}=\cos 2x + i \sin 2x \, </math> and
+:<math>e^{-2 x i}=\cos 2x - i \sin 2x \, </math>
+:so the response is equal to
-:<math>\dfrac{t e^{-2 i x} + t e^{2 i x}}{2} = t\cos 2t </math>
+:<math>\cos 2t </math>