Difference between revisions of "HW4.3 Jungu (Joe) Choi ECE301Fall2008mboutin" - Rhea

Revision as of 17:21, 26 September 2008

   This is only a preview; changes have not yet been saved! (????)

h(t) = e^{-t}u(t)

H(jw) = \int_0^{\infty} e^{-\tau}e^{-jw{\tau}}\,d{\tau}

= [-{1 \over 1 + jw}e^{-\tau}e^{-jwr} ]^{\infty}_0

= {1 \over 1+ jw}

Rewriting the periodic signal in Question 1,

x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4})

x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) [e^{j 2w_0 t}]+ {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) [e^{-j 2w_0 t}]

x(t) = \sum_{i=-2}^2 a_ie^{jkw_0 t}

@@ Line 9: / Line 9: @@
 == CT LTI system Part b ==
-<math>y(t) = \sum_{i=m}^n x_i = x_m + x_{m+1} + x_{m+2} +\dots+ x_{n-1} + x_n. </math>
+Rewriting the periodic signal in Question 1, <br><br>
+:<math> x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4})  </math><br><br>
+:<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) [e^{j 2w_0 t}]+ {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) [e^{-j 2w_0 t}]</math><br><br>
+:<math>x(t) = \sum_{i=-2}^2 a_ie^{jkw_0 t} </math>