(→CT LTI system Part b) |
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:<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) = 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_{ | + | :<math>x(t) = \sum_{k=-2}^2 a_ke^{jkw_0 t} </math><br><br> |
| + | Now, calculating y(t) <br><br> | ||
| + | :<math> y(t) = \sum_{k={-2}}^2 b_ke^{jk w_0 t} </math><br><br> | ||
| + | with <math> b_k = a_k H(jkw_0) </math>, so that<br><br> | ||
| + | |||
| + | <math> b_0 = 1 </math><br><br><math> b_1 = {1 \over 2j}({1 \over 1+jw_0}) </math><br><br><math> b_{-1} = {-1 \over 2j}({1 \over 1-jw_0})</math><br><br><math> b_2 = </math><br><br><math> b_{2} = </math><br><br> | ||
Revision as of 16:28, 26 September 2008
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CT LTI system Part a
- $ 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} $
- $ = [-{1 \over 1 + jw}e^{-\tau}e^{-jwr} ]^{\infty}_0 $
CT LTI system Part b
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_{k=-2}^2 a_ke^{jkw_0 t} $
Now, calculating y(t)
- $ y(t) = \sum_{k={-2}}^2 b_ke^{jk w_0 t} $
with $ b_k = a_k H(jkw_0) $, so that
$ b_0 = 1 $
$ b_1 = {1 \over 2j}({1 \over 1+jw_0}) $
$ b_{-1} = {-1 \over 2j}({1 \over 1-jw_0}) $
$ b_2 = $
$ b_{2} = $
