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| − | == | + | [[Category:problem solving]] |
| − | + | [[Category:ECE301]] | |
| + | [[Category:ECE]] | ||
| + | [[Category:Fourier series]] | ||
| + | [[Category:signals and systems]] | ||
| + | == Example of Computation of Fourier series of a CT SIGNAL == | ||
| + | A [[Signals_and_systems_practice_problems_list|practice problem on "Signals and Systems"]] | ||
| + | ---- | ||
== CT Signal == | == CT Signal == | ||
| − | :<math> x(t) = 1 + sin(w_0 t) + 3cos( | + | :<math> x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4}) </math><br><br> |
This is a signal with period <math>T = {2\pi \over w_0}</math><br><br> | This is a signal with period <math>T = {2\pi \over w_0}</math><br><br> | ||
| − | :<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j | + | :<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j 2w_0 t + {\pi \over 4})}+e^{-(j 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} [e^{j | + | :<math> x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}]</math><br><br> |
| + | :<math>e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} </math><br> | ||
| + | :<math>e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} </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} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\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> | ||
| + | <br> | ||
| + | Thus, the Fourier series coefficients for this are:<br><br> | ||
| + | <math> a_0 = 1 </math><br><br><math> a_1 = {1 \over 2j}</math><br><br><math> a_{-1} = {-1 \over 2j}</math><br><br><math> a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}})</math><br><br><math> a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) </math><br><br><math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:57, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
CT Signal
- $ x(t) = 1 + sin(w_0 t) + 3cos(2w_0 t + {\pi \over 4}) $
This is a signal with period $ T = {2\pi \over w_0} $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t} - e^{-j w_0 t}] + {3 \over 2}[e^{(j 2w_0 t + {\pi \over 4})}+e^{-(j 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} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\pi \over 4}}] $
- $ e^{j {\pi \over 4}} = {1 \over \sqrt{2}} + j{1 \over \sqrt{2}} $
- $ e^{-j{\pi \over 4}} = {1 \over \sqrt{2}} - j{1 \over \sqrt{2}} $
- $ x(t) = 1 + {1 \over 2j}[e^{j w_0 t}] + ({-1 \over 2j})e^{-j w_0 t} + {3 \over 2} [e^{j 2w_0 t}e^ {j{\pi \over 4}}]+ {3 \over 2}[e^{-j 2w_0 t} e^{-j{\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}] $
Thus, the Fourier series coefficients for this are:
$ a_0 = 1 $
$ a_1 = {1 \over 2j} $
$ a_{-1} = {-1 \over 2j} $
$ a_2 = {3 \over 2}({1 \over \sqrt{2}} + j{1 \over \sqrt{2}}) $
$ a_{-2} = {3 \over 2}({1 \over \sqrt{2}} - j{1 \over \sqrt{2}}) $
$ ---- [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] $
