(New page: ==Periodic CT Signal== The first signal that comes to mind when i think of a periodic CT signal is one involving sines and cosines, so let's work with one of those. Let <math>x(t) = sin(...) |
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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"]] | ||
| + | ---- | ||
| + | |||
==Periodic CT Signal== | ==Periodic CT Signal== | ||
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This can also be expressed as | This can also be expressed as | ||
| − | <math>x(t) = e^{14jt} | + | <math>x(t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2}</math> |
| + | |||
| + | I contend that the <math>\omega_0=2</math> since both functions are periodic based on it. | ||
| + | |||
| + | <math>a_7=\frac{1}{2j}</math> | ||
| + | |||
| + | <math> a_{-7} = \frac{-1}{2j}</math> | ||
| + | |||
| + | <math> a_1 = \frac{1+3j}{2} </math> | ||
| + | |||
| + | <math> a_{-1} = \frac{1+3j}{2} </math> | ||
| + | |||
| + | We can write this as a sum: | ||
| + | |||
| + | <math> x(t)=\sum^{\infty}_{k = -\infty} a_k e^{j2k}\,</math> | ||
| + | |||
| + | Where | ||
| + | |||
| + | <math> a_1=a_{-1}=\frac{1+3j}{2} </math> | ||
| + | |||
| + | <math> a_7 = -a_{-7} = \frac{1}{2j} </math> | ||
| + | |||
| + | <math> a_k = 0, k \neq 1, -1, 7, -7 </math> | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 09:56, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Periodic CT Signal
The first signal that comes to mind when i think of a periodic CT signal is one involving sines and cosines, so let's work with one of those.
Let $ x(t) = sin(14t)+(1+3j)cos(2t) $
This can also be expressed as
$ x(t) = \frac{e^{14jt}}{2j}-\frac{e^{-14jt}}{2j} + \frac{(1+3j)*e^{2jt}}{2} + \frac{(1+3j)e^{-2jt}}{2} $
I contend that the $ \omega_0=2 $ since both functions are periodic based on it.
$ a_7=\frac{1}{2j} $
$ a_{-7} = \frac{-1}{2j} $
$ a_1 = \frac{1+3j}{2} $
$ a_{-1} = \frac{1+3j}{2} $
We can write this as a sum:
$ x(t)=\sum^{\infty}_{k = -\infty} a_k e^{j2k}\, $
Where
$ a_1=a_{-1}=\frac{1+3j}{2} $
$ a_7 = -a_{-7} = \frac{1}{2j} $
$ a_k = 0, k \neq 1, -1, 7, -7 $
