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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"]] | ||
| + | ---- | ||
| + | |||
== Fourier sum definition == | == Fourier sum definition == | ||
The function as defined by summing fourier coefficients <math>\,a_k</math> is defined as: | The function as defined by summing fourier coefficients <math>\,a_k</math> is defined as: | ||
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Now splitting up: | Now splitting up: | ||
| − | <math>x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{ | + | <math>x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t}</math> |
choose <math>\,\omega_0</math> as <math>\,\pi</math>, the smallest period between the two parts. | choose <math>\,\omega_0</math> as <math>\,\pi</math>, the smallest period between the two parts. | ||
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<math>a_1=\frac{1+j}{2}</math> | <math>a_1=\frac{1+j}{2}</math> | ||
| − | <math>a_-1=\frac{1+j}{2}</math> | + | <math>a_{-1}=\frac{1+j}{2}</math> |
<math>a_2=\frac{1+j}{2j}</math> | <math>a_2=\frac{1+j}{2j}</math> | ||
| − | <math>a_-2=\frac{-1-j}{2j}</math> | + | <math>a_{-2}=\frac{-1-j}{2j}</math> |
All other <math>\,a_k</math> values are zero. | All other <math>\,a_k</math> values are zero. | ||
| + | ---- | ||
| + | [[Signals_and_systems_practice_problems_list|Back to Practice Problems on Signals and Systems]] | ||
Latest revision as of 10:55, 16 September 2013
Example of Computation of Fourier series of a CT SIGNAL
A practice problem on "Signals and Systems"
Fourier sum definition
The function as defined by summing fourier coefficients $ \,a_k $ is defined as:
$ x(t)=\sum^{\infty}_{k=-\infty} a_k e^{jk\omega_0 t}\, $
Example of a periodic CT signal
The following is a periodic signal:
$ \,x(t)=(1+j)cos(\pi t)+sin(2\pi t) $
Using Eulers formula, we can interpret this function in terms of exponentials which can then be used to compute the $ \,a_k $ values for a Fourier series:
$ \,x(t)=(1+j)\frac {e^{j\pi t}+e^{-j \pi t}}{2} + \frac {e^{j2 \pi t}-e^{-j2 \pi t}}{2j} $
Now splitting up:
$ x(t)=\frac{1+j}{2}e^{j\pi t}+\frac{1+j}{2}e^{-j\pi t}+\frac{1+j}{2j}e^{j2\pi t}+\frac{-1-j}{2j}e^{-j2\pi t} $
choose $ \,\omega_0 $ as $ \,\pi $, the smallest period between the two parts.
so this function becomes:
$ x(t)=\sum^{\infty}_{k=-\infty} a_k e^{jk\pi t}\, $
Which very nearly matches our function, we only need solve or point out our $ \,a_k $ values.
$ a_1=\frac{1+j}{2} $
$ a_{-1}=\frac{1+j}{2} $
$ a_2=\frac{1+j}{2j} $
$ a_{-2}=\frac{-1-j}{2j} $
All other $ \,a_k $ values are zero.
