Difference between revisions of "HW4.1 Tyler Houlihan ECE301Fall2008mboutin" - Rhea

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.

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