Difference between revisions of "HW4.1 Ben Laskowski ECE301Fall2008mboutin" - Rhea

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"

The Signal

Consider the signal $ 7\sin(2t)+(1+j)\cos(3t) $.

The Formulae

Recall the Fourier Series formulae for the continuous time signal case:

$ x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0t} $

and

$ a_k=\frac{1}{T}\int_0^Tx(t)e^{-jk\omega_0t}dt $.

Finding the Series

It will be helpful -- necessary even -- to find the fundamental period of the signal. In our case, the period of the overall signal is $ 2\pi $, so $ \omega_0 $ will be $ \frac{2\pi}{2\pi}=1 $.

A good place to start is the calculation of $ a_0 $, which is the average of the signal. Plotting the signal makes it look like the average is 0, but we can integrate to check.

$ a_0=\frac{1}{T}\int_0^T[7\sin(2t)+(1+j)\cos(3t)]e^{-jk\omega_0t}dt $

$ =\frac{7}{2\pi}\int_0^{2\pi}\sin(2t)dt + \frac{1+j}{2\pi}\int_0^{2\pi}\cos(3t)dt $

$ =\frac{-7}{4\pi}\cos(2t)|_0^{2\pi}+\frac{1+j}{6\pi}\sin(3t)|_0^{2\pi} $

$ =\frac{-7}{4\pi}(\cos(4\pi)-\cos(0))+\frac{1+j}{6\pi}(\sin(6\pi)-\sin(0))=0 $

After this point, integrating becomes quite tedious, so I'll revert to using complex exponential identities to continue the solution. Our signal then becomes

$ x(t)=\frac{7}{2j}(e^{2j}-e^{-2j})+\frac{1+j}{2}(e^{3j}+e^{-3j}) $

$ =\frac{7}{2j}(e^2e^j-e^{-2}e^j)+\frac{1+j}{2}(e^3e^j+e^{-3}e^j) $

Now the Fourier coefficients should be fairly obvious.

$ a_{-3}=a_3=\frac{1+j}{2} $

$ a_{-2}=\frac{-7}{2j} $

$ a_2=\frac{7}{2j} $

All other $ a_k=0 $.

Checking the Results

I checked my Fourier coefficients against what MATLAB thinks they should be. I am off by a factor of $ \pi $ from MATLAB's result, but I can't seem to to figure out why. Can anyone lend a hand?

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