Difference between revisions of "HW4.1 Daniel Morris ECE301Fall2008mboutin" - Rhea

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Example of Computation of Fourier series of a CT SIGNAL

A practice problem on "Signals and Systems"

CT Signal

Choose the CT signal to be $ x(t)=4cos(4t)+(2+12j)sin(12t) $

Forier Series Equations

$ x(t) = \sum^{\infty}_{k = -\infty} a_ke^{jK{w_0}t}\! $

where $ a_k = \frac{1}{T}\int_{0}^{T} x(t)e^{-jK{w_0}t}dt\! $

Finding The Fourier Series Coefficients

$ x(t)=4cos(4t)+(2+12j)sin(12t)=4(\frac{e^{j4t}+e^{-j4t}}{2})+(2+12j)(\frac{e^{j12t}-e^{-j12t}}{2j}) $

$ x(t)=2e^{j4t}+2e^{-j4t}+\frac{2+12j}{2j}e^{j12t}-\frac{2+12j}{2j}e^{-j12t}=2e^{j4t}+2e^{-j4t}+(6-j)e^{j12t}-(6-j)e^{-j12t} $

$ x(t)=2e^{1(j4t)}+2e^{-1(j4t)}+(6-j)e^{3(j4t)}+(j-6)e^{-3(j4t)} $

x(t) is written as a sum of exponential functions, so take the coefficients of those.

$ a_1 $ is the coefficient of $ 2e^{j4t} $ which is 2.

$ a_{-1} $ is the coefficient of $ 2e^{-j4t} $ which is 2.

$ a_3 $ is the coefficient of $ (6-j)e^{3(j4t)} $ which is 6-j.

$ a_{-3} $ is the coefficient of $ (j-6)e^{-3(j4t)} $ which is j-6.

$ a_0 $ is the average value of the function over one period. $ a_0= \frac{1}{T}\int_{0}^{T} x(t)e^{0}dt\! $.

$ w_0 $ is 4, so the period is $ T=\frac{2\pi}{4}=\frac{\pi}{2} $

$ a_0= \frac{1}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} 4cos(4t)+(2+12j)sin(12t)dt\! $

$ \frac{2}{\pi}(sin(4t)|_{0}^{\frac{\pi}{2}}-(\frac{1}{6}+j)cos(12t)|_{0}^{\frac{\pi}{2}})\! $

$ \frac{2}{\pi}(sin(2\pi)-sin(0)-(\frac{1}{6}+j)(cos(6\pi)-cos(0)))\! $

$ a_0=0 $

$ a_k=0 $ for all other k

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