(New page: ==CT Signal== Choose the CT signal to be <math>x(t)=\frac(1 j)cos(4\pit)+5sin(5\pit)</math>) |
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==CT Signal== | ==CT Signal== | ||
| − | Choose the CT signal to be <math>x(t)=\frac(1 j) | + | Choose the CT signal to be <math>x(t)=4cos(4t)+(2+12j)sin(12t)</math> |
| + | |||
| + | ==Forier Series Equations== | ||
| + | |||
| + | <math> x(t) = \sum^{\infty}_{k = -\infty} a_ke^{jK{w_0}t}\!</math> | ||
| + | |||
| + | where <math> a_k = \frac{1}{T}\int_{0}^{T} x(t)e^{-jK{w_0}t}dt\!</math> | ||
| + | |||
| + | ==Finding The Fourier Series Coefficients== | ||
| + | |||
| + | <math>x(t)=4cos(4t)+(2+12j)sin(12t)=4(\frac{e^{j4t}+e^{-j4t}}{2})+(2+12j)(\frac{e^{j12t}-e^{-j12t}}{2j})</math> | ||
| + | |||
| + | <math>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}</math> | ||
| + | |||
| + | <math>x(t)=2e^{1(j4t)}+2e^{-1(j4t)}+(6-j)e^{3(j4t)}+(j-6)e^{-3(j4t)}</math> | ||
| + | |||
| + | x(t) is written as a sum of exponential functions, so take the coefficients of those. | ||
| + | |||
| + | <math>a_1</math> is the coefficient of <math>2e^{j4t}</math> which is 2. | ||
| + | |||
| + | <math>a_{-1}</math> is the coefficient of <math>2e^{-j4t}</math> which is 2. | ||
| + | |||
| + | <math>a_3</math> is the coefficient of <math>(6-j)e^{3(j4t)}</math> which is 6-j. | ||
| + | |||
| + | <math>a_{-3}</math> is the coefficient of <math>(j-6)e^{-3(j4t)}</math> which is j-6. | ||
| + | |||
| + | <math>a_0</math> is the average value of the function over one period. <math>a_0= \frac{1}{T}\int_{0}^{T} x(t)e^{0}dt\!</math>. | ||
| + | |||
| + | <math>w_0</math> is 4, so the period is <math>T=\frac{2\pi}{4}=\frac{\pi}{2}</math> | ||
| + | |||
| + | <math>a_0= \frac{1}{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}} 4cos(4t)+(2+12j)sin(12t)dt\!</math> | ||
| + | |||
| + | <math>\frac{2}{\pi}(sin(4t)|_{0}^{\frac{\pi}{2}}-(\frac{1}{6}+j)cos(12t)|_{0}^{\frac{\pi}{2}})\!</math> | ||
| + | |||
| + | <math>\frac{2}{\pi}(sin(2\pi)-sin(0)-(\frac{1}{6}+j)(cos(6\pi)-cos(0)))\!</math> | ||
| + | |||
| + | <math>a_0=0</math> | ||
Revision as of 08:57, 25 September 2008
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 $
