(New page: <math>X(\omega)=cos{(6\omega)+ (\pi/6)}</math> Start by guessing the solution: <math>\,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math> <math>\,\mathcal{X}(\omeg...)
 
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<math>X(\omega)=cos{(6\omega)+ (\pi/6)}</math>
+
<math>X(\omega)=\cos{(6\omega + \pi/6)}</math>
  
 
Start by guessing the solution:
 
Start by guessing the solution:
  
 +
<math>X(t)=(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6)</math>
 +
 +
Then take the fourier transform of the guessed solution to make sure it's right...
  
  
 
<math>\,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math>
 
<math>\,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\,</math>
  
−
<math>\,\mathcal{X}(\omega)=\int_{0}^{+\infty}e^{-5t}cos{(2t)}e^{-j\omega t}\,dt,</math>
 
  
−
<math>\,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-5t}e^{2jt}e^{-j\omega t}\,dt + 1/2\int_{0}^{+\infty}e^{-5t}e^{-2jt}e^{-j\omega t}\,dt\,</math>
+
<math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}[(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6)]e^{-j\omega t}\,dt,</math>
 +
 
 +
 
 +
<math>\,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}(1/2)e^{-j(\pi/6)}\delta(t-6)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}(1/2)e^{j(\pi/6)}\delta(t+6)e^{-j\omega t}\,dt\,</math>
 +
 
 +
<math>\,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t-6)e^{-j\omega t}\,dt + (1/2)e^{j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t+6)e^{-j\omega t}\,dt\,</math>
  
−
<math>\,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-t(5-2j+j\omega)}\,dt + 1/2\int_{0}^{+\infty}e^{-t(5+2j+j\omega)}\,dt\,</math>
+
<math>\,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}e^{-6jt} + (1/2)e^{j(\pi/6)}e^{6jt}</math>
  
−
<math>\,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\,</math>
 
  
−
<math>\,\mathcal{X}(\omega)=\left. 1/2\frac{e^{-t(5-2j+j\omega)}}{-(5-2j+j\omega)}\right]_{0}^{+\infty} + \left. 1/2\frac{e^{-t(5+2j+j\omega)}}{-(5+2j+j\omega)}\right]_{0}^{+\infty}</math>
+
<math>\,\mathcal{X}(\omega)=(1/2)e^{-j(6t + \pi/6)} + (1/2)e^{j(6t + \pi/6)}</math>
  
  
−
<math>\,\mathcal{X}(\omega)=\frac{1}{2(5-2j+j\omega)} + \frac{1}{2(5+2j+j\omega)}</math>
+
<math>\,\mathcal{X}(\omega)=\cos{(6\omega + \pi/6)} </math>

Revision as of 17:30, 8 October 2008

$ X(\omega)=\cos{(6\omega + \pi/6)} $

Start by guessing the solution:

$ X(t)=(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6) $

Then take the fourier transform of the guessed solution to make sure it's right...


$ \,\mathcal{X}(t)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $


$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}[(1/2)e^{-j(\pi/6)}\delta(t-6)+(1/2)e^{j(\pi/6)}\delta(t+6)]e^{-j\omega t}\,dt, $


$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}(1/2)e^{-j(\pi/6)}\delta(t-6)e^{-j\omega t}\,dt + \int_{-\infty}^{+\infty}(1/2)e^{j(\pi/6)}\delta(t+6)e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t-6)e^{-j\omega t}\,dt + (1/2)e^{j(\pi/6)}\int_{-\infty}^{+\infty}\delta(t+6)e^{-j\omega t}\,dt\, $

$ \,\mathcal{X}(\omega)=(1/2)e^{-j(\pi/6)}e^{-6jt} + (1/2)e^{j(\pi/6)}e^{6jt} $


$ \,\mathcal{X}(\omega)=(1/2)e^{-j(6t + \pi/6)} + (1/2)e^{j(6t + \pi/6)} $


$ \,\mathcal{X}(\omega)=\cos{(6\omega + \pi/6)} $

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Basic linear algebra uncovers and clarifies very important geometry and algebra.

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