(Computing the Inverse Fourier Transform)
(Computing the Inverse Fourier Transform)
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The inverse Fourier transform is defined as:
 
The inverse Fourier transform is defined as:
  
<math> x(t) = \int_{-infty}^{infty} \frac{X(w)}{2 \pi} e^{jwt} dw </math>
+
<math> x(t) = \int_{-\infty}^{\infty} \frac{X(w)}{2 \pi} e^{jwt} dw </math>
 +
 
 +
Using this formula to determine the signal:
 +
 
 +
<math>\ x(t) = \frac{8 \pi}{2 \pi} \int_{-\infty}^{\infty} w e^{jwt} \delta(w-9) dw + \frac{2}{2 \pi} \int_{-\infty}^{\infty}w^{3} \delta(w-4 \pi) e^{jwt} dw </math>
 +
 
 +
Now using the sifting property of the delta function we find that the signal is
 +
 
 +
<math>\ x(t) = 36 e^{j9t} + 64 \pi^{2} e^{j4\pi t} </math>

Revision as of 18:31, 8 October 2008

Computing the Inverse Fourier Transform

$ \ X(\omega)= 8 \pi w \delta(w-9) + 2 \pi w^{3} \delta(w-4 \pi) $

The inverse Fourier transform is defined as:

$ x(t) = \int_{-\infty}^{\infty} \frac{X(w)}{2 \pi} e^{jwt} dw $

Using this formula to determine the signal:

$ \ x(t) = \frac{8 \pi}{2 \pi} \int_{-\infty}^{\infty} w e^{jwt} \delta(w-9) dw + \frac{2}{2 \pi} \int_{-\infty}^{\infty}w^{3} \delta(w-4 \pi) e^{jwt} dw $

Now using the sifting property of the delta function we find that the signal is

$ \ x(t) = 36 e^{j9t} + 64 \pi^{2} e^{j4\pi t} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010