Math Help - Fourier Transform

1. Fourier Transform

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I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

Thanks

2. Originally Posted by hammer
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I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

Thanks
I presume the problem is in simplifying $X(\omega)$?

$X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n}$

Define $m = n + 4$. Then
$X(\omega) = \sum_{n = -4}^{\infty} \left ( \frac{1}{4} \right )^ne^{-j \omega n} = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^{m - 4}e^{-j \omega (m - 4)}$

$= \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m \cdot \left ( \frac{1}{4} \right )^{-4}e^{-j \omega m }e^{-(-4)j \omega }$

So
$X(\omega) = \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }4^4e^{4j \omega }$

Now, note that the last two factors are independent of m:
$X(\omega) = 4^4e^{4j \omega } \cdot \sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m }$

Note also that the summation is the summation of an infinite geometric series. The sum of any geometric series is
$\sum_{m = 0}^{\infty} ar^m = \frac{a}{1 - r}$

Thus
$\sum_{m = 0}^{\infty} \left ( \frac{1}{4} \right )^m e^{-j \omega m } = \sum_{m = 0}^{\infty} \left ( \frac{e^{-j \omega}}{4} \right )^m = \frac{1}{1 - \frac{e^{-j \omega}}{4}}$

So finally we get:
$X(\omega) = 4^4e^{4j \omega } \cdot \frac{1}{1 - \frac{e^{-j \omega}}{4}}$

$X(\omega) = \frac{4^4e^{4j \omega }}{1 - \frac{e^{-j \omega}}{4}}$

-Dan

3. Originally Posted by hammer
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I was asked to find the fourier transform of x(n). Could someone explain whats happening in this answer?

Thanks
This is a Discrete Fourier Transform, $X(\omega)$ (on the second line) is the transform of $x(n)$
by definition of the DFT (and that $u(n)=0$ for $n<0$, and $1$ otherwise).

The rest is just simplification using the sum of a geometric series formula.

ZB