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This gives up a magnitude and phase graphs. Something noteworthy is that this function is periodic with an <math>{\omega} = 2\pi\,</math> | This gives up a magnitude and phase graphs. Something noteworthy is that this function is periodic with an <math>{\omega} = 2\pi\,</math> | ||
| + | |||
---- | ---- | ||
Example 2 | Example 2 | ||
---- | ---- | ||
| + | |||
Let's try something a little more challenging..... | Let's try something a little more challenging..... | ||
| + | |||
| + | |||
| + | Lets let : <math>x[n] = (\frac{1}{2})^{n-1}u[n-1]\,</math> | ||
| + | |||
| + | Using the Fourier Transform Equation: | ||
| + | |||
| + | <math>X(e^{j\omega}) = \sum^{\infty}_{n = -\infty} u[n] e^{-j\omega n}\,</math> | ||
| + | |||
| + | Transforming our equation into the Fourier Transform Equation we get | ||
| + | |||
| + | <math>X(e^{j\omega}) = \sum^{\infty}_{n = 1} (\frac{1}{2})^{n-1}e^{-j\omega n}\,</math> | ||
| + | |||
| + | <math>X(e^{j\omega}) = \sum^{\infty}_{n = 0} (\frac{1}{2})^{n}e^{-j\omega (n+1)}\,</math> | ||
| + | |||
| + | <math>X(e^{j\omega}) = e^{-j\omega}(\frac{1}{1-(\frac{1}{2})e^{-j\omega}})\,</math> | ||
| + | |||
| + | ---- | ||
| + | Example 3 | ||
| + | ---- | ||
| + | |||
| + | Let's try to get the inverse Fourier Transform in this next one..... | ||
| + | |||
| + | Suppose <math>X_1(e^{-j\omega}) = \sum^{\infty}_{k = -\infty}[2\pi\delta(\omega-2\pi k) + \pi\delta(\omega - (\frac{\pi}{2}) - 2\pi k) + \pi\delta(\omega + (\frac{\pi}{2}) - 2\pi k)]\,</math> | ||
| + | |||
| + | Using the Fourier Transform Synthesis Equation: | ||
| + | |||
| + | <math>x_1[n] = (\frac{1}{2\pi})\int^{\pi}_{-\pi} X_1(e^{-j\omega})e^{j\omega n}d\omega \,</math> | ||
| + | |||
| + | Transforming our equation into the Inverse Fourier Transform Synthesis Equation | ||
| + | |||
| + | <math>x_1[n] = (\frac{1}{2\pi})\int^{\pi}_{-\pi} [2\pi\delta(\omega) + \pi\delta(\omega - (\frac{\pi}{2})) + \pi\delta(\omega + (\frac{\pi}{2}))]e^{-j\omega n}\,</math> | ||
| + | |||
| + | <math>e^{j0} + (\frac{1}{2}e^{j(\frac{\pi}{2}n)}) + (\frac{1}{2}e^{-j(\frac{\pi}{2}n)}) \,</math> | ||
Revision as of 11:36, 24 October 2008
Hello, This is my Homework 7 Contribution.
I am having some trouble still with the process of doing Fourier Transforms so I thought it would be a good idea to do some examples of how to do a Fourier Transform to help clarify the process.
Example 1
Lets take a simple example to start.
Lets let : $ x[n] = a^nu[n], |a| < 1\, $
Converting to $ X(e^{j\omega})\, $ notation we get
$ X(e^{j\omega}) = \sum^{\infty}_{n = -\infty} a^n u[n] e^{-j\omega n}\, $
Simplifying we get...
$ X(e^{j\omega}) = \sum^{\infty}_{n = 0} (ae^{-j\omega})^n\, $
$ X(e^{j\omega}) = \frac{1}{1-a e^{-j\omega}}\, $
This gives up a magnitude and phase graphs. Something noteworthy is that this function is periodic with an $ {\omega} = 2\pi\, $
Example 2
Let's try something a little more challenging.....
Lets let : $ x[n] = (\frac{1}{2})^{n-1}u[n-1]\, $
Using the Fourier Transform Equation:
$ X(e^{j\omega}) = \sum^{\infty}_{n = -\infty} u[n] e^{-j\omega n}\, $
Transforming our equation into the Fourier Transform Equation we get
$ X(e^{j\omega}) = \sum^{\infty}_{n = 1} (\frac{1}{2})^{n-1}e^{-j\omega n}\, $
$ X(e^{j\omega}) = \sum^{\infty}_{n = 0} (\frac{1}{2})^{n}e^{-j\omega (n+1)}\, $
$ X(e^{j\omega}) = e^{-j\omega}(\frac{1}{1-(\frac{1}{2})e^{-j\omega}})\, $
Example 3
Let's try to get the inverse Fourier Transform in this next one.....
Suppose $ X_1(e^{-j\omega}) = \sum^{\infty}_{k = -\infty}[2\pi\delta(\omega-2\pi k) + \pi\delta(\omega - (\frac{\pi}{2}) - 2\pi k) + \pi\delta(\omega + (\frac{\pi}{2}) - 2\pi k)]\, $
Using the Fourier Transform Synthesis Equation:
$ x_1[n] = (\frac{1}{2\pi})\int^{\pi}_{-\pi} X_1(e^{-j\omega})e^{j\omega n}d\omega \, $
Transforming our equation into the Inverse Fourier Transform Synthesis Equation
$ x_1[n] = (\frac{1}{2\pi})\int^{\pi}_{-\pi} [2\pi\delta(\omega) + \pi\delta(\omega - (\frac{\pi}{2})) + \pi\delta(\omega + (\frac{\pi}{2}))]e^{-j\omega n}\, $
$ e^{j0} + (\frac{1}{2}e^{j(\frac{\pi}{2}n)}) + (\frac{1}{2}e^{-j(\frac{\pi}{2}n)}) \, $
