Difference between revisions of "Thomas Wroblewski - HW7 Contribution ECE301Fall2008mboutin" - Rhea

Latest revision as of 12:03, 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}\, $

$ x_1[n] = e^{j0} + (\frac{1}{2}e^{j(\frac{\pi}{2}n)}) + (\frac{1}{2}e^{-j(\frac{\pi}{2}n)}) \, $

$ x_1[n] = 1 + cos(\frac{\pi n}{2})\, $

Example 4

Ok, Lets try one using Table's 5.1 and 5.2 from the text

Lets determine the numerical value for A when $ A = \sum^{\infty}_{n = 0} n(\frac{1}{2})^{n}\, $

From Table 5.2 we know that $ (\frac{1}{2})^n u[n] <--FT--> \frac{1}{1 - (\frac{1}{2})e^{-j\omega}}\, $

Using the "Differentiation in Frequency" Property from Table 5.1, we can transform our equation..

$ x[n] = n(\frac{1}{2})^{n} <--FT--> X(e^{j\omega}) = j\frac{d}{d\omega} (\frac{1}{1-(\frac{1}{2})e^{-j\omega}}) \, $

$ X(e^{j\omega}) = \frac{(\frac{1}{2})e^{-j\omega}}{(1-(\frac{1}{2})e^{-j\omega})^{2}}\, $

So.....

$ \sum^{\infty}_{n = 0}n(\frac{1}{2})^{n} = \sum^{\infty}_{n = -\infty}x[n] = X(e^{j0}) = 2 \, $

Hope this helps - Thomas Wroblewski