Kevin.Starkey at 04:25, 3 November 2009

2009-11-03T04:25:12Z

@@ Line 5: / Line 5: @@
 /T <math>\Longrightarrow df \!</math> <br>
 n/T <math>\Longrightarrow df \!</math> <br>
-<math> \sum_{n=-\infty} ^\infty
+<math> \sum_{n=-\infty} ^\infty 1/T \Longrightarrow \int_{-\infty}^\infty ()df \!</math> <br>

Kevin.Starkey: New page: Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that $\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...$

2009-11-03T03:24:07Z

New page: Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that <math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...

New page

Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that
<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^')e^{j2\pi nt/T},\!</math> where 
<math>\ 1/T\int_{-T/2}^{T/2} x(t^')e^{-j2\pi nt^'/T}dt^' = \alpha_n \!</math> 
first we need to remove the restiction x(t) = x(t + T) by following these steps. 
1/T <math>\Longrightarrow df \!</math> 
n/T <math>\Longrightarrow df \!</math> 
<math> \sum_{n=-\infty} ^\infty

3 - The Game Simplified - Revision history

Kevin.Starkey at 04:25, 3 November 2009

Kevin.Starkey: New page: Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that \lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...

Kevin.Starkey: New page: Lets look at what happens if our signals are not periodic. We can achieve this by setting our period T to infinity such that $\lim_{T \to \infty}\sum_{n=-\infty}^\infty (1/T\int_{-T/2...$