Assignment - Revision history

Joshua.Sarris at 21:58, 15 October 2009

2009-10-15T21:58:14Z

@@ Line 5: / Line 5: @@
 Begining with a Fourier Series:
-<math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math>
+(1) <math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math>
 where
@@ Line 13: / Line 13: @@
 We then take the limit of a Fourier series as its period T approaches infinity:
-<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^')e^{\frac{j2\pi nt}{T}}\!</math>
+(2)<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^')e^{\frac{j2\pi nt}{T}}\!</math>
 In order to evaluate this limit we need the following relationships:
@@ Line 54: / Line 54: @@
 We can now write out the following:
-<math>x(t) = \lim_{T \to \infty} \left[ \sum_{n=- \infty}^{\infty} \left(\frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{- \frac{j2 \pi nt'}{T}} \,dt' \right) e^{\frac{j2 \pi nt}{T}} \right] </math>
+(3)<math>x(t) = \lim_{T \to \infty} \left[ \sum_{n=- \infty}^{\infty} \left(\frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{- \frac{j2 \pi nt'}{T}} \,dt' \right) e^{\frac{j2 \pi nt}{T}} \right] </math>
 which can also be written as:
-<math>x(t) =\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math>
+(4)<math>x(t) =\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math>
 using,
@@ Line 66: / Line 66: @@
 we now have
-<math>\,X(f) =  \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt'
+(5)<math>\,X(f) =  \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt'
 </math>

Joshua.Sarris at 21:47, 15 October 2009

2009-10-15T21:47:06Z

← Older revision		Revision as of 14:47, 15 October 2009
Line 16:		Line 16:

	In order to evaluate this limit we need the following relationships:		In order to evaluate this limit we need the following relationships:
			<table>
			<tr>
			<td width=100>
			<math>\frac{1}{T}</math>
			</td>
			<td width=50>
			<math>\rightarrow</math>
			</td>
			<td widt=100>
			<math>\,df</math>
			</td>
			</tr>
			<tr>
			<td>
			<math>\frac{n}{T}</math>
			</td>
			<td>
			<math>\rightarrow</math>
			</td>
			<td>
			<math>\,f</math>
			</td>
			</tr>
			<tr>
			<td>
			<math>\sum_{n=- \infty}^{\infty} \frac{1}{T}</math>
			</td>
			<td>
			<math>\rightarrow</math>
			</td>
			<td>
			<math>\int_{-\infty}^{\infty}(\mbox{ })\,df</math>
			</td>
			</tr>
			</table>

			We can now write out the following:

			<math>x(t) = \lim_{T \to \infty} \left[ \sum_{n=- \infty}^{\infty} \left(\frac{1}{T} \int_{-\frac{T}{2}}^{\frac{T}{2}} x(t')e^{- \frac{j2 \pi nt'}{T}} \,dt' \right) e^{\frac{j2 \pi nt}{T}} \right] </math>

			which can also be written as:

			<math>x(t) =\int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt' \right) e^{j2 \pi tf} \,df</math>

			using,

			<math>\alpha_n\rightarrow X(f) \!</math>

			we now have

			<math>\,X(f) = \int_{-\infty}^{\infty} \,x(t') e^{-j2 \pi ft'} \,dt'

			</math>

Joshua.Sarris: New page: Summery of the class notes from Oct. 5: What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can ...

2009-10-15T21:35:10Z

New page: Summery of the class notes from Oct. 5: What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can ...

New page

Summery of the class notes from Oct. 5:

What if a periodic signal had an infinite period? We would no longer be able to tell the difference between it and a non periodic signal. We can use this property to look at signals that do not have a period (an observable one at least).

Begining with a Fourier Series:

<math>x(t) = x(t+T) = \sum_{n=- \infty}^{\infty} \alpha _n e^{\frac{j2 \pi nt}{T}}</math>

where

<math>\alpha_n\!=\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^'\!</math>

We then take the limit of a Fourier series as its period T approaches infinity:

<math>\lim_{T \to \infty}\sum_{n=-\infty}^\infty (\frac{1}{T}\int_{-\frac{T}{2}}^{\frac{T}{2}}x(t^')e^{\frac{-j2\pi nt^'}{T}}dt^')e^{\frac{j2\pi nt}{T}}\!</math>

In order to evaluate this limit we need the following relationships: