Homework Four - Revision history

Nicholas.Christman at 16:36, 8 November 2009

2009-11-08T16:36:39Z

← Older revision		Revision as of 09:36, 8 November 2009
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	[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br><br>		[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br><br>
	1. '''Find <math>\mathcal{F} \left[ ~~\int_{- \infty}^{\infty}10^{t}~~g(t)e^{j2 \pi ~~ft_~~{0}} ~~\,dt~~ \right] </math><br/>'''		1. '''Find <math>\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] </math><br/>'''

	~~To begin, we know that~~<br/>		This is a fairly straightforward property and is known as ''complex modulation''<br/>

	<math>		<math>
	\mathcal{F} \left[~~\int_{- \infty}^{\infty}10^{t}~~g(t)e^{j2 \pi ~~ft_0~~} ~~\,dt~~ \right]		\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = \int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt
	= \int_{- \infty}^{\infty} \left~~( \int_{-\infty}^{\infty}10^{t}~~g(t)e^{j2 \pi ~~ft_0~~} ~~\,dt~~ \right) e^{-j2 \pi ft}\,dt
	</math>		</math>
	~~<br/>~~

	~~After some factoring and combinting <math> \Longleftarrow </math> "combining-Kevin" of like~~ terms we get:		Combining terms, we get:

	~~<br/>~~
	<math>		<math>
	~~\mathcal{F} \left[~~ \int_{- \infty}^{\infty}~~10^{t}~~g(t)e^{j2 \pi ~~ft_0~~} ~~\,dt~~ \right]		\int_{- \infty}^{\infty} \left[ g(t)e^{j2 \pi f_{0}t} \right] e^{-j2 \pi ft} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt
	~~= \int_~~{- \~~infty~~}^{\~~infty} \left(~~ \int_{- \infty}^{\infty~~} 10^{t~~}g(t~~) \,dt \right~~) e^{j2 \pi f~~(t_0~~-t)}\,dt
	</math>		</math>
	<br/>		<br/>

	~~But recall that~~		Now let's make the following substitution <math> \displaystyle \theta = f-f_{0}</math>
	<math>\~~int_{- \infty}^{\infty}e^{j2~~ \pi f~~(t_{0}-t)} \,dt \equiv \delta (t_{0}-t) \mbox{ or } \delta (t~~-t_{0})</math>

			This now gives us a surprisingly familiar function:
	~~Because of this definition (and some "math magic") our problem has been simplified significantly~~: ~~<br/>~~

	<math>		<math>
	~~\mathcal{F} \left[~~ \int_{- \infty}^{\infty~~}10^{t~~}g(t)e^{j2 \pi ~~ft_0~~} \,dt ~~\right]~~		\int_{- \infty}^{\infty} g(t)e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(t)e^{-j2 \pi \theta t} \,dt
	= \int_{- \infty}^{\infty~~}10^{t~~}g(t) \~~delta (~~t~~-t_{0~~})\,dt ~~= 10^{t_0}g(t_0)~~		</math>
	</math>
	<br/>		<br/>

	~~Therefore,~~		This looks just like <math> \displaystyle G(\theta )</math>!

			We can now conclude that:
			<br/>

	<math> \mathcal{F} \left[ ~~\int_{- \infty}^{\infty}10^{t}~~g(t)e^{j2 \pi ~~ft_0~~} ~~\,dt~~ \right] = ~~10^~~{~~t_0~~}~~g(t_0~~) </math>		<math>
			\mathcal{F} \left[ g(t)e^{j2 \pi f_{0}t} \right] = G(\theta ) = G(f-f_{0})
			</math>
	<br>		<br>
	~~Other than the one typo looks~~ good - Kevin
			Looks good - Kevin

Kevin.Starkey at 04:41, 3 November 2009

2009-11-03T04:41:47Z

← Older revision		Revision as of 21:41, 2 November 2009
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	<br/>		<br/>

	After some factoring and combinting of like terms we get:		After some factoring and combinting <math> \Longleftarrow </math> "combining-Kevin" of like terms we get:

	<br/>		<br/>
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	<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>		<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>
			<br>
			Other than the one typo looks good - Kevin

Nicholas.Christman at 20:41, 31 October 2009

2009-10-31T20:41:33Z

← Older revision		Revision as of 13:41, 31 October 2009
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	----		----

	'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''		[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br><br>
			1. '''Find <math>\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_{0}} \,dt \right] </math><br/>'''

	To begin, we know that<br/>		To begin, we know that<br/>

	<math> \mathcal{F}[~~10^{t}g(t)e^{j2 \pi ft_0}] =~~ \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{~~-j2~~ \~~pi ft~~}\~~,dt =~~ \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ~~f(t_~~{0}-t)}\,dt</math>		<math>
			\mathcal{F} \left[\int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
			= \int_{- \infty}^{\infty} \left( \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right) e^{-j2 \pi ft}\,dt
			</math>
			<br/>

	~~But recall that <math>\int_{-\infty}^{\infty}e^{j2 \pi f(t_{0}-t)}df \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>~~		After some factoring and combinting of like terms we get:

	~~The following needs to be fixed, because the previous thing (just above this) which we just fixed wasn't an identity. Hint:~~ <math>10^{t}~~</math> is related to <math>~~e^{t}</math>		<br/>
			<math>
			\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
			= \int_{- \infty}^{\infty} \left( \int_{- \infty}^{\infty} 10^{t}g(t) \,dt \right) e^{j2 \pi f(t_0-t)}\,dt
			</math>
			<br/>

	~~Because of this definition~~, ~~our problem has now been simplified significantly:~~ <br/>		But recall that
			<math>\int_{- \infty}^{\infty}e^{j2 \pi f(t_{0}-t)} \,dt \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>

	<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> <br/>
			Because of this definition (and some "math magic") our problem has been simplified significantly: <br/>

			<math>
			\mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right]
			= \int_{- \infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)
			</math>
			<br/>

	Therefore,		Therefore,

	<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>		<math> \mathcal{F} \left[ \int_{- \infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0} \,dt \right] = 10^{t_0}g(t_0) </math>

Frohro at 22:10, 19 October 2009

2009-10-19T22:10:13Z

← Older revision		Revision as of 15:10, 19 October 2009
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	<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>		<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>

	But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>		But recall that <math>\int_{-\infty}^{\infty}e^{j2 \pi f(t_{0}-t)}df \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>

			The following needs to be fixed, because the previous thing (just above this) which we just fixed wasn't an identity. Hint: <math>10^{t}</math> is related to <math>e^{t}</math>

	Because of this definition, our problem has now been simplified significantly: <br/>		Because of this definition, our problem has now been simplified significantly: <br/>

Nicholas.Christman at 22:16, 16 October 2009

2009-10-16T22:16:27Z

← Older revision		Revision as of 15:16, 16 October 2009
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	[[Fourier Transform Properties\|Fourier Transform Properties]]		[[Fourier Transform Properties\|<b>Fourier Transform Properties</b>]]

			----

	'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''		'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''

Nicholas.Christman at 22:15, 16 October 2009

2009-10-16T22:15:34Z

← Older revision		Revision as of 15:15, 16 October 2009
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	[[~~Nick Christman~~\|~~<b><u>Nick Christman</u></b>~~]]~~<br><br>~~		[[Fourier Transform Properties\|Fourier Transform Properties]]

	'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''		'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math><br/>'''

Nicholas.Christman: New page: Nick Christman

'''Find $\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]$
''' To begin, we know that
$\mathcal{F}[10^{t}g(t)e^{...$

2009-10-16T22:14:36Z

New page: Nick Christman '''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math> ''' To begin, we know that <math> \mathcal{F}[10^{t}g(t)e^{...

New page

[[Nick Christman|Nick Christman]] 
'''Find <math>\mathcal{F}[10^{t}g(t)e^{j2 \pi ft_{0}}]</math> '''

To begin, we know that 

<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi ft_0}e^{-j2 \pi ft}\,dt = \int_{-\infty}^{\infty}10^{t}g(t)e^{j2 \pi f(t_{0}-t)}\,dt</math>

But recall that <math>e^{j2 \pi f(t_{0}-t)} \equiv \delta (t_{0}-t) \mbox{ or } \delta (t-t_{0})</math>

Because of this definition, our problem has now been simplified significantly: 

<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = \int_{-\infty}^{\infty}10^{t}g(t) \delta (t-t_{0})\,dt = 10^{t_0}g(t_0)</math> 

Therefore,

<math> \mathcal{F}[10^{t}g(t)e^{j2 \pi ft_0}] = 10^{t_0}g(t_0) </math>