Fourier Transform Property review - Revision history

Max.Woesner at 02:58, 21 October 2009

2009-10-21T02:58:41Z

← Older revision		Revision as of 19:58, 20 October 2009
Line 55:		Line 55:
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>
	Or rather <br>		Or rather <br>
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{2}j[G(f-f_0)+ G(f+f_0)]\!</math>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{2}j[G(f+f_0)- G(f-f_0)]\!</math>

Max.Woesner at 02:55, 21 October 2009

2009-10-21T02:55:51Z

← Older revision		Revision as of 19:55, 20 October 2009
Line 54:		Line 54:
	So we now have the identity,<br>		So we now have the identity,<br>
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>
			Or rather <br>
			<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{2}j[G(f-f_0)+ G(f+f_0)]\!</math>

Max.Woesner at 02:36, 21 October 2009

2009-10-21T02:36:41Z

← Older revision		Revision as of 19:36, 20 October 2009
Line 54:		Line 54:
	So we now have the identity,<br>		So we now have the identity,<br>
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>
	~~Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>~~

Max.Woesner at 06:01, 20 October 2009

2009-10-20T06:01:01Z

← Older revision		Revision as of 23:01, 19 October 2009
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	<math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math><br>		<math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math><br>
	The derivation should be as follows.<br><br>		The derivation should be as follows.<br><br>
	'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br>'''		'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br><br>'''
	Recall <math> w_0 = 2\pi f_0\!</math>, so expanding we have, <math>\mathcal{F}[sin(w_0t)g(t)] = \mathcal{F}[sin(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>		Recall <math> w_0 = 2\pi f_0\!</math>, <br>
	Also recall <math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math>, so we can convert to exponentials.<br>		so expanding we have, <br>
			<math>\mathcal{F}[sin(w_0t)g(t)] = \mathcal{F}[sin(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br>
			Also recall <math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math>, <br>
			so we can convert to exponentials.<br>
	<math>\int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>		<math>\int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math><br>
	Now integrating gives us,<br>		Now integrating gives us,<br>
	<math>\int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt-\frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt = \frac{1}{j2}G(f-f_0)- \frac{1}{j2}G(f+f_0)\!</math><br>		<math>\int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt-\frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt = \frac{1}{j2}G(f-f_0)- \frac{1}{j2}G(f+f_0)\!</math><br>
	So we now have the identity,		So we now have the identity,<br>
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>
	Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>		Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>

Max.Woesner at 04:24, 20 October 2009

2009-10-20T04:24:48Z

← Older revision		Revision as of 21:24, 19 October 2009
Line 4:		Line 4:
	=== Homework #5 - Reveiew a Fourier Transform Property ===		=== Homework #5 - Reveiew a Fourier Transform Property ===

	[[Joshua Sarris\|~~<b><u>~~Joshua Sarris~~</u></b>~~]] derived the following Fourier Transform Property [[Fourier Transform Properties\|here]] for Homework #4.<br><br>		[[Joshua Sarris\|Joshua Sarris]] derived the following Fourier Transform Property [[Fourier Transform Properties\|here]] for Homework #4.<br><br>
	'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br>'''		'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br>'''

Line 51:		Line 51:
	So we now have the identity,		So we now have the identity,
	<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>		<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math><br>
	Your answer looks correct, but I don't know how you got if from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>		Your answer looks correct, but I don't know how you got it from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>

Max.Woesner: New page: == Max Woesner == Back to my Home Page === Homework #5 - Reveiew a Fourier Transform Property === Joshua Sarris derived the following Four...

2009-10-20T01:04:01Z

New page: == Max Woesner == Back to my Home Page === Homework #5 - Reveiew a Fourier Transform Property === Joshua Sarris derived the following Four...

New page

== Max Woesner ==
[[Max Woesner|Back to my Home Page]]

=== Homework #5 - Reveiew a Fourier Transform Property ===

[[Joshua Sarris|Joshua Sarris]] derived the following Fourier Transform Property [[Fourier Transform Properties|here]] for Homework #4. 
'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math> '''

Recall
<math> w_0 = 2\pi f_0\!</math>,

so expanding we have,

<math>\mathcal{F}[sin(w_0t)g(t)] = \mathcal{F}[sin(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math> 

Also recall
<math> sin(\theta) = \frac{1}{j2}(e^{j\theta} + e^{-j\theta})\!</math>,

so we can convert to exponentials.

<math>\int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math> 

Now integrating gives us,

<math>\int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}+e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt+\frac{1}{2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt = \frac{1}{j2}G(f-f_0)+ \frac{1}{j2}G(f+f_0)\!</math> 

So we now have the identity,

<math>\mathcal{F}[cos(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)+ G(f+f_0)]\!</math>

orr rather

<math>\mathcal{F}[cos(w_0t)g(t)] =\frac{1}{2}j[G(f-f_0)- G(f+f_0)]\!</math>

----

====My Review====

Josh, Josh, it appears that you copied my code and forgot to change some necessary elements so it is works for your equation, such as the cosines on lines 4, 11, and 13. 
Also, you forgot a <math> j \!</math> in the second term of the second equation on line 9, and your identity for <math> sin(\theta)\!</math> has a sign error. It should be: 
<math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math> 
The derivation should be as follows. 
'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math> '''
Recall <math> w_0 = 2\pi f_0\!</math>, so expanding we have, <math>\mathcal{F}[sin(w_0t)g(t)] = \mathcal{F}[sin(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math> 
Also recall <math> sin(\theta) = \frac{1}{j2}(e^{j\theta} - e^{-j\theta})\!</math>, so we can convert to exponentials. 
<math>\int_{-\infty}^{\infty}sin(2\pi f_0t)g(t)e^{-j2\pi ft}dt = \int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt\!</math> 
Now integrating gives us, 
<math>\int_{-\infty}^{\infty} \frac{1}{j2}[e^{j2\pi f_0t}-e^{-j2\pi f_0t}]g(t)e^{-j2\pi ft}dt = \frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f-f_0)t}g(t)dt-\frac{1}{j2}\int_{-\infty}^{\infty}e^{-j2\pi (f+f_0)t}g(t)dt = \frac{1}{j2}G(f-f_0)- \frac{1}{j2}G(f+f_0)\!</math> 
So we now have the identity,
<math>\mathcal{F}[sin(w_0t)g(t)] = \frac{1}{j2}[G(f-f_0)- G(f+f_0)]\!</math> 
Your answer looks correct, but I don't know how you got if from the equation above it. Your operator between the <math> G \!</math> terms changed from a <math> + \!</math> to a <math> - \!</math>