Fourier Transform Properties - Revision history

Max.Woesner at 06:26, 1 December 2009

2009-12-01T06:26:19Z

Brandon.price at 01:08, 30 November 2009

2009-11-30T01:08:24Z

← Older revision		Revision as of 18:08, 29 November 2009
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	==[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>==		==[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>==

			'''Look carefully at the signs in section 2 for <math>f^'</math> and <math>f^{''}</math> after the * operation. I think the signs are backwards but it ends up working out just fine because <math>\delta (f^{''}-f^')=\delta (f^'-f^{''})</math> -Brandon'''


	1. '''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br><br>'''		1. '''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br><br>'''
	Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br><br>		Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br><br>

Joshua.Sarris: /* Joshua Sarris */

2009-11-09T03:22:03Z

Joshua Sarris

← Older revision		Revision as of 20:22, 8 November 2009
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	We begin by finding the Fourier of x(t).		We begin by finding the Fourier of x(t).

	<math>\mathcal{F}[\frac{d}{dt} X(f)] = \frac{d}{dt} [ \int_{-\infty}^{\infty} X(f)e^{-j2 \pi f t} df]</math>		<math>\mathcal{F}[\frac{d}{dt} x(t)] = \frac{d}{dt} [ \int_{-\infty}^{\infty} x(t)e^{-j2 \pi f t} df]</math>

	We can then pull the derivitive into the integral and carry out its opperation.		We can then pull the derivitive into the integral and carry out its opperation.


	<math> \int_{-\infty}^{\infty} x(f)\frac{d}{dt} e^{-j2 \pi ft} df = \int_{-\infty}^{\infty}-j2\pi f x(f) e^{j2\pi f t} df</math>		<math> \int_{-\infty}^{\infty} x(t)\frac{d}{dt} e^{-j2 \pi ft} dt = -j2\pi f \int_{-\infty}^{\infty} x(t) e^{-j2\pi f t} dt</math>

			Since we know <math>\int_{-\infty}^{\infty} x(t) e^{-j2\pi f t} dt</math> is X(f) we can simplify.

			<math>\mathcal{F}[\frac{d}{dt} x(t)]= -j2\pi f X(f)</math>

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Joshua.Sarris at 03:16, 9 November 2009

2009-11-09T03:16:56Z

Joshua.Sarris at 03:00, 9 November 2009

2009-11-09T03:00:48Z

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 Also reviewed by [[Nick Christman]]
 -- Looks good.
 ----

Nicholas.Christman at 00:31, 9 November 2009

2009-11-09T00:31:57Z

← Older revision		Revision as of 17:31, 8 November 2009
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	<math>		<math>
	\int_{- \infty}^{\infty} g(t-t_{0})e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt		\int_{- \infty}^{\infty} g(t-t_{0})e^{-j2 \pi (f-f_{0})t} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,d \lambda
	</math>		</math>

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	<math>		<math>
	\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,dt = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt		\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})(\lambda + t_{0})} \,d \lambda = \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,d \lambda
	</math>		</math>

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	<math>		<math>
	\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,dt = e^{-j2 \pi (f-f_{0})t_{0}} \left[ \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } \,dt \right]		\int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } e^{-j2 \pi (f-f_{0})t_{0}} \,d \lambda = e^{-j2 \pi (f-f_{0})t_{0}} \left[ \int_{- \infty}^{\infty} g(\lambda )e^{-j2 \pi (f-f_{0})\lambda } \,d \lambda \right]
	</math>		</math>

Nicholas.Christman at 00:30, 9 November 2009

2009-11-09T00:30:53Z

← Older revision		Revision as of 17:30, 8 November 2009
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	<br/>		<br/>

	'''~~PLEASE ENTER PEER REVIEW HERE~~'''		'''Reviewed by [[Kevin Starkey]] --> add <math>\, d \lambda</math> above... other than that it looks good.'''

	<br/>		<br/>

Joshua.Sarris at 00:08, 9 November 2009

2009-11-09T00:08:06Z

← Older revision		Revision as of 17:08, 8 November 2009
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	<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, ~~''(<math> ** </math> I believe you are missing 'j' in the denominator of the second term)''~~		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><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>


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	Also reviewed by [[Nick Christman]]		Also reviewed by [[Nick Christman]]
	-- Looks good. ~~I found one typo (I think), see <math> ** </math>. Good job Josh! <br> <br> <br>~~		-- Looks good.

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Nicholas.Christman at 16:52, 8 November 2009

2009-11-08T16:52:46Z

← Older revision		Revision as of 09:52, 8 November 2009
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	1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>		1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>
	First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt </math> <br>		First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt </math> <br>
	We also know that <math> \mathcal{F}\left[s(t)\right] = S(f) and \int_{-\infty}^ \infty e^{-j2\pi ft} dt = \delta(f) </math> <br>		We also know that <math> \mathcal{F}\left[s(t)\right] = S(f) \mbox{ and } \int_{-\infty}^ \infty e^{-j2\pi ft} dt = \delta(f) </math> <br>
	Which gives us <math> \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{j2\pi ft} dt = \int_{-\infty}^ \infty S(f) \delta (f)df </math> <br>		(+) Which gives us <math> \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt = \int_{-\infty}^ \infty S(f) \delta (f)df </math> <br>
	Since <math> \int_{-\infty}^ \infty \delta (f) df </math> is only non-zero at f = 0 this yeilds <br>		Since <math> \int_{-\infty}^ \infty \delta (f) df </math> is only non-zero at f = 0 this yeilds <br>
	<math> \int_{-\infty}^ \infty S(f) \delta (f)df = S(0) </math> <br>		<math> \int_{-\infty}^ \infty S(f) \delta (f)df = S(0) </math> <br>
	So <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = S(0)</math> <br><br>		So <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = S(0)</math> <br><br>

	''~~'PLEASE ENTER PEER REVIEW HERE'''~~ <br><br><br>		Reviewed by [[Nick Christman]]
			-- I fixed one typo (needed a minus sign in the exponential). I'm not sure about the step (+). I would like to believe it, but I'm just not sure that it works... if you are sure it works, maybe add a little comment to explain it a little better. Other than that, it looks good! <br><br><br>

	2. Find <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] </math><br>		2. Find <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] </math><br>

Kevin.Starkey at 16:19, 8 November 2009

2009-11-08T16:19:09Z

← Older revision		Revision as of 23:26, 30 November 2009
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	==[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>==		==[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>==

	'''Look carefully at the signs in section 2 for <math>f^'</math> and <math>f^{''}</math> after the * operation. I think the signs are backwards but it ends up working out just fine because <math>\delta (f^{''}-f^')=\delta (f^'-f^{''})</math> -Brandon'''		'''Look carefully at the signs in section 2 for <math>f^'</math> and <math>f^{''}</math> after the * operation. I think the signs are backwards but it ends up working out just fine because <math>\delta (f^{''}-f^')=\delta (f^'-f^{''})</math> -Brandon'''<br>
			'''Corrected -Max'''


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	Similarly, <math> h(t)= \mathcal{F}^{-1}[H(f)] = \int_{-\infty}^{\infty}H(f)e^{j2\pi ft}df\!</math><br>		Similarly, <math> h(t)= \mathcal{F}^{-1}[H(f)] = \int_{-\infty}^{\infty}H(f)e^{j2\pi ft}df\!</math><br>
	So <math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^(t) dt\bigg] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}G(f^')e^{j2\pi f^'t}df^' \Bigg(\int_{-\infty}^{\infty}H(f^{''})e^{j2\pi f^{''}t}df^{''}\Bigg)^ dt \!</math><br>		So <math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^(t) dt\bigg] = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}G(f^')e^{j2\pi f^'t}df^' \Bigg(\int_{-\infty}^{\infty}H(f^{''})e^{j2\pi f^{''}t}df^{''}\Bigg)^ dt \!</math><br>
	Now <math>\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}G(f^')e^{j2\pi f^'t}df^' \Bigg(\int_{-\infty}^{\infty}H(f^{''})e^{j2\pi f^{''}t}df^{''}\Bigg)^* dt = \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^*(f^{''})\int_{-\infty}^{\infty}e^{j2\pi (f^{''}-f^')t}dt df^{''} df^' \!</math><br><br>		Now <math>\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}G(f^')e^{j2\pi f^'t}df^' \Bigg(\int_{-\infty}^{\infty}H(f^{''})e^{j2\pi f^{''}t}df^{''}\Bigg)^* dt = \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^*(f^{''})\int_{-\infty}^{\infty}e^{j2\pi (f^'-f^{''})t}dt df^{''} df^' \!</math><br><br>
	Note that <math>\int_{-\infty}^{\infty}e^{j2\pi (f^{''}-f^')t}dt = \delta (f^{''}-f^') \!</math><br><br>		Note that <math>\int_{-\infty}^{\infty}e^{j2\pi (f^'-f^{''})t}dt = \delta (f^'-f^{''}) \!</math><br><br>
	<i>Added step per Nick's suggestion</i><br><br>		<i>Added step per Nick's suggestion</i><br><br>
	Substituting gives us <math>\int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\int_{-\infty}^{\infty}e^{j2\pi (f^{''}-f^')t}dt df^{''} df^' = \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\delta (f^{''}-f^') df^{''} df^' \!</math><br><br>		Substituting gives us <math>\int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\int_{-\infty}^{\infty}e^{j2\pi (f^'-f^{''})t}dt df^{''} df^' = \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\delta (f^'-f^{''}) df^{''} df^' \!</math><br><br>
	And <math> \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\delta (f^{''}-f^') df^{''} df^' = \int_{-\infty}^{\infty}G(f^')H^(f^')df^' \!</math><br><br>		And <math> \int_{-\infty}^{\infty}G(f^')\int_{-\infty}^{\infty}H^(f^{''})\delta (f^'-f^{''}) df^{''} df^' = \int_{-\infty}^{\infty}G(f^')H^(f^')df^' \!</math><br><br>
	Since <math> f^' \!</math> is a simply a dummy variable, we can conclude that: <br><br>		Since <math> f^' \!</math> is a simply a dummy variable, we can conclude that: <br><br>
	<math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^(t) dt\bigg] = \int_{-\infty}^{\infty}G(f)H^(f)df \!</math><br><br>		<math>\mathcal{F}\bigg[\int_{-\infty}^{\infty}g(t) h^(t) dt\bigg] = \int_{-\infty}^{\infty}G(f)H^(f)df \!</math><br><br>

← Older revision		Revision as of 20:16, 8 November 2009
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	[http://cnx.org/content/m0045/latest/ Some properties to choose from if you are having difficulty....]		[http://cnx.org/content/m0045/latest/ Some properties to choose from if you are having difficulty....]

	[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>		==[[Max Woesner\|<b><u>Max Woesner</u></b>]]<br><br>==
	1. '''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br><br>'''		1. '''Find <math>\mathcal{F}[cos(w_0t)g(t)]\!</math><br><br>'''
	Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br><br>		Recall <math> w_0 = 2\pi f_0\!</math>, so <math>\mathcal{F}[cos(w_0t)g(t)] = \mathcal{F}[cos(2\pi f_0t)g(t)] = \int_{-\infty}^{\infty}cos(2\pi f_0t)g(t)e^{-j2\pi ft}dt\!</math><br><br>
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	----		----

	[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br><br>		==[[Nick Christman\|<b><u>Nick Christman</u></b>]]<br><br>==
	Note: After scratching my head for a couple of hours, I decided that I would try a different Fourier Property. In fact, I chose a property that would need to be defined in order to show my second property.		Note: After scratching my head for a couple of hours, I decided that I would try a different Fourier Property. In fact, I chose a property that would need to be defined in order to show my second property.

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	[[Joshua Sarris\|<b><u>Joshua Sarris</u></b>]]<br><br>		==[[Joshua Sarris\|<b><u>Joshua Sarris</u></b>]]<br><br>==
	'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br>'''		'''Find <math>\mathcal{F}[sin(w_0t)g(t)]\!</math><br>'''

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	'''Find <math>\mathcal{F}[\frac{d}{dt} x(t)] \!</math><br>'''		'''Find <math>\mathcal{F}[\frac{d}{dt} x(t)] \!</math><br>'''

	We begin ~~our proof~~ by finding the ~~inverse Fourer~~ of X(f) ~~in terms of t~~.		We begin by finding the Fourier of x(t).

	<math>\mathcal{F~~}^{-1~~}[\frac{d}{dt} X(f)] = \frac{d}{dt} [ \int_{-\infty}^{\infty} X(f)e^{j2\pi f t} df]</math>		<math>\mathcal{F}[\frac{d}{dt} X(f)] = \frac{d}{dt} [ \int_{-\infty}^{\infty} X(f)e^{-j2 \pi f t} df]</math>

	We can then pull the derivitive into the integral and carry out its opperation.		We can then pull the derivitive into the integral and carry out its opperation.


	<math> \int_{-\infty}^{\infty} X(f)\frac{d}{dt} e^{j2\pi ~~f t~~} df = \int_{-\infty}^{\infty}j2\~~pif X~~(f) e^{j2\pi f t} df</math>		<math> \int_{-\infty}^{\infty} x(f)\frac{d}{dt} e^{-j2 \pi ft} df = \int_{-\infty}^{\infty}-j2\pi f x(f) e^{j2\pi f t} df</math>

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	[[Kevin Starkey\|<b><u>Kevin Starkey</u></b>]] <br> <br>		==[[Kevin Starkey\|<b><u>Kevin Starkey</u></b>]] <br> <br>==
	1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>		1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>
	First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt </math> <br>		First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt </math> <br>

← Older revision		Revision as of 09:19, 8 November 2009
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	[[Kevin Starkey\|<b><u>Kevin Starkey</u></b>]] <br> <br>		[[Kevin Starkey\|<b><u>Kevin Starkey</u></b>]] <br> <br>
	1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>		1. Find <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right]</math> <br>
	First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{j2\pi ft} dt </math> <br>		First we know that <math> \mathcal{F}\left[\int_{-\infty}^ \infty s(t)dt\right] = \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{-j2\pi ft} dt </math> <br>
	We also know that <math> \mathcal{F}\left[s(t)\right] = S(f) and \int_{-\infty}^ \infty e^{j2\pi ft} dt = \delta(f) </math> <br>		We also know that <math> \mathcal{F}\left[s(t)\right] = S(f) and \int_{-\infty}^ \infty e^{-j2\pi ft} dt = \delta(f) </math> <br>
	Which gives us <math> \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{j2\pi ft} dt = \int_{-\infty}^ \infty S(f) \delta (f)df </math> <br>		Which gives us <math> \int_{-\infty}^\infty\left(\int_{-\infty}^ \infty s(t)dt\right)e^{j2\pi ft} dt = \int_{-\infty}^ \infty S(f) \delta (f)df </math> <br>
	Since <math> \int_{-\infty}^ \infty \delta (f) df </math> is only non-zero at f = 0 this yeilds <br>		Since <math> \int_{-\infty}^ \infty \delta (f) df </math> is only non-zero at f = 0 this yeilds <br>
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	2. Find <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] </math><br>		2. Find <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] </math><br>
	First <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] = \int_{- \infty}^{\infty}e^{j2\pi f_0t}s(t)e^{-j2\pi ft} </math><br>		First <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] = \int_{- \infty}^{\infty}e^{j2\pi f_0t}s(t)e^{-j2\pi ft} </math><br>
	or rearranging we get <math> \int_{- \infty}^{\infty}e^{j2\pi f_0t}s(t)e^{-j2\pi ft}dt = \int_{- \infty}^{\infty}s(t)e^{j2\pi t(f_0 -f)}dt</math><br>		or rearranging we get <math> \int_{- \infty}^{\infty}e^{j2\pi f_0t}s(t)e^{-j2\pi ft}dt = \int_{- \infty}^{\infty}s(t)e^{-j2\pi t(f -f_0)}dt</math><br>
	Which leads to <math> \int_{- \infty}^{\infty}s(t)e^{j2\pi t(f_0 -f)}dt = S(f-f_0)</math><br>		Which leads to <math> \int_{- \infty}^{\infty}s(t)e^{-j2\pi t(f -f_0)}dt = S(f-f_0)</math><br>
	So <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] = S(f-f_0) </math><br><br><br>		So <math>\mathcal{F}\left[e^{j2\pi f_0t}s(t)\right] = S(f-f_0) </math><br><br><br>

	'''PLEASE ENTER PEER REVIEW HERE'''<BR><BR><BR>		'''PLEASE ENTER PEER REVIEW HERE'''<BR><BR><BR>