<?xml version="1.0" encoding="UTF-8"?><rss version="2.0"
	xmlns:content="http://purl.org/rss/1.0/modules/content/"
	xmlns:dc="http://purl.org/dc/elements/1.1/"
	xmlns:atom="http://www.w3.org/2005/Atom"
	xmlns:sy="http://purl.org/rss/1.0/modules/syndication/"
		>
<channel>
	<title>Comments on: The Trouble with White Spaces</title>
	<atom:link href="http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/feed/" rel="self" type="application/rss+xml" />
	<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/</link>
	<description>A regular old blog</description>
	<lastBuildDate>Fri, 04 Sep 2009 23:51:11 +0000</lastBuildDate>
	<generator>http://wordpress.org/?v=2.9.1</generator>
	<sy:updatePeriod>hourly</sy:updatePeriod>
	<sy:updateFrequency>1</sy:updateFrequency>
		<item>
		<title>By: Brett Glass</title>
		<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/comment-page-1/#comment-427467</link>
		<dc:creator>Brett Glass</dc:creator>
		<pubDate>Sat, 22 Nov 2008 00:22:15 +0000</pubDate>
		<guid isPermaLink="false">http://bennett.com/blog/?p=5065#comment-427467</guid>
		<description>Richard: Yes, non-cooperating neighbors -- without equipment that cooperates automatically -- is the biggest weakness of Part 15. I&#039;ve long been an advocate of spectrum etiquettes. However, the guys who just want all spectrum to be exclusively owned have done their best to get the FCC not to require them. Why? Because without etiquettes on the unlicensed bands, bedlam ensues and their licensed spectrum is more valuable.

Jason: Part 15 contains no requirement that minimum power be used. It will after the final rules for the white spaces are codified -- and that will be ONLY for the white spaces. However, there is an inherent problem in it: to determine the minimum power required to communicate, you have to turn the power down regularly so as to cause communications to fail. Broadband providers&#039; equipment cannot do this; it needs to work perfectly, 24x7. Another reason why the release of the white spaces as unlicensed is a horrible idea.</description>
		<content:encoded><![CDATA[<p>Richard: Yes, non-cooperating neighbors &#8212; without equipment that cooperates automatically &#8212; is the biggest weakness of Part 15. I&#8217;ve long been an advocate of spectrum etiquettes. However, the guys who just want all spectrum to be exclusively owned have done their best to get the FCC not to require them. Why? Because without etiquettes on the unlicensed bands, bedlam ensues and their licensed spectrum is more valuable.</p>
<p>Jason: Part 15 contains no requirement that minimum power be used. It will after the final rules for the white spaces are codified &#8212; and that will be ONLY for the white spaces. However, there is an inherent problem in it: to determine the minimum power required to communicate, you have to turn the power down regularly so as to cause communications to fail. Broadband providers&#8217; equipment cannot do this; it needs to work perfectly, 24&#215;7. Another reason why the release of the white spaces as unlicensed is a horrible idea.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Richard Bennett</title>
		<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/comment-page-1/#comment-427466</link>
		<dc:creator>Richard Bennett</dc:creator>
		<pubDate>Fri, 21 Nov 2008 23:21:49 +0000</pubDate>
		<guid isPermaLink="false">http://bennett.com/blog/?p=5065#comment-427466</guid>
		<description>Yes, I&#039;m talking about non-cooperating neighbors, the model for both Wi-Fi and the Google whitespaces model.</description>
		<content:encoded><![CDATA[<p>Yes, I&#8217;m talking about non-cooperating neighbors, the model for both Wi-Fi and the Google whitespaces model.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jason</title>
		<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/comment-page-1/#comment-427465</link>
		<dc:creator>Jason</dc:creator>
		<pubDate>Fri, 21 Nov 2008 23:07:22 +0000</pubDate>
		<guid isPermaLink="false">http://bennett.com/blog/?p=5065#comment-427465</guid>
		<description>&lt;blockquote&gt; On a shared channel, only one frame can be successfully
transmitted at a time, so the length of time my neighbors occupy the
channel directly affects my ability to use the channel.
&lt;/blockquote&gt;

Besides being not true (c.f. any non-TDMA system like CDMA or FDMA)
the implicit model being discussed is non-cooperating users. That is,
other users of the channel are external interferers.

&lt;blockquote&gt;
You donâ€™t
need Shannonâ€™s Law to understand that, itâ€™s simple arithmetic.
&lt;/blockquote&gt;

Like I said in the previous post this has nothing to do with Shannon&#039;s
law (the premise used is the rate-robustness tradeoff which I do not
dispute).</description>
		<content:encoded><![CDATA[<blockquote><p> On a shared channel, only one frame can be successfully<br />
transmitted at a time, so the length of time my neighbors occupy the<br />
channel directly affects my ability to use the channel.
</p></blockquote>
<p>Besides being not true (c.f. any non-TDMA system like CDMA or FDMA)<br />
the implicit model being discussed is non-cooperating users. That is,<br />
other users of the channel are external interferers.</p>
<blockquote><p>
You donâ€™t<br />
need Shannonâ€™s Law to understand that, itâ€™s simple arithmetic.
</p></blockquote>
<p>Like I said in the previous post this has nothing to do with Shannon&#8217;s<br />
law (the premise used is the rate-robustness tradeoff which I do not<br />
dispute).</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Richard Bennett</title>
		<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/comment-page-1/#comment-427460</link>
		<dc:creator>Richard Bennett</dc:creator>
		<pubDate>Fri, 21 Nov 2008 10:53:11 +0000</pubDate>
		<guid isPermaLink="false">http://bennett.com/blog/?p=5065#comment-427460</guid>
		<description>Sorry, Jason, but this is nonsense:

&lt;blockquote&gt;All else being equal the rate at which other users transmit has no impact on the performance on a given userâ€™s communication. &lt;/blockquote&gt;

On a shared channel, only one frame can be successfully transmitted at a time, so the length of time my neighbors occupy the channel directly affects my ability to use the channel. You don&#039;t need Shannon&#039;s Law to understand that, it&#039;s simple arithmetic.</description>
		<content:encoded><![CDATA[<p>Sorry, Jason, but this is nonsense:</p>
<blockquote><p>All else being equal the rate at which other users transmit has no impact on the performance on a given userâ€™s communication. </p></blockquote>
<p>On a shared channel, only one frame can be successfully transmitted at a time, so the length of time my neighbors occupy the channel directly affects my ability to use the channel. You don&#8217;t need Shannon&#8217;s Law to understand that, it&#8217;s simple arithmetic.</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Jason</title>
		<link>http://bennett.com/blog/2008/10/the-trouble-with-white-spaces/comment-page-1/#comment-427459</link>
		<dc:creator>Jason</dc:creator>
		<pubDate>Fri, 21 Nov 2008 02:52:32 +0000</pubDate>
		<guid isPermaLink="false">http://bennett.com/blog/?p=5065#comment-427459</guid>
		<description>&lt;blockquote&gt;
The corollary of Shannonâ€™s Law is that lower data rates require less bandwidth, Jason. Work the problem of low signaling rates in a shared channel and I think youâ€™ll see Brettâ€™s point. If itâ€™s not immediately obvious, consider that many of our popular wireless systems use redundant coding to increase robustness, simply adding or subtracting redundancy at constant modulation.
&lt;/blockquote&gt;

I understand perfectly well that you can trade robustness for rate. I
was disputing that this is a consequence of the channel capacity
theorem. My opinion is that working with capacity and interference is
not informative: rate and BER are meaningful. But this unimportant
in order to make the following point:

All else being equal the rate at which other users transmit has no
impact on the performance on a given user&#039;s communication. The
rate-race-to-the-bottom assertion is not valid.

&lt;blockquote&gt;
Jason, you should read the Part 15 regulations. They absolutely do not say that a device should transmit with no more power than necessary
&lt;/blockquote&gt;

The whitespaces order says:

&lt;blockquote&gt;
All devices must include adaptable power control so
that they use the minimum power necessary to accomplish communications.
&lt;/blockquote&gt;</description>
		<content:encoded><![CDATA[<blockquote><p>
The corollary of Shannonâ€™s Law is that lower data rates require less bandwidth, Jason. Work the problem of low signaling rates in a shared channel and I think youâ€™ll see Brettâ€™s point. If itâ€™s not immediately obvious, consider that many of our popular wireless systems use redundant coding to increase robustness, simply adding or subtracting redundancy at constant modulation.
</p></blockquote>
<p>I understand perfectly well that you can trade robustness for rate. I<br />
was disputing that this is a consequence of the channel capacity<br />
theorem. My opinion is that working with capacity and interference is<br />
not informative: rate and BER are meaningful. But this unimportant<br />
in order to make the following point:</p>
<p>All else being equal the rate at which other users transmit has no<br />
impact on the performance on a given user&#8217;s communication. The<br />
rate-race-to-the-bottom assertion is not valid.</p>
<blockquote><p>
Jason, you should read the Part 15 regulations. They absolutely do not say that a device should transmit with no more power than necessary
</p></blockquote>
<p>The whitespaces order says:</p>
<blockquote><p>
All devices must include adaptable power control so<br />
that they use the minimum power necessary to accomplish communications.
</p></blockquote>
]]></content:encoded>
	</item>
</channel>
</rss>

<!-- Dynamic Page Served (once) in 0.284 seconds -->
