<?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"
	>
<channel>
	<title>Comments on: ORDERS V TIME - COMPARING OVERALL LEVELS</title>
	<atom:link href="http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/feed/" rel="self" type="application/rss+xml" />
	<link>http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/</link>
	<description>Notes, tutorials, news and articles on digital signal capture, processing, techniques and applications</description>
	<pubDate>Mon, 13 Oct 2008 04:53:20 +0000</pubDate>
	<generator>http://wordpress.org/?v=2.5</generator>
		<item>
		<title>By: Colin Mercer</title>
		<link>http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-390</link>
		<dc:creator>Colin Mercer</dc:creator>
		<pubDate>Thu, 13 Mar 2008 11:31:13 +0000</pubDate>
		<guid isPermaLink="false">http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-390</guid>
		<description>What is an order?   This is a fundamental point I did not include!   An  order is that frequency directly proportional to the rotational speed.  So if we have a rotation speed of R rpm then first order is (R/60)Hz, second order is (2R/60)Hz and so on.  Suppose we have a single cylinder 4 stroke engine  then there is a firing pulse every two revolutions so we would expect excitation to happen every half order.   If the engine had four cylinders  then we would get two firing pulses every revolution So we expect excitation at every second order, that is at orders 2, 4, 6 and so on.

Colin Mercer</description>
		<content:encoded><![CDATA[<p>What is an order?   This is a fundamental point I did not include!   An  order is that frequency directly proportional to the rotational speed.  So if we have a rotation speed of R rpm then first order is (R/60)Hz, second order is (2R/60)Hz and so on.  Suppose we have a single cylinder 4 stroke engine  then there is a firing pulse every two revolutions so we would expect excitation to happen every half order.   If the engine had four cylinders  then we would get two firing pulses every revolution So we expect excitation at every second order, that is at orders 2, 4, 6 and so on.</p>
<p>Colin Mercer</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Muthuraman</title>
		<link>http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-386</link>
		<dc:creator>Muthuraman</dc:creator>
		<pubDate>Tue, 11 Mar 2008 12:41:47 +0000</pubDate>
		<guid isPermaLink="false">http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-386</guid>
		<description>hi,

could you explain me what is order and more on 2nd 4th and 6th order in engine..</description>
		<content:encoded><![CDATA[<p>hi,</p>
<p>could you explain me what is order and more on 2nd 4th and 6th order in engine..</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Colin</title>
		<link>http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-57</link>
		<dc:creator>Colin</dc:creator>
		<pubDate>Mon, 10 Dec 2007 16:54:28 +0000</pubDate>
		<guid isPermaLink="false">http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-57</guid>
		<description>Hi
 There is not usually a problem with differentiating signals in the time domain. What I am often asked about is integration when it is vital to remove any dc offset (eg reduce to zero mean first or use a high pass filter or both!)

In the frequency domain if your displacement signal had a spectrum X(w) then the corresponding velocity spectrum  is given by i*w*X(w) where w = 2*PI*f. That is differentiating is like a high pass filter with a frequency response iw.  As a result if you have a lot of high frequency noise in your displacement signal then this will dominate the calculated velocity. 

I would suggest low pass filtering your displacement signal, but look at the spectrum first. For example suppose you are interested up to 100Hz and have some noise at 1000Hz then this noise will be "magnified" ten times more than the region of interest.  Another thing to think about is the sample rate as sometimes this is too high for the region you are really concerned about!  No matter what differentiation formula is used it will involve dividing by the time between samples. It we have a very small value due to a high sample rate then  that may introduce more error; this is another manifestation of amplifying to higher frequent noise. So sometimes it pays to &lt;a href="http://blog.prosig.com/2001/06/06/data-decimation-what-do-i-do/" rel="nofollow"&gt;decimate the signal&lt;/a&gt; (doing it properly with the necessary filtering step of course).</description>
		<content:encoded><![CDATA[<p>Hi<br />
 There is not usually a problem with differentiating signals in the time domain. What I am often asked about is integration when it is vital to remove any dc offset (eg reduce to zero mean first or use a high pass filter or both!)</p>
<p>In the frequency domain if your displacement signal had a spectrum X(w) then the corresponding velocity spectrum  is given by i*w*X(w) where w = 2*PI*f. That is differentiating is like a high pass filter with a frequency response iw.  As a result if you have a lot of high frequency noise in your displacement signal then this will dominate the calculated velocity. </p>
<p>I would suggest low pass filtering your displacement signal, but look at the spectrum first. For example suppose you are interested up to 100Hz and have some noise at 1000Hz then this noise will be &#8220;magnified&#8221; ten times more than the region of interest.  Another thing to think about is the sample rate as sometimes this is too high for the region you are really concerned about!  No matter what differentiation formula is used it will involve dividing by the time between samples. It we have a very small value due to a high sample rate then  that may introduce more error; this is another manifestation of amplifying to higher frequent noise. So sometimes it pays to <a href="http://blog.prosig.com/2001/06/06/data-decimation-what-do-i-do/" rel="nofollow">decimate the signal</a> (doing it properly with the necessary filtering step of course).</p>
]]></content:encoded>
	</item>
	<item>
		<title>By: Dr ShiPing Yao</title>
		<link>http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-39</link>
		<dc:creator>Dr ShiPing Yao</dc:creator>
		<pubDate>Wed, 28 Nov 2007 10:20:15 +0000</pubDate>
		<guid isPermaLink="false">http://blog.prosig.com/2007/10/23/orders-v-time-comparing-overall-levels/#comment-39</guid>
		<description>Dear Colin,

If I process the displacement time domain data to get velocity and acceleration, what points will I need to pay attention? It seemed I could not get reasonable velocity/acceleration signals by differentiate the time domain signal of displacement by using DATS software from Prosig. Please advise. Thanks.

S. Yao</description>
		<content:encoded><![CDATA[<p>Dear Colin,</p>
<p>If I process the displacement time domain data to get velocity and acceleration, what points will I need to pay attention? It seemed I could not get reasonable velocity/acceleration signals by differentiate the time domain signal of displacement by using DATS software from Prosig. Please advise. Thanks.</p>
<p>S. Yao</p>
]]></content:encoded>
	</item>
</channel>
</rss>
