Path: news.mathworks.com!not-for-mail From: <HIDDEN> Newsgroups: comp.soft-sys.matlab Subject: Re: Bacterial specific growth rate calculation. Date: Wed, 14 Mar 2012 08:11:12 +0000 (UTC) Organization: The MathWorks, Inc. Lines: 16 Message-ID: <jjpjr0$qe3$1@newscl01ah.mathworks.com> References: <jjpg8m$gej$1@newscl01ah.mathworks.com> Reply-To: <HIDDEN> NNTP-Posting-Host: www-01-blr.mathworks.com Content-Type: text/plain; charset=UTF-8; format=flowed Content-Transfer-Encoding: 8bit X-Trace: newscl01ah.mathworks.com 1331712672 27075 172.30.248.46 (14 Mar 2012 08:11:12 GMT) X-Complaints-To: news@mathworks.com NNTP-Posting-Date: Wed, 14 Mar 2012 08:11:12 +0000 (UTC) X-Newsreader: MATLAB Central Newsreader 1187260 Xref: news.mathworks.com comp.soft-sys.matlab:760885 "smith" wrote in message <jjpg8m$gej$1@newscl01ah.mathworks.com>... > I need to calculate the bacterial specific growth rate for a batch process i.e I want to calculate the maximum growth rate at exponential growing condition. I have to calculate for a large set of varied cases. > In the plot of log of biomass versus time,I have been taking the mean of slopes at each point (x2-x1)/(t2-t1) from time t=0 to the end of exponential growth phase. This method is not accurate,gives a reduced growth rate. For each case one has to take the mean of the slopes those are at higher range and exclude the lower values of slope. > Is there any other way to get this? - - - - - - - - - - If your time intervals are all equal you could get a second order approximation to the derivative using matlab's 'gradient' function. With unequal intervals you can use the formula (x2-x1)/(t2-t1)*(t3-t2)/(t3-t1) + (x3-x2)/(t3-t2)*(t2-t1)/(t3-t1) as a second order approximation to the derivative at (t2,x2) with (t1,x1) and (t3,x3) as points on either side. Second order approximation here means the slope of a parabola at (t2,x2) which runs through all three points. Both methods assume that your plotted data is relatively noise-free. Roger Stafford