Polyfit on semilogy. Straight line equations and intercept point

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Emanuele Sortino
Emanuele Sortino 2017-1-17
评论： Emanuele Sortino 2017-1-18
Hi all. I'm trying to fit this two lines on matlab. I take the last 30 second of my experiment, the sampling is 2 data per second so i take the last 60 points. 40 before the peak and 20 after it.
I'd like to do a linear regression on the first 20 datas and then make a straight line. (Blue line) But using polyfit I've a curve on it. How can I make a straight line in a semilogy? I can't understand the equation to use in this case. I don't know where to put the 10 or the e. Also I'd like to consider the tangent from the peak to the second before. But if i'm able to do it for two point i can't elongate the red line to the blue line. Then i'd like to find the intercept of this two line. How can I do it mathematically? Anyone can help? this is the code i'm using
time=all_data(:,1);
voltage=all_data(:,3);
current=all_data(:,4);
power=current.*voltage/(0.7*2*4);
logpower=log(power);
maxt=find(power==max(power));
timebef=time(maxt-40:maxt+20);
powerbef=power(maxt-40:maxt+20);
close all
fig=figure(j);
set(fig,'units','normalized','outerposition',[0 0 1 1])
semilogy(time,power,'LineWidth',2,'Color',[0 0 0]);
xlim([min(timebef),max(timebef)]);
xlabel('time (s)');
ylabel('Log Power density mW/mm^3');
set(gca,'FontSize',16);
ylim([10,10000]);
ylabel('Pow Den mW/mm^3');
xlabel('time (s)');
title(lista(j,:));
hold on
slope=polyfit(time((maxt-40):(maxt-20)),power((maxt-40):(maxt-20)),1);
Y1=(slope(1).*time((maxt-40):(maxt+20)))+slope(2);
semilogy(time((maxt-40):(maxt+20)),Y1,'LineWidth',2,'Color',[0 0 1])
X2=[time(maxt-2) time(maxt)];
Y2=[power(maxt-2) power(maxt)];
semilogy(X2,Y2,'LineWidth',2,'Color',[1 0 0])

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采纳的回答

David Goodmanson
David Goodmanson 2017-1-17
Hello Emanuele, when you are looking at a semilogy plot you are interested in log(y) vs. x. So you need to go into log-land and do all the fits on the logpower variable you created. The fit
polyfit(t,logpower(t),1)
gives you a linear relationship of the form log(power) = at + b (for a and b you have slope(1) and slope(2) respectively). The actual fit is on the log, but the inverse of this relation is power = exp(at +b), so if for the blue line you do
slope = polyfit(time((maxt-40):(maxt-20)),logpower((maxt-40):(maxt-20)),1);
Y1 = exp( (slope(1).*time((maxt-40):(maxt+20)))+slope(2) );
semilogy(time((maxt-40):(maxt+20)),Y1,'LineWidth',2,'Color',[0 0 1])
then the blue line represents an exponential for the actual power and a straight line on a log plot. Same for the red line, you can do a fit on logpower to find a and b, then exponentiate and expand your x array to make a line that crosses the blue one.
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David Goodmanson
David Goodmanson 2017-1-17
编辑：David Goodmanson 2017-1-17
The idea is to create the line in log-land first, and exponentiate it afterwards. This parallels the blue line code so it should work:
Y1 = exp( PEAK(2) + PEAK(1).*X2 );
Emanuele Sortino
Emanuele Sortino 2017-1-18
with blue line i'm fine. It's the red that give me problem, goes to infinite, but in a semilog scale is not. what I'm missing on it?

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