- You should have log(sigma)^2, not log(sigma^2).
 - Don't forget the "dx" part when integrating the curve.
 
Discretising a size distribution function and area under the curve
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I have some parameters for some data and I am plotting a size distribution function. The probability function I am using is given below:

Currently I have the following code which plots the function:
clear
clc
close all
mu = 0.015;  % geometric mean radius
sigma = 1.6; % geometric standard deviation
Ntot = 850;  % total number concentration
nbins = 200; % number of bins
rMin = (mu/(10*sigma)); % determine the minimum radius
rMax = (mu*10*sigma);   % determine the maximum radius
rs = logspace(log10(rMin),log10(rMax),nbins+1); % vector containing rs
for i = 1:length(rs)
    sizedist = Ntot/((sqrt(2*pi))*log(sigma))*exp(-(log(rs./mu).^2)./(2*log(sigma^2)));
end
% plot the size distribution function 
semilogx(rs,sizedist,'Linewidth',2);
xlim([10e-4 10e1]);
xlabel('Particle Radius, \mum');
ylabel('Number Concentration, cm^3');
 The first problem I have occurs here. Shouldn't the area under the curve be equal to the total number concentration specified? As this doesn't seem to be the case.
For the model I am coding it is necessary for the aerosol population to be discretised into bins. 

Now suppose I wish to discretise this function into bins, currently the vector rs is the bin edges. I would like to know how to find the number concentration in each bin, but as stated above the area under the curve doesn't seem to equal Ntot. Can anyone see whether i'm doing anything wrong? 
What I would like to do is to determine the Number Concentration in each bin, which when added together should total the value of Ntot.
Hopefully someone can help!!! 
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Antworten (2)
  Alan Stevens
      
      
 am 26 Feb. 2021
        mu = 0.015;  % geometric mean radius
sigma = 1.6; % geometric standard deviation
Ntot = 850;  % total number concentration
nbins = 200; % number of bins
rMin = (mu/(10*sigma)); % determine the minimum radius
rMax = (mu*10*sigma);   % determine the maximum radius
rs = logspace(log10(rMin),log10(rMax),nbins+1); % vector containing rs
dx = (log(rMax)-log(rMin))/nbins;
sizedist = Ntot/(sqrt(2*pi)*log(sigma))*exp(-log(rs./mu).^2./(2*log(sigma)^2));
disp(sum(sizedist)*dx)   % This should display the area under the curve
% plot the size distribution function 
semilogx(rs,sizedist,'Linewidth',2);
xlim([10e-4 10e1]);
xlabel('Particle Radius, \mum');
ylabel('Number Concentration, cm^3');
4 Kommentare
  Alan Stevens
      
      
 am 27 Feb. 2021
				
      Bearbeitet: Alan Stevens
      
      
 am 27 Feb. 2021
  
			You could do the following
mu = 0.015;  % geometric mean radius
sigma = 1.6; % geometric standard deviation
Ntot = 850;  % total number concentration
nbins = 200; % number of bins
rMin = (mu/(10*sigma)); % determine the minimum radius
rMax = (mu*10*sigma);   % determine the maximum radius
rs = logspace(log10(rMin),log10(rMax),nbins+1); % vector containing rs
dx = (log(rMax)-log(rMin))/nbins;
sizedist = Ntot/(sqrt(2*pi)*log(sigma))*exp(-log(rs./mu).^2./(2*log(sigma)^2));
disp(sum(sizedist)*dx)   % This should display the area under the curve
nbins2 = 21;
rs2 = logspace(log10(rMin),log10(rMax),nbins2+1); % vector containing rs2
sizedist2 = Ntot/(sqrt(2*pi)*log(sigma))*exp(-log(rs2./mu).^2./(2*log(sigma)^2));
x = []; y = []; bw = rs2(1)/2;
for i = 1:nbins2
   x = [x rs2(i) rs2(i) rs2(i+1) rs2(i+1)];
   yav = (sizedist2(i+1)+sizedist2(i))/2;
   y = [y 0 yav yav 0];
end
% plot the size distribution function 
semilogx(rs,sizedist,'Linewidth',2);
hold on
semilogx(x,y)
xlabel('Particle Radius, \mum');
ylabel('Number Concentration, cm^3');
which produces

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