How to plot This graph in matlab?

x=[1.0:0.1:3.0];
%Composition

%Outlet Specification [PFAD(4)]
y=-2.03^9-3*x^2)+5.76^(-3x)+1;
z=-1.11^-2*x^2+6.49^-2*x+9.99;

plot(x,y,'-b',x,z,'-r')
title('Distillate composition vs time');
xlabel('time'); 
ylabel('xd'); 
legend('a) FFA','b)TAG');

Answer 1

[DGM]

1.00E-2 is shorthand for scientific notation 1.00*10^-2.

% define x
x = 1.0:0.05:3.0;

% YData as row vectors in a matrix
y = zeros(3,numel(x)); % preallocate
y(1,:) = -2.03E-3*x.^2 + 5.76E-3*x + 1E2;
y(2,:) = -1.11E-2*x.^2 + 6.49E-2*x + 9.99E1;
y(3,:) = -2.73E-2*x.^3 + 1.05E-1*x.^2 - 1.42E-1*x + 1E2;

% plot all y vs x
hp = plot(x,y);

% set colors using short names or direct tuples
% set line and marker styles as needed
linecolors = {'k',[0.8 0.5 0],[0.8 0.8 0]};
for k = 1:3
    hp(k).Color = linecolors{k};
    hp(k).Marker = 'o';
    hp(k).MarkerFaceColor = linecolors{k};
    hp(k).MarkerSize = 3;
end

title('Distillate composition vs time');
xlabel('time'); 
ylabel('xd'); 
legend(hp,{'a) FFA','b) TAG','c) something'});

If you want to add equations to the plot, you can use annotation()

Why aren't the curves in the same location? The equations in the annotations contain constants which are rounded. It's also likely that the curves are measured data and the equations are a polynomial fit, so they might not actually be identical.

[Itqan Ismail]

Thank you however I couldnt run the coding above. It shown 'Unable to perform assignment because the size of the left side is 1-by-21 and the size of the right side is 1-by-41.'

[DGM]

You've probably defined some variable called y in some other part of the code.

I've edited the example to explicitly preallocate y. This should ensure that the assignment works, but it'll overwrite anything else called y.

I can't know what you wrote, so you'll have to figure out how things are named and sort that out.

[DGM]

To try to better approximate what the graph shows while staying within the rounded constants given:

% define x
x = 1.0:0.05:3.0;

% YData as row vectors in a matrix
y = zeros(3,numel(x)); % preallocate
% using polyval() is the same as writing a polynomial with the specified coefficients
p1 = [-2.025E-3 5.755E-3 99.996];
p2 = [-1.115E-2 6.495E-2 99.883];
p3 = [-2.728E-2 1.051E-1 -1.423E-1 100.066];
y(1,:) = polyval(p1,x);
y(2,:) = polyval(p2,x);
y(3,:) = polyval(p3,x);

% plot all y vs x
hp = plot(x,y);

% set colors using short names or direct tuples
% set linestyles as needed
linecolors = {'k',[0.8 0.5 0],[0.8 0.8 0]};
for k = 1:3
    hp(k).Color = linecolors{k};
    hp(k).Marker = 'o';
    hp(k).MarkerFaceColor = linecolors{k};
    hp(k).MarkerSize = 3;
end

title('Distillate composition vs time');
xlabel('time'); 
ylabel('xd'); 
legend(hp,{'a) FFA','b) TAG','c) something'});

Otherwise, you'll have to find a better source of information than a picture.

[DGM]

... or if we assume the plotted data are measured values and not the polynomial fit, we could approximate the measured values instead of trying to refine the fit.

% define x
x = 1.0:0.05:3.0;

% YData as row vectors in a matrix
S = load('y.mat');

% plot all y vs x
hp = plot(x,S.y);

% set colors using short names or direct tuples
% set line and marker styles as needed
linecolors = {'k',[0.8 0.5 0],[0.8 0.8 0]};
for k = 1:3
    hp(k).Color = linecolors{k};
    hp(k).Marker = 'o';
    hp(k).MarkerFaceColor = linecolors{k};
    hp(k).MarkerSize = 3;
end

title('Distillate composition vs time');
xlabel('time'); 
ylabel('xd'); 
legend(hp,{'a) FFA','b) TAG','c) something'});

Note that this is a better approximation of what the image itself shows.

Answer 2

[KSSV]

I have shown one plot for you reference. You may extend the same to others.

x=1.0:0.1:3.0;
y=-2.03*10^-3*x.^2+5.76*10^-3*x+1*10^2 ; 

plot(x,y,'-b')
title('Distillate composition vs time');
xlabel('time'); 
ylabel('xd'); 
legend('a) FFA');

[Itqan Ismail]

Thank you for your response