How to plot This graph in matlab?
显示 更早的评论
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');
采纳的回答
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'});
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.
4 个评论
Itqan Ismail
2023-2-1
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.
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.
... 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.
更多回答（1 个）
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');
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