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How to find the maximum value

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sophp
sophp am 6 Feb. 2018
Kommentiert: Jan am 7 Feb. 2018
Below is a 3D graph.
T=600:1:850;
t=0.2:0.1:20;
[tm,Tm] = meshgrid(t, T);
k1 = 10^7.*exp(-12700./Tm);
k2 = 5*10^4.*exp(-10800./Tm);
k3 = 7*10^7.*exp(-15000./Tm);
Y_B = (k1.*tm)./(((k2.*tm)+1).*(1+(tm.*(k1+k3))));
mesh(t,T,Y_B)
view(40, 45)
grid on
How do I find the maximum value of Y_B for t=0.2:0.1:20 over the range of T from 600 to 850 and plots on a double y-axis graph the value of T and Y_Bmax with t?

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Jan
Jan am 6 Feb. 2018
Bearbeitet: Jan am 6 Feb. 2018
t = 0.2:0.1:20;
T = 600:50:850;
for i = 1:numel(T)
k1 = 1e7 .* exp(-12700 ./ T(i));
k2 = 5e4 .* exp(-10800 ./ T(i));
k3 = 7e7 .* exp(-15000 ./ T(i));
for j = 1:numel(t)
Y_B(i,j) = (k1.*t(j)) ./ (((k2.*t(j))+1).*(1+(t(j).*(k1+k3))));
% Prefer:
% Y_B(i,j) = k1 .* t(j) ./ ((k2 .* t(j) + 1) .* (1 + t(j) .* (k1 + k3)));
end
end
plot(t, max(Y_B, [], 1), '.-');
k1, k2, k3 are not changed inside the inner loop, so move them to the outer one. I would not overdo it with the parentheses.
By the way, you can even omit the loops:
t = 0.2:0.1:20;
T = (600:50:850).';
k1 = 1e7 .* exp(-12700 ./ T);
k2 = 5e4 .* exp(-10800 ./ T);
k3 = 7e7 .* exp(-15000 ./ T);
Y_B = k1 .* t ./ ((k2 .* t + 1) .* (1 + t .* (k1 + k3))); % >= R2016
plot(t, max(Y_B, [], 1), '.-');
This uses auto expanding, which was introduced in Matlab R2016b. Your meshgrid approach was fine also, while the loops suggested by Birdman have no advantage here.
  2 Kommentare
sophp
sophp am 7 Feb. 2018
HI Jan thank you for this, it works perfectly! I am trying to plot a graph to show the relationship between residence time t and the operating temperature T(y_B max). However, one is row vector where as the other is a column vector. HOw do I do this?
Jan
Jan am 7 Feb. 2018
@sophp: By transposing one of the two vectors?

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Weitere Antworten (1)

Rik
Rik am 6 Feb. 2018
You can specify the dimension max should operate on, so you can insert your matrix. See the documentation for instructions.
For a double y-axis, see yyaxis, or search the File Exchange for one of many examples.
  1 Kommentar
sophp
sophp am 6 Feb. 2018
Hi Rik,
I should have been more specific with my question. But, I am new to MATLAB and cannot find anything that tells me to find the maximum for each loop of i, form an array and plot a graph. This is the code I have
t = [0.2:0.1:20];
T = [600:50:850];
hold on;
for i=1:numel(T)
for j=1:numel(t)
k1(i) = 1e7 .* exp(-12700 ./ T(i));
k2(i) = 5e4 .* exp(-10800 ./ T(i));
k3(i) = 7e7 .* exp(-15000 ./ T(i));
Y_B(i,j) = (k1(i).*t(j))./(((k2(i).*t(j))+1).*(1+(t(j).*(k1(i)+k3(i)))));
end
[val,idt] = max(Y_B(i,j))
end
hold on
plot(t(idt),Y_B(idt))
However the correct graph does not appear.

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