how can i plot this function?
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FRANCESCA CANALE
am 19 Jun. 2022
Kommentiert: Walter Roberson
am 21 Jun. 2022
I have to plot this formula:

on matlab:
%Meyer-Peter Muller
taus=0.01:1;
tausc1=0.047;
fi1=8*(taus-tausc1).^(3/2);
and I have write like this:
subplot (2,2,1) %Meyer-Peter Muller
hold on
plot(taus,fi1,'LineWidth',3)
grid on
axis([0.01 1 0.001 10])
set(gca,'XScale','log');
set(gca,'YScale','log');
xlabel('\tau_*')
ylabel('\phi_o')
but there is this error:
Warning: Imaginary parts of complex X and/or Y arguments ignored.
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Akzeptierte Antwort
Sam Chak
am 20 Jun. 2022
Because that is not an ordinary plot, but a plot using base-10 logarithmic scale on the x-axis and the y-axis. I have made changes in 3 lines. See comments.
% Meyer-Peter Muller
taus = linspace(0.048, 10, 4977); % ← Changes in this line
tausc1=0.047;
fi1=8*(taus-tausc1).^(3/2);
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Figures
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% subplot (2,2,1) % Meyer-Peter Muller
% hold off
loglog(taus,fi1,'LineWidth',3) % ← Changes in this line
grid on
axis([1e-2 1e0 1e-4 1e1]) % ← Changes in this line
set(gca,'XScale','log');
set(gca,'YScale','log');
xlabel('\tau_*')
ylabel('\phi_o')
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Weitere Antworten (1)
Walter Roberson
am 19 Jun. 2022
0.01 minus 0.047 is negative. negative raised to 3/2 is complex. Your x are non-negative real but your fi1 are a mix of real and complex values.
You also ask for log scale for both axes. log of a complex value would be complex, so it is not obvious what you want plotted.
0.01:1 is the single value 0.01. Remember that the default increment is 1
What do you expect a log plot of complex values to look like?
2 Kommentare
Walter Roberson
am 21 Jun. 2022
I plotted the formula as given. I also used the general form of the formula, and read off some readings from the graph, and solve simultaneous equations to get approximate solutions.
Notice that the mulitiplier found through reading from the graph is fairly close to the 4 you already have, and that the exponent found through approximation is fairly close to the 3/2 you already have. The value supportraced has a notable difference though.
I had to choose representative values from the graph; it is possible that I did not pick consistent values.
syms a b c x
f(x,a,b,c) = a * (x-b)^c
eqn1 = f(10^-1.75, a, b, c) == 10^-4
eqn2 = f(10^0, a, b, c) == 10^0.6
eqn3 = f(10^-1, a, b, c) == 10^-1
sol = solve([eqn1, eqn2, eqn3], a, b, c)
f0 = f(x, 4, 0.047, 3/2)
f1 = subs(f(x,a,b,c), sol)
vpa(f1, 10)
fplot([f0, f1], [0.05, 1])
set(gca, 'XScale', 'log', 'YScale', 'log')
legend({'Meyer-Peter Muller', 'Approximated from graph'})
xlim([10^-2, 10^0])
ylim([10^-4, 10^1])
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