Smooth decay of a function
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I have a function which dies of quickely. I need this function to smoothly decay off to 0 . Is there any way in Matlab to make that happen?
4 Commenti
Alan Stevens
il 26 Lug 2020
What's the function?
Saurav
il 26 Lug 2020
Image Analyst
il 26 Lug 2020
Please give values for everything in that alphabet soup. And include a screenshot of what you got, and what you'd like to achieve, but only after you read this link.
T = linspace(-5, 5, 500);
Tc = 3;
y = 2;
a = 7;
b = 6;
m = ((-b + (((b*b)-(4*y*a*(T-Tc))).^0.5))/(2*y)).^0.5;
subplot(2, 1, 1);
plot(T, real(m), 'b-', 'LineWidth', 2);
title('Real(m) vs. T', 'FontSize', 20);
grid on;
xlabel('T', 'FontSize', 20);
ylabel('real(m)', 'FontSize', 20);
% For T>Tc, m(T) becomes complex so m(T>Tc)=0
% Hence I want m(T) to smoothly decay to 0 for T>Tc
subplot(2, 1, 2);
plot(T, imag(m), 'b-', 'LineWidth', 2);
title('Imag(m) vs. T', 'FontSize', 20);
grid on;
xlabel('T', 'FontSize', 20);
ylabel('imag(m)', 'FontSize', 20);

For the random values I chose, it does look like the curve approaches the Tc point smoothly. What does it look like for you, with your values???
Saurav
il 27 Lug 2020
Risposte (1)
Alan Stevens
il 26 Lug 2020
Modificato: Alan Stevens
il 26 Lug 2020
How about writing a function like:
function mfn = mvalue(T,Tc,a,b,y)
disc = b^2 - 4*y*a*(T - Tc);
quad = -b + disc^0.5/(2*y);
if disc>0 && quad>0
mfn = quad^0.5;
else
mfn = 0;
end
end
and then calling
m(i) = mvalue(T(i), Tc, a, b, y);
5 Commenti
Saurav
il 26 Lug 2020
Alan Stevens
il 26 Lug 2020
The value of m could be complex before T reaches Tc (if quad is negative). The mvalue function simply returns zero whenever it would otherwise have returned a compex number.
Saurav
il 26 Lug 2020
Alan Stevens
il 26 Lug 2020
Modificato: Alan Stevens
il 26 Lug 2020
Impossible to tell without knowing about the system you are modelling. Mathematically you could probably multiply m by some sort of exponential decay term. However, would that make sense in the context of your model?
Saurav
il 27 Lug 2020
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