# Finding gradient of a part of a graph

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Yen Tien Yap le 9 Avr 2022
Modifié(e) : Tala le 9 Avr 2022
May I know how to find the gradient of the first linear part? What function should I use?
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### Réponse acceptée

Sam Chak le 9 Avr 2022
Modifié(e) : Sam Chak le 9 Avr 2022
Technically, if is plotted by a function f(x) with a uniform step size h, then you can use the nabla = gradient(f)/h to compute the slope of f(x).
To find the gradient of the transient response, you need to pick a point in that region, for example, , and then find the index idx that is nearest to, or exactly at this point.
You can try @Faraz Hedayati's code or this code. Both are good learning experiences for you
h = 0.01;
x = 0:h:10;
y = 1 - exp(-x/sqrt(2)).*(cos(x/sqrt(2)) + sin(x/sqrt(2)));
plot(x, y)
hold on
[M, idx] = max(nabla);
plot(x(idx), y(idx), 'o', 'linewidth', 1.5)
m = nabla(idx) % slope at point p
c = y(idx) - m*x(idx) % y-intercept
z = m*x + c; % line equation at point p
plot(x, z, 'linewidth', 1.5)
hold off
grid on
xlabel('x')
ylabel('y')
title('y = f(x) and the tangent line at the steepest slope')
legend('function f(x)', 'the point at f(p)', 'tangent line', 'location', 'best')
Result:
Note: If the spread of the data points are not uniform, then you probably need to use the interpolation technique, interp1().
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### Plus de réponses (1)

Tala le 9 Avr 2022
Modifié(e) : Tala le 9 Avr 2022
A=your signal; dx=epsilon(2)-epsilon(1); % assuming the sampling frequency is constant
[M,I] = max(A);
B=diff(A(1:I))./dx;
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