Calculate dV/dQ to plot differential voltage analysis curve?

69 views (last 30 days)
I have all the required data but do not know how to differentiate my voltage with respect to discharge/charge capacity given.

Sign in to comment.

Accepted Answer

Alberto Cuadra Lara
Alberto Cuadra Lara on 3 Jun 2022
Hello Ekagra,
I guess you have numerical values of the voltage as a function of the discharge/charge capacity, right? In this case, you can compute the first derivative numerically, e.g., using finite central differences as follows, where x and y will be Q and V, respectively.
% Definitions
x = linspace(0, 2*pi);
y = sin(x);
% Compute first derivative
dydx = compute_first_derivative(y, x);
% Plot
figure; hold on;
plot(x, y);
plot(x(2:end), dydx);
xlabel('x', 'interpreter', 'latex')
ylabel('y', 'interpreter', 'latex')
legend({'y(x)', 'dy(x)/dx'}, 'interpreter', 'latex', 'location', 'northeastoutside');
function dxdy = compute_first_derivative(x, y)
% Compute first central derivate using a non-uniform grid
% Args:
% x (float): Values for the corresponding grid
% y (float): Grid values
% Returns:
% dxdy (float): Value of the first derivate for the given grid and its corresponding values
% Author: Alberto Cuadra-Lara
h = y(2:end) - y(1:end-1);
hmax = max(h);
mu = h / hmax;
dxdy = zeros(1, length(h));
dxdy(1) = ((x(2) - x(1)) ./ h(1));
for i = 2:length(mu)-1
dxdy(i) = (mu(i)^2 * x(i+1) - (mu(i)^2 - mu(i+1)^2) * x(i) - mu(i+1)^2 * x(i-1)) / ((mu(i)^2 * mu(i+1) + mu(i) * mu(i+1)^2) * hmax);
% Direct method
% dxdy(2:end-1) = (mu(2:end-1).^2 .* x(3:end-1) - (mu(2:end-1).^2 - mu(3:end).^2) .* x(2:end-2) - mu(3:end).^2 .* x(1:end-3)) ./ ((mu(2:end-1).^2 .* mu(3:end) + mu(2:end-1) .* mu(3:end).^2) * hmax);
dxdy(end) = ((x(end) - x(end-1)) ./ h(end));

More Answers (0)


Find more on Creating and Concatenating Matrices in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by