Calculate dV/dQ to plot differential voltage analysis curve?

51 ビュー (過去 30 日間)
ekagra gupta
ekagra gupta 2022 年 6 月 3 日
回答済み: Alberto Cuadra Lara 2022 年 6 月 3 日
I have all the required data but do not know how to differentiate my voltage with respect to discharge/charge capacity given.
  5 件のコメント
Alberto Cuadra Lara
Alberto Cuadra Lara 2022 年 6 月 3 日
My mistake, I thought I posted it as an answer. Thanks @John D'Errico!



Alberto Cuadra Lara
Alberto Cuadra Lara 2022 年 6 月 3 日
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));

その他の回答 (0 件)


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