Calculate dV/dQ to plot differential voltage analysis curve?

Question

ekagra gupta 2022-6-3

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1732985-calculate-dv-dq-to-plot-differential-voltage-analysis-curve

回答： Alberto Cuadra Lara 2022-6-3

I have all the required data but do not know how to differentiate my voltage with respect to discharge/charge capacity given.

5 个评论
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Alberto Cuadra Lara 2022-6-3

编辑：Alberto Cuadra Lara 2022-6-3

在 MATLAB Online 中打开

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');

% SUB-PASS FUNCTION

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);

end

dxdy(end) = ((x(end) - x(end-1)) ./ h(end));

end

John D'Errico 2022-6-3

@Alberto Cuadra Lara Comments cannot be accepted as answers. If you want credit for an answer, then you need to post it as an answer.

Alberto Cuadra Lara 2022-6-3

My mistake, I thought I posted it as an answer. Thanks @John D'Errico!

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Answer 1

Alberto Cuadra Lara 2022-6-3

1
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1732985-calculate-dv-dq-to-plot-differential-voltage-analysis-curve#answer_977825

在 MATLAB Online 中打开

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');

% SUB-PASS FUNCTION

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);

end

% 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));

end