Calculate dV/dQ to plot differential voltage analysis curve?

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

[Alberto Cuadra Lara]

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

[ekagra gupta]

Thank you for your time. This code works.

[Alberto Cuadra Lara]

Great! if you don't mind, please mark as accepted answer ;)

[John D'Errico]

@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]

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

Answer 1

[Alberto Cuadra Lara]

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