How to solve the following equations for Vc(t)?

How can I solve the following problem? The capacitor voltage.

Vc(t) = 1/c * Int (Ic(t))dt + Vc(0)

function [Ic,Vc] = fcn(S1,S2,S3,S4,Is1)

Ic = (S1*S4+S2*S3)*Is1; %capacitor current - it´s working

Vc= 1/0.01*int(Ic,t) + 2000;

I have the capacitor current flowing, it will depend on which switch is on. So I will have my chain as a function of time. But to know the value of the capacitor voltage in each time it is necessary to integrate the capacitor current as a function of time

Answer 1

[Alan Stevens]

Can you do it using the differential form of the relationship between Vc and Ic? Something like the following (obviously, you would need to replace my arbitrary data/functions with your own):

% dVc/dt = (1/c)Ic  with  Vc(0)
Is1 = @(t) exp(-t);

tspan = [0 3];
Vc0 = 2000;
[t, Vc] = ode45(@rate,tspan, Vc0,[],Is1);

plot(t,Vc)
xlabel('time'),ylabel('Vc')

function dVcdt = rate(t,~,Is1)
        c = 0.01;
        dVcdt = Sfn(t).*Is1(t)/c;
end

function S = Sfn(t)
         if t<0.5
             S1 = 1; S2 = 0; S3 = 0; S4 = 1;
         elseif t < 1
             S1 = 0; S2 = 0; S3 = 0; S4 = 0;
         else
             S1 = 0; S2 = 1; S3 = 1; S4 = 0;
         end
         S = S1*S4 + S2*S3;
end

[Yelena]

Thanks! But than, in the end, I´ll have my current, and not the voltage as a function of the current in the capacitor, like this equation below:

where the current i = (S1*S4+S2*S3)*Is1 in function time.

[Alan Stevens]

It's a trivial addition to plot Vc against Ic

Is1 = @(t) 0.01*exp(-t);

tspan = [0 3];
Vc0 = 2000;
[t, Vc] = ode45(@rate,tspan, Vc0,[],Is1);

c = 0.01;
Ic = Sfn(t).*Is1(t)/c;
plot(Ic,Vc)
xlabel('Ic'),ylabel('Vc')

function dVcdt = rate(t,~,Is1)
        c = 0.01;
        dVcdt = Sfn(t).*Is1(t)/c;
end

function S = Sfn(t)
         if t<0.5
             S1 = 1; S2 = 0; S3 = 0; S4 = 1;
         elseif t < 1
             S1 = 0; S2 = 0; S3 = 0; S4 = 0;
         else
             S1 = 0; S2 = 1; S3 = 1; S4 = 0;
         end
         S = S1*S4 + S2*S3;
end

[Yelena]

Thank you!

I was thinking to do something like this:

function [Ic, Vc] = fcn(S1,S2,S3,S4,Is1,k) %k=sampling time
Ic = (S1*S4+S2*S3)*Is1;         %every time that S1S4==1 ans S2S3==1 will have current 
Ic(k)= Ic*((k+1)-(k));          %current value for each time
V0 = 2000                       %Initial value
t= [0 k];                       %Max and Min value
V(k) = Integral(Ic(k),t);  
C=0.01;
Vc(k) = 1/C*V(k)+V0;
V0=Vc(k);
end

[Alan Stevens]

Does it work for you? (Note - you will need integer values of k doing it that way).

[Yelena]

No, but k is my time so it´ll be integer values. Or not?

[Alan Stevens]

A little more like this, perhaps:

Is1 = @(t) exp(-t);
Icfn = @(t) Sfn(t).*Is1(t);
N = 100;                        % Nbr of timesteps
T = 3;                          % End time
C = 0.01;                   
V0 = 2000;
Vc = zeros(1,N); Vc(1) = V0;
Ic = zeros(1,N); Ic(1) = Is1(0);
dt = T/N;
t = 0:dt:T-dt;

for i = 2:numel(t)
    Ic(i) = Icfn(t(i));
    Vc(i) = integral(Icfn,t(i-1),t(i))*dt/C +Vc(i-1);
end

subplot(3,1,1)
plot(t,Ic),ylabel('Ic')
xlabel('time'),ylabel('Ic')
subplot(3,1,2)
plot(t,Vc)
xlabel('time'),ylabel('Vc')
subplot(3,1,3)
plot(Ic,Vc)
xlabel('Ic'),ylabel('Vc')

function S = Sfn(t)
         if t<0.5
             S1 = 1; S2 = 0; S3 = 0; S4 = 1;
         elseif t < 1
             S1 = 0; S2 = 0; S3 = 0; S4 = 0;
         else
             S1 = 0; S2 = 1; S3 = 1; S4 = 0;
         end
         S = S1*S4 + S2*S3;
end