I wan to find the branch currents using mathematical expression

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Zeeshan
Zeeshan 2024-9-6
评论： Sam Chak 2024-9-8
Hello
from the given differential equation i want to find the brach current and branch voltage. My question is how i can solve these first order differential equations. if anyone who work on the MMC fault current calculation can help in this regards
  2 个评论
Zeeshan
Zeeshan 2024-9-6
This is the paper i am trying to implement
"Y. Yuan, J. Li, P. Lyu, Z. Qian, Y. Jiang and J. Wang, "Modified Fault Current Calculation Scheme Combined With MMC Control for AC/DC Distribution Networks," in IEEE Access, vol. 12, pp. 106822-106831, 2024, doi: 10.1109/ACCESS.2024.3436667."
Star Strider
Star Strider 2024-9-6
I would use the Symbolic Math Toolbox.
The easiest way would likely be to convert theme to Laplace space using the laplace function, solve them in Laplace space for your variables of interest, then if necesary use ilaplace to transform them back to time domaiin functions.

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Sam Chak
Sam Chak 2024-9-6
Hi @Zeeshan
The system appears to be linear; however, it is a 9th-order system (see Abel's Impossibility Theorem). Therefore, you may not be able to obtain a complete analytical solution in terms of the coefficients {, , } of the ordinary differential equations. I suggest using a numerical method to obtain the solutions. The code snippet is provided below:
tstart = ...; % begin of simulation time
tfinal = ...; % end of simulation time
tspan = [tstart, tfinal];
x0 = [..., ..., ..., ..., ..., ..., ..., ..., ...]; % initial values
[t, x] = ode45(@RLC_circuit, tspan, x0);
%% Plot result
plot(t, x)
%% RLC
function dxdt = RLC_circuit(t, x)
%% declarations
i12 = x(1);
i25 = x(2);
i30 = x(3);
i34 = x(4);
i50 = x(5);
u1 = x(6);
u2 = x(7);
u3 = x(8);
u4 = x(9);
%% defined parameters
R12 = ...;
Rc1 = ...;
Rc2 = ...;
...
L12 = ...;
Lc1 = ...;
Lc2 = ...;
...
C1 = ...;
C2 = ...;
C3 = ...;
C4 = ...;
%% designs
is1 = ...;
is2 = ...;
is3 = ...;
is4 = ...;
%% derived functions on the right-hand side
f1 = ;
f2 = ;
f3 = ;
f4 = ;
f5 = ;
f6 = ;
f7 = ;
f8 = ;
f9 = ;
%% differential equations
di12 = f1;
di25 = f2;
di30 = f3;
di34 = f4;
di50 = f5;
du1 = f6;
du2 = f7;
du3 = f8;
du4 = f9;
%% dxdt
dxdt = [di12
di25
di30
di34
di50
du1
du2
du3
du4];
end
  2 个评论
Zeeshan
Zeeshan 2024-9-8
Thank you for your answer
i try to use this formate but the current values are exceeding. Secondlay, this is a 4 terminal MTDC network equation and one of the converter is voltage control. so how can we add the source current effect in the fault current
Sam Chak
Sam Chak 2024-9-8
Hi @Zeeshan
I am unfamiliar with the four-terminal MTDC network, converter, source current effects, and fault current. Because Equations (28)–(36) are ordinary differential equations, it was natural for me to suggest that you place all parameters and ODEs within a single function (RLC_circuit(t, x)) and then solve it using the ode45() solver with the known initial values.
If you encounter any difficulties running the simulation, please post the full code here by clicking the indentation icon .

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