使用Matlab在VB中使用Laplace变压器 [英] Laplace transformer in VB using matlab

问题描述

你好:

我想使用MatLab制造的dll在vb中计算Laplace变压器，但是在vb中运行它会出现自动化错误.该功能在Matlab中可以正常工作.

是:

在matlab中

hello:

I want to calculate Laplace transformer in vb using a dll made in MatLab, but to run it in vb I get an automation error. The function works fine in Matlab.

is:

in matlab

function transf = lap(f)

syms x s

transf=laplace(f,x,s);

end

现在在VB中

Now in VB

Dim lapla As New Laplace.Transformada
Dim resultado As variant


Call lapla.lap(1, resultado, 3 * x)

请帮助.. !!!
我快疯了！

坦克. :confused:

plase help..!!!
I''m going crazy!

tanks. :confused:

推荐答案

using System;
using System.Collections.Generic;
using System.Text;
namespace ComportMath
{
    class Laplace
    {
        public delegate double FunctionDelegate(double t);
        static double[] V;       //  Stehfest coefficients
        static double ln2;       //  log of 2
        const int DefaultStehfestN = 14;
        static Laplace()
        {
            InitStehfest(DefaultStehfestN);
        }
        public static double Transform(FunctionDelegate F, double s)
        {
            const int DefaultIntegralN = 5000;
            double du = 0.5 / (double)DefaultIntegralN;
            double y =  - F(0) / 2.0;
            double u = 0;
            double limit = 1.0 - 1e-10;
            while (u < limit)
            {
                u += du;
                y += 2.0 * Math.Pow(u, s - 1) * F(-Math.Log(u));
                u += du;
                y += Math.Pow(u, s - 1) * F(-Math.Log(u));
            }
            return 2.0 * y * du / 3.0;
        }
        public static double InverseTransform(FunctionDelegate f, double t)
        {
            double ln2t = ln2 / t;
            double x = 0;
            double y = 0;
            for (int i = 0; i < V.Length; i++)
            {
                x += ln2t;
                y += V[i] * f(x);
            }
            return ln2t * y;
        }
        public static double Factorial(int N)
        {
            double x = 1;
            if (N > 1)
            {
                for (int i = 2; i <= N; i++)
                    x = i * x;
            }
            return x;
        }

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