Changing sampling rate using quadratic regression

I want to change sampling rate of a signal from 333Hz to 5000Hz. To get a smoother result, I have implemented second order hold system instead of popular zero order hold (ZOH) system. This approach can also be used for down sampling. Another possible application is to filter out noise without introducing phase delay. And it can be useful for real-time application.

Approach 1: Quadratic regression
Quadratic function is defined as

$$f=w_0+w_1 x + w_2 x^2$$

Then, cost function to be minimized is defined as

$$J(w)=\frac{1}{2}\sum_{i=1}^{n}(y_i-f_i)^2$$
$$J(w)=\frac{1}{2}\sum_{i=1}^{n}(y_i-w_0+w_1 x_i + w_2 x_i^2)^2$$

Optimal weights can be found by differentiating the cost function and setting them to zero.

$$\frac{\partial J(w)}{\partial w_0}=0$$
$$-\sum_{i=1}^{n}(y_i-w_0+w_1 x_i + w_2 x_i^2)=0$$
$$w_0\sum_{i=1}^{n}1+w_1\sum_{i=1}^{n}x_i+w_2\sum_{i=1}^{n}x_i^2=\sum_{i=1}^{n}y_i$$

Similarly, differentiating with w₁ and w₂ gives,

$$w_0\sum_{i=1}^{n}x_i+w_1\sum_{i=1}^{n}x_i^2+w_2\sum_{i=1}^{n}x_i^3=\sum_{i=1}^{n}y_i x_i$$
$$w_0\sum_{i=1}^{n}x_i^2+w_1\sum_{i=1}^{n}x_i^3+w_2\sum_{i=1}^{n}x_i^4=\sum_{i=1}^{n}y_i x_i^2$$

When these three equations are written in matrix form.

$$ \begin{bmatrix} \sum_{i=1}^{n}1 & \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}x_i^2 \\ \sum_{i=1}^{n}x_i & \sum_{i=1}^{n}x_i^2 & \sum_{i=1}^{n}x_i^3 \\ \sum_{i=1}^{n}x_i^2 & \sum_{i=1}^{n}x_i^3 & \sum_{i=1}^{n}x_i^4 \end{bmatrix} \begin{bmatrix} w_0 \\ w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} \sum_{i=1}^{n}y_i \\ \sum_{i=1}^{n}y_i x_i \\ \sum_{i=1}^{n}y_i x_i^2 \end{bmatrix} $$
$$\mathbf{A}\mathbf{W}=\mathbf{B}$$
$$\mathbf{W}=\mathbf{A}^{-1}\mathbf{B}$$

Approach 2: Three equations
In our case, we can use only the last three points and we can get three equations from these three points.

$$y_1=w_0+w_1 x_1 + w_2 x_1^2$$
$$y_2=w_0+w_1 x_2 + w_2 x_2^2$$
$$y_3=w_0+w_1 x_3 + w_2 x_3^2$$
$$ \begin{bmatrix} 1 & x_1 & x_1^2 \\ 1 & x_2 & x_2^2 \\ 1 & x_3 & x_3^2 \end{bmatrix} \begin{bmatrix} w_0 \\ w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} y_1 \\ y_2 \\ y_3 \end{bmatrix} $$
$$\mathbf{A}\mathbf{W}=\mathbf{B}$$
$$\mathbf{W}=\mathbf{A}^{-1}\mathbf{B}$$

Approach 3: Two equations
If we define x₁=-1, x₂=0, and x₃=1, y₂ is equal to w₀. And,

$$y_1=y_2-w_1+ w_2$$
$$y_3=y_2+w_1+ w_2$$
$$ \begin{bmatrix} -1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} y_1-y_2 \\ y_3-y2 \end{bmatrix} $$
$$ \begin{bmatrix} w_1 \\ w_2 \end{bmatrix} = \begin{bmatrix} -0.5 & 0.5 \\ 0.5 & 0.5 \end{bmatrix} \begin{bmatrix} y_1-y_2 \\ y_3-y2 \end{bmatrix} $$

Then, we have

$$w_0=y_2$$
$$w_1=-0.5(y_1-y_2)+0.5(y_3-y2)$$
$$w_2=0.5(y_1-y_2)+0.5(y_3-y2)$$

I have tested the above three approaches using MatLab and it is shown below.

%-------------------------------------------------------------------------
clc;
close all;
clear all;
%-------------------------------------------------------------------------
% y= w0 + w1*x + w2* x^2;
%-------------------------------------------------------------------------
%Got x and y
x=[-1 0 1]';
Wo=[4 3 2]';
y=Wo(1)+Wo(2)*x+Wo(3).*x.*x;

%-------------------------------------------------------------------------
%Approach 1
%Polynomial regression of order 2
%For n=3
S1=3;
Sx=sum(x);
Sx2=sum(x.*x);
Sx3=sum(x.*x.*x);
Sx4=sum(x.*x.*x.*x);
Sy=sum(y);
Syx=sum(y.*x);
Syx2=sum(y.*x.*x);

P=[S1 Sx Sx2; Sx Sx2 Sx3; Sx2 Sx3 Sx4];
B=[Sy Syx Syx2]';
%P1=P^(-1);
W1=P\B

%-------------------------------------------------------------------------
%Approach 2
%Linear equations
A=[1 x(1) x(1)*x(1);1 x(2) x(2)*x(2); 1 x(3) x(3)*x(3)];
W2=A\y
%-------------------------------------------------------------------------
%Approach 3
%Only 2 linear equations
w0=y(2);
w1=-0.5*( y(1)- y(2))+0.5*( y(3)- y(2));
w2=0.5*( y(1)- y(2))+0.5*(y(3)- y(2));
W3=[w0 w1 w2]'
%-------------------------------------------------------------------------

The following figure shows the result of using this method (blue color plot) compare to ordinary zero order hold (black color plot). This method gives much more smoother result but it should be noted that it introduces one sample delay.

The implementation of this method in LabVIEW using C code is shown in the following figure.

The first two approaches involve finding inverse of a 3x3 matrix and I have developed a C program as shown below.

#include
#include
main()
{
    float M[3][3]={{3,0,2},{0,2,0},{2,0,2}}; //initialize a 3x3 matrix
    float N[3][3]={{0,0,0},{0,0,0},{0,0,0}}; //allocate for inverse
    int i,j;
    float d;
    //-------------------------------------------------------------------------
    N[0][0]=(M[1][1]*M[2][2]-M[2][1]*M[1][2]);
    N[1][0]=-(M[1][0]*M[2][2]-M[2][0]*M[1][2]);
    N[2][0]=(M[1][0]*M[2][1]-M[1][1]*M[2][0]);
    d=M[0][0]*N[0][0]+M[0][1]*N[1][0]+M[0][2]*N[2][0];
    N[0][0]/=d;
    N[1][0]/=d;
    N[2][0]/=d;
    N[0][1]=-(M[0][1]*M[2][2]-M[0][2]*M[2][1])/d;
    N[1][1]=(M[0][0]*M[2][2]-M[0][2]*M[2][0])/d;
    N[2][1]=-(M[0][0]*M[2][1]-M[0][1]*M[2][0])/d;
    N[0][2]=(M[0][1]*M[1][2]-M[0][2]*M[1][1])/d;
    N[1][2]=-(M[0][0]*M[1][2]-M[0][2]*M[1][0])/d;
    N[2][2]=(M[0][0]*M[1][1]-M[0][1]*M[1][0])/d;
    //-------------------------------------------------------------------------
    //print 3x3 matrix
    for(i=0;i<3;i++)
    {
        for(j=0;j<3;j++) printf("%3.4f ",N[i][j]);
        printf("\n");
    }
    getch();
    return 0;
}