How can I fit linear regression model with the measured data with a known intercept?

I am working on close-in reference path loss model in MATLAB, where the code is given below:

PL_dB= file(:,2); dis_inch=file(:,1); dis_m= dis_inch * 0.0254

c = 3*10^8; %in light speed in free space fc = 2.4*10^9; lamda=c/fc; do =0.0254; % free space refernce distance for 5.8Ghz 1 inch=0.3048m PL_do = 20*log10(4*pi*do/lamda);% y intercept dis_dB = 10* log10 (dis_m/do); % slope of the regression line/ path loss exponent, n % find the z-score of measurement values z_dis = dis_dB- mean(dis_dB)/sqrt(var(dis_dB)); z_PL = PL_dB- mean(PL_dB)/sqrt(var(PL_dB)); r= sum(z_dis.* z_PL)/ length(z_PL) % correlation coefficient

In this case, the intercept is known. How can I find the linear regression plot based on y and x-axis data?

Answer 1

[Star Strider]

You have not stated what ‘x’ and ‘y’ are in your code. However, knowing that ‘PL_do’ is the desired y-intercept, I would do:

b = x(:) \ [y(:)-PL_do];                                % Linear Regression

then to estimate a linear regression line to plot with the desired x-data:

y_fit = x(:)*b + PL_do;

You can then plot ‘y_fit’ as a function of ‘x’.

Answer 2

[Mahfuza Khatun]

Then, b will provide another intercept which is not similar to PL_do. PL_dB is the y-axis value and dis_dB is the x-axis value in the code.

Thanks for your reply!

[Star Strider]

As always, my pleasure!

The revised code (using your variables) are:

b = dis_dB(:) \ [PL_dB(:)-PL_do];                                % Linear Regression
y_fit = dis_dB(:)*b + PL_do;

The regression that calculates ‘b’ will have a 0 intercept. Adding ‘PL_do’ to the evaluated line will produce the linear regression with the y-intercept you want.

Answer 3

[mer potter]

Hello. Do you still happen to have a code for this problem, cause I have a very simillar asignment and i'm completely lost. Thank you in advance.