How to plot a linear function?

Question

Jenni 2024-2-7

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2079636-how-to-plot-a-linear-function

评论： William Rose 2024-2-9

采纳的回答： William Rose

Hey!

Here is my function and I am trying to plot a linear function in form:

y = b_0 + b_1x.

How could I plot this function using the result values b_0 and b_1?

Here is my function code called "sovittaja":

function [b, bci] = sovittaja(x,y,delta_y)

%%Helper variables:

s_1 = sum(1./(delta_y.^2));

s_2 = sum((x.*y)./(delta_y.^2));

s_3 = sum(x./(delta_y.^2));

s_4 = sum(y./(delta_y.^2));

s_5 = sum(x.^2./(delta_y.^2));

%%Defining D:

D = (s_1*s_5)-(s_3)^2;

%%Analyyttiset lausekkeet:

b_1 = (1/D)*((s_1)*(s_2)-(s_3)*(s_4));

b_0 = (1/D)*((s_5)*(s_4)-(s_3)*(s_2));

%%Margin of errors:

delta_b_1 = sqrt((1/D)*(s_1));

delta_b_0 = sqrt((1/D)*(s_5));

%%Indexs of matrix bci :

bci_11 = b_0 - delta_b_0;

bci_12 = b_1 - delta_b_1;

bci_21 = b_0 + delta_b_0;

bci_22 = b_1 + delta_b_1;

%% Return values:

b = [b_0, b_1];

bci = [bci_11,bci_12;bci_21,bci_22];

And here is what I want to do with the function "sovittaja":

clear all

close all

% Measurement:

i=[1,2,3,4,5]';

% Frequency:

x=[519,549,688,740,821]';

% Voltage:

y=[1.0,1.2,1.9,2.4,2.3]';

% Margin of Error:

delta_y=[0.15,0.40,0.20,0.30,0.10];

sovittaja(x,y,delta_y)

So, how could I plot my result to a linear function y = b_0 + b_1x?

And additional question: How could I do the same thing by using command fitlm?

Thank you for the help!

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John D'Errico 2024-2-7

编辑：John D'Errico 2024-2-7

Why would you want to use fitlm to plot a line? It is not a tool that creates plots. So what are you asking? I think you want to know a tool that can fit a line through your points, using some given weights at those points. I am not entirely sure if the "margin of error" is a set of weights, or it is a set of hard constraints that the line is not allow to pass.

Plotting a line is easy enough, even trivial. Choose two points on the line. Plot them, connecting them with a line.