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# How to build up linear functions in MATLAB and plot the?

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Hi, I am new in MATLAB and I want to build a code for these linear functions and plot them to use it in my research paper.

The general shape of the linear fuctions are attached.

The idea from these functions that if I have values for example from (0 to 0.04999) in x axis and i substitute them in the first linear function, I want the output to be from (0-0.25) ...

in the second linear function the input from (0.05 to 0.24999) so the output that I want to be are (0.3 - 0.55).

in the third linear function the input from (0.25 to 0.6) so the output that I want to be are (0.6 - 1).

I dont intrest in exact value of the slope at this stage, I just want general functions to achieve what i want.

Thanks in advance

##### 0 Comments

### Answers (3)

Star Strider
on 3 Aug 2022

The lilnes are easy enough to plot —

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), ym(k,:), '-k')

end

hold off

What do you want to do with them after that?

.

##### 21 Comments

M
on 3 Aug 2022

Edited: M
on 3 Aug 2022

@Star Strider, thanks but I want the functions itself to substitute any number on it not plot the numbers only! and then plot the functions..

I mean i need those three functions

Star Strider
on 3 Aug 2022

I am still not certain what you want.

To get one line with a slope common to all the points listed:

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

[xq,ix] = sort(xm(:));

yq = ym(ix);

DM = [xq ones(size(xq))];

B = DM \ yq % [Slope; Intercept]

B = 2×1

1.5000
0.1500

yi = DM * B;

figure

plot(xq, yi, '.-')

grid

.

M
on 3 Aug 2022

@Star Strider, I was wrong when I said to you that the three function should have the same slop.. sorry.

How can I change the slop for each function Please?

M
on 3 Aug 2022

Edited: M
on 3 Aug 2022

@Star Strider, I want to use these functions in fuzzy logic problem .. thanks

Star Strider
on 3 Aug 2022

In my original Answer, the first row of the ‘B’ matrix are the sloopes of the individual line segments. You can change that to any values you want.

To calculate the resulting lines, and using my original code (and slopes) as a starting point:

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

B

B = 2×3

5.0010 1.2501 1.1429
0 0.2375 0.3143

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-k')

end

hold off

B(1,:) = rand(size(B(1,:)))

B = 2×3

0.5312 0.9381 0.2751
0 0.2375 0.3143

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-k')

end

hold off

This simply demonstrates how to set the slopes. They can be anything you want.

.

M
on 3 Aug 2022

Star Strider
on 3 Aug 2022

I am not certain how to define a function for that.

Try this —

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

B

B = 2×3

5.0010 1.2501 1.1429
0 0.2375 0.3143

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-k')

end

hold off

NewSlopes = [5.2 3.4 0.6]; % Define Slopes Vector

B(1,:) = NewSlopes % Assign Slopes Vector To Create New Linear Functions

B = 2×3

5.2000 3.4000 0.6000
0 0.2375 0.3143

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-k')

end

hold off

.

M
on 3 Aug 2022

@Star Strider sorry for bothering, but you didnt get me well.

I mean for example if there is input value (x value) 0.12 and i want to see it's output (y value), it's output should be in the second function range.

so I have to substitute in a certain function, where is that function?

Star Strider
on 3 Aug 2022

That requirement is cedrtainly not obvious from the previous descriptions!

Try this —

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

xq = 0.12; % Supplied Value For 'x'

xm_row = @(x) find(xm(:,1) <= x & xm(:,2) >= x); % Function To Determine Segment Membership

Check = xm_row(xq) % Check Result

Check = 2

yq = [xq 1] * B(:,xm_row(xq)) % Call Function & Find 'y'

yq = 0.3875

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-', 'DisplayName',sprintf('Segment %d',k))

end

plot(xq,yq,'rs', 'DisplayName','(xq,yq)')

hold off

legend('Location','best')

That seems to meet the current requirements. (It would not work if ‘x’ (or ‘xq’) is not within any of those limits.)

.

Sam Chak
on 3 Aug 2022

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

m = B(1, 1:3) % Slopes

m = 1×3

5.0010 1.2501 1.1429

c = B(2, 1:3) % Intercepts

c = 1×3

0 0.2375 0.3143

Having that you just need to write the line functions, where you may want to use them in your Fuzzy Logic System

x1 = linspace(xm(1), xm(2), 101);

x2 = linspace(xm(3), xm(4), 101);

x3 = linspace(xm(5), xm(6), 101);

y1 = m(1)*x1 + c(1); % line #1: y = 5.0010*x

y2 = m(2)*x2 + c(2); % line #2: y = 1.2501*x + 0.2375

y3 = m(3)*x3 + c(3); % line #3: y = 1.1429*x + 0.3143

plot(x1, y1, x2, y2, x3, y3)

M
on 3 Aug 2022

Edited: M
on 3 Aug 2022

@Sam Chak, this is what I want exactly ... now do you know how can I make these as memebrship functions? general membership functions not specifically to use it in Sugeno..

I mean I have many values and I want to substitute them in these functions and save their outputs.. can you help plesae in this? Thanks

Star Strider
on 3 Aug 2022

M
on 3 Aug 2022

Edited: M
on 3 Aug 2022

@Star Strider yes thanks. I didnt notice that.

can you help me in formulating for loop for "xq" .. to substitute all values that I have please and save their outputs?

dpb
on 3 Aug 2022

"... I have many values and I want to substitute them in these functions and see save their outputs.."

Show us why what I gave you to begin with will not let you do precisely that...I showed you precisely that result as the first plot where the vector yh was computed from a whole bunch of x values; the x can be totally arbitrary in number and value (although they ought to be within the range of the defined breakpoints range, obviously).

Illustrate again --

>> yNew=fnY(pi/10) % save a new single value at some arbitrary point, x

yNew =

0.6733

>>

or, with a vector of points...

>> xvals=rescale(rand(1,7),0,0.6) % 7 random points between 0 and 0.6

xvals =

0 0.6000 0.4549 0.5091 0.2921 0.4800 0.0824

>> yvals=fnY(xvals) % and the function values to go with 'em...

yvals =

0 1.0000 0.8341 0.8961 0.6481 0.8629 0.3405

>>

and just to show that they satisfy the criteria we'll put 'em on that original figure yet again...

>> plot(xvals,yvals,'or')

which again shows they all are identically on the given line segments as defined...

You've yet to show us why this won't work for you.

Star Strider
on 3 Aug 2022

I do not know what other values you have, so I created some — .

xm = [0 0.04999; 0.05 0.24999; 0.25 0.6];

ym = [0 0.25; 0.3 0.55; 0.6 1];

for k = 1:size(xm,1)

B(:,k) = [xm(k,:); 1 1].' \ ym(k,:).'; % Slopes & Intercepts

end

xq = [0.12 rand(1,9)*0.6]; % Supplied Values For 'x'

for k = 1:numel(xq)

xm_row = @(x) find(xm(:,1) <= x & xm(:,2) >= x); % Function To Determine Segment Membership

yq(k) = [xq(k) 1] * B(:,xm_row(xq(k))); % Call Function & Find 'y'

end

Points = [xq; yq]

Points = 2×10

0.1200 0.1433 0.5412 0.4145 0.2087 0.4961 0.3901 0.5859 0.0221 0.4086
0.3875 0.4166 0.9328 0.7880 0.4983 0.8813 0.7601 0.9839 0.1103 0.7813

figure

hold on

for k = 1:size(xm,1)

plot(xm(k,:), [xm(k,:); 1 1].'*B(:,k), '-', 'DisplayName',sprintf('Segment %d',k))

end

plot(xq,yq,'rs', 'DisplayName','(xq,yq)')

hold off

legend('Location','best')

The points are random, however they all appear to plot correctly.

.

dpb
on 3 Aug 2022

dpb
on 3 Aug 2022

% the lookup table/interpolant function...

y=[0 0.25 0.30 0.55 0.6 1];

x=[0 0.05-eps(0.05) 0.05 0.25-eps(0.25) 0.25 0.6];

% let's use it...

xh=[0:0.01:0.6]; % sample points to evaluate

yh=interp1(x,y,xh); % functional values

plot(xh,yh,'x') % see what it looks like...

results in

plot() by default would draw the line between the breakpoints; hence the markers only here to illustrate...

##### 8 Comments

dpb
on 3 Aug 2022

You can evaluate the function at any single point you need -- and do anything you want with the result.

I just plotted the results to show you what the above gives over the range of values of the independent variable -- that doesn't prohibit you from using it as desired -- you can encapsulate it as an anonymous function

What, specifically, do you need with the functions that can't be done evaluating one expression instead? Show us a place in your code the above won't work to accomplish the objective.

>> fnY=@(v)interp1(x,y,v); % an anonymous function to use

>> y=fnY(pi/10) % an arbitrary point selected

y =

0.6733

>> hold on

>> plot(pi/10,y,'or') % show it on top of previous plot

>> xh=[0 x(2)]; % the range of the function first breakpoint

>> plot(xh,fnY(xh),'g-') % show it in full over that range, too...

>>

What, specifically, that you need doesn't that let you do?

dpb
on 3 Aug 2022

Do what, specifically?

Why can't you pass/use the function I defined? Have you even tried it?

Show us code that you've tried and that doesn't function; don't just complain.

Alternatively, you could do something like

>> bkpt=reshape(x,2,[]).'

bkpt =

0 0.0500

0.0500 0.2500

0.2500 0.6000

>> resp=reshape(y,2,[]).'

resp =

0 0.2500

0.3000 0.5500

0.6000 1.0000

>> b=arrayfun(@(i)polyfit(bkpt(i,:),resp(i,:),1),1:size(bkpt,1),'UniformOutput',0);

>> for i=1:numel(b)

fnY(i)={@(x)polyval(b{i},x)};

end

>> whos fnY

Name Size Bytes Class Attributes

fnY 1x3 408 cell

>> fnY{3}(pi/10)

ans =

0.6733

>>

gives a handle array of the three line segments defined above that could be passed as separate array elemnts but it's not clear what advantage this has over simply using the one global function wherever it is needed.

I'll 'fess that I don't have and have never used the fuzzy logic TB so there may be something I'm missing, but we would need to see why that solution doesn't work to figure out what might do instead or to correct whatever is wrong with it (or in your use thereof).

dpb
on 3 Aug 2022

Edited: dpb
on 4 Aug 2022

The lookup table aspects of @doc:interp1 can be extremely handy tool for shortening user code such as here -- using it for reverse lookup is also something not to be overlooked.

The only disadvantage here I can see is that one does, indeed, have to use values within each segment only to draw a continuous line segment that will show the discontinuity without a line segment drawn between sections. But, the other solutions also all have to have some other way to handle it that is at least as much code/logic as would be using the segment breakpoints in a loop similarly as to @Star Strider

xq = [0.12 rand(1,9)*0.6]; % Supplied Values For 'x'

yq=fnY(xq); % 'y'

Points = [xq; yq]

figure

hold on

xm=reshape(x,2,[]).'; % convert to S-S's variable from mine

for k = 1:size(xm,1)

plot(xm(k,:), fnY(xm(k,:)), '-', 'DisplayName',sprintf('Segment %d',k))

end

plot(xq,yq,'rs', 'DisplayName','(xq,yq)')

hold off

legend('Location','best')

gives

which is same plot with slightly different set of randomized xq,yq points...I was just too lazy to go to the extra trouble so just dumped the points out to simulate the lines.

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