Help with piecewise function? Can't use else/if?

Question

AJ 2014-2-18

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https://ww2.mathworks.cn/matlabcentral/answers/116532-help-with-piecewise-function-can-t-use-else-if

评论： Carlos Guerrero García 2022-11-14

采纳的回答： Karan Gill

在 MATLAB Online 中打开

Hi, I am having trouble on a piece-wise homework problem I am having.

This is my code. I plot it and in the middle of the graph from negative pi to positive pi where y should be the cosine of x isn't right. I just have a straight line at y=-1 across the graph. I know I need to make it a vector or a loop it so it doesn't use if/else and skip the middle cosine function.

Whenever I try to get a function command I get this: Function definitions are not permitted in this context.

code so far:

clear all
close all
clc
x=-2*pi:.01:2*pi
if (x<-pi)
    y=-1;
elseif(x>=-pi&x<=pi)
    y=cos(x);
else(x>pi)
    y=-1;
plot(x,y)
end

This is what I'm getting: The middle part (cosine function) is wrong. I was told it's just graphing the part after else.

Help please?

Thank you.

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Matt Tearle 2014-2-18

Homework problem clearly stated as being a homework problem
Own effort and work shown
Problem explained and results shown

Congratulations on writing a great (homework) question!

M. Y. Najjar 2016-12-28

Don't you need a double '&' sign in the If statement?

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Answer 1

Karan Gill 2016-12-1

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编辑：Karan Gill 2017-10-17

If you have R2016b, you can just use the piecewise function: https://www.mathworks.com/help/symbolic/piecewise.html.

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Sarah Palfreyman 2016-12-14

编辑：Sarah Palfreyman 2016-12-14

Here is the example.

john Snori 2021-9-1

The second method involves the use of if-else statements along with for loop. In this method, we’ll define all the sub-functions along with the constraints using if-else statements and then we will plot the piecewise function.

Source : https://www.entechin.com/how-to-plot-a-piecewise-function-in-matlab/

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Answer 2

AJ 2014-2-18

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If there any possible way anyone can give me the exact code?

I looked at your answer Image Analyst and I appreciate it but I still can't figure out and I think I will understand it after seeing how it's supposed to be done. I've been trying for 5 hours now and I have other work to do :(

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Jos (10584) 2014-2-18

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Take a look at this:

    % Step 1: initialise x and y
    x = -10:2:10
    y = zeros(size(x))
    % Step 2: selective processing
    tf = x < -5
    y(tf) = -5
    tf = x > -5 & x < 5
    y(tf) = -2 + 2 * x(tf) 
    tf = x > 5
    y(tf) = 5
    % step 3: visualisation
    plot(x,y,'bo-')

burak ergocmen 2016-12-1

it really works . thanks for the codes.

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Answer 3

Image Analyst 2014-2-18

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编辑：Image Analyst 2014-2-18

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Use two &:

elseif x(k) >=-pi && x(k) <= pi

and you need to make an array out of y=-1:

y(k) = -1;

otherwise it's just a single number, not an array of -1s. Plus you need to have it in a loop over k like I mentioned

for k = 1 : length(x)

then everything inside the look has a (k) index. Of course there is a vectorized way to do it, if you want that.

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Jenik Skapa 2019-3-26

编辑：Jenik Skapa 2019-3-26

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I would use this construction without the "for" loop:

x = -2*pi : pi/5 : 2 * pi;
y = NaN * ones(size(x));
y(x < -pi) = -1;
y((x >= -pi) & (x < pi)) = cos(x((x >= -pi) & (x < pi)));
y(x >= pi) = -1;
plot(x, y, 'r*-'); grid on;