Plotting Functions using For Loop and If Statements?

Hey all,
I'm attempting to write a program using a for loop and conditional statements for the following:
I thought this was a very simple assignment but however when I try to implement my code my plot comes up completely blank. This is my first time working with MATLAB so apologies if this is trivial to most of you. Could someone help me understand what I'm doing wrong? Thanks in advance.
Below is my code:
t=0;
x1= ((t.^2)-(2*t)+3);
x2 = (4*cos((2*pi*t)-(pi/8))+3*sin(2*pi*t));
x3 = (sinc(t));
figure(1);
hold on
for i= -4:0.1:4
if (i> - 4) && (i< -2)
t = i;
y = x1;
elseif (i >-2) && (i < 2)
t = i;
y = x2;
elseif (i > 2) && (i< 4)
t = i;
y = x3;
end
plot(t,y)
end

 Respuesta aceptada

KALYAN ACHARJYA
KALYAN ACHARJYA el 17 de Feb. de 2021
Editada: KALYAN ACHARJYA el 17 de Feb. de 2021
t=-4:0.1:4
x1= (t.^2)-(2*t)+3;
x2 = 4*cos((2*pi*t)-(pi/8))+3*sin(2*pi*t);
x3 = sinc(t);
y=zeros(1,length(t));
for i=1:length(t)
if (t(i)>-4) && (t(i)< -2)
y(i)= x1(i);
elseif (t(i)>-2) && (t(i) < 2)
y(i)= x2(i);
else
y(i)= x3(i);
end
end
figure,plot(t,y);
Important: Here you can avoid loop and if else condition (recommended), please try once, we are here to help you.

4 comentarios

KALYAN ACHARJYA
KALYAN ACHARJYA el 17 de Feb. de 2021
Editada: KALYAN ACHARJYA el 17 de Feb. de 2021
Here: Without Loop and if else Statement
t=-4:0.1:4;
x1= (t.^2)-(2*t)+3;
x2 = 4*cos((2*pi*t)-(pi/8))+3*sin(2*pi*t);
x3 = (sinc(t));
y=x3;
id1=t>-4 & t<-2;
id2=t>-2 & t<2;
y(id1)=x1(id1);
y(id2)=x2(id2);
figure,plot(t,y);
Thank you for your help! This worked perfectly. Could you explain why you put the code below in or have a link where I can read more about this? I'd like to understand why it is done this way.
t(i)>-4 && t(i)<-2
%specifically the "t(i)" portion
This is logical indexing, Here the links
https://blogs.mathworks.com/loren/2013/02/20/logical-indexing-multiple-conditions/
https://in.mathworks.com/help/matlab/matlab_prog/techniques-for-improving-performance.html
Thank you for your help. I really appreciate it.

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Versión

R2019b

Preguntada:

el 17 de Feb. de 2021

Editada:

el 17 de Feb. de 2021

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