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I am getting an error as "Operands to the || and && operators must be convertible to logical scalar values. Error in xd (line 8) if (t>=t1)&&(t<t2)".

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my code is
t1 = 0;
t2 = 10;
t3 = 20;
t4 = 30;
t = 0:50;
if (t>=t1)&&(t<t2)
r = 0.6;
if (t>=t2)&&(t<t3)
r = 0.3;
if (t>=t3)&&(t<t4)
r = 0.3;
if (t>=t4)
r = 0.6;
end
end
end
end
plot(t,r)

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Accepted Answer

Omer Yasin Birey
Omer Yasin Birey on 17 Jan 2019
Edited: Omer Yasin Birey on 17 Jan 2019
You cannot compare a vector t (which is 1x51 double) with a single value for an if statement, with the logical operators. Even if you could, the plot at the end won't give anything. I believe you want to do this.
t1 = 0;
t2 = 10;
t3 = 20;
t4 = 30;
t = 0:50;
r = zeros(length(t),1);
for i = 1:length(t)
if (t(i)>=t1)&&(t(i)<t2)
r(i) = 0.6;
elseif (t(i)>=t2)&&(t(i)<t3)
r(i) = 0.3;
elseif (t(i)>=t3)&&(t(i)<t4)
r(i) = 0.3;
else
r(i) = 0.6;
end
end
plot(t,r)
axis([0 60 0.2 0.7])

  2 Comments

Stephen Cobeldick
Stephen Cobeldick on 17 Jan 2019
"You cannot compare a vector t (which is 1x51 double) with a single value for an if statement"
Actually you can: comparing a vector against a scalar value is permitted, and providing a non-scalar condition to if is also permitted (but you should read the if documentation very carefully before using this):
>> if [1,2,3]>0, disp('oh, it works!'), end
oh, it works!
>>
Personally I would recommend against non-scalar if or while conditions, but they certainly are possible in MATLAB.

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More Answers (2)

Stephen Cobeldick
Stephen Cobeldick on 17 Jan 2019
Edited: Stephen Cobeldick on 17 Jan 2019
Learn to use logical indexing
Logical indexing is a very basic MATLAB concept that will make your code simpler and more efficient:
>> t1 = 0;
>> t2 = 10;
>> t3 = 20;
>> t4 = 30;
>> t = 0:50;
>> r = nan(size(t));
>> r((t>=t1)&(t<t2)) = 0.6;
>> r((t>=t2)&(t<t3)) = 0.3;
>> r((t>=t3)&(t<t4)) = 0.3;
>> r((t>=t4)) = 0.6;
>> plot(t,r)

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madhan ravi
madhan ravi on 17 Jan 2019
Edited: madhan ravi on 17 Jan 2019
&& and || for scalars for vectors use & and | , logical indexing is efficient:
t1 = 0;
t2 = 10;
t3 = 20;
t4 = 30;
t = 0:50;
r = zeros(size(t));
id1=(t>=t1)&(t<t2)|(t>=t4);
r(id1)=0.6
id2=(t>=t2)&(t<t3);
r(id2) = 0.3;
id3=(t>=t3)&(t<t4)
r(id3) = 0.3;
plot(t,r)
axis([0 60 0 1])

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