Function ranges in Matalb
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Hello everybody,
I am trying to use Gui matlab to determine the output of a function in two different ranges which are compared with user-entered values. I set the axis Z as
Z=0.000001:500;
The I used the entered values by user in calculations:
Z03= (4.*E.^1.6)/(r3.*0.8862269);% E and r3 are given by the user
Z04= (4.*E.^1.6)/(r4.*0.8862269);% E and r4 are given by the user
delat3=0;
delat4=0;
p3=-exp((-(Z-delat3)./Z03).^2);
p4=-exp((-(Z-delat4)./Z04).^2);
After that I defined the function's ranges as follows:
for i=1:numel(Z)
if Z(i)<=d3 % d3 is given by the user
P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user
P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
end
end
and plot the results by
plot(Z, P2L,'linewidth',2)
When I press 'Plot' I got this error
Operands to the || and && operators must be convertible to logical scalar values.
Would you please help me to slove this error?
Thanks in advance,
Ahmed
6 Comments
junzhuo ren
on 17 Aug 2020
After the first condition in the for loop,matlab automatically regard z(i)>d3,so it makes no sense to show it again.
Ahmed Elsherif
on 17 Aug 2020
Edited: Ahmed Elsherif
on 17 Aug 2020
dpb
on 17 Aug 2020
for i=1:numel(Z)
if Z(i)<=d3 % d3 is given by the user
P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user
P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
end
end
in
elseif (Z(i) > d3) && (Z(i) < d4) % d4 is given by the user
can't get there unless Z(i)>d3 because all cases of Z(i)<=d3 are in the other branch already...so, j r is pointing out the code is same as
for i=1:numel(Z)
if Z(i)<=d3 % d3 is given by the user
P2L=(p3.*(2*(Z-delat3)./Z03))./((Z03).^2);
else
if Z(i)<d4
P2L=(p4.*(2*(Z-delat4)./Z04))./((Z04).^2);
end
end
end
to separate out the two ... or, of course, you can leave the elseif in and just remove the superfluous test.
The error as you wrote is owing to the use of '&&' and that d3 which is not defined in the supplied code snippet must have been a vector, not a single element...observe at the command line if write:
>> v0 % a variable happened to be in my current workspace for convenience...
v0 =
10.00 11.00 12.00 13.00 14.00 15.00 16.00 17.00 18.00 19.00 20.00
>> v0(1) && 3 % short-circuit && and a constant
ans =
logical
1
>> v0(1) && [3 4] % short-circuit && and a vector
Operands to the || and && operators must be convertible to logical scalar values.
>>
NB: the latter goes "boom"...
Ahmed Elsherif
on 17 Aug 2020
Well, whenever it was that the code posted before was run, the error message is clear that wasn't a scalar.
Use the debugger to see in context.
But, as written there's no need for the & operator anyways...and it is "&" you would want here, not "&&" if did need the compound test.
Ahmed Elsherif
on 21 Aug 2020
Answers (1)
Walter Roberson
on 21 Aug 2020
for i=1:numel(Z)
if Z(i)<=d3 % d3 is given by the user
P2L(i)=(p3.*(2*(Z(i)-delat3)./Z03))./((Z03).^2);
elseif (Z(i) < d4) % d4 is given by the user
P2L(i)=(p4.*(2*(Z(i)-delat4)./Z04))./((Z04).^2);
else
P2L(i) = nan; %undefined outside the range given
end
end
1 Comment
Ahmed Elsherif
on 21 Aug 2020
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