How can i use the 'mod' function in an Embedded Matlab function?

I have the following function. I want to use this algorithm in an embedded Matlab function in simulink.

function y   = fcn( u )

y=[0;0;0;0;0;0;0;0;0;0];
newu=0;
nr=0;
s=false;

if u<0 s=true;
end

while u~=0 
  newu=newu*10+mod(u,10);
  u=(u/10);
  nr=nr+1;
end;

for i=nr:0   
  y(i)=mod(newu,10);
  newu=(newu/10);
end;

for i=0:nr 
y = y+65;
end

end

After compilation it gives me the error :'Function 'mod' is not defined for values of class 'embedded.fi'. 'u'(input value) being of class 'embedded.fi'. What do i have to change to get my output value?

Thank you, Zoli

[Walter Roberson]

Note:

for i=nr:0

is not going to execute any loops when nr > 0.

Note: if u < 0 then you set s, but you do not do anything with s and you do not change the sign of u; are you sure you are going to get the right answer? Perhaps you want nr:-1:0 ? But then how are you going to store a value at y(0) ?

Note: y = y + 65 is going to add 65 to every member of y. And you have that in a loop...

[Azzi Abdelmalek]

I think he wants for i=nr:-1:0

[Walter Roberson]

Probably wants to count backwards, yes, but still has problems at y(0)

[Azzi Abdelmalek]

 y(i+1)=mod(newu,10);

[Walter Roberson]

nr is the number of digits, but nr:-1:0 is looping nr+1 times.

[Zoltan]

yes, i want to count backwards (downto- in pascal) and i don't know the equivalent in Matlab.

[Walter Roberson]

Syntax is

for i = nr: -1: 0

but please re-check whether you want to stop at 0 or at 1.

[Zoltan]

at 0...but my question is, can i use mod with the u variable which is of class'embedded.fi'? Because it doesn't let me use it. Can i store u in another type of variable, changing the class?

[Zoltan]

*correction

for i = nr: -1: 0 
y(i)=mod(newu,10);
  newu=(newu/10);
end;

for i=0:nr
if y(i)==0 then y(i)=65; 
end;
if y(i)==1 then y(i)=66;
end;
if y(i)==2 then y(i)=67;
end;
if y(i)==3 then y(i)=68;
end;
if y(i)==4 then y(i)=69;
end;
if y(i)==5 then y(i)=70;
end;
if y(i)==6 then y(i)=71;
end; 
if y(i)==7 then y(i)=72;
end;
if y(i)==8 then y(i)=73;
end;
if y(i)==9 then y(i)=74;
end;
end

[Walter Roberson]

Look at your code again, what you show above. In the two "for" loops, where is the data being stored when "i" is 0 ?

[Zoltan]

oh...you mean i can't use 0 as an index?

[Walter Roberson]

Correct. You cannot use 0 as an index. But you probably don't want to: you are only manipulating nr values, not nr+1 values, so 1:nr and nr:-1:1 would seem to be appropriate.

[Zoltan]

OK, so i understand how to use syntax correctly. I wouldn't need this algorithm if i would be able to send values through RS-232 to my microcontroller...but for some reason i get the same sent value back only in the 64..127 range. for example, if i send 0, i receive -192 on the scope in my simulink model. why is this? thanks

[Walter Roberson]

Double-check the parity of the line. But even more so, double check how the heck an 8 bit result is generating a negative number below -128, as 8 bit signed values can only be in the range -128 to +127 . Check your scope settings.

[Zoltan]

thanks

Answer 1

[Azzi Abdelmalek]

In the below part of your code

while u~=0 
  newu=newu*10+mod(u,10);
  u=(u/10);
  nr=nr+1;
end;

newu increases rapidly to infinity. that's what causes a problem when

 y(i)=mod(newu,10);

[Zoltan]

why does it increase to infinity?

[Azzi Abdelmalek]

 newu=newu*10+mod(u,10);

[Azzi Abdelmalek]

try this

u=1,newu=0;nr=0
while u~=0 
  newu=newu*10+mod(u,10);
  u=(u/10);
  nr=nr+1;
end;
newu

[Walter Roberson]

Remember, Azzi, the input is fixed point, so the division is going to truncate at some point.

[Azzi Abdelmalek]

Yes, but after 309 iterations, 10^309=inf

In the above example, nr=324

[Walter Roberson]

This is only a problem if the fi object has a representation for infinity and if the input is infinity. Otherwise abs(u) is diminishing in each step and there would be a finite end to the process. It could potentially overflow the buffer of 10 y locations, but that would depend upon the maximum value possible for that particular fi object.