Value Exceeds while dividing

I have 2 matrices

A ranges between -32768 to 32767 ,it is 100x1 matrix

class(A) is int16

B matrix ranges from -1473 to 1570,it is 100x1 matrix

class(B) is double

asuming i divide manually the values ,max(B)/max(A), i get correct answer , but while runnning in m file , the value exceeds

The maximum value must be .049 ,but i get 1.73,

I converted both A and B to double,still i get same error ,

please assist

Answer 1

[Walter Roberson]

What happens if you break it up and display the values:

ma = max(A)
class(ma)
size(ma)
mb = max(B)
class(mb)
size(mb)
mb / ma
mb ./ ma

[Pat]

class of ma is int16

class of B is double 

size same for A,B-100x1

max(A) is 32767,B is 1540

mb/ma-

??? Error using ==> mrdivide
Linear algebra is not supported for
integer data types.

mb./ma

??? Error using ==> rdivide
Class of operand is not supported.

After converting to double

double(max(B))./double(max(A))

ans =

      0.0470

[Walter Roberson]

So the problem has gone away?

[Pat]

it is manula calculation in command window,but if i perform in m file i get error

[Walter Roberson]

What happens if you paste the commands into the .m file right after your current calculation ?

ma = max(A)
class(ma)
size(ma)
mb = max(B)
class(mb)
size(mb)
double(mb) ./ double(ma)

and what is the exact line you are using for calculation at the moment?

[Pat]

same answer

0.0470

Walter i mean if i divide the max values ,the value is 0.04,y do i get larger values than that ,while dividing others

[Walter Roberson]

I don't know: you do not show the other values you are dividing.

If you are dividing elements of A and B pairwise, then the maximum possible result is reached if max(B(B>0)) is at the same location as min(A(A>=0)), or if min(B(B<0)) is at the same location as min(A(A<0)); and the maximum can be infinite because 0 is a valid member of A; NaN is also a possible output if a 0 in B happens to be at the same location as a 0 in A.

[Pat]

Walter suppose am dividing a 14bit value/16 bit value,will i get 5 bit value or larger than that

[Walter Roberson]

Any non-zero value divided by 0 gives an infinite result, which is going to be more than 5 bits.

If you have a 14 bit value divided by 1 (which is a valid 16 bit value) then you are going to get a 14 bit value output.

Any 14 bit value divided by a value that could be represented (without compression) in 8 or fewer binary digits, is going to give a result that requires more than 5 bits.

64 (7 bits) / 2 (2 bits) = 32 (6 bits) so the numbers represented as 14 bits can be pretty small and you would still have problems.

[Pat]

Thanks walter