Numerictype for fimath output

When adding two fixed point numbers, I am not understanding the output I am getting. I think it may have to do with a subtlety I am missing with fimath properties. Here is an example:

FMval = fimath;

FMval.SumMode = 'KeepMSB';

FMval.SumWordLength = bitDepthd;

FMval.ProductMode = 'KeepMSB';

FMval.ProductWordLength = bitDepthd;

FMval.OverflowAction = 'Saturate';

FMval.RoundingMethod = 'Nearest';

A = fi(3, 0, 4, 0, FMval);

B = fi(5, 0, 4, 0, FMval);

I would think that A+B = 8, but when I add them together:

>> C = A+B

C =

    10

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Unsigned
            WordLength: 4
        FractionLength: -1

Why is the answer not 8? Why did the FractionLength go to -1? There is no overflow in this addition, and both variables have a fractionlength of 0.

I understand that adding two four bit numbers requires 5 bits, and then I could maybe understand an answer of, say, sixteen (moving the fractionlength to -1), but 5(0011 bin) + 3(0011 bin) should not equal 10 any way I can figure.

Another example, this time with a 4-bit overflow:

D = fi(12, 0, 4, 0, FMval); E = fi(7, 0, 4, 0, FMval);

>> F = E+D

F =

    20

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Unsigned
            WordLength: 4
        FractionLength: -1

What am I missing here? Is there a way to set the fimath properties to get the answers I'm expecting?

Answer 1

[Shivansh]

Hello Megan,

As you mentioned above, adding two 4-bits number will need 5 bits, but you have set the "fimath" to work with 4-bits and as a result, you are getting unexpected results. A possible approach to get the desired result can be to change the "SumMode" to "FullPrecision".

bitDepthd = 4; 
FMval = fimath;
FMval.SumMode = 'FullPrecision';
FMval.SumWordLength = bitDepthd;
FMval.ProductMode = 'KeepMSB';
FMval.ProductWordLength = bitDepthd;
FMval.OverflowAction = 'Saturate';
FMval.RoundingMethod = 'Nearest';
A = fi(3, 0, 4, 0, FMval);
B = fi(5, 0, 4, 0, FMval);
C=A+B

C = 
     8

          DataTypeMode: Fixed-point: binary point scaling
            Signedness: Unsigned
            WordLength: 5
        FractionLength: 0

        RoundingMethod: Nearest
        OverflowAction: Saturate
           ProductMode: FullPrecision
               SumMode: FullPrecision

C.storedInteger

ans = uint8 8

I hope it clarifies the issue!