wsobhan
Newbie level 3
Hello all, newbie to these boards. I have a question regarding absolute value in terms of binary. For instance (a and b are 4 bits long):
|a - b| + 1
My thoughts are do it like:
- for b, negate all the bits
- pass the negated 'b' into a full adder with 'a' as the other input. Cin of the full adder is permanently set to 1 to complete the 2's complement negation of b.
- if the Cout of this adder is 1, you know your result was negative from a-b
This is where I get fuzzy. I know how I can implement the rest of the circuit, I just don't think it's the most efficient.
- have Cout control as the select of a mux. When Cout=1, I1 of the mux gets selected.
- what feeds into this I1 input of the mux is the sum of the first full adder negated and plus 1 to convert it to it's positive version.
- what feeds into the I0 input of the mux is the just the sum of the first full adder. If Cout=0, you know this value is positive.
- another adder on the output of the mux is used to add the final + 1 as indicated in the original equation.
Can someone help me simply this so it uses less adders?? RIght now it's using three. I think two is possible but can't get it in my head, much appreciated!
|a - b| + 1
My thoughts are do it like:
- for b, negate all the bits
- pass the negated 'b' into a full adder with 'a' as the other input. Cin of the full adder is permanently set to 1 to complete the 2's complement negation of b.
- if the Cout of this adder is 1, you know your result was negative from a-b
This is where I get fuzzy. I know how I can implement the rest of the circuit, I just don't think it's the most efficient.
- have Cout control as the select of a mux. When Cout=1, I1 of the mux gets selected.
- what feeds into this I1 input of the mux is the sum of the first full adder negated and plus 1 to convert it to it's positive version.
- what feeds into the I0 input of the mux is the just the sum of the first full adder. If Cout=0, you know this value is positive.
- another adder on the output of the mux is used to add the final + 1 as indicated in the original equation.
Can someone help me simply this so it uses less adders?? RIght now it's using three. I think two is possible but can't get it in my head, much appreciated!