Question 4 of 6: Two-bit subtractor and HS/FS cascade

Question 4: Two-bit subtractor and HS/FS cascade (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Minuend $N_1=AB$, subtrahend $N_2=CD$ (each 2-bit unsigned, $A,C$ the MSBs); output $XYZ$ is the difference in 3-bit 2’s complement, value $=-4X+2Y+Z$.

Find. The truth table of $X,Y,Z$, their minimal SoP, and the wiring of the supplied HS + FS into a 2-bit subtractor.

Approach. Compute $(2A+B)-(2C+D)$ for all 16 input combinations, encode each signed value in 3-bit 2’s complement, minimise each output on a K-map, then recognise the ripple-borrow structure that the HS and FS implement.

(a) Truth table

(b) K-map minimisation

$$X = \overline{A}\,C + \overline{A}\,\overline{B}\,D + \overline{B}\,C\,D.$$

$$Z = B\overline{D} + \overline{B}D = B\oplus D.$$

The middle bit is a three-way exclusive-OR, $Y=A\oplus C\oplus(\overline{B}D)$, whose checkerboard map does not merge; written flat as a minimal SoP it is

$$Y = \overline{A}\,\overline{B}\,\overline{C}\,D + \overline{A}BC + A\overline{B}CD + \overline{A}C\overline{D} + A\overline{C}\,\overline{D} + AB\overline{C}.$$

(c) HS + FS cascade

The 2-bit subtraction ripples a borrow from the LSB stage to the MSB stage. The half subtractor forms the least-significant bit and its borrow; the full subtractor forms the next bit, its borrow-in being the LSB borrow, and its borrow-out is the 2’s-complement sign bit:

• $Z = \text{HS.Diff} = B\oplus D$, and $\text{HS.Bout}=\overline{B}D$ (borrow when $B<D$).
• $Y = \text{FS.Diff} = A\oplus C\oplus(\overline{B}D)$.
• $X = \text{FS.Bout} = \overline{A}C + \overline{A}\,\overline{B}D + \overline{B}CD$ — a borrow out of the top stage is the sign bit (it weights $-4$), so no extra inverter is needed.