Design of Combinational Circuit for 2's Complement of a 4-bit Number

Concept

The 2's complement of a binary number is obtained by:

1. Taking the 1's complement (inverting all bits), then
2. Adding 1 to the result

For a 4-bit input A = A₃A₂A₁A₀, the output is S = S₃S₂S₁S₀

Truth Table

The 2's complement of N is: S = 2⁴ - N = 16 - N

Logical Design Approach

The copy-and-invert rule

There is a well-known pattern for the 2's complement:

Copy all bits from the right up to and including the first 1, then invert every bit above it.

The rule follows from the definition. Complementing every bit gives $\overline{A_i}$, and adding 1 to that propagates a carry only while the complemented bits are 1, that is only while the original bits are 0. So with $B_i = \overline{A_i}$ and an initial carry $C_0 = 1$:

$$S_i = B_i \oplus C_i = \overline{A_i} \oplus C_i$$

$$C_{i+1} = B_i \cdot C_i = \overline{A_i} \cdot C_i$$

The complement must appear in BOTH equations. Using $A_i$ rather than $\overline{A_i}$ builds an incrementer that computes $A + 1$, not the 2's complement.

Boolean Expressions

Carry Equations:

$$C_0 = 1$$

$$C_1 = \overline{A_0} \cdot C_0 = \overline{A_0}$$

$$C_2 = \overline{A_1} \cdot C_1 = \overline{A_1} \cdot \overline{A_0} = \overline{A_1 + A_0}$$

$$C_3 = \overline{A_2} \cdot C_2 = \overline{A_2 + A_1 + A_0}$$

Output Equations:

Since $\overline{x} \oplus \overline{y} = x \oplus y$, each output simplifies neatly:

$$\boxed{S_0 = \overline{A_0} \oplus 1 = A_0}$$

$$\boxed{S_1 = \overline{A_1} \oplus \overline{A_0} = A_1 \oplus A_0}$$

$$\boxed{S_2 = A_2 \oplus (A_1 + A_0)}$$

$$\boxed{S_3 = A_3 \oplus (A_2 + A_1 + A_0)}$$

Read together these are exactly the copy-and-invert rule: $S_0$ is a plain copy of $A_0$, and each higher bit is inverted precisely when some lower bit is 1, which is when the first 1 has already been passed.

Circuit Diagram

A0 ──────────────────────────────────────────────── S0   (direct connection)

A1 ──┐
     XOR ─────────────────────────────────────────── S1
A0 ──┘

A1 ──┐
     OR ── P1 = A1 + A0 ──┐
A0 ──┘                    XOR ───────────────────── S2
A2 ───────────────────────┘

A2 ──┐
     OR ── P2 = A2 + P1 ──┐
P1 ──┘                    XOR ───────────────────── S3
A3 ───────────────────────┘

Gates required: two OR gates (the second reuses the output of the first) and three XOR gates. $S_0$ needs no gate at all, so seven gate inputs in total realise the whole circuit.

Verification

Take A = 0110 (decimal 6). Then $A_0 = 0$, $A_1 = 1$, $A_2 = 1$, $A_3 = 0$, so

$$S_0 = A_0 = 0$$

$$S_1 = A_1 \oplus A_0 = 1 \oplus 0 = 1$$

$$S_2 = A_2 \oplus (A_1 + A_0) = 1 \oplus 1 = 0$$

$$S_3 = A_3 \oplus (A_2 + A_1 + A_0) = 0 \oplus 1 = 1$$

The output is S = 1010 (decimal 10), and $16 - 6 = 10$, which agrees with the truth table row for 0110.

Checking the two extremes as well: for $A = 0000$ every OR term is 0 and every output is 0, which is correct because the 2's complement of zero is zero; for $A = 1111$ the outputs are $1, 0, 0, 0$ read from $S_0$, giving $S = 0001$, and $16 - 15 = 1$.

Conclusion

The 2's complement circuit does not need a full adder chain. Because the carry out of the increment propagates only while the complemented bits are 1, each output reduces to $S_i = \overline{A_i} \oplus C_i$ with $C_{i+1} = \overline{A_i} \cdot C_i$ and $C_0 = 1$, which flattens to a copy of $A_0$ and an XOR of each higher bit with the OR of all bits below it. Two OR gates and three XOR gates are enough.

Convert $(1011.110)_2$ to Decimal and Hexadecimal

Step 1: Extract Given Data

Given data:

• Binary number: $1011.110_2$
• Integer part: $1011$
• Fractional part: $110$

Required conversions: to decimal (base 10) and to hexadecimal (base 16).

Step 2: Solve

Part A: Binary to Decimal

Integer part $1011$:

$$1\times2^3 + 0\times2^2 + 1\times2^1 + 1\times2^0 = 8 + 0 + 2 + 1 = 11$$

Fractional part $.110$:

$$1\times2^{-1} + 1\times2^{-2} + 0\times2^{-3} = 0.5 + 0.25 + 0 = 0.75$$

Combine:

$$(1011.110)2 = 11 + 0.75 = (11.75){10}$$

Part B: Binary to Hexadecimal

Group into blocks of 4 bits from the binary point outward, padding with zeros.

Integer part: $1011$ (already 4 bits)

$$1011_2 = 8+2+1 = 11 = B$$

Fractional part: $110 \rightarrow$ pad right to $1100$

$$1100_2 = 8+4 = 12 = C$$

Result:

$$(1011.110)2 = (B.C){16}$$

Check via decimal: $11.75_{10} = B + 0.75$. Fraction $0.75 = \frac{12}{16} = 0.C_{16}$. ✓

Summary