Exercises — Comparators

The outputs are ==$G$== ($A>B$), ==$E$== ($A=B$), and ==$L$== ($A<B$). Exactly one is high at any instant, so $G+E+L=1.$ What this means: the three signals partition every possible input — there is no fourth case, and no two can be true at once.

Why: read each row as a number comparison. $0=0$ → $E$. $0<1$ → $L$. $1>0$ → $G$. $1=1$ → $E$. Notice every row has exactly one $1$ — that is $G+E+L=1$ in action.

Equality is XNOR: $E=\overline{A\oplus B}$. XOR outputs $1$ when the bits differ, which is the opposite of equal. Negating XOR gives "no difference" = equal. See XOR and XNOR Gates.

• $G=A\bar{B}=1\cdot \overline{0}=1\cdot 1=1$.
• $E=\overline{A\oplus B}=\overline{1\oplus 0}=\overline{1}=0$.
• $L=\bar{A}B=\overline{1}\cdot 0=0\cdot 0=0$.

So $(G,E,L)=(1,0,0)$: $A>B$, correct since $1>0$. Sum $=1$. ✔

Here $A_1=1,A_0=0$ and $B_1=0,B_0=1$.

• First term: $A_1{\bar{B}}_1=1\cdot \overline{0}=1\cdot 1=1$.
• Because the first term is already $1$, $G=1$ regardless of the second term.

Decimal check: $A=2$, $B=1$, indeed $2>1$, so $G=1$. ✔ The MSB alone settled it — exactly the "MSB rules" idea.

$A_1=0,A_0=1$, $B_1=0,B_0=1$.

• $x_1=\overline{A_1\oplus B_1}=\overline{0\oplus 0}=\overline{0}=1$ (MSBs match).
• $x_0=\overline{A_0\oplus B_0}=\overline{1\oplus 1}=\overline{0}=1$ (LSBs match).
• $E=x_1{x}_0=1\cdot 1=1$.
• $G$ term 1: $A_1{\bar{B}}_1=0\cdot 1=0$; term 2: $x_1{A}_0{\bar{B}}_0=1\cdot 1\cdot \overline{1}=1\cdot 1\cdot 0=0$ → $G=0$.
• $L$ term 1: ${\bar{A}}_1{B}_1=1\cdot 0=0$; term 2: $x_1{\bar{A}}_0{B}_0=1\cdot 0\cdot 1=0$ → $L=0$.

$(G,E,L)=(0,1,0)$: equal. Decimal: $1=1$. ✔

Take $A=01_2$, $B=10_2$ (so $A=1$, $B=2$; truth is $A<B$).

• Correct $L={\bar{A}}_1{B}_1+x_1{\bar{A}}_0{B}_0$: ${\bar{A}}_1{B}_1=\overline{0}\cdot 1=1$ → $L=1$. Good.
• Buggy $G=A_1{\bar{B}}_1+A_0{\bar{B}}_0=0\cdot \overline{1}+1\cdot \overline{0}=0+1=1$.

The buggy circuit reports $G=1$ and $L=1$ simultaneously — it claims $A>B$ and $A<B$. The unguarded LSB term $A_0{\bar{B}}_0$ fired even though the MSB already ruled $A<B$. This violates $G+E+L=1$. The $x_1$ guard exists precisely to silence lower bits once a higher bit has decided.

Scan from the MSB (bit 3) down, stopping at the first position where the two numbers differ.

• Bit 3: $A_3=1$, $B_3=0$. They differ here, at the very top.
• Rule: at the highest differing bit, $A=1,B=0$ ⇒ $A>B$, and nothing below matters.

The three $1$s in $0111$ live in bits 2,1,0 — all less significant than the decision. Their combined weight is $4+2+1=7$, but bit 3 alone is worth $8$. So $8>7$. This is why comparators scan MSB-first, unlike a Ripple Carry Adder which propagates carries LSB-first.

$G+E+L=A\bar{B}+\overline{A\oplus B}+\bar{A}B$. Expand XNOR: $\overline{A\oplus B}=\bar{A}\bar{B}+AB$. So the sum is $A\bar{B}+\bar{A}\bar{B}+AB+\bar{A}B.$ Group by $\bar{B}$ and $B$: $(A+\bar{A})\bar{B}+(A+\bar{A})B=1\cdot \bar{B}+1\cdot B=\bar{B}+B=1.$ Every one of the four minterms $\{\bar{A}\bar{B},\bar{A}B,A\bar{B},AB\}$ appears exactly once, which is the whole truth-table space — so the sum is a tautology. See Truth Tables and Minterms.

Per-bit equality flags: $x_2=\overline{A_2\oplus B_2}$, $x_1=\overline{A_1\oplus B_1}$. The first differing bit (from the top) with $A_i=1,B_i=0$ decides "greater": $G=A_2{\bar{B}}_2+x_2{A}_1{\bar{B}}_1+x_2{x}_1{A}_0{\bar{B}}_0.$ Reading it: term 1 = MSB of $A$ wins outright. Term 2 = MSBs tied ($x_2$), bit 1 breaks it. Term 3 = top two bits tied ($x_2{x}_1$), the LSB breaks it. Each lower term is guarded by the equality of all bits above it. Equality is $E=x_2{x}_1{x}_0$.

$(A>B{)}_{out}=G_H+E_H\cdot (A>B{)}_{in}=0+1\cdot 1=1.$ Interpretation: the high nibble didn't decide (its bits all matched, $E_H=1$), so it passes through the low chip's verdict. The low chip says $A>B$, so the whole 8-bit result is $A>B$. This is the human "first differing digit" rule lifted from single bits to whole 4-bit chips.

If every bit of both 8-bit numbers is equal, each chip has $E_{local}=1$ and $G_{local}=L_{local}=0$. The low chip then computes $(A=B{)}_{out}=E_{local}\cdot (A=B{)}_{in}=1\cdot 1=1$ — but only because we forced $(A=B{)}_{in}=1$ at the bottom of the chain. That tie value ripples up: the high chip sees $(A=B{)}_{in}=1$ from below and, with $E_H=1$, outputs $(A=B{)}_{out}=1$. Tie the bottom greater/less inputs to $0$ so they never falsely trigger. Rule of thumb: the very bottom of any cascade must default to "equal" so that "all bits match" correctly bubbles to the top.

Every bit matches ($A_i=B_i=0$), so each $x_i=\overline{0\oplus 0}=\overline{0}=1$. Then $E=x_3{x}_2{x}_1{x}_0=1\cdot 1\cdot 1\cdot 1=1.$ In $G$, every term contains a factor $A_i{\bar{B}}_i=0\cdot \overline{0}=0$, so $G=0$. Likewise $L$ has factors ${\bar{A}}_i{B}_i=1\cdot 0=0$, so $L=0$. Result $(G,E,L)=(0,1,0)$ — zero equals zero, and $G+E+L=1$ holds even in this degenerate all-zero case.

The very first term is $A_3{\bar{B}}_3=1\cdot \overline{0}=1$. Since $G$ is a sum (OR) of terms, one true term makes $G=1$; the guard chain below is irrelevant. Decimal: $15>0$. ✔ Equality $E=x_3\dots$ dies immediately because $x_3=\overline{1\oplus 0}=\overline{1}=0$ zeroes the product. So $(G,E,L)=(1,0,0)$.

The plain magnitude comparator is correct for unsigned numbers — it never claimed to handle signs. As unsigned, $1000_2=8>7=0111_2$, so $(A>B)$ is right. The "failure" is only if you mis-interpret the MSB.

Fix for signed: the MSB is a sign bit, not a magnitude bit. Swap its meaning — a $1$ in the MSB means smaller (negative), not larger. Concretely, invert both MSBs before feeding them in, or XOR the MSB comparison. After inverting the sign bits, $A=0000$-flavoured, $B=1111$-flavoured, and the same comparator now reports $A<B$ correctly. This is exactly the sign-bit trick used in Binary Subtraction.

Numeric check: unsigned $8>7$ (comparator output), signed $-8<7$ (after sign fix).

Which bit decides magnitude first? ::: The MSB — the first differing bit from the top. Why guard lower-bit terms with $x_i$? ::: So they only fire when all higher bits tied; prevents contradictory $G$ and $L$. Cascade default at the bottom of a 7485 chain? ::: $(A=B{)}_{in}=1$, others $0$, so a full tie propagates up. All-zero input $(G,E,L)$? ::: $(0,1,0)$ — equality holds, sum still $1$. Fix for signed comparison? ::: Invert the MSBs (sign bits) before comparing.