Combinational Circuits

Q1 (12)

(a) $A=1011$ ($A_3A_2A_1A_0$), $B=0110$, $C_0=1$.

(2 marks for g/p table)

Carry equations:

• $C_1 = g_0 + p_0C_0 = 0 + 1\cdot1 = 1$
• $C_2 = g_1 + p_1C_1 = 1 + 0\cdot1 = 1$
• $C_3 = g_2 + p_2C_2 = 0 + 1\cdot1 = 1$
• $C_4 = g_3 + p_3C_3 = 0 + 1\cdot1 = 1$

(4 marks — 1 each)

Sum $S_i = p_i \oplus C_i$:

• $S_0 = 1\oplus1 = 0$
• $S_1 = 0\oplus1 = 1$
• $S_2 = 1\oplus1 = 0$
• $S_3 = 1\oplus1 = 0$

So $S = 0010$ with $C_4=1$ → result $10010_2 = 18$. Check: $11+6+1=18$. ✓ (2 marks)

(b) CLA: g/p ready at 1 delay; all carries produced with one AND-OR level = 2 gate delays total for carries (sums add 1 more XOR delay). Ripple-carry: carry ripples through 4 full adders at 2 delays each = 8 gate delays. CLA is ~4× faster. (4 marks: 2 for CLA=2, 2 for RCA=8)

Q2 (12)

(a) Truth table (X = don't care):

(6 marks — priority correctly captured via don't-cares)

(b) Let index of highest active input decide bits.

• $V = D_0 + D_1 + D_2 + D_3 + D_4 + D_5$
• $Y_2 = D_4 + D_5$ (codes 4 and 5 have MSB set)
• $Y_1 = \overline{D_5}\,\overline{D_4}(D_3 + D_2)$ (bit1 set only for codes 2,3, and only if 4,5 inactive)
• $Y_0 = D_5 + \overline{D_4}\,\overline{D_3}\,D_2 \cdot 0 \dots$ — evaluate bit0 = codes 1,3,5:
  • $Y_0 = D_5 + \overline{D_4}D_3 + \overline{D_4}\,\overline{D_3}\,\overline{D_2}D_1$

(6 marks: 1.5 each; equivalent minimal forms accepted)

Q3 (12)

(a) Group minterms by $ABC$ (rows), $D$ (column). Minterm number $= 8A+4B+2C+D$.

Minterms present: 1,3,4,6,9,11,12,14.

So $I_0=D,\ I_1=D,\ I_2=\overline{D},\ I_3=\overline{D},\ I_4=D,\ I_5=D,\ I_6=\overline{D},\ I_7=\overline{D}$. (8 marks)

(b) Pattern: $I = D$ when $C=0$, $I=\overline{D}$ when $C=1$. So $F = \overline{C}\,D + C\,\overline{D} = C \oplus D$.

Check minterm 6 ($C=1,D=0$): $1\oplus0=1$ ✓; minterm 1 ($C=0,D=1$): $0\oplus1=1$ ✓.

$\boxed{F = C \oplus D}$ (4 marks)

Q4 (12)

(a) K-map ($A$ rows, $BC$ columns), $F=\overline A C + A\overline B$:

$\overline A C$ covers cells $(A{=}0,C{=}1)$: minterms 1,3. $A\overline B$ covers $(A{=}1,B{=}0)$: minterms 4,5. (3 marks — loops identified)

(b) Both terms are 1 in the region where the two loops are adjacent but not overlapping. Consider $A:0\to1$ with $B=0, C=1$:

• Before ($A=0,B=0,C=1$, m1): $\overline A C=1$, $A\overline B=0$ → $F=1$ via term1.
• After ($A=1,B=0,C=1$, m5): $\overline A C=0$, $A\overline B=1$ → $F=1$ via term2.

During the transition, $\overline A C$ turns off before $A\overline B$ turns on (due to inverter delay on $A$), so $F$ momentarily = 0 → static-1 hazard. The changing variable is $A$ (with $B=0,C=1$ held). (5 marks)

(c) Consensus of $\overline A C$ and $A\overline B$ (eliminate $A$): $C\cdot\overline B = \overline B C$. Hazard-free: $F = \overline A C + A\overline B + \overline B C$ This term stays 1 across the transition ($B=0,C=1$), covering the glitch. Cost: original = 4 literals, hazard-free = 6 literals (+2). (4 marks)

Q5 (12)

(a) With $x_i = \overline{A_i\oplus B_i}$ (bit-equal): $A>B = A_3\overline{B_3} + x_3 A_2\overline{B_2} + x_3 x_2 A_1\overline{B_1} + x_3 x_2 x_1 A_0\overline{B_0}$ (higher bits must be equal before a lower bit decides). (6 marks)

(b) $A=1001$, $B=1010$.

Terms:

• $A_3\overline{B_3}=0$
• $x_3A_2\overline{B_2}=1\cdot0=0$
• $x_3x_2A_1\overline{B_1}=1\cdot1\cdot0=0$
• $x_3x_2x_1A_0\overline{B_0}=1\cdot1\cdot0\cdot1=0$

$A>B = 0$. Since bit1: $A_1=0,B_1=1$ (first inequality from MSB), $A<B$. Check: $1001_2=9$, $1010_2=10$, so $9<10$ → $A<B$. ✓ (6 marks)

[
{"claim":"Q1 CLA sum: 11+6+1 = 18 = 10010b","code":"A=0b1011; B=0b0110; C0=1; s=A+B+C0; result = (s==18 and format(s,'05b')=='10010')"},
{"claim":"Q1 all carries C1..C4 equal 1","code":"g=[0,1,0,0]; p=[1,0,1,1]; C=[1]; \nfor i in range(4):\n    C.append(g[i] or (p[i] and C[i]))\nresult = C[1:]==[1,1,1,1]"},
{"claim":"Q3 F = C xor D matches minterms 1,3,4,6,9,11,12,14","code":"mins=set([1,3,4,6,9,11,12,14]); ok=True\nfor n in range(16):\n    A=(n>>3)&1;B=(n>>2)&1;Cc=(n>>1)&1;D=n&1\n    f=Cc^D\n    if f!=(1 if n in mins else 0): ok=False\nresult=ok"},
{"claim":"Q4 hazard-free F equals original F over all inputs","code":"ok=True\nfor n in range(8):\n    A=(n>>2)&1;B=(n>>1)&1;Cc=n&1\n    f1=((1-A)&Cc)|(A&(1-B))\n    f2=f1|((1-B)&Cc)\n    if f1!=f2: ok=False\nresult=ok"},
{"claim":"Q5 A=9,B=10 gives A<B (A>B expression = 0)","code":"A=0b1001;B=0b1010; result = (A<B) and not (A>B)"}
]