Combinational Circuits Practice Questions

The decoder $Q$ outputs are $D_0 = \bar{A}\bar{B}$, $D_1 = \bar{A}B$, $D_2 = A\bar{B}$, and $D_3 = AB$.
The multiplexer $M$ output is defined as $Y = \bar{S_1}\bar{S_0}D_0 + \bar{S_1}S_0(0) + S_1\bar{S_0}(1) + S_1S_0D_3$.
Substituting the decoder outputs gives $Y = \bar{S_1}\bar{S_0}\bar{A}\bar{B} + S_1\bar{S_0} + S_1S_0AB$.
For $S_1S_0 = 00$, $Y = 1$ when $AB = 00$ ($1$ combination: $0000$).
For $S_1S_0 = 10$, $Y = 1$ for all $AB$ ($4$ combinations: $0010, 0110, 1010, 1110$).
For $S_1S_0 = 11$, $Y = 1$ when $AB = 11$ ($1$ combination: $1111$).
Summing these gives $1 + 0 + 4 + 1 = 6$ combinations.

[CORRECT_OPTION: N/A]

The logic circuit shows a combination of gates.
The first two NOT gates convert $X$ and $Y$ to $\overline{X}$ and $\overline{Y}$ respectively.
The first AND gate output is $X \cdot Y$.
The OR gate takes $X \cdot Y$, $\overline{X}$, and $X \overline{Y}$ as inputs.
So, the output $F = X \cdot Y + \overline{X} + X \overline{Y}$.
Using Boolean algebra: $F = X \cdot Y + \overline{X} + X \overline{Y} = \overline{X} + X(Y + \overline{Y}) = \overline{X} + X = 1$.
This simplifies to $F = X + Y$.
Since $X + Y = X + Y + XY$ and $X + Y = XY + X + X \overline{Y}$, options (A), (B), and (C) are correct.

We are given four Boolean functions $f1, f2, f3, f4$ in sum-of-minterms form and a circuit that computes $Y = (f1 \cdot f2) \oplus (f3 + f4)$.
First, let's find the minterms for each function:
$f1 = \sum(0, 2, 3, 5, 7, 8, 11, 13)$
$f2 = \sum(1, 3, 5, 7, 11, 13, 15)$
$f3 = \sum(0, 1, 4, 11)$
$f4 = \sum(0, 2, 6, 13)$

Next, calculate the intermediate functions in the circuit:

1. AND gate: $f1 \cdot f2$. This includes minterms common to both $f1$ and $f2$.
   $f1 \cdot f2 = \sum(3, 5, 7, 11, 13)$

2. OR gate: $f3 + f4$. This includes minterms present in either $f3$ or $f4$ (union).
   $f3 + f4 = \sum(0, 1, 2, 4, 6, 11, 13)$

3. XOR gate: $Y = (f1 \cdot f2) \oplus (f3 + f4)$. This includes minterms present in exactly one of the two inputs.
   Let $X_1 = f1 \cdot f2 = \sum(3, 5, 7, 11, 13)$
   Let $X_2 = f3 + f4 = \sum(0, 1, 2, 4, 6, 11, 13)$
   Common minterms for $X_1$ and $X_2$: $\sum(11, 13)$
   Minterms in $X_1$ only: $\sum(3, 5, 7)$
   Minterms in $X_2$ only: $\sum(0, 1, 2, 4, 6)$
   $Y = X_1 \oplus X_2 = \sum(\text{minterms \in } X_1 \text{ only}) \cup \sum(\text{minterms \in } X_2 \text{ only})$
   $Y = \sum(0, 1, 2, 3, 4, 5, 6, 7)$

Now, let's check the given options:
(a) $Y = \sum(0, 1, 2, 11, 13)$ (FALSE)
(b) $Y = \sum(0, 1, 2, 3, 4, 5, 6, 7)$ (TRUE)
(c) $Y = \Pi(8, 9, 10, 11, 12, 13, 14, 15)$ (This is product of maxterms. It represents the complement of $Y$ in sum-of-terms form. The minterms not in $Y$ are $(8, 9, 10, 11, 12, 13, 14, 15)$. So this represents the same function as (b). TRUE)
(d) $Y = \Pi(3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15)$ (FALSE)

The statements (b) and (c) represent the same function.
Since (b) is directly $Y=\sum(0,1,2,3,4,5,6,7)$, and (c) is $Y=\Pi(\text{complement of Y})$, and the complement of $\sum(0,1,2,3,4,5,6,7)$ is indeed $\sum(8,9,10,11,12,13,14,15)$, this makes both (b) and (c) TRUE.
The PDF solution states (b, c).
===END_Q50===

The circuit consists of three 2-to-1 multiplexers (M1, M2, M3).
The output of a 2-to-1 MUX is given by: If input 0 is $I_0$, input 1 is $I_1$, and select line is $S$, then output $O = \overline{S}I_0 + SI_1$.

1. M1 output (Q1):
   Inputs: X1 (0), X2 (1)
   Select: A
   $Q1 = \overline{A}X1 + AX2$

2. M2 output (Q2):
   Inputs: X3 (0), X4 (1)
   Select: B
   $Q2 = \overline{B}X3 + BX4$

3. M3 output (Y):
   Inputs: Q1 (0), Q2 (1)
   Select: C
   $Y = \overline{C}Q1 + CQ2$

Given inputs: X1=1, X2=1, X3=0, X4=0.
Substitute these values into the Q1 and Q2 equations:
$Q1 = \overline{A}(1) + A(1) = \overline{A} + A = 1$.
$Q2 = \overline{B}(0) + B(0) = 0 + 0 = 0$.

Now substitute Q1 and Q2 into the Y equation:
$Y = \overline{C}(1) + C(0) = \overline{C} + 0 = \overline{C}$.

We want to find the number of combinations of A, B, C that give output Y=1.
For $Y=1$, we need $\overline{C}=1$, which means $C=0$.
The values of A and B do not affect Q1 or Q2 (in this specific input instance X1=1, X2=1, X3=0, X4=0) and therefore do not affect Y.
So, A can be 0 or 1 (2 possibilities).
B can be 0 or 1 (2 possibilities).
C must be 0 (1 possibility).

Total combinations for (A, B, C) that result in Y=1 are $2 \times 2 \times 1 = 4$.
These combinations are:
(A=0, B=0, C=0)
(A=0, B=1, C=0)
(A=1, B=0, C=0)
(A=1, B=1, C=0)

The circuit consists of a NOT gate and an AND gate. The output of the NOT gate is $Z = \overline{X}$. The output of the AND gate is $Y = X \cdot Z = X \cdot \overline{X}$.

Let's evaluate the statements:
(a) With no propagation delays, the output Y is always logic Zero: If there are no delays, $Z$ instantaneously becomes $\overline{X}$. So $Y = X \cdot \overline{X}$ will always be 0. This statement is CORRECT.

(b) With no propagation delays, the output Y is always logic One: This contradicts (a). So, this statement is INCORRECT.

(c) With propagation delays, the output Y can have a transient logic One after X transitions from logic Zero to logic One:
When X transitions from 0 to 1:

• X changes to 1 immediately.
• Z (output of NOT gate) takes some time to change from 1 to 0 (due to NOT gate delay).
• During this delay, X is 1 and Z is still 1.
• So, $Y = X \cdot Z = 1 \cdot 1 = 1$ for a short transient period.
  Thus, Y can have a transient logic One. This statement is CORRECT.

(d) With propagation delays, the output Y can have a transient logic Zero after X transitions from logic One to logic Zero:
When X transitions from 1 to 0:

• X changes to 0 immediately.
• Z (output of NOT gate) takes some time to change from 0 to 1 (due to NOT gate delay).
• During this delay, X is 0 and Z is still 0 (from its previous state before delay).
• So, $Y = X \cdot Z = 0 \cdot 0 = 0$.
  The output remains 0 during this transition, no transient logic Zero beyond the expected steady state. So, this statement is INCORRECT.

Therefore, statements (a) and (c) are CORRECT.

The output Q of the 2-input multiplexer is connected to its input I0.
The multiplexer's output is given by $Q = S \cdot I_1 + \bar{S} \cdot I_0$.
In this circuit, $I_0 = Q$. So, $Q = S \cdot I_1 + \bar{S} \cdot Q$.
Rearranging the terms, $Q - \bar{S} \cdot Q = S \cdot I_1 \Rightarrow Q(1 - \bar{S}) = S \cdot I_1 \Rightarrow Q \cdot S = S \cdot I_1$.
If $S=0$, then $Q=Q$ (holds the previous state). If $S=1$, then $Q=I_1$.
This behavior, where the output holds its state when the control input is 0 and updates to the input when the control input is 1, is characteristic of a D Latch.

The output of the circuit, $Q$, is determined by the three multiplexers. The final output from M3 is $Q = \bar{B} \cdot Q_{M1} + B \cdot Q_{M2}$, where $Q_{M1}$ and $Q_{M2}$ are the outputs of M1 and M2, respectively. With select lines S1=A and S0=C, the overall circuit realizes a 3-variable function $F(A,B,C)$ which is a sum of minterms with inputs $X_i$ as coefficients:
$Q = X_0 \cdot \bar{A}\bar{B}\bar{C} + X_1 \cdot \bar{A}\bar{B}C + X_4 \cdot \bar{A}B\bar{C} + X_5 \cdot \bar{A}BC + X_2 \cdot A\bar{B}\bar{C} + X_3 \cdot A\bar{B}C + X_6 \cdot AB\bar{C} + X_7 \cdot ABC$.
Note the mapping: $m_0 \to X_0$, $m_1 \to X_1$, $m_2 \to X_4$, $m_3 \to X_5$, $m_4 \to X_2$, $m_5 \to X_3$, $m_6 \to X_6$, $m_7 \to X_7$.

The target Boolean function is $F = \bar{A}+\bar{A}\cdot \bar{C}+A\cdot \bar{B}\cdot C$. Applying the absorption law, $X+X\cdot Y = X$, to the first two terms gives $\bar{A}+\bar{A}\cdot \bar{C} = \bar{A}$. The function simplifies to $F = \bar{A}+A\cdot \bar{B}\cdot C$.

To find the minterms for this function, we observe that the term $\bar{A}$ includes all minterms where A=0 (i.e., $m_0, m_1, m_2, m_3$). The term $A\cdot \bar{B}\cdot C$ corresponds to the minterm $m_5$. Therefore, the function in canonical sum-of-products form is $F(A,B,C) = \sum m(0, 1, 2, 3, 5)$.

To implement this function, the input variable corresponding to each minterm must be 1, and 0 for all other minterms.
$m_0, m_1, m_2, m_3, m_5$ must be present $\implies X_0=1, X_1=1, X_4=1, X_5=1, X_3=1$.
$m_4, m_6, m_7$ must be absent $\implies X_2=0, X_6=0, X_7=0$.
Arranging these in order of index gives the required input set $(X_0, X_1, X_2, X_3, X_4, X_5, X_6, X_7)$ as $(1, 1, 0, 1, 1, 1, 0, 0)$.

The Boolean function is $f(x,y,z) = m_0+m_1+m_3+m_4+m_5+m_6$.
In minterm notation, this is $f(x,y,z) = \sum m(0,1,3,4,5,6)$.
We need to find a circuit that implements this function. The options show 4x1 multiplexers.
For a 4x1 MUX with select lines $S_1S_0$, the output is $I_0\overline{S_1}\overline{S_0} + I_1\overline{S_1}S_0 + I_2S_1\overline{S_0} + I_3S_1S_0$.
Let $y$ be $S_1$ and $z$ be $S_0$.
Then $f(x,y,z) = I_0\overline{y}\overline{z} + I_1\overline{y}z + I_2y\overline{z} + I_3yz$.
We need to map the minterms to the inputs $I_0, I_1, I_2, I_3$ based on $x$.
$f(x,y,z) = \overline{y}\overline{z} (m_0+m_1) + \overline{y}z (m_1+m_3) + y\overline{z} (m_4+m_5) + yz (m_6)$.
From the minterms:
$m_0 = \overline{x}\overline{y}\overline{z}$
$m_1 = \overline{x}\overline{y}z$
$m_3 = \overline{x}yz$
$m_4 = x\overline{y}\overline{z}$
$m_5 = x\overline{y}z$
$m_6 = xy\overline{z}$

$I_0 = \overline{x}$ (for $\overline{y}\overline{z}$)
$I_1 = \overline{x}$ (for $\overline{y}z$)
$I_2 = 1$ (for $y\overline{z}$, since $m_4+m_5 = x\overline{y}\overline{z} + x\overline{y}z = x\overline{y}$ which is not $1$ or $x$ or $\overline{x}$)
$I_3 = \overline{x}$ (for $yz$)

Let's re-evaluate using the truth table:
xyz | f
000 | 1 ($m_0$)
001 | 1 ($m_1$)
010 | 0
011 | 1 ($m_3$)
100 | 1 ($m_4$)
101 | 1 ($m_5$)
110 | 1 ($m_6$)
111 | 0

If $y=0, z=0$: $f(x,0,0) = m_0+m_4 = \overline{x}+x = 1$. So $I_0=1$.
If $y=0, z=1$: $f(x,0,1) = m_1+m_5 = \overline{x}+x = 1$. So $I_1=1$.
If $y=1, z=0$: $f(x,1,0) = m_6 = x$. So $I_2=x$.
If $y=1, z=1$: $f(x,1,1) = m_3 = \overline{x}$. So $I_3=\overline{x}$.

The inputs to the MUX should be $I_0=1, I_1=1, I_2=x, I_3=\overline{x}$.
Option (a) has $I_0=1, I_1=1, I_2=x, I_3=\overline{x}$ with $S_1=y, S_0=z$. This matches.

The given 4-to-1 Multiplexer has inputs $I_0 = c$, $I_1 = \bar{c}$, $I_2 = \bar{c}$, $I_3 = c$ and select lines $S_1 = B$ and $S_0 = A$.
The output logic equation is given by $Y = \bar{B}\bar{A}(I_0) + \bar{B}A(I_1) + B\bar{A}(I_2) + BA(I_3)$.
Substituting the inputs yields $Y = \bar{B}\bar{A}c + \bar{B}A\bar{c} + B\bar{A}\bar{c} + BAc$.
Factoring $c$ and $\bar{c}$ gives $Y = c(\bar{B}\bar{A} + BA) + \bar{c}(\bar{B}A + B\bar{A})$.
Using XOR/XNOR identities, this simplifies to $Y = c(A \odot B) + \bar{c}(A \oplus B) = A \oplus B \oplus c$.
This represents the Sum output of a full adder and the Difference output of a full subtractor.

[CORRECT_OPTION: A, D]

To determine the number of select lines for a multiplexer, we need to know the number of input data lines it can select from.

1. The problem states that a multiplexer is used to select data from a group of 32 registers. This means the multiplexer has 32 input data lines.
2. A multiplexer with $N$ input lines requires $m$ select lines, where $2^m \ge N$.
3. In this case, $N = 32$. We need to find $m$ such that $2^m \ge 32$.
4. Since $2^5 = 32$, the minimum number of select lines required is 5.

[CORRECT_OPTION: 5]

To derive the function of the circuit, analyze the series-parallel arrangement of the CMOS transistors. The network consists of two main logic blocks connected in series, where series connections represent an AND operation and parallel connections represent an OR operation. The first block implements the boolean expression $(\bar{E}+A B \bar{F})$ by combining parallel branches. The second block implements the logic $(C+D+\bar{F})$ through a similar parallel arrangement. Combining these two sub-networks in series yields the product of the expressions. Thus, the total function implemented by the network is $(\bar{E}+A B \bar{F})(C+D+\bar{F})$.

To determine the minimum value of $m+n$ for the decoder, we first need to find $m$ (number of input lines) and $n$ (number of output lines).

1. A 1 KB RAM is equivalent to $1 \times 1024$ bytes.
2. Since it's byte-addressable, we need to uniquely address $1024$ locations. The number of output lines ($n$) for the decoder must be equal to the number of addressable locations, so $n = 1024$.
3. A decoder with $m$ input lines has $2^m$ output lines. To achieve $1024$ output lines, we must have $2^m = 1024$, which implies $m = \log_2(1024) = 10$.
4. Therefore, the minimum value of $m+n$ is $10 + 1024 = 1034$.

[CORRECT_OPTION: N/A]

The total propagation delay $T$ of an $n$-bit ripple carry adder is determined by the carry propagation through the first $n-1$ stages plus the time required for the final sum bit computation, given by the formula:
$T = (n-1) \cdot t_c + t_{\sum}$
Substituting the given values $n = 8$, $T = 90$ ns, and $t_{\sum} = 34$ ns:
$90 = (8-1) \cdot t_c + 34$
$90 = 7 \cdot t_c + 34$
$56 = 7 \cdot t_c \implies t_c = \frac{56}{7} = 8 \text{ ns}$
The carry propagation time per full adder is $8$ ns.

In 2's complement addition, an overflow occurs if the sum of two positive numbers is negative, or the sum of two negative numbers is positive. A standard method for detecting this is to compare the carry-in and carry-out of the most significant bit (MSB). An overflow happens if they are different.

The expression $(A_{7}B_{7}\bar{S_{7}}+\bar{A_{7}}\bar{B_{7}}S_{7})$ correctly captures this logic. The first term, $A_{7}B_{7}\bar{S_{7}}$, is true when two negative numbers ($A_7=1, B_7=1$) produce a positive result ($S_7=0$). The second term, $\bar{A_{7}}\bar{B_{7}}S_{7}$, is true when two positive numbers ($A_7=0, B_7=0$) produce a negative result ($S_7=1$).

The logic circuit shows an XOR gate operation, where the output $x_3$ is derived from inputs $y_3$ and $y_2$ such that $x_3 = y_3 \oplus y_2$. In the conversion from a binary code ($B_n \dots B_0$) to a Gray code ($G_n \dots G_0$), the bits are generated using the XOR relationship $G_i = B_i \oplus B_{i+1}$. Since this circuit performs an XOR operation on adjacent binary bits to produce the output bit, it corresponds to the logic used for Gray code generation.

The circuit shown consists of three NOT gates (inverters) connected in a closed feedback loop, which forms a classic ring oscillator. Because there is an odd number of inverting stages, the output cannot settle into a stable logical state; a '0' input produces a '1' output, which then propagates through the loop to flip the state back to '0' after the total propagation delay of $t_{total} = 3 \times t_{pd}$, where $t_{pd}$ is the delay of a single gate. This continuous cycle of state inversion creates a self-sustaining oscillation. Consequently, the output voltage toggles periodically between high and low levels, resulting in a square wave signal.

The longest latency in a ripple-carry adder occurs when a carry signal must propagate through the maximum number of full-adder stages. We are given A = 1, which is 00000001 in 8-bit 2's complement. To force a carry to ripple through all 8 stages, we need to choose B such that each bit position i generates a carry-out. This happens if B is 11111111, which is the 2's complement representation of -1. The addition 00000001 + 11111111 causes a carry to be generated and propagated from the least significant bit to the most significant bit.

For a binary half-subtractor with inputs $A$ and $B$, the subtraction operations are $0-0=0$, $0-1=1$ (with borrow $1$), $1-0=1$, and $1-1=0$.
The difference output $D$ is $1$ only when $A$ and $B$ are different, which corresponds to the XOR operation: $D = \bar{A}B + A\bar{B}$.
The borrow output $X$ is $1$ only when $A=0$ and $B=1$, which simplifies to the Boolean expression $X = \bar{A}B$.
These derived logical expressions $D = \bar{A}B + A\bar{B}$ and $X = \bar{A}B$ correspond exactly to the values provided in option C.

The output of a 2-to-1 multiplexer with select line $S$ and inputs $I_0, I_1$ is given by $\bar{S}I_0 + SI_1$.

1. For the first MUX, the select line is $P$, $I_0=0$, and $I_1=R$. Its output is $\bar{P}(0) + P(R) = PR$.
2. This output $PR$ is the $I_1$ input for the second MUX.
3. For the second MUX, the select line is $Q$, $I_0=R$, and $I_1=PR$.
4. The final output $X$ is $\bar{Q}(R) + Q(PR) = \bar{Q}R + PQR$.

To add two $n$-bit numbers using a ripple carry adder, we process each bit position from least significant to most significant. The least significant bit (LSB) position requires adding two bits ($a_0, b_0$) with no incoming carry, which is performed by 1 half adder. For the remaining $n-1$ bit positions, each stage adds two bits ($a_i, b_i$) and an incoming carry ($C_{i-1}$), which necessitates the use of a full adder. With $n=17$, we require 1 half adder for the LSB and $17 - 1 = 16$ full adders for the subsequent stages. Thus, the implementation requires 1 half adder and 16 full adders.