Unit 3 - Practice Quiz

To convert decimal 10 to binary, we use division by 2: 10/2 gives a quotient of 5 and remainder 0. 5/2 gives a quotient of 2 and remainder 1. 2/2 gives a quotient of 1 and remainder 0. 1/2 gives a quotient of 0 and remainder 1. Reading the remainders from bottom to top gives 1010.

The NOT gate is called an inverter because it inverts the input signal. A HIGH (1) input becomes a LOW (0) output, and a LOW (0) input becomes a HIGH (1) output.

This is the complement law. If A is 0, is 1, so . If A is 1, is 0, so . In both cases, the result is 1.

In BCD, each decimal digit is represented by its 4-bit binary equivalent. The binary equivalent of the decimal number 7 is 0111.

A group of two adjacent cells containing '1' in a K-map is called a pair. It eliminates one variable from the resulting term.

The expression is a sum of several product terms (AB and A'C). This format is known as the Sum of Products (SOP) form.

To convert 1101 to decimal: .

NAND and NOR gates are called universal gates because any other logic gate (like AND, OR, NOT) can be constructed using only NAND gates or only NOR gates.

De Morgan's first theorem states that the complement of a sum is equal to the product of the complements. Thus, .

The Excess-3 code is found by adding 3 to the decimal digit and then converting the result to its 4-bit binary form. For digit 5, . The 4-bit binary for 8 is 1000.

A K-map is a graphical method used to simplify Boolean algebra expressions without having to use Boolean algebra theorems and laws.

The AND gate functions like a logical 'multiplication'. Its output is 1 only if all of its inputs are 1. If any input is 0, the output will be 0.

The hexadecimal system uses digits 0-9 and letters A-F. The letter F represents the highest value, which is 15 in the decimal system.

A minterm is a product term (AND term) in a Boolean expression that includes all the variables of the function. It corresponds to a single '1' in the function's truth table.

To convert binary to Gray code: The first bit (MSB) remains the same. Then, XOR each binary bit with the bit to its left. So, for 1011: G3=1; G2=1 XOR 0=1; G1=0 XOR 1=1; G0=1 XOR 1=0. The result is 1110.

The eXclusive-OR (XOR) gate gives a HIGH output only when its inputs are different. For a 2-input gate, this means one input is 0 and the other is 1, which means there is an odd number of HIGH inputs (one).

The Commutative law of addition states that the order of variables in an OR operation does not affect the result, so is the same as .

To convert from hexadecimal to binary, convert each hex digit to its 4-bit binary equivalent. '2' is 0010 and 'A' (which is 10 in decimal) is 1010. Combining them gives 00101010.

To convert Gray to binary: The MSB remains the same. Then, XOR the current binary bit with the next Gray code bit. For 1101: B3=1; B2=B3 XOR G2 = 1 XOR 1=0; B1=B2 XOR G1 = 0 XOR 0=0; B0=B1 XOR G0 = 0 XOR 1=1. The result is 1001.

A NOR gate is an OR gate followed by a NOT gate. The OR gate output is 0 only when all inputs are 0. The inverter then changes this 0 to a 1. For all other input combinations, the OR output is 1, which the NOR inverts to 0.

To convert to decimal, we use place values: \n \n \n.

The correct option follows directly from the given concept and definitions.

The correct option follows directly from the given concept and definitions.

To convert Gray to binary: \n1. The Most Significant Bit (MSB) of the binary number is the same as the MSB of the Gray code. So, Binary MSB = 1. \n2. XOR the current binary bit with the next Gray code bit to get the next binary bit. \n \n \n \n. \nSo, the binary number is 1001.

To convert binary to Gray: \n1. The MSB of the Gray code is the same as the MSB of the binary number. So, Gray MSB = 1. \n2. XOR each binary bit with the next binary bit to its right. \n \n \n \n \n. \nSo, the Gray code is 11101.

The XOR function is . This can be implemented using NAND gates as . This circuit configuration requires exactly 4 NAND gates. It's a standard and optimized implementation.

A 3-input NOR gate produces the output . When this output is passed through an inverter (NOT gate), the result is , which simplifies to . This is the function of a 3-input OR gate.

This expression is an example of the Consensus Theorem: . Here, X=A, Y=B, Z=C. The term is redundant. So, the expression simplifies to . \nExpanding this: . \nApplying consensus again on this result: .

To find the complement of , we use De Morgan's theorem. \n \n \n \n.

We expand each term to include all variables: \n1. . Then, . These are minterms . \n2. . These are minterms . \nCombining and removing duplicates, we get minterms 1, 4, 5, 6, 7. So, .

The POS form is derived from the maxterms, which correspond to the input combinations for which the function is 0. For a 3-variable function, the possible minterm indices are 0 to 7. The given minterms (where F=1) are 0, 2, 3, 5, 7. The missing indices are 1, 4, 6. These are the maxterm indices. Therefore, the canonical POS expression is .

Placing the minterms on a 3-variable K-map reveals two groups: \n1. A quad (group of 4) can be formed with minterms . This group simplifies to . \n2. A pair (group of 2) can be formed with minterms . This group simplifies to . \nCombining these gives the simplified expression .

When the minterms are plotted on a 4-variable K-map, two main groups can be identified: \n1. A group of four consisting of minterms . This group simplifies to the term . \n2. A group of four consisting of minterms . This group simplifies to the term . \nThe final simplified SOP expression is the sum of these terms: , which is the XNOR function of B and D.

To find the POS expression, we group the zeros on the K-map. The minterms where F=1 are 0, 1, 2, 4, 5. Therefore, the zeros are at minterm locations 3, 6, 7. \nOn the K-map: \n1. Group the zeros at (011) and (111). This pair simplifies to the SOP term . The corresponding maxterm is . \n2. Group the zeros at (110) and (111). This pair simplifies to the SOP term . The corresponding maxterm is . Wait, this is not optimal. Let's regroup.\n1. Group zeros at and . This gives SOP term , so POS term is . \n2. Group zeros at and . Let's try and . The group is . Let's try and . Wait, zeros are at 3,6,7. and simplifies to . This gives POS term . The zero at is grouped with , which gives . This gives POS term . This gives . This is not an option. Let's re-examine the minterms. Zeroes are 3,6,7. K-map for zeroes: Minterm 3 (A'BC), 6 (ABC'), 7 (ABC). Grouping 3 and 7 gives BC. Grouping 6 and 7 gives AB. So F' = AB + BC. Then F = (F')' = (AB+BC)' = (A'+B')(B'+C'). This is a correct derivation.

The output of the first XNOR gate is . The output of the second XNOR gate is . These two outputs are the inputs to the AND gate. Therefore, the final output expression is the logical AND of the two XNOR outputs, which is .

This is a reduction theorem, not the absorption law. It can be proven: \n (Distributive Law) \nSince , the expression becomes: \n. This rule, , is a fundamental identity used for simplification.

We convert the integer and fractional parts separately. \nInteger Part (43): \n remainder $3$ \n remainder $5$ \nReading remainders from bottom up gives . \nFractional Part (0.375): \n \nReading the integer part of the product gives . \nCombining both parts, we get .

BCD code uses 4 bits to represent decimal digits from 0 to 9. The bit patterns for 0 (0000) to 9 (1001) are valid. The bit pattern 1100 corresponds to decimal 12. Since this is greater than 9, it is an invalid representation for a single decimal digit in BCD.

We can expand the POS expression using the distributive law: \n \n \nSince and , the expression simplifies to: \n. This is the SOP expression for the XOR function.

The correct option follows directly from the given concept and definitions.