EEE 120 FINAL EXAM QUESTIONS ANSWERED CORRECTLY LATEST UPDATE 2026
AND gate truth table - Answers Output is 1 ONLY when ALL inputs are 1. (0,0→0; 0,1→0; 1,0→0;
1,1→1)
OR gate truth table - Answers Output is 1 when AT LEAST ONE input is 1. (0,0→0; 0,1→1; 1,0→1;
1,1→1)
NOT gate (Inverter) truth table - Answers Output is the complement of the input. (0→1; 1→0)
NAND gate truth table - Answers Output is 0 ONLY when ALL inputs are 1. (0,0→1; 0,1→1; 1,0→1;
1,1→0) — complement of AND
NOR gate truth table - Answers Output is 1 ONLY when ALL inputs are 0. (0,0→1; 0,1→0; 1,0→0;
1,1→0) — complement of OR
XOR gate truth table - Answers Output is 1 when inputs are DIFFERENT. (0,0→0; 0,1→1; 1,0→1;
1,1→0) — Y = A'B + AB'
XNOR gate - Answers Output is 1 when inputs are the SAME. Complement of XOR. (0,0→1; 0,1→0;
1,0→0; 1,1→1)
NAND gate — alternate symbol (DeMorgan) - Answers OR gate with INVERTED (bubbled) inputs.
Equivalent to (AB)'
NOR gate — alternate symbol (DeMorgan) - Answers AND gate with INVERTED (bubbled) inputs.
Equivalent to (A+B)'
Bubble on a gate input - Answers Represents inversion (NOT) at that input. A bubble on the output
also inverts the output.
Universal gate - Answers A gate that can implement any Boolean function by itself. NAND and NOR
are both universal gates.
DeMorgan's Law #1 - Answers (A · B)' = A' + B' — The complement of an AND is the OR of the
complements
DeMorgan's Law #2 - Answers (A + B)' = A' · B' — The complement of an OR is the AND of the
complements
How to apply DeMorgan's Law - Answers Break the bar, change the operator (AND↔OR).
Complement each literal.
NAND expressed via DeMorgan - Answers (AB)' = A' + B'
NOR expressed via DeMorgan - Answers (A+B)' = A' · B'
Idempotency rule - Answers A · A = A and A + A = A — combining identical inputs eliminates one gate
OR with 1 rule - Answers A + 1 = 1 — any OR with a constant 1 produces 1
AND with 0 rule - Answers A · 0 = 0 — any AND with a constant 0 produces 0
Complement rule - Answers A · A' = 0 and A + A' = 1
Double complement rule - Answers (A')' = A
Minterm - Answers A product (AND) term in which every variable appears exactly once
(complemented or uncomplemented). Row where output = 1.
Maxterm - Answers A sum (OR) term in which every variable appears exactly once. Row where
output = 0.
Sum of Minterms (shorthand) - Answers Σm(#, #, ...) — lists the row numbers where the output is 1
Product of Maxterms (shorthand) - Answers ΠM(#, #, ...) — lists the row numbers where the output is
0
Sum of Products (SOP) form - Answers Boolean expression written as OR of AND terms (minterms).
Example: Y = AB' + A'B + AB
Product of Sums (POS) form - Answers Boolean expression written as AND of OR terms (maxterms).
Example: Y = (A+B)(A'+C)
Canonical SOP - Answers SOP where every product term is a minterm (contains all variables)
Canonical POS - Answers POS where every sum term is a maxterm (contains all variables)
How to convert truth table → SOP - Answers Write one minterm (AND term) for every row where
output = 1; OR all minterms together
How to convert truth table → POS - Answers Write one maxterm (OR term) for every row where
output = 0; AND all maxterms together
Karnaugh Map (K-map) - Answers A graphical tool for minimizing Boolean functions by grouping
adjacent 1s (for SOP) or 0s (for POS)
K-map grouping rule - Answers Groups must be rectangular, contain only 1s (for SOP), and have a size
that is a power of 2 (1,2,4,8,...). No diagonals.
, K-map: eliminating a variable - Answers When a variable changes within a group, it is eliminated from
the resulting product term
Prime implicant - Answers The largest possible grouping of 1s in a K-map that cannot be absorbed
into a larger group
Essential prime implicant - Answers A prime implicant that is the ONLY group covering a particular
minterm — must be included in the minimal expression
How to get minimum SOP from K-map - Answers 1) Circle groups of 1s (sizes = powers of 2). 2) Each
group = one product term (drop changing variables). 3) OR all product terms.
How to get minimum POS from K-map - Answers 1) Circle groups of 0s. 2) Each group = one sum term
using INVERTED variables that don't change. 3) AND all sum terms.
Don't-care conditions in K-map - Answers Marked as X; can be treated as 0 or 1 — choose whichever
makes the grouping larger
K-map wrap-around - Answers The edges of a K-map are adjacent — top/bottom rows and left/right
columns can be grouped together
Binary (base 2) - Answers Number system using only digits 0 and 1. Each position is a power of 2.
Hexadecimal (base 16) - Answers Number system using digits 0-9 and A-F (A=10 ... F=15). Each hex
digit = 4 bits.
Octal (base 8) - Answers Number system using digits 0-7. Each octal digit = 3 bits.
Convert binary → decimal - Answers Multiply each bit by 2^(position) and sum. Example: 1011₂ =
8+0+2+1 = 11₁₀
Convert decimal → binary - Answers Repeatedly divide by 2; remainders (read bottom-up) give the
binary representation
Convert hex → binary - Answers Replace each hex digit with its 4-bit binary equivalent. Example: A3₁₆
= 1010 0011₂
Convert binary → hex - Answers Group bits into 4-bit nibbles from the right; convert each nibble to a
hex digit
Most Significant Bit (MSB) - Answers The leftmost (highest-weight) bit in a binary number
Least Significant Bit (LSB) - Answers The rightmost (lowest-weight) bit in a binary number
Sign bit (signed binary) - Answers The MSB in a signed binary representation: 0 = positive, 1 =
negative
One's complement - Answers Bitwise NOT of a binary number — flip every 0 to 1 and every 1 to 0
Two's complement (operation) - Answers Flip all bits (one's complement) then add 1. Used to get the
negative of a number for subtraction.
Two's complement representation - Answers The standard way to represent signed integers. Positive
numbers start with 0; negative numbers start with 1.
Two's complement subtraction - Answers A - B = A + (two's complement of B). Convert B, add, ignore
carry out of the MSB.
Overflow condition (signed addition) - Answers Overflow occurs when the carry INTO the MSB
(C_MSB) and the carry OUT OF the MSB (C_out) are DIFFERENT
Overflow: positive + positive = negative - Answers Overflow! Two positive numbers produced a
negative result — the sign bit flipped incorrectly
Overflow: negative + negative = positive - Answers Overflow! Two negative numbers produced a
positive result — the sign bit flipped incorrectly
Range of n-bit two's complement - Answers From -2^(n-1) to +2^(n-1) - 1. Example: 4-bit → -8 to +7
How to read a negative two's complement number - Answers The MSB contributes -2^(n-1). Add the
remaining bit values. Example: 1101₂ = -8+4+1 = -3
Half adder - Answers Adds two 1-bit inputs A and B. Outputs: Sum = A XOR B, Carry = A AND B. No
carry-in input.
Full adder - Answers Adds three 1-bit inputs A, B, Cin. Sum = A XOR B XOR Cin. Cout = AB + ACin +
BCin.
Ripple carry adder - Answers Multiple full adders chained together; the carry out of each stage feeds
into the carry in of the next
Adder/subtractor circuit - Answers Uses a full adder with XOR gates on one operand and a mode bit
(M). M=0 → add; M=1 → subtract (XOR inverts B, Cin=1 adds 1 for two's complement)
Multiplexer (MUX) - Answers Selects ONE of 2^n data inputs to pass to the output based on n select
(control) lines
AND gate truth table - Answers Output is 1 ONLY when ALL inputs are 1. (0,0→0; 0,1→0; 1,0→0;
1,1→1)
OR gate truth table - Answers Output is 1 when AT LEAST ONE input is 1. (0,0→0; 0,1→1; 1,0→1;
1,1→1)
NOT gate (Inverter) truth table - Answers Output is the complement of the input. (0→1; 1→0)
NAND gate truth table - Answers Output is 0 ONLY when ALL inputs are 1. (0,0→1; 0,1→1; 1,0→1;
1,1→0) — complement of AND
NOR gate truth table - Answers Output is 1 ONLY when ALL inputs are 0. (0,0→1; 0,1→0; 1,0→0;
1,1→0) — complement of OR
XOR gate truth table - Answers Output is 1 when inputs are DIFFERENT. (0,0→0; 0,1→1; 1,0→1;
1,1→0) — Y = A'B + AB'
XNOR gate - Answers Output is 1 when inputs are the SAME. Complement of XOR. (0,0→1; 0,1→0;
1,0→0; 1,1→1)
NAND gate — alternate symbol (DeMorgan) - Answers OR gate with INVERTED (bubbled) inputs.
Equivalent to (AB)'
NOR gate — alternate symbol (DeMorgan) - Answers AND gate with INVERTED (bubbled) inputs.
Equivalent to (A+B)'
Bubble on a gate input - Answers Represents inversion (NOT) at that input. A bubble on the output
also inverts the output.
Universal gate - Answers A gate that can implement any Boolean function by itself. NAND and NOR
are both universal gates.
DeMorgan's Law #1 - Answers (A · B)' = A' + B' — The complement of an AND is the OR of the
complements
DeMorgan's Law #2 - Answers (A + B)' = A' · B' — The complement of an OR is the AND of the
complements
How to apply DeMorgan's Law - Answers Break the bar, change the operator (AND↔OR).
Complement each literal.
NAND expressed via DeMorgan - Answers (AB)' = A' + B'
NOR expressed via DeMorgan - Answers (A+B)' = A' · B'
Idempotency rule - Answers A · A = A and A + A = A — combining identical inputs eliminates one gate
OR with 1 rule - Answers A + 1 = 1 — any OR with a constant 1 produces 1
AND with 0 rule - Answers A · 0 = 0 — any AND with a constant 0 produces 0
Complement rule - Answers A · A' = 0 and A + A' = 1
Double complement rule - Answers (A')' = A
Minterm - Answers A product (AND) term in which every variable appears exactly once
(complemented or uncomplemented). Row where output = 1.
Maxterm - Answers A sum (OR) term in which every variable appears exactly once. Row where
output = 0.
Sum of Minterms (shorthand) - Answers Σm(#, #, ...) — lists the row numbers where the output is 1
Product of Maxterms (shorthand) - Answers ΠM(#, #, ...) — lists the row numbers where the output is
0
Sum of Products (SOP) form - Answers Boolean expression written as OR of AND terms (minterms).
Example: Y = AB' + A'B + AB
Product of Sums (POS) form - Answers Boolean expression written as AND of OR terms (maxterms).
Example: Y = (A+B)(A'+C)
Canonical SOP - Answers SOP where every product term is a minterm (contains all variables)
Canonical POS - Answers POS where every sum term is a maxterm (contains all variables)
How to convert truth table → SOP - Answers Write one minterm (AND term) for every row where
output = 1; OR all minterms together
How to convert truth table → POS - Answers Write one maxterm (OR term) for every row where
output = 0; AND all maxterms together
Karnaugh Map (K-map) - Answers A graphical tool for minimizing Boolean functions by grouping
adjacent 1s (for SOP) or 0s (for POS)
K-map grouping rule - Answers Groups must be rectangular, contain only 1s (for SOP), and have a size
that is a power of 2 (1,2,4,8,...). No diagonals.
, K-map: eliminating a variable - Answers When a variable changes within a group, it is eliminated from
the resulting product term
Prime implicant - Answers The largest possible grouping of 1s in a K-map that cannot be absorbed
into a larger group
Essential prime implicant - Answers A prime implicant that is the ONLY group covering a particular
minterm — must be included in the minimal expression
How to get minimum SOP from K-map - Answers 1) Circle groups of 1s (sizes = powers of 2). 2) Each
group = one product term (drop changing variables). 3) OR all product terms.
How to get minimum POS from K-map - Answers 1) Circle groups of 0s. 2) Each group = one sum term
using INVERTED variables that don't change. 3) AND all sum terms.
Don't-care conditions in K-map - Answers Marked as X; can be treated as 0 or 1 — choose whichever
makes the grouping larger
K-map wrap-around - Answers The edges of a K-map are adjacent — top/bottom rows and left/right
columns can be grouped together
Binary (base 2) - Answers Number system using only digits 0 and 1. Each position is a power of 2.
Hexadecimal (base 16) - Answers Number system using digits 0-9 and A-F (A=10 ... F=15). Each hex
digit = 4 bits.
Octal (base 8) - Answers Number system using digits 0-7. Each octal digit = 3 bits.
Convert binary → decimal - Answers Multiply each bit by 2^(position) and sum. Example: 1011₂ =
8+0+2+1 = 11₁₀
Convert decimal → binary - Answers Repeatedly divide by 2; remainders (read bottom-up) give the
binary representation
Convert hex → binary - Answers Replace each hex digit with its 4-bit binary equivalent. Example: A3₁₆
= 1010 0011₂
Convert binary → hex - Answers Group bits into 4-bit nibbles from the right; convert each nibble to a
hex digit
Most Significant Bit (MSB) - Answers The leftmost (highest-weight) bit in a binary number
Least Significant Bit (LSB) - Answers The rightmost (lowest-weight) bit in a binary number
Sign bit (signed binary) - Answers The MSB in a signed binary representation: 0 = positive, 1 =
negative
One's complement - Answers Bitwise NOT of a binary number — flip every 0 to 1 and every 1 to 0
Two's complement (operation) - Answers Flip all bits (one's complement) then add 1. Used to get the
negative of a number for subtraction.
Two's complement representation - Answers The standard way to represent signed integers. Positive
numbers start with 0; negative numbers start with 1.
Two's complement subtraction - Answers A - B = A + (two's complement of B). Convert B, add, ignore
carry out of the MSB.
Overflow condition (signed addition) - Answers Overflow occurs when the carry INTO the MSB
(C_MSB) and the carry OUT OF the MSB (C_out) are DIFFERENT
Overflow: positive + positive = negative - Answers Overflow! Two positive numbers produced a
negative result — the sign bit flipped incorrectly
Overflow: negative + negative = positive - Answers Overflow! Two negative numbers produced a
positive result — the sign bit flipped incorrectly
Range of n-bit two's complement - Answers From -2^(n-1) to +2^(n-1) - 1. Example: 4-bit → -8 to +7
How to read a negative two's complement number - Answers The MSB contributes -2^(n-1). Add the
remaining bit values. Example: 1101₂ = -8+4+1 = -3
Half adder - Answers Adds two 1-bit inputs A and B. Outputs: Sum = A XOR B, Carry = A AND B. No
carry-in input.
Full adder - Answers Adds three 1-bit inputs A, B, Cin. Sum = A XOR B XOR Cin. Cout = AB + ACin +
BCin.
Ripple carry adder - Answers Multiple full adders chained together; the carry out of each stage feeds
into the carry in of the next
Adder/subtractor circuit - Answers Uses a full adder with XOR gates on one operand and a mode bit
(M). M=0 → add; M=1 → subtract (XOR inverts B, Cin=1 adds 1 for two's complement)
Multiplexer (MUX) - Answers Selects ONE of 2^n data inputs to pass to the output based on n select
(control) lines
