DIC201 Progress Test 1 Practice - Ch1-4

Q1 (Ch1): Which statement correctly distinguishes an analog quantity from a digital quantity? A) Analog has discrete values; digital has continuous values B) Analog has continuous values; digital has a discrete set of values C) Both have continuous values, just different voltage ranges D) Digital quantities can only be represented in decimal

B) Analog has continuous values; digital has a discrete set of values

Q2 (Ch1): A digital waveform has a period of 4 microseconds and a pulse width (HIGH time) of 1 microsecond. What are its frequency and duty cycle? A) 400 kHz, 25% B) 250 kHz, 40% C) 250 kHz, 25% D) 4 MHz, 75%

C) 250 kHz, 25%

Q3 (Ch1): What distinguishes a periodic digital waveform from a nonperiodic one? A) Periodic waveforms only occur in analog circuits B) A periodic waveform repeats at a fixed time interval; a nonperiodic one does not C) Nonperiodic waveforms always have a 50% duty cycle D) Periodic waveforms cannot contain pulses, only sine waves

B) A periodic waveform repeats at a fixed time interval; a nonperiodic one does not

Q4 (Ch1): Which statement about the three basic logic functions is correct? A) An AND gate's output is HIGH when any input is HIGH B) An OR gate's output is HIGH only when all inputs are HIGH C) A NOT gate's output is the complement of its single input D) An OR gate's output is LOW when any input is HIGH

C) A NOT gate's output is the complement of its single input

Q5 (Ch1): What is the key difference between combinational and sequential logic? A) Combinational logic has memory; sequential logic does not B) Sequential logic output depends only on the current inputs C) Combinational output depends only on current inputs (no memory); sequential output depends on current inputs plus previous state (has memory) D) There is no functional difference, only naming convention

C) Combinational output depends only on current inputs (no memory); sequential output depends on current inputs plus previous state (has memory)

Q6 (Ch2): Convert 1011011 (binary) to decimal, hexadecimal, and octal. A) 91, 5B(16), 133(8) B) 91, 5A(16), 133(8) C) 83, 5B(16), 123(8) D) 91, 5B(16), 143(8)

A) 91, 5B(16), 133(8)

Q7 (Ch2): Convert decimal 173 to binary using the sum-of-weights method. A) 10101101 B) 10101110 C) 11001101 D) 10110101

A) 10101101

Q8 (Ch2): Represent decimal -45 as an 8-bit signed binary number using 2's complement. A) 11010010 B) 11010011 C) 10101101 D) 11001101

B) 11010011

Q9 (Ch2): Add these signed 8-bit 2's complement numbers: 01111000 + 00010100. What's the result, and does overflow occur? A) 10001100, overflow occurs B) 10001100, no overflow C) 01001100, overflow occurs D) 10001101, no overflow

A) 10001100, overflow occurs

Q10 (Ch2): Convert decimal 39 to BCD (8421 code). A) 0011 1001 B) 0100 1001 C) 0011 1000 D) 0010 1001

A) 0011 1001

Q11 (Ch2): Convert binary 1101 to Gray code. A) 1011 B) 1101 C) 1110 D) 0111

A) 1011

Q12 (Ch2): Why is Gray code preferred over pure binary in devices like rotary shaft encoders? A) It uses fewer bits than binary for the same range B) Only one bit changes between adjacent values, reducing the risk of an erroneous reading during a transition C) Gray code values are always even D) It's easier to convert directly to decimal

B) Only one bit changes between adjacent values, reducing the risk of an erroneous reading during a transition

Q13 (Ch3): For a 3-input NOR gate with inputs A, B, C, when is the output HIGH? A) When at least one input is HIGH B) When all three inputs are HIGH C) Only when all three inputs are LOW (A=B=C=0) D) When exactly two inputs are HIGH

C) Only when all three inputs are LOW (A=B=C=0)

Q14 (Ch3): Why is the NAND gate called a "universal gate"? A) It has more inputs than any other gate B) Any logic function (AND, OR, NOT, etc.) can be implemented using only NAND gates C) It's the fastest logic gate available D) It can only be used in combinational circuits, not sequential ones

B) Any logic function (AND, OR, NOT, etc.) can be implemented using only NAND gates

Q15 (Ch3): For a 2-input XOR gate, when is the output HIGH? A) When both inputs are the same (both 0 or both 1) B) When the inputs are different (exactly one is HIGH) C) When both inputs are HIGH D) When both inputs are LOW

B) When the inputs are different (exactly one is HIGH)

Q16 (Ch3): Input A is HIGH for the first half of a clock period; input B is HIGH for the entire period. What does the output of an AND gate driven by A and B look like? A) HIGH for the whole period B) LOW for the whole period C) HIGH for the first half, LOW for the second half D) HIGH for the second half only

C) HIGH for the first half, LOW for the second half

Q17 (Ch3): What is the dual (equivalent) symbol for a NAND gate, and when would you use it in a logic diagram? A) A NOR symbol; use it when the gate functions as a NOR B) A negative-OR symbol (OR shape with input bubbles); use it when the gate is functioning as a negative-OR, to make the diagram easier to read C) A plain AND symbol with no bubble; used interchangeably at all times D) There is no dual symbol for NAND, only for NOR

B) A negative-OR symbol (OR shape with input bubbles); use it when the gate is functioning as a negative-OR, to make the diagram easier to read

Q18 (Ch3): For an alarm system where the output should be HIGH only when door sensor D is HIGH AND the system is armed S is HIGH, which single gate implements this? A) OR gate B) NAND gate C) AND gate D) XOR gate

C) AND gate

Q19 (Ch4): Simplify F = AB + AB'. A) F = A B) F = B C) F = AB D) F = 1

A) F = A

Q20 (Ch4): Simplify F = A + A'B. A) F = A B) F = A + B C) F = AB D) F = B

B) F = A + B

Q21 (Ch4): Apply DeMorgan's theorem to fully expand (A' + BC)'. A) A'B'C' B) A' + B'C' C) AB' + AC' D) A'BC

C) AB' + AC'

Q22 (Ch4): Apply DeMorgan's theorem to fully expand (AB'C)'. A) A' + B + C' B) A'.B.C' C) A + B' + C D) A' + B' + C'

A) A' + B + C'

Q23 (Ch4): For F = A + BC (3 variables A, B, C), how many minterms (rows where F=1) are in its truth table? A) 3 B) 4 C) 5 D) 6

C) 5

Q24 (Ch4): Given a 3-variable K-map with minterms 1, 3, 5, 7 = 1 (all others 0), find the simplified SOP. A) F = A B) F = C C) F = A'C D) F = B

B) F = C

Q25 (Ch4): Given a 4-variable K-map with minterms 4, 5, 6, 7, 12, 13, 14, 15 = 1 (all others 0), find the simplified SOP. A) F = A B) F = B C) F = AB D) F = C

B) F = B

Q26 (Ch4): Simplify F = A'B'CD + A'B'CD' + AB'CD + AB'CD' using a K-map. A) F = B'C B) F = B'D C) F = A'C D) F = BC'

A) F = B'C