McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Electrical Engineering , Bachelor of Science in Computer Science FY (BSc CS) , Bachelor of Computer Applications (BCA) , Bachelor of Science in Computer Science (BSc CS) .
| 51. |
Which of the following circuit parameters would be most likely to limit the maximum operating frequency of a flip-flop? |
| A. | setup and hold time |
| B. | clock pulse high and low time |
| C. | propagation delay time |
| D. | clock transition time |
| Answer» C. propagation delay time | |
| 52. |
What is the result of taking more samples during the quantization process? |
| A. | more errors in the analog-to-digital conversion |
| B. | more bit requirements |
| C. | more accurate signal representation |
| D. | more bit requirements and more accurate signal representation |
| Answer» D. more bit requirements and more accurate signal representation | |
| 53. |
Which A/D conversion method has a fixed conversion time? |
| A. | single-slope analog-to-digital converter |
| B. | dual-slope analog-to-digital converter |
| C. | digital-ramp analog-to-digital converter |
| D. | successive-approximation analog-to-digital converter |
| Answer» D. successive-approximation analog-to-digital converter | |
| 54. |
Which is a typical application of digital signal processing? |
| A. | noise elimination |
| B. | music signal processing |
| C. | image processing |
| D. | all of the above |
| Answer» D. all of the above | |
| 55. |
If a DAC has a full-scale, or maximum, output of 12 V and accuracy of 0.1%, then the maximum error for any output voltage is ________. |
| A. | 12 v |
| B. | 120 mv |
| C. | 12 mv |
| D. | 0 v |
| Answer» C. 12 mv | |
| 56. |
What do we call the manipulation of an analog signal in a digital domain? |
| A. | analog-to-digital conversion |
| B. | digital-to-analog conversion |
| C. | digital signal processing |
| D. | signal filtering |
| Answer» B. digital-to-analog conversion | |
| 57. |
How many address bits are needed to select all memory locations in the 2118 16K × 1 RAM? |
| A. | 8 |
| B. | 10 |
| C. | 14 |
| D. | 16 |
| Answer» C. 14 | |
| 58. |
The check sum method of testing a ROM: |
| A. | indicates if the data in more than one memory location is incorrect. |
| B. | provides a means for locating and correcting data errors in specific memory locations. |
| C. | allows data errors to be pinpointed to a specific memory location. |
| D. | simply indicates that the contents of the rom are incorrect. |
| Answer» D. simply indicates that the contents of the rom are incorrect. | |
| 59. |
Refer to the given figures (a) and (b). A logic analyzer is used to check the circuit in figure (a) and displays the waveforms shown in figure (b). The actual analyzer display shows all four data outputs, Q0-Q3. The analyzer's cursor is placed at position X and all four of the data output lines show a LOW level output. What is wrong, if anything, with the circuit? |
| A. | nothing is wrong, according to the display. the outputs are in the open state and should show zero output voltage. |
| B. | the circuit is in the read mode and the outputs, q0-q3, should reflect the contents of the memory at that address. the chip is defective; replace the chip. |
| C. | the circuit is in the mode and should be writing the contents of the selected address to q0–q3. |
| D. | the q0–q3 lines can be either low or high, since the chip is in the tristate mode in which case their level is unpredictable. |
| Answer» D. the q0–q3 lines can be either low or high, since the chip is in the tristate mode in which case their level is unpredictable. | |
| 60. |
What is the meaning of RAM, and what is its primary role? |
| A. | readily available memory; it is the first level of memory used by the computer in all of its operations. |
| B. | random access memory; it is memory that can be reached by any sub- system within a computer, and at any time. |
| C. | random access memory; it is the memory used for short-term temporary data storage within the computer. |
| D. | resettable automatic memory; it is memory that can be used and then automatically reset, or cleared, after being read from or written to. |
| Answer» C. random access memory; it is the memory used for short-term temporary data storage within the computer. | |
| 61. |
The storage element for a static RAM is the ________. |
| A. | diode |
| B. | resistor |
| C. | capacitor |
| D. | flip-flop |
| Answer» D. flip-flop | |
| 62. |
In a DRAM, what is the state of R/W during a read operation? |
| A. | low |
| B. | high |
| C. | hi-z |
| D. | none of the above |
| Answer» B. high | |
| 63. |
The condition occurring when two or more devices try to write data to a bus simultaneously is called ________. |
| A. | address decoding |
| B. | bus contention |
| C. | bus collisions |
| D. | address multiplexing |
| Answer» B. bus contention | |
| 64. |
The difference between a PLA and a PAL is: |
| A. | the pla has a programmable or plane and a programmable and plane, while the pal only has a programmable and plane. |
| B. | the pal has a programmable or plane and a programmable and plane, while the pla only has a programmable and plane. |
| C. | the pal has more possible product terms than the pla. |
| D. | pals and plas are the same thing. |
| Answer» A. the pla has a programmable or plane and a programmable and plane, while the pal only has a programmable and plane. | |
| 65. |
ALM is the acronym for ________. |
| A. | array logic matrix |
| B. | arithmetic logic module |
| C. | asynchronous local modulator |
| D. | adaptive logic module |
| Answer» D. adaptive logic module | |
| 66. |
The GAL16V8 has: |
| A. | 16 dedicated inputs. |
| B. | 8 special function pins. |
| C. | 8 pins that are used as inputs or outputs. |
| D. | all of the above |
| Answer» C. 8 pins that are used as inputs or outputs. | |
| 67. |
PALs tend to execute ________ logic. |
| A. | sap |
| B. | sop |
| C. | pla |
| D. | spd |
| Answer» B. sop | |
| 68. |
How many pins are in an EDF10K70 package? |
| A. | 70 |
| B. | 140 |
| C. | 240 |
| D. | 532 |
| Answer» C. 240 | |
| 69. |
Convert hexadecimal value 16 to decimal. |
| A. | 2210 |
| B. | 1610 |
| C. | 1010 |
| D. | 2010 |
| Answer» A. 2210 | |
| 70. |
Convert the following decimal number to 8-bit binary. |
| A. | 101110112 |
| B. | 110111012 |
| C. | 101111012 |
| D. | 101111002 |
| Answer» A. 101110112 | |
| 71. |
Convert binary 111111110010 to hexadecimal. |
| A. | ee216 |
| B. | ff216 |
| C. | 2fe16 |
| D. | fd216 |
| Answer» B. ff216 | |
| 72. |
Convert the binary number 1001.00102 to decimal. |
| A. | 90.125 |
| B. | 9.125 |
| C. | 125 |
| D. | 12.5 |
| Answer» B. 9.125 | |
| 73. |
One hex digit is sometimes referred to as a(n): |
| A. | byte |
| B. | nibble |
| C. | grouping |
| D. | instruction |
| Answer» B. nibble | |
| 74. |
Which of the following is the most widely used alphanumeric code for computer input and output? |
| A. | gray |
| B. | ascii |
| C. | parity |
| D. | ebcdic |
| Answer» B. ascii | |
| 75. |
If a typical PC uses a 20-bit address code, how much memory can the CPU address? |
| A. | 20 mb |
| B. | 10 mb |
| C. | 1 mb |
| D. | 580 mb |
| Answer» C. 1 mb | |
| 76. |
Convert 59.7210 to BCD. |
| A. | 111011 |
| B. | 01011001.01110010 |
| C. | 1110.11 |
| D. | 0101100101110010 |
| Answer» B. 01011001.01110010 | |
| 77. |
Convert 8B3F16 to binary. |
| A. | 35647 |
| B. | 011010 |
| C. | 1011001111100011 |
| D. | 1000101100111111 |
| Answer» D. 1000101100111111 | |
| 78. |
Which is typically the longest: bit, byte, nibble, word? |
| A. | bit |
| B. | byte |
| C. | nibble |
| D. | word |
| Answer» D. word | |
| 79. |
Assign the proper odd parity bit to the code 111001. |
| A. | 1111011 |
| B. | 1111001 |
| C. | 0111111 |
| D. | 0011111 |
| Answer» B. 1111001 | |
| 80. |
Convert decimal 64 to binary. |
| A. | 01010010 |
| B. | 01000000 |
| C. | 00110110 |
| D. | 01001000 |
| Answer» B. 01000000 | |
| 81. |
Convert hexadecimal value C1 to binary. |
| A. | 11000001 |
| B. | 1000111 |
| C. | 111000100 |
| D. | 111000001 |
| Answer» A. 11000001 | |
| 82. |
The given hexadecimal number (1E.53)16 is equivalent to |
| A. | (35.684)8 |
| B. | (36.246)8 |
| C. | (34.340)8 |
| D. | (35.599)8 |
| Answer» B. (36.246)8 | |
| 83. |
The octal number (651.124)8 is equivalent to |
| A. | 16 |
| B. | (1b0.10)16 |
| C. | (1a8.a3)16 |
| D. | (1b0.b0)16 |
| Answer» A. 16 | |
| 84. |
The octal equivalent of the decimal number (417)10 is |
| A. | (641)8 |
| B. | (619)8 |
| C. | (640)8 |
| D. | (598)8 |
| Answer» A. (641)8 | |
| 85. |
Convert the hexadecimal number (1E2)16 to decimal: |
| A. | 480 |
| B. | 483 |
| C. | 482 |
| D. | 484 |
| Answer» C. 482 | |
| 86. |
(170)10 is equivalent to |
| A. | (fd)16 |
| B. | (df)16 |
| C. | (aa)16 |
| D. | (af)16 |
| Answer» C. (aa)16 | |
| 87. |
Convert the binary number (01011.1011)2 into decimal: |
| A. | (11.6875)10 |
| B. | (11.5874)10 |
| C. | (10.9876)10 |
| D. | (10.7893)10 |
| Answer» A. (11.6875)10 | |
| 88. |
1011)2 = (11.6875)10 |
| A. | (111101)2 |
| B. | (010100)2 |
| C. | (111100)2 |
| D. | (101010)2 |
| Answer» B. (010100)2 | |
| 89. |
On addition of +38 and -20 using 2’s complement, we get |
| A. | 11110001 |
| B. | 100001110 |
| C. | 010010 |
| D. | 110101011 |
| Answer» C. 010010 | |
| 90. |
On addition of -46 and +28 using 2’s complement, we get |
| A. | 00101110 |
| B. | 0101110 |
| C. | 00101111 |
| D. | 1001111 |
| Answer» B. 0101110 | |
| 91. |
On subtracting +28 from +29 using 2’s complement, we get |
| A. | 11111010 |
| B. | 111111001 |
| C. | 100001 |
| D. | 1 |
| Answer» D. 1 | |
| 92. |
The decimal number 10 is represented in its BCD form as |
| A. | 10100000 |
| B. | 01010111 |
| C. | 00010000 |
| D. | 00101011 |
| Answer» C. 00010000 | |
| 93. |
When numbers, letters or words are represented by a special group of symbols, this process is called |
| A. | decoding |
| B. | encoding |
| C. | digitizing |
| D. | inverting |
| Answer» B. encoding | |
| 94. |
Carry out BCD subtraction for (68) – (61) using 10’s complement method. |
| A. | 00000111 |
| B. | 01110000 |
| C. | 100000111 |
| D. | 011111000 |
| Answer» A. 00000111 | |
| 95. |
How many bits would be required to encode decimal numbers 0 to 9999 in straight binary codes? |
| A. | 12 |
| B. | 14 |
| C. | 16 |
| D. | 18 |
| Answer» B. 14 | |
| 96. |
The decimal equivalent of the excess-3 number 110010100011.01110101 is |
| A. | 970.42 |
| B. | 1253.75 |
| C. | 861.75 |
| D. | 1132.87 |
| Answer» A. 970.42 | |
| 97. |
In boolean algebra, the OR operation is performed by which properties? |
| A. | associative properties |
| B. | commutative properties |
| C. | distributive properties |
| D. | all of the mentioned |
| Answer» D. all of the mentioned | |
| 98. |
The expression for Absorption law is given by |
| A. | a + ab = a |
| B. | a + ab = b |
| C. | ab + aa’ = a |
| D. | a + b = b + a |
| Answer» A. a + ab = a | |
| 99. |
According to boolean law: A + 1 = ? |
| A. | 1 |
| B. | a |
| C. | 0 |
| D. | a’ |
| Answer» A. 1 | |
| 100. |
The involution of A is equal to |
| A. | a |
| B. | a’ |
| C. | 1 |
| D. | 0 |
| Answer» A. a | |
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