For the 8 - bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory?
| A. | 10011001 |
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| B. | 00011001 |
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| C. | 00111000 |
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| D. | 11000110 |
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Option: B Explanation :
The Hamming Word initially calculated was:
Doing an exclusive-OR of 0111 and 1101 yields 1010 indicating an error in bit 10 of the Hamming Word. Thus, the data word read from memory was 00011001. Click on Discuss to view users comments. |
Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that this microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor?
| A. | 8x106 bytes/sec |
| B. | 4x106 bytes/sec |
| C. | 16x106 bytes/sec |
| D. | 4x109 bytes/sec |
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Option: B Explanation :
Since minimum bus cycle duration = 4 clock cycles and bus clock = 8 MHz Click on Discuss to view users comments. |
The RST 7 instruction in 8085 microprocessor is equivalent to :
| A. | CALL 0010 H |
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| B. | CALL 0034 H |
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| C. | CALL 0038 H |
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| D. | CALL 003C H |
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Option: C Explanation : The Table shows the vector addresses of all interrupts.
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The equivalent hexadecimal notation for octal number 2550276 is :
| A. | FADED |
| B. | AEOBE |
| C. | ADOBE |
| D. | ACABE |
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Option: C Explanation : binary equivalent of 2550276 is 010101101000010111110. now writing in pair of 4 bits and converting to hexadecimal it becomes AD0BE Click on Discuss to view users comments. |
The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on an average for executing an instruction. The fifty percent of the cycles use memory bus. A memory read/ write employs one machine cycle. For execution of the programs, the system utilizes 90 percent of the CPU time. For block data transfer, an IO device is attached to the system while CPU executes the background programs continuously. What is the maximum IO data transfer rate if programmed IO data transfer technique is used?
| A. | 500 Kbytes/sec |
| B. | 2.2 Mbytes/sec |
| C. | 125 Kbytes/sec |
| D. | 250 Kbytes/sec |
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Option: D Explanation :
CPU speed: 10^6 instructions/sec In status Check status Branch R/W data in memory Max. data transfer rate = CPU Speed / 4 = 10^6 / 4 = 250 Kbytes/sec Click on Discuss to view users comments. |
