- State the purpose of registers in a microprocessor. (1)
- Registers are temporary storage locations on the processor which can be used to store data or instructions required by the CPU.
- Describe the purpose of two signals found on the control bus. (2)
- Read – signals a memory read operation is to take place
- Write – signals a memory write operation is to take place
- Clock – synchronises the signals being passed in and out of the processor
- Reset – clears all registers and busses of data
- Interrupt – tells the processor to save current process, run interrupt process until completed, then resume its activities (can be ignored)
- NMI – tells the processor to save current process, run interrupt process until completed, then resume its activities (cannot be ignored)
- An IT manager believes that a Pegasus computer with a 700MHz processor is faster than a Hercules computer with a 650MHz processor. Suggest two reasons why he may be incorrect. (2)
- Hercules computer may:
- have larger data bus (so can transfer more bits per clock cycle)
- have more RAM so can store more programs in fast-access memory
- What is the purpose of a control unit in a CPU? (1)
- A control unit manages all the other parts of the processor and makes sure that the program instructions of the computer are carried out in the correct order
- If a processor needs an instruction from memory, a read operation is carried out. Describe the steps of the memory read operation with reference to the processor, memory and buses. (4)
-
- Address to be read stored in MAR
- Contents of MAR transferred to Address bus
- Read control line activated
- Contents of memory at location from Address bus placed on Data bus
- Contents of Data bus transferred to MDR
- Processor clock speed is one factor which affects system performance. Name one other factor and describe how it affects system performance. (2)
- Data bus size – the larger the data bus the more bits that can be transferred in one clock cycle
- RAM – more fast-access memory improves system performance as the processor does not need to use slow-access Virtual RAM
- Graphics card with VRAM – takes away complex graphic calculations from processor, leaving it free to carry out other operations
- Calculate the maximum amount of addressable memory that a processor with 24-bit address bus and 16-bit data bus could access. Express your answer in the appropriate units, (3)
- Addressable memory (bits) =
- data bus (bits) x 2 ^ address bus (bits)
- 16 x 2^24 = 268435456 bits
- /8 = 33554432 bytes
- / 1024 = 32768 Kb
- / 1024 = 32 Mb
comments nothing to do with this post but its the only way i can think of to contact you sir as my glow mail still doesnt work, but could i tweet you @chspticm for help? the final part of my rgb program doesnt work
Sorry Darryn, come in at lunch time tomorrow and we can get to the bottom of it. Make sure that you have prepared your DFD & pseudocode for tomorrow.