Data phases - computer architecture, Computer Engineering

Assignment Help:

Data phases:

After the address phase (particularly, starting with the cycle that DEVSEL# goes low) comes a burst of one or more data phases. In all the cases, initiator drives active-low byte choose signals on the C/BE[3:0]# lines, however the data on the AD[31:0] may be driven by the initiator (on case of writes) or target (in case of reads).

During data phases, the C/BE [3:0] # lines are interpreted as active-low byte enables. In case of a write, the asserted signals indicate which of the four bytes on the AD bus are to be written to the addressed location. In the case of a read, they specify which bytes the initiator is interested in. For reads, it is always permissible to ignore the byte enable signals and simply return all 32 bits; cacheable memory resources are needed to always return 32 valid bits. The byte enables are chiefly useful for I/O space accesses where reads have side effects.

A data phase with all 4 C/BE# lines deserted is explicitly allowed by the PCI standard, and must have no effect on the target (other than to advance the address in the burst access in progress).

The data phase continues till both parties are ready to complete the transfer and continue to the next data phase. The initiator asserts the IRDY# (initiator is ready) when it no longer required to wait, whereas the target asserts TRDY# (target ready). Whichever side is providing the data have to drive it on the AD bus before asserting its ready signal.

Once one of the contributors asserts its ready signal, this cannot become un-ready or otherwise alter its control signals till the end of the data phase. The data recipient have to latch the AD bus each cycle till it sees IRDY# and TRDY#  both asserted, which marks the end of the current data phase and mention that the just -latched data is the word to be transferred.

To maintain complete burst speed, the data sender then has half a clock cycle after seeing TRDY# and IRDY# both asserted to drive the next word onto the AD bus.

It continues the address cycle shown above, supposing a single address cycle having medium DEVSEL, so the target responds for clock 3in time. Though, at that time, neither side is ready to transfer data. For clock 4, initiator is ready to transfer, but the target is not ready. On clock 5, both are ready, and a data transfer takes place (as mention by the vertical lines). For clock 6, the target is ready to transfer, but the initiator is not ready. On clock 7, the initiator becomes ready, and then data is transferred. For clocks 8 and 9, both sides remain ready to transfer data and transferred it at the maximum possible rate (32 bits per clock cycle).

In particular case of a read, clock 2 is reserved for turning around the AD bus, so the target is not allowed to drive data on the bus even if it is capable of fast DEVSEL.

Fast DEVSEL# on reads

A target that supports fast DEVSEL could in theory start responding to a read the cycle after the address is existing. However, this particular cycle is reserved for AD bus turnaround. Therefore, a target may not drive the AD bus (and therefore may not assert TRDY#) on the second cycle of a transaction. Notice that most of the targets will not be this type of fast and will not require any special logic to enforce this condition.


Related Discussions:- Data phases - computer architecture

Goals - artificial intelligence, Goals - artificial intelligence: One ...

Goals - artificial intelligence: One desirable way to make perfect an agent's performance is to enable it to have some details of what it is trying to complete. If it is given

Determine the concepts of object oriented analysis, Determine the concepts ...

Determine the concepts of Object Oriented Analysis  In OOA the initial focus is on identifying objects from the application domain, after that fitting those procedures around

How many address bits are needed to show a 32 K memory, How many address bi...

How many address bits are required to represent a 32 K memory ? Ans. 32K = 25 x 210 = 215, Hence 15 address bits are needed; Only 16 bits can address this.

What are the values of the slack or surplus variables, Consider the followi...

Consider the following linear programming problem: Minimize:        70M + 40N Subject to:           3M + 7N ≥ 233                             10M + 2N ≥ 254

Define about anchor tag, Q. Define about Anchor Tag? Anchor tag is used...

Q. Define about Anchor Tag? Anchor tag is used to create links between various objects such as HTML pages, web sites, files etc. It is introduced by characters and termi

Explain the while loop in c, Explain The while loop in C The while loop...

Explain The while loop in C The while loop keeps repeating an action until an associated test returns false. This is useful where the programmer does not know in advance how ma

Describe real and protected mode, Describe Real and protected mode: Op...

Describe Real and protected mode: Operation of Real mode interrupt:   When microprocessor completes executing the current instruction, it concludes whether an interrupt is act

Explain the high level Language - computer programming, Explain the High Le...

Explain the High Level Language? The programming language such as FORTRAN, C, or Pascal that enables a programmer to write programs those are more or less independent of a parti

Sorting algorithms, Two way merge sort for 84,83,78,90,23,123,98,159,8,200

Two way merge sort for 84,83,78,90,23,123,98,159,8,200

Determine the abstraction mechanisms for modelling, Determine the abstracti...

Determine the abstraction mechanisms for modelling The object orientation conceptual structure helps in providing abstraction mechanisms for modelling, that includes: Cl

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd