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

Explain about heterogeneous linked list, If you are using C language to imp...

If you are using C language to implement the heterogeneous linked list, what pointer type will you use? The heterogeneous linked list having different data types in its nodes a

Slot machine, A special Slot Machine is introduced in a new casino; this ma...

A special Slot Machine is introduced in a new casino; this machine has a digital interface of a 3X3 grid. An interface button named as SPIN for making the machine spin. There are f

Explain about transmission media, Q. Explain about Transmission Media? ...

Q. Explain about Transmission Media? These, also known as physical channels, can be either bounded or unbounded. Bounded media, in which signals representing data are confined

What is a multitape tm, What is a multitape TM?  A  multi-tape  Turing ...

What is a multitape TM?  A  multi-tape  Turing  machine  having  of  a  finite  control  with  k-tape  heads  and k-tapes each tape is infinite in both directions. On a one mov

Difference among java beans & servlets, Java bean is a reusable component, ...

Java bean is a reusable component, where as the servlet is the java program which extends the server capability.

Explain scientific applications and image processing, Scientific Applicatio...

Scientific Applications/Image processing Major of concurrent processing applications from science and other academic disciplines, are mostly have based on numerical simulations

Software Engineering, explanation of the difference between syntax and sema...

explanation of the difference between syntax and semantic errors

Evaluate fixed execution time, The statement of Gustafson's law can be desc...

The statement of Gustafson's law can be described with the help of an illustration. Let us take a problem, say P, which has to be solved using a parallel computer. Let Ts be the ti

Illustrate about compact disk rewritable, Q. Illustrate about Compact Disk ...

Q. Illustrate about Compact Disk Rewritable? CD-RW optical disk can be repeatedly written and overwritten same as with a magnetic disk. Even though a number of techniques have

Mode counter, desing mode level counter starting at 0011 using D flipflop

desing mode level counter starting at 0011 using D flipflop

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