Memories and the Memory Subsystem; The Memory Hierarchy; Caching; ROM

Post on 01-Jan-2016

71 Views

Category:

Documents

1 Downloads

Preview:

Click to see full reader

DESCRIPTION

Memories and the Memory Subsystem; The Memory Hierarchy; Caching; ROM. Memory: Some embedded systems require large amounts; others have small memory requirements Often must use a hierarchy of memory devices Memory allocation may be static or dynamic Main concerns [in embedded systems]: - PowerPoint PPT Presentation

Transcript

Memories and the Memory Subsystem;The Memory Hierarchy;

Caching;ROM

Memory:Some embedded systems require large amounts; others have small memory requirements

Often must use a hierarchy of memory devices

Memory allocation may be static or dynamic

Main concerns [in embedded systems]:make sure allocation is safeminimize overhead

Memory types:

RAMDRAM—asynchronous; needs refreshingSRAM—asynchronous; no refreshingSemistatic RAMSDRAM—synchronous DRAM

ROM—read only memoryPROM—one timeEPROM—reprogram (uv light)EEPROM—electrical reprogrammingFLASH—reprogram without removing from

circuit

Altera chips: http://www.altera.com/literature/hb/cyc/cyc_c51007.pdfmemory blocks with parity bit (supports error checking)synchronous, can emulate asynchronouscan be used as:

single port—nonsimultaneous read/writesimple dual-port—simultaneous read/write“true” dual-port (bidirectional)—2 read; 2 write; one read, one write

at different frequenciesshift registerROMFIFO

flash memory:

http://www.altera.com/literature/hb/cfg/cfg_cf52001.pdf?GSA_pos=10&WT.oss_r=1&WT.oss=program%20flash%20memory%20up3%20board

fig_04_01

Standard memory configuration:

Memory a “virtual array”

Address decoder

Signals: address, data, control

fig_04_02

ROM—usually read-only (some are programmable-”firmware”)

transistor (0 or 1)

fig_04_04

SRAM—similar to ROMIn this example—6 transistors per cell (compare to flipflop?)

fig_04_05

SRAM read: precharge bi and not bi to a value halfway between 0 and 1; word line drives bi’s to stored value

SRAM write: R/W line is driven low

fig_04_06

Dynamic RAM: only 1 transistor per cellREAD causes transistor to discharge; it must be restored each timerefresh cycle time determined by part specification

fig_04_08

DRAM read and write timing:

fig_04_08

Comparison—SRAM / DRAM

fig_04_09

Memory: typical organization (SRAM or DRAM):

fig_04_11

Two important time intervals: access time and cycle time

fig_04_12

Terminology:

Block: logical unit of transfer

Block size

Page—logical unit; a collection of blocks

Bandwidth—word transition rate on the I/O bus (memory can be organized in bits, bytes, words)

Latency—time to access first word in a sequence

Block access time—time to access entire block

“virtual” storage

Memory interface:

Restrictions which must be dealt with:

Size of RAM or ROM

width of address and data I/O lines

fig_04_14

Memory example:

4K x 16 SRAM

Uses 2 8-bit SRAMs (to achieve desired word size)

Uses 4 1K blocks (to achieve desired number of words)

Address:10 bits within a block2 bits to specify block— CS (chip select)

fig_04_16

Write: 8-bit bus, two cycles per word

fig_04_17

Read: choose upper or lower bits to put on the bus

If insufficient I/O lines: must multiplex signals and store in registers until data is accumulated (common in embedded system applications)

Requires MAR / MDR configuration typically

DRAM:

Variations available: EDO, SDRAM, FPM—basically DRAMs trying to accommodate ever faster processors

Techniques:--synchronize DRAM to system clock--improve block accessing--allow pipelining

As with SRAM, typically there are insufficient I/O pins and multiplexing must be used

fig_04_19

Terminology:

RAS—row address strobeCAS—column address strobe

note either leading or trailing edge can capture address

RAS cycle time

RAS to CAS delay

Refresh period

fig_04_20

Example:

EDO—extended data output: one row, multiple columns

fig_04_21

Refreshing the DRAM: overhead; must refresh all rows, not just those accessed;Will this be controlled internally or externally?

Example: refresh one row at a time, external refresh

4M memory: 4K rows, 1K columns: 22 address bits12 I/O pins, 10 shared between row and columnRefresh each row every 64 ms

2-phase clock (for greater flexibility), 50 MHz source

fig_04_22

Refresh timing: must refresh one row every 16 musec

Use a 9-bit counter incremented from either phase of the 25 MHz clock, refresh at count 384 (15.36musec)—this provides some timing margin

fig_04_23

Refresh address—12 bit binary counter—increment following the completion of each row refresh

fig_04_24

Address selection:

Read / write

refresh

fig_04_25

Refresh arbitration: avoid R/W conflicts

1. If normal R/W operation starts, allow it to complete

2. If refresh operation has started, remember R/W operation

3. For a tie, normal R/W operation has priority

Required signals:R/W—has been initiatedRefresh interval—has elapsedNormal request—by arbitration logicRefresh request—by arbitration logicRefresh grant—by arbitration logicNormal active—R/W has startedRefresh active—refresh has started

fig_04_25

Row and column addresses (column only 10 bits):

Generate row, column addresses on phase 1, RAS, CAS on phase 2:

fig_04_27

Arbitration circuit: request portion followed by grant portion

fig_04_28

Complete system:

fig_04_29

R/W cycles:

fig_04_30

Memory organization:

Typical “Memory map”

For power loss

Issue in embedded systems design: stack overflow

Example: should recursion be used?

Control structures: “sequential” +:

“Primitive” “Structured programming”GOTO choice (if-else, case)Cond GOTO iteration (pre, post-test)

?recursion?

[functions, macros: how do these fit into the list of control structures?]

fig_04_31

Memory hierarchy

fig_04_32

Paging / Caching

Why it typically works:

locality of reference

(spatial/temporal)

“working set”

Note: in real-time embedded systems, behavior may be atypical; but caching may still be a useful technique

fig_04_33

Typical memory system with cache: hit rate (miss rate) important

Basic caching strategies:

Direct-mapped Associative

Block-set associative questions:

what is “associative memory”?

what is overhead?

what is efficiency (hit rate)?

is bigger cache better?

Associative memory: storage location related to data storedExample—hashing:--When software program is compiled or assembled, a symbol table must be created to link addresses with symbolic names--table may be large; even binary search of names may be too slow--convert each name to a number associated with the name, this number will be the symbol table indexFor example, let a = 1, b = 2, c = 3,…Then “cab” has value 1 + 2 + 3 = 6 “ababab” has value 3 *(1 + 2) = 9And “vvvvv” has value 5*22 = 110Address will be modulo a prime p, if we expect about 50 unique identifiers, can take p = 101 (make storage about twice as large as number of items to be stored, reduce collisions)Now array of names in symbol table will look like:0—>1—>2--->…6--->cab…9--->ababab--->vvvvv…Here there is one collision, at address 9; the two items are stored in a linked listAccess time for an identifier <= (time to compute address) + 1 + length of longest linked list ~ constant

Caching: the basic process—note OVERHEAD for each task--program needs information M that is not in the CPU--cache is checked for M how do we know if M is in the cache?--hit: M is in cache and can be retrieved and used by CPU--miss: M is not in cache (M in RAM or in secondary memory) where is M?

* M must be brought into cache* if there is room, M is copied into cache how do we know if there is room?* if there is no room, must overwrite some info M’ how do we select M’?

++ if M’ has not been modified, overwrite it how do we know if M’ has been modified?++ if M’ has been modified, must save changeshow do we save changes to M’?

fig_04_34

Example: direct mapping32-bit words, cache holds 64K words, in 128 0.5K blocksMemory addresses 32 bitsMain memory 128M words; 2K pages, each holds 128 blocks (~ cache)

fig_04_35

fig_04_36

2 bits--byte; 9 bits--word address;7 bits—block address (index); 11 (of 15)—tag (page block is from)

Tag table: 128 entries (one for each block in the cache). Contains:Tag: page block came fromValid bit: does this block contain data write-through: any change propagated immediately to main memorydelayed write: since this data may change again soon, do not propagate change to main memory immediately—this saves overhead; instead, set the dirty bitIntermediate: use queue, update periodicallyWhen a new block is brought in, if the valid bit is true and the dirty bit is true, the old block must first be copied into main memoryReplacement algorithm: none; each block only has one valid cache location

fig_04_37

Problem with direct mapping: two frequently used parts of code can be in different “Block0’s”—so repeated swapping would be necessary; this can degrade performance unacceptably, especially in realtime systems (similar to “thrashing” in operating system virtual memory system)Another method: associative mapping: put new block anywhere in the cache; now we need an algorithm to decide which block should be removed, if cache is full

fig_04_38

Step 1: locate the desired block within the cache; must search tag table, linear search may be too slow; search all entries in parallel or use hashingStep 2: if miss, decide which block to replace.a.Add time accessed to tag table info, use temporal locality:Least recently used (LRU)—a FIFO-type algorithmMost recently used (MRU)—a LIFO-type algorithmb. Choose a block at random

Drawbacks: long search timesComplexity and cost of supporting logicAdvantages: more flexibility in managing cache contents

fig_04_39

Intermediate method: block-set associative cacheEach index now specifies a set of blocksMain memory: divided into m blocks organized into n groupsGroup number = m mod n Cache set number ~ main memory group numberBlock from main memory group j can go into cache set jSearch time is less, since search space is smallerHow many blocks: simulation answer (one rule of thumb: doubling associativity ~ doubling cache size, > 4-way probably not efficient)

Two-way set-associative scheme

Example: 256K memory-64 groups, 512 blocks Block Group (m mod

64)0 64 128 . . . 384 448 01 65 129 . . . 385 449 12 66 130 . . . 386 450 2. . .63 127 192 . . . 447 511 63

fig_04_40

Dynamic memory allocation “virtual storage”):--for programs larger than main memory--for multiple processes in main memory--for multiple programs in main memory

General strategies may not work well because of hard deadlines for real-time systems in embedded applications—general strategies are nondeterministic

Simple setup:Can swap processes/programsAnd their contexts--Need storage (may be infirmware)--Need small swap time comparedto run time--Need determinismEx: chemical processing, thermal control

fig_04_41

Overlays (“pre-virtual storage”):Seqment program into one main section and a set of overlays (kept in ROM?)Swap overlaysChoose segmentation carefully to prevent thrashing

fig_04_42

Multiprogramming: similar to paging

Fixed partition size: Can get memory fragmentationExample:If each partition is 2K and we have 3 jobs: J1 = 1.5K, J2 = 0.5K, J3 = 2.1KAllocate to successive partitions (4)J2 is using only 0.5 K J3 is using 2 partitions, one of size 0.1KIf a new job of size 1K enters system, there is no place for it, even though there is actually enough unused memory for it

Variable size:Use a scheme like pagingInclude compactionChoose parameters carefully to prevent thrashing

fig_04_43

Memory testing: Components and basic architecture

fig_04_45

Faults to test: data and address lines; stuck-at and bridging(if we assume no internal manufacturing defects)

fig_04_49

ROM testing: stuck-at faults, bridging faults, correct data stored Method: CRC (cyclic reduncancy check) or signature analysisUse LFSR to compress a data stream into a K-bit pattern, similar to error checking(Q: how is error checking done?)ROM contents modeled as N*M-bit data stream, N= address size, M = word size

Error checking: simple examples1.Detect one bit error: add a parity bit2.Correct a 1-bit error: Hamming codeExample: send m message bits + r parity bitsThe number of possible error positions ism + r + 1, we need 2r >= m + r + 1If m = 8, need r = 4; ri checks parity of bits with i in binary representationPattern:Bit #: 1 2 3 4 5 6 7 8 9 10 11 12Info: r0 r1 m1 r2 m2 m3 m4 r3 m5 m6 m7 m8

--- --- 1 --- 1 0 0 --- 0 1 1 1Set parity = 0 for each groupr0: bits 1 + 3 + 5 + 7 + 9 + 11 = r0 + 1 + 1 + 0 + 0 + 1 r0 = 1r1: bits 2 + 3 + 6 + 7 + 10 + 11 = r1 + 1 + 0 + 0 + 1 + 1 r1 = 1r2: bits 4 + 5 + 6 + 7 + 12 = r2 + 1 + 0 + 1 r2 = 0r3: bits 8 + 9 + 10 + 11 + 12 = r3 + 0 + 1 + 1 + 1 r3 = 1Exercise: suppose message is sent and 1 bit is flipped in received messageCompute the parity bits to see which bit is incorrect

Addition: add an overall parity bit to end of message to also detect two errors

Note: a.this is just one example, a more general formulation of Hamming codes using the finite field arithmetic can also be given b. this is one example of how error correcting codes can be obtained, there are many more complex examples, e.g., Reed-Solomon codes used in CD players

top related