XIB-E【PDF 40/64 页】BOARD INTERFACE ETHERNET

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Response packet delay

As a packet propagates through the repeater network, if any node receives the data and

generates a quick response, the response needs to be delayed so as not to collide with

subsequent retransmissions of the original packet. To reduce collisions, both repeater and end

node radios in a repeater network will delay transmission of data shifted in the serial port to allow

any repeaters within range to complete their retransmissions.

The time for this delay is computed by the formula:

Maximum Delay (ms) = L * DS

DS = ((-41-(-100))/10)*RN)+RN+1

Where L is the length of the transmitted packet in milliseconds, DS is the number of delay slots

to wait, RSSI is the received signal strength in dBm, and RN is the value of the RN register.

Use Case - Broadcast Repeater Network

Consider modules R1 through R10 each communicating to a PLC using the ModBus protocol and

spaced evenly in a line. All ten nodes are configured as destinations & repeaters within the

scope of Basic Broadcast Communications (MD=3, AM, DT=0xFFFF, PK=0x100, RO=0x03,

RB=0x100, RN=1). The Base Host (BH) shifts payload that is destined for R10 to R1. R1

initializes RF communication and transmits payload to nodes R2 through R5 which are all within

range of R1. Modules R2 through R5 receive the RF packet and retransmit the packet

simultaneously. They also send the data out the serial ports, to the PLC's.

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Command

Binary

Command

AT Command Name

Range

# Bytes

Returned

Factory

Default

0x3A (58d)

Auto-set MY

0x00 (0d)

Destination Address

0 0xFFFF

0x3C (60d)

RF Mode

3 - 4

0x2A (42d)

Source Address

0 0xFFFF

0xFFFF

0x19 (25d)

Delay Slots

0 0xFF [slots]

0x08 (8d)

Write

Bandwidth Considerations

Using broadcast repeaters in a network reduces the overall network data throughput as each

repeater must buffer an entire packet before retransmitting it. For example: if the destination is

within range of the transmitter and the packet is 32 bytes long, the transmission will take 72ms

on a 9600 baud XStream Module (much faster modules are available). If that same packet has to

propagate through two repeaters, it will take 72ms to arrive at the first repeater, another 72 ms

to get to the second and a final 72ms to get to the destination for a total of 216ms. Taking into

account UART transfer times (~1ms/byte at 9600 baud), a server to send a 32 byte query and

receive a 32 byte response is ~200ms, allowing for 5 polls per second. With the two repeaters in

the path, the same query/response sequence would take about 500ms for 2 polls per second.

To summarize, this system is sending and receiving 64 bytes 5 times per second for a throughput

of 320 bytes per second with no repeaters and 128 bytes per second with 2 repeaters. Generally,

the network throughput will decrease by a factor of 1/(R+1), with R representing the number of

repeaters between the source and destination.

Note that these numbers are absolutely worst case to illustrate how the system would perform in

a typical, low bandwidth system. As a counter example the 115kbps 9XTend radio can transfer

the same 32 byte packet in 12 ms for a round trip with UART transfer times of ~30ms or 33 polls

per second (1066 bytes per second) with no repeaters. With two repeaters the time would be

~100ms round trip time for 10 polls per second or 320 bytes per second network throughput with

two repeaters.

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