Response Time

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The calculation for response time given below assumes ideal conditions on the radio channel, i.e. that there are no collisions or repeating of packets. A description of the capacity of the radio channel is given here.

The response time is taken to be the time from sending the data from the control centre system (application) through the Morse network to the end station and back.

The overall time consists of:

  1. Transmission time over the serial (wire) interface between the control centre and the connected radio modem (Twc)
  2. Transmission time over the radio channel (Tr)
  3. Transmission time from the radio modem over the serial interface to the control system (PLC) at the end station (Twe)
  4. Processing time in the PLC (Tplc)
  5. Transmission time from the PLC to the radio modem over the serial interface (Twe)
  6. Transmission time over the radio channel (Tr)
  7. Transmission time from the radio modem over the serial interface to the control centre (Twc)

T = Twc + Tr + Twe + Tplc + Twe + Tr + Twc

1. Transmission time over the serial interface (Tw)

The transmission time over the serial interface Tw.. is given by the signalling rate and the amount of data transmitted. Upon receiving data to the radio modem the transmission time is extended by a so-called idle value time which is the time over which no data arrives. After this time the buffer is closed and sent for further processing. The idle value is typically set to 5 bytes in the Morse network.

For a 100 Byte long packet and signalling rate of 19.200 b/s

Tw = ((100+5) . 10.5)/19.2 = 57.42 ms

where:
100 – length of data in the byte
5 – idle value
10.5 – corresponding number of bits to one byte (1 start bit, 8 bits, 1.5 stop bits, no parity)
19.2 – signalling rate in kbps

2. Transmission time over the radio channel (Tr)

2.1 Transmission time on the radio link between two radio modems

Consists of two basic components

2.1.1 Actual transmission over the radio channel (Tt)

The length of the packet transmitted over the radio channel is not the same as the length of data received over the serial channel.

The size of the packet is reduced by:

  • Non-transmitted headers _ depends on the method of implementation of the specific protocol on the serial interface. In the worst case the full size of the packet is transmitted.

The size of the packet is increased by:

  • >Synchronising sequence – 18 bytes (includes 2ms switching time)
  • MORSE protocol header – 12 bytes
  • CRC – 4 bytes
  • Coefficient of data extension due to the influence of FEC – 1.375

Tt = 8 . (18 + 1.375 . (100 + 12 + 4))/21.68 = 65.5 ms

where:
8 – number of bits in a byte
18 – synchronising sequence (18 bytes)
1.375 – coefficient of data extension due to the influence of FEC
100 – size of user data after processing by the protocol at the interface
12 – MORSE protocol header (12 bytes)
4 – CRC (4 bytes)
21.68 – modulation signalling rate kbps

2.1.2 Access to the radio channel (Ta)

The time for accessing the radio channel is a random number given by the algorithm, the initial conditions of which can be set within wide boundary limits. The settings need to be optimised for the considered type of operation and topology of the network.

The time needed to hardware switch the radio modem (2ms) is not considered in the calculation because the beginning of transmitting the synchronising sequence starts together with the request for switching the radio modem to transmitting mode.

The time of access to the channel consists of a fixed and randomly calculated time:

Ta = 15 + 1.5 * 15 = 37.5 ms

where:
15 – fixed time
1.5 – average number of timeslots (randomly 0-3)
15 – length of time slot

Total transmission time over the radio channel on the link between two radio modems:

Tl = Tt + Ta

Tl = 65.5 + 37.5 = 103 ms

2.2 Routing

In the case of routing one or more times it is necessary to calculate the time needed to send an ACK packet:

Tack = 8 . (18 + 1.375 . (12 + 4))/21.6 = 14.8 ms

where:
8 – number bits in a byte
18 – synchronising sequence (18 bytes)
1.375 – coefficient of data extension due to the influence of interleaving
12 – MORSE protocol header (12 bytes)
4 – CRC (4 bytes)
21.68 – modulation signalling rate kbps

If the data is received correctly the ACK packet is sent immediately. Software access to the radio channel is not used.

Each routing means a prolongation of the transmission time over the radio channel by 2 x Tack + 2x Tl.

3. Total response time

Examples of the total response time for various cases are given in the following table.

Sig. rate on RS232 ports
Control centre/substation
Data length
Control centre/substation
Number of retransmissions
1 2 3 4 5
19200/9600 20/20 407 561 716 870 1025
300/20 1008 1305 1601 1898 2194
115200/38400 20/20 343 497 652 806 961
300/20 587 884 1180 1477 1773

The reaction time of the PLC for generating Tplc responses is estimated to be 200ms.

From the above calculations it is obvious that the speed of communication over the serial connection between the radio modem and the control system (at the control centre and at the end point) has negligible impact on the response time. Logically, as the packets get longer their influence increases. For this reason it is necessary to set communication parameters on the serial interface to the highest possible signalling rate.

An MS Excel program for calculating response time is available for downloading here.

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