Hivertec motionCAT HCPCI-MNT720M, Руководство пользователя

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1.5 Calculating the Communication Time
The communication time is classified into:
⚫
Amount of time for a cyclic communication
⚫
Amount of time for one data communication
⚫
Total amount of time for a cyclic communication including data communications
<Abbreviations>
N: Number of connected local devices
B: Number of bytes of transmitted data (in case of two-byte data transmission: B = 2)
K: Transmission speed coefficient
Communication speed (Mbps) K
20 1
10 2
5 4
2.5 8
Table 1.5-1 motionCAT communication times
(1) Amount of time for a cyclic communication
Communication time for one local device CT = 7.7

K
Amount of time for a cyclic communication (μs) = (CT + 7.4)

N
Example: Where the transmission speed = 20 Mbps, The number of local devices = 30
A cyclic communication time = (7.7

1 + 7.4)

30 = 453μs
(2) Amount of time for one data communication
Data communications are classified into the following two types:
⚫
When data is present in a response from the local device (variable-length data)
⚫
When data is absent in a response from the local device
Send time ST = (B

0.6 + 3.25)

K
Respond time: with response data JT = (B

0.6 + 5.65)

K
Respond time: without response data JT = 5.05

K
Amount of tim
e for one data communication (μs) = ST + JT + 7.4
Example 1: Time required for writing data to one register of a motion device, where the transmission
speed = 20 Mbps:
A write command (2 bytes) and register data (4 bytes), totaling 6 bytes, are transmitted to write the
data to the motion device register. There is no response data. (A response frame only)
Data communication time = (6

0.6+3.25)

1 + 5.05

1 + 7.4 = 19.
3μs
Example 2: Time required for reading data from one register of a motion device, where the transmission
speed = 20 Mbps:
A read command (2 bytes) is transmitted to write the data to the motion device register.
Response data consists of a read command (2 bytes) and register data (4 bytes), totaling 6 bytes.
A data communication time = (2

0.6 + 3.25)

1 + (6

0.6 + 5.65)

1 + 7.4 = 21.1μs