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AWR2544LOPEVM: What is the difference between ethPktRdyCnt in 1xmode and 2xmode in AWR2544?

Jack Nong
Prodigy 60 points
Part Number: AWR2544LOPEVM
Other Parts Discussed in Thread: AWR2544

Tool/software:

Hi,TI Team：

We can see "profile_3d_3Azim_1ElevTx_DDM_awr2544_1Xmode.cfg" and "profile_3d_3Azim_1ElevTx_DDM_awr2544_2Xmode.cfg" in the profiles file of 2544. The "ethPktRdyCnt" in the "procChainCfg" CLI parameter uses different values, one is set to 530, the other is 6(Figure 1), and the two values are quite different. I have learned from relevant documents that ethPktRdyCnt in 1xmode should be calculated according to the formula in the following figure(Figure 2). However, the documentation does not yet see how this value should be calculated in 2xmode. Can you provide the relevant calculation method of this value in 2xmode?

Figure 1:

Figure 2:

Best Regards

Jack

  • 0
    TI__Guru*** 141644 points

    Hi,

    We will be able to get back to you early next week

    thank you

    Cesar

  • 0 in reply to Cesar
    TI__Genius 12170 points

    Hi Jack,

    You should be able to use the same formula irrespective of 1x Mode/2x Mode.

    Regards,

    Samhitha

  • 0 in reply to Samhitha Bobba
    Prodigy 60 points

    Hi，Samhitha,

    Can you help to analyze how "ethPktRdyCnt = 6" is calculated in 2x Mode configuration?

    Thank you very much！

    Best Regards

    Jack

  • 0 in reply to Jack Nong
    TI__Genius 12170 points

    Hi Jack,

    In case of configuration in 2xmode, PacketsPerChirp will be 3. This is because size of compressed radar cube is 3072 bytes. So, this is sent in three packets, each of size 1120 (1024+96) bytes.

    N_c = 768, T_c = 26.5 us, PacketSize = 1120 (1024+96) bytes, PacketsPerChirp = 3, EthernetSpeed = 1Gbps

    Regards,

    Samhitha

  • 0 in reply to Samhitha Bobba
    Prodigy 60 points

    Hi，Samhitha,

    Some details about the calculation method of ethPktRdyCnt would like to consult again:

    1. In the Demo program, is the EthernetSpeed value 900Mbps or 1000Mbps? For example, in 1xmode, N_c = 768, T_c = 27.81 us, PacketSize = 1120 (1024+96) bytes, PacketsPerChirp = 1. If EthernetSpeed is 900, ethPktRdyCnt is 493.07. If EthernetSpeed is 1000, ethPktRdyCnt is 247.44. ethPktRdyCnt is 530 in the Demo Profile. It seems that EthernetSpeed = 900 is closer to the actual application.

    2. According to the description in the document, the actual value of ethPktRdyCnt must be greater than the calculated value. Are there any constraints associated with this greater range? For example, calculated ethPktRdyCnt = 493.07, I use the value of ethPktRdyCnt with 893 can it?

    3, in 2xmode, according to the parameters you provide, the calculated value is negative?(-11)?

    Best Regards

    Jack

  • 0 in reply to Jack Nong
    TI__Genius 12170 points

    Hi Jack,

    Jack Nong said:
    In the Demo program, is the EthernetSpeed value 900Mbps or 1000Mbps? For example, in 1xmode, N_c = 768, T_c = 27.81 us, PacketSize = 1120 (1024+96) bytes, PacketsPerChirp = 1. If EthernetSpeed is 900, ethPktRdyCnt is 493.07. If EthernetSpeed is 1000, ethPktRdyCnt is 247.44. ethPktRdyCnt is 530 in the Demo Profile. It seems that EthernetSpeed = 900 is closer to the actual application.

    In the demo, maximum speed of the ethernet is 1Gbps. Please recheck your calculation. The value 247.44 that is obtained might be Nc - ethPktRdyCnt. In that case, ethPktRdyCnt will be 768 - 247.44 = 520.56.

    Also consider using 1kB = 1024 bytes instead of 1kB = 1000 bytes while computing the value of ethPktRdyCnt.

    Jack Nong said:
    According to the description in the document, the actual value of ethPktRdyCnt must be greater than the calculated value. Are there any constraints associated with this greater range?

    Maximum value depends on the DSS_L3 size (i.e maximum data that can be stored on L3 before the trigger).

    Jack Nong said:
    in 2xmode, according to the parameters you provide, the calculated value is negative?(-11)?

    As mentioned earlier, consider using 1kB = 1024 bytes instead of 1kB = 1000 bytes while computing the value of ethPktRdyCnt.

    Regards,

    Samhitha

  • 0 in reply to Samhitha Bobba
    Prodigy 60 points

    Hi，Samhitha,

    Thank you very much for your patient answer, which makes my first two questions have been well solved. I just wanted to get some details on the third question. In 2x mode, N_c = 768, T_c = 26.5 us, PacketSize = 1120 (1024+96) bytes, PacketsPerChirp = 3, EthernetSpeed = 1024. According to the formula, ethPktRdyCnt = 7.245 (as shown in the figure below). According to the manual (figure below), the value used should be >7.245. But the program is set to 6. Can you help me analyze this problem?
    Thank you very much!

    Best Regards

    Jack

  • +1 in reply to Jack Nong
    TI__Genius 12170 points

    Hi Jack,

    We cannot get the accurate value using the formula. So, it has to be based on the runtime observations. We considered the value to be 6 after few experiments. You can take a value greater than 7.245 for ethPktRdyCnt.

    Regards,

    Samhitha

  • 0 in reply to Samhitha Bobba
    Prodigy 60 points

    Hi，Samhitha,

    Thank you very much, I have solved the problem, wish you a happy life！

    Best Regards

    Jack