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IWR6843ISK: 如何计算点云大小?

Part Number:IWR6843ISK

使用的是IWR6843ISK ES2.0。

在3D_people_counting_demo_implementation_guide.pdf中的4.5.2.1节,提到了点云大小DOA_OUTPUT_MAXPOINTS最大是750个点。

在CCS的3D_people_count_68xx_dss工程中找到了这一项,定义DOA_OUTPUT_MAXPOINTS=MAX_DYNAMIC_CFAR_PNTS*4+MAX_STATIC_CFAR_PNTS

请问这几项,MAX_DYNAMIC_CFAR_PNTS是多普勒维CFAR检测点数,4表示接收天线数,MAX_STATIC_CFAR_PNTS是距离维CFAR检测点数,是这样的吗?

使能3TX 4RX的情况下,IWR6843ISK在方位角有8个虚拟天线,俯仰角有2个虚拟天线(是2个吧?我不太确定),为什么这里用4呢?

另外,为什么MAX_STATIC_CFAR_PNTS这一项是相加而不是相乘呢?

rockzero:

还有我想请问一下IWR6843ISK所支持的单帧点云最大是多少个点呢?

,

Shine:

已把您的问题升级到英文e2e论坛,请关注下面帖子的回复。https://e2e.ti.com/support/sensors-group/sensors/f/sensors-forum/1191014/iwr6843-calculation-about-doa_output_maxpoints

,

rockzero:

谢谢!

,

Shine:

不客气~

,

Shine:

请看下面e2e工程师的回复。

I believe that factor of 4 is there to encourage more dynamic points than static points because they're often more relevant to the user. The maximum number of points is 750 for this application based off the code you've provided. I am confident that you could increase or decrease this limit depending on the amount of processing you do on-device. It is likely a timing challenge, where if there are too many points, the UART will become overwhelmed or the processor may not have enough time to complete its operations before the next frame begins, so as the code becomes more custom, the limits will have to be tested.

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rockzero:

也就是说,这个4只是一个系数,与天线数没有关系,它可以根据情况改成3或5或者其他的数。这个MAX_DYNAMIC_CFAR_PNTS和MAX_STATIC_CFAR_PNTS,实际上所代表的是通过速度维的结果进行划分,速度为0的划分到MAX_STATIC_CFAR_PNTS,速度不为0的划分到MAX_DYNAMIC_CFAR_PNTS。最终输出点云的大小要看UART的传输速率对吗?

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Shine:

是的,可以增加或者减少,实际情况还是要测试,保证在下一帧开始前能有足够的事件处理当前帧。

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