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Original question:

TDA4VM: graph

TDA4VM: graph

jialin xie
Intellectual 260 points

Part Number: TDA4VM

SDK is 8.1
In the my demo, Capture->LDC->SRV->MOSAIC->DISPLAY. When I ready to rendering, I print the the local time in srv kernel.
In the meanwhile, after vxGraphParameterDequeueDoneRef(), I queried the timestamp of reference of capture and csitx, and used vxQueryGraph() to know the overall performance of the graph. My question is below:
1. The execution time of all nodes is ~60ms, so the timestamps of sink node should be more than the timestamps of source node. why the timestamp1 is equal to timestamp2?

2. Using vxGraphParameterEnqueueReadyRef() 4 times, it means capture frame1, capture frame2, capture frame3 and capture frame4 are enqueued. The srv node begin to work . I want to know why enqueue reference 4 times, srv node begin to work? And the first frame used by srv node is capture frame1 or capture frame3 ?

3. Using vxQueryGraph(), what does graph_perf.num means?

code :

result:

  • 0
    Intellectual 260 points

    Appreciate your guidance here.

  • 0 in reply to jialin xie
    TI__Guru* 87451 points

    Hi

    Sorry for the delay in response

    jialin xie said:
    The execution time of all nodes is ~60ms, so the timestamps of sink node should be more than the timestamps of source node. why the timestamp1 is equal to timestamp2?

    You had mentioned that Capture->LDC->SRV->MOSAIC->DISPLAY is the flow of your graph, May i know where is the csi-tx node being used here?

    jialin xie said:
    Using vxGraphParameterEnqueueReadyRef() 4 times, it means capture frame1, capture frame2, capture frame3 and capture frame4 are enqueued. The srv node begin to work

    As the capture node is a source node, it is designed such that a minimum of 3 buffers are required by it before it starts dequeuing the buffer.

    jialin xie said:
    Using vxQueryGraph(), what does graph_perf.num means?

    graph_perf.num holds the number of measurements taken.

    Regards,
    Nikhil

  • 0 in reply to Nikhil Dasan
    Intellectual 260 points

    1. Capture->LDC->SRV->MOSAIC->DISPLAY is the flow of my graph, and DISPLAY node is csi-tx node.

    2.  I want to know why enqueue reference 4 times, srv node begin to work? And the first frame used by srv node is capture frame1 or capture frame3 ?

    Appreciate your guidance here.

  • +1 in reply to jialin xie
    TI__Guru* 87451 points

    Hi,

    Nikhil Dasan said:
    As the capture node is a source node, it is designed such that a minimum of 3 buffers are required by it before it starts dequeuing the buffer

    As mentioned above, the capture node requires minimum of 3 buffers to begin transmitting frames to the next node, which is why we are enqueueing 4 frames initially.

    jialin xie said:
    And the first frame used by srv node is capture frame1 or capture

    The frames from the capture node to the SRV node will be in order, i.e., the first frame used by srv node is capture frame 1.

    Regards,
    Nikhil