Can UDMA still be applied to C7 core?
I need to do some image copying with DSP, UDMA is faster.
But when I turn on this switch, it will prompt that the channel has failed to open. It doesn't seem to support it.
How should I use it?
For example, the appUdmaCopy2D interface .
SDK version11.0
Hi,
This is the debugging information I added, only for DMA copying
int32_t App_udmaMemcpy2D( const img_select_2d_copy_params_t *params)
{
int32_t retVal = UDMA_SOK;
uint64_t pDesc;
uint32_t trRespStatus;
uint64_t trpdMemPhy;
uint64_t current_time = 0;
current_time = tivxPlatformGetTimeInUsecs();
/* Initialize TRPD for 2D copy */
App_udmaTrpdInit2D(g_udmaChHandle, g_udmaTrpdMem, params);
trpdMemPhy = (uint64_t) Udma_defaultVirtToPhyFxn(g_udmaTrpdMem, 0U, NULL);
/* Submit TRPD to channel */
retVal = Udma_ringQueueRaw(Udma_chGetFqRingHandle(g_udmaChHandle), trpdMemPhy);
if (retVal != UDMA_SOK)
{
VX_PRINT(VX_ZONE_ERROR,"Failed to submit TRPD to UDMA channel\r\n");
return retVal;
}
/* Wait for completion */
SemaphoreP_pend(&g_udmaDoneSem, SystemP_WAIT_FOREVER);
retVal = Udma_ringDequeueRaw(Udma_chGetCqRingHandle(g_udmaChHandle), &pDesc);
if (retVal != UDMA_SOK)
{
VX_PRINT(VX_ZONE_ERROR,"Failed to dequeue from completion ring\r\n");
return retVal;
}
/* Check TR response status */
CacheP_inv(g_udmaTrpdMem, UDMA_TEST_TRPD_SIZE, CacheP_TYPE_ALLD);
trRespStatus = UdmaUtils_getTrpdTr15Response(g_udmaTrpdMem, 1U, 0U);
if (CSL_UDMAP_TR_RESPONSE_STATUS_COMPLETE != trRespStatus)
{
VX_PRINT(VX_ZONE_ERROR,"UDMA 2D transfer failed with status: %d\r\n", trRespStatus);
retVal = UDMA_EFAIL;
return retVal;
}
VX_PRINT(VX_ZONE_ERROR,"App_udmaMemcpy2D time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
return retVal;
}
VX_PRINT(VX_ZONE_ERROR,"App_udmaMemcpy2D time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
Regard
/mcu_plus_sdk_j722s_11_00_00_12/examples/drivers/udma/udma_memcpy_interrupt/udma_memcpy_interrupt.c
This is the source code for your example. He is a copy of 1D.
I have done 2D configuration, everything else is the same.
1D is slower than 2D。
Hi,
int32_t App_udmaMemcpy2D( const img_select_2d_copy_params_t *params)
{
int32_t retVal = UDMA_SOK;
uint64_t pDesc;
uint32_t trRespStatus;
uint64_t trpdMemPhy;
uint64_t current_time = 0;
current_time = tivxPlatformGetTimeInUsecs();
/* Initialize TRPD for 2D copy */
App_udmaTrpdInit2D(g_udmaChHandle, g_udmaTrpdMem, params);
VX_PRINT(VX_ZONE_ERROR,"App_udmaTrpdInit2D time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
trpdMemPhy = (uint64_t) Udma_defaultVirtToPhyFxn(g_udmaTrpdMem, 0U, NULL);
/* Submit TRPD to channel */
retVal = Udma_ringQueueRaw(Udma_chGetFqRingHandle(g_udmaChHandle), trpdMemPhy);
if (retVal != UDMA_SOK)
{
VX_PRINT(VX_ZONE_ERROR,"Failed to submit TRPD to UDMA channel\r\n");
return retVal;
}
VX_PRINT(VX_ZONE_ERROR,"Udma_ringQueueRaw time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
/* Wait for completion */
SemaphoreP_pend(&g_udmaDoneSem, SystemP_WAIT_FOREVER);
VX_PRINT(VX_ZONE_ERROR,"Transfer time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
retVal = Udma_ringDequeueRaw(Udma_chGetCqRingHandle(g_udmaChHandle), &pDesc);
if (retVal != UDMA_SOK)
{
VX_PRINT(VX_ZONE_ERROR,"Failed to dequeue from completion ring\r\n");
return retVal;
}
/* Check TR response status */
CacheP_inv(g_udmaTrpdMem, UDMA_TEST_TRPD_SIZE, CacheP_TYPE_ALLD);
trRespStatus = UdmaUtils_getTrpdTr15Response(g_udmaTrpdMem, 1U, 0U);
if (CSL_UDMAP_TR_RESPONSE_STATUS_COMPLETE != trRespStatus)
{
VX_PRINT(VX_ZONE_ERROR,"UDMA 2D transfer failed with status: %d\r\n", trRespStatus);
retVal = UDMA_EFAIL;
return retVal;
}
VX_PRINT(VX_ZONE_ERROR,"App_udmaMemcpy2D time : %d\r\n", (int)(tivxPlatformGetTimeInUsecs() - current_time)/1000 );
return retVal;
}
Hi,
Which pdk example are you using for this BCDMA?
These are the throughput times which I got from pdk documentation.
Regards,
Sivadeep
Hi,
Could you please check the time between submission and wait for transfer completion like the below.
Regards,
Sivadeep
Summary of CPU load, ==================== CPU: mpu1_0: TOTAL LOAD = 30.65 % ( HWI = 1.97 %, SWI = 0.29 % ) CPU: mcu2_0: TOTAL LOAD = 1.25 % ( HWI = 0. 0 %, SWI = 0. 0 % ) CPU: c7x_1: TOTAL LOAD = 0.46 % ( HWI = 0. 0 %, SWI = 0. 0 % ) CPU: c7x_2: TOTAL LOAD = 0.13 % ( HWI = 0. 0 %, SWI = 0. 0 % ) HWA performance statistics, =========================== HWA: LDC : LOAD = 7.35 % ( 39 MP/s ) HWA: GPU : LOAD = 72. 0 % ( 92 MP/s ) DDR performance statistics, =========================== DDR: READ BW: AVG = 595 MB/s, PEAK = 4999 MB/s DDR: WRITE BW: AVG = 667 MB/s, PEAK = 6091 MB/s DDR: TOTAL BW: AVG = 1262 MB/s, PEAK = 11090 MB/s GRAPH: avm_graph (#nodes = 3, #executions = 347) NODE: CAPTURE1: capture_node: avg = 33042 usecs, min/max = 47 / 44608 usecs, #executions = 347 NODE: VPAC_LDC1: ldc_node: avg = 9184 usecs, min/max = 8676 / 14326 usecs, #executions = 347 NODE: MPU-0: OpenGL_APS_SRV_Node: avg = 26002 usecs, min/max = 23403 / 90819 usecs, #executions = 347 NODE: DSP_C7-1: ImgSelectNode: avg = 20351 usecs, min/max = 15204 / 24047 usecs, #executions = 347 NODE: CSITX: CsitxNode: avg = 16657 usecs, min/max = 16623 / 16750 usecs, #executions = 347
This is all the information of the NODE, as well as some resource information。
Remove TIDL. The time has decreased, around 19-20.
Hi,
How to calculate this time? can you teach me？Thank you
7.2M /0.015sec = 480M/sec。
Does not match the content of the image
