I2S/TDM ADX Demux

[jonathanh]

Hi Chris,

Looking at the ADX driver I see it does a …

memset(adx->map, 0, sizeof(adx->map));

… and writes the entire map to the actual RAM map. So I believe you should not have to define any unused entries at the end of the frame.

Yes the ADX support S32_LE as well.

Cheers
Jon

[chris.duncan]

Hi Jon,

Apologies. This is incorrect…

I can confirm that this does not work. It needs the unused entries which perhaps is a bug. I will debug the ADX driver when i get some time.

It does indeed work as described. My mistake was the adx-slot-size being incorrect.

Thanks for your time, Jon. Once i have finally tidied everything up i will share all the drivers etc.

Regards,
Chris

Hi Chris,

I have been looking at this a bit more, and what you have looks correct to me, but could be simplified as …

nvidia,adx-slot-size = <8>;
nvidia,adx-slot-map = <
TDM_SLOT_MAP(0, 1, 0)
TDM_SLOT_MAP(0, 1, 1)
TDM_SLOT_MAP(1, 1, 0)
TDM_SLOT_MAP(1, 1, 1)
TDM_SLOT_MAP(2, 1, 0)
TDM_SLOT_MAP(2, 1, 1)
TDM_SLOT_MAP(3, 1, 0)
TDM_SLOT_MAP(3, 1, 1)>;

So each TDM_SLOT_MAP entry maps an input byte in the frame to an output byte on one of the ADX outputs. In your case each ADX output only has 1 channel and is made up of two bytes. So looking at the above, where there are 4 channels each with 16 bits and so a total of 8 bytes, we have …

Input frame byte 0 → slot map entry 0 → ADX1-1 output channel 1 byte 0
Input frame byte 1 → slot map entry 1 → ADX1-1 output channel 1 byte 1
Input frame byte 2 → slot map entry 2 → ADX1-2 output channel 1 byte 0
Input frame byte 3 → slot map entry 3 → ADX1-2 output channel 1 byte 1
Input frame byte 4 → slot map entry 4 → ADX1-3 output channel 1 byte 0
Input frame byte 5 → slot map entry 5 → ADX1-3 output channel 1 byte 1
Input frame byte 6 → slot map entry 6 → ADX1-4 output channel 1 byte 0
Input frame byte 7 → slot map entry 7 → ADX1-4 output channel 1 byte 1

Let me know if this helps.

Regards,
Jon

[jonathanh]

Hi Chris,

Thanks for confirming!

Regards,
Jon