CC2640R2F: some questions about CC2640R2F AOA IQ

Hello,

cindy wang89 said:

1. as shown following, is the IQ samples packet sent during the advertising time or in the connected time? How fast is one packet sent? Is this based on the red or the blue arrow value we set in the code? 

It is in the advertising time. 100ms (RED)

cindy wang89 said:

2. Could we reduce this value to faster the packet rate? what's the mininum value should we set ?

10ms x channel

cindy wang89 said:

3. How many IQ samples are there in one packet? 

Only 1 tone per packet

cindy wang89 said:

4. If the IQ packet is sent in the advertising time , which channel is it using ? if it advertise using all the 3 channels, are the IQ samples in each advertising channel the same ? or if the IQ samples are not the same , what's the relationship between them ?

37, 38,39 and the tone is the same on each channel. Please review

static void transmitterInitAdvData(uint8_t *advData, uint8_t packetId)

  Best regards,

    David

Hello,

cindy wang89 said:

1. About the tone, what's the byte size of a tone?

20 Bytes - #define AOA_ADV_TONE_LEN     20

cindy wang89 said:

2.You mean the IQ samples packet are sent while advertising, and not sent when connecting.  Am i right?

The tone is being sent using the advertising channels, 37, 38 and 39. The example supports both cases, when you are connected and connectionless.

cindy wang89 said:

3. In question 2 you answered as upper, the lowest value for advertising interval can be set to 10ms, it's enough for sending one tone(also for one packet ), is it ?

Correct

cindy wang89 said:

4. Is it possible that we want to receive  10*512 IQ samples in 1 second? and if we want to receive different IQ samples at each second, how should we config in sourcecode?

The IQ samples can be captured at a rate of up to 4 MHz. So for each tone received we store 2KB IQ samples (32 bits per IQ pair) and this takes 128us. So yes this could possible, let me try to explain:

1. We have 32 antenna patterns and we switch antenna pattern every 4us therefore  (32 * 4 =) 128us is time that takes to go over the antenna pattern.

2. Now, the radio capture rate is 4MHz, therefore for 1us [1us / (1 / 4MHz) =] we get 4 samples and each sample is 32 bits (IQ) 4 * 32 =  128 bits

3. In 128us we can capture 2048 bytes ( = 128 * 128 / 8). And that is the maximum size that we can capture in one tone as is the size of the buffer.

  Hopefully this helps.

    David