AWR1843: How changing the "number of ADC samples" justifies, without changing the "ADC sampling time" and sampling frequency

Palguna Kumar Reddy Gopireddy

Prodigy 141 points

Part Number: AWR1843

Tool/software:

This is the chirp cycle time and the values in profileCfg given while using AWR1843

As given in the figure

ADC sampling time = number of ADC samples / Sampling Rate

I am using AWR1843 with DCA1000 to get raw data. So while using mmWaveStudio application t in the sensorCofig tab we put the profileCfg values.

Here are the values I have given

Frequency slope = 29.982 MHz/uS = 29.982*10^12 Hz/S

sampling rate = 10000 ksps (kilo samples per second)

ADC start time = 6 us

Ramp end time = 60 us

number of ADC samples = 256

From above values

ADC sampling time = 60 - 6 = 54 us

number of ADC samples = 54 us*10000 ksps = 540 samples

So my questions are

1) if choosing the number of ADC samples as 256 justifies this. What about the remaining 540-256 = 284 samples? (In the profileCfg we have the provision to change the number of samples to any values. What about the remaining samples)

2) How should I calculate the range resolution, IF frequency resolution? from the page 12 in mmwaveSensing-FMCW-offlineviewing_0.pdf (ti.com.cn) both are related. IF frequency resolution = (Frequency slope*2*range resolution)/speed of light.

From page 12 in mmwaveSensing-FMCW-offlineviewing_0.pdf (ti.com.cn), the range resolution = speed of light / (2*bandwidth)

here bandwidth = frequency slope * ADC sampling time = 29.982*10^12*54*10^-6 = 1.619028 GHz

so range resolution = 3*10^8/(2*1.619028*10^9) = 0.0926481815014935 meters

Hence, IF frequency resolution = (29.982*10^12 *2*0.0926481815014935)/(3*10^8) = 18,518.5185 Hz

generally, frequency resolution is calculated as Sampling rate/ Number of ADC samples,

Here if the Number of ADC samples is taken as 540, then frequency resolution = 10000*10^3/540 = 18,518.518 Hz

But if I consider number of samples as 256 as given in profileCfg, frequency resolution = 10000*10^3/256 = 39,062.5 Hz.

from IF frequency resolution = (Frequency slope*2*range resolution)/speed of light., if I calculate the range resolution it is coming as (39,062.5 *3*10^8)/(2*29.982*10^12) = 0.1954297578547128 m

So which of the 0.0926 and 0.1954 to take as range resolution, which of 18,518.5185 and 39,062.5 to take as IF frequency resolution?

over 1 year ago

0 Kaushik Gowda over 1 year ago

TI__Mastermind 20605 points

Hi Palguna,

Thank you for reaching out on e2e!

Please find my responses as follows:

The number of samples denotes after how many samples to stop the sampling of the IF signal. This will be the effective number of samples that you have captured or the number of samples that will be available in the ADC buffer (256 in your case). The maximum number of samples that can be set would have to be less than the possible number of samples that can be acquired within the sampling window following the sampling rate (This appears to be 540 samples in your case).
Range resolution is equal to (C/2*B) where C is speed of light and B is the effective bandwidth of your IF signal. I recommend you go through the following document to understand more about the same.
1. www.ti.com/.../swra553

Regards,

Kaushik

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

Hi Kaushik,

the first question is resolved.

Coming to the second question, I have seen swra553a.pdf (ti.com). It did not solve it as no mentioning of frequency resolution.

I again clarify this doubt to you. I basically got this doubt of frequency resolution, range resolution while preparing the range grid when plotting the plots like range doppler, range angle plots etc.

For suppose if I prepare range grid like

rng_res = c/(2*BW); %0.0926

rng_grid = (0:fft_Rang-1).'*rng_res;

I get range grid values as 0, 0.0926, 0.1852, 0.2778, 0.3703, 0.4629, 0.5555, 0.6481, 0.7407, 0.8333, -------

if i prepare range grid like

freq_res = sampling_rate/fft_Rang; %39,062.5

freq_grid = (0:fft_Rang-1).'*freq_res;

rng_grid = freq_grid*c/frequency_slope/2;

I get range grid values as 0, 0.1953, 0.3906, 0.5859, 0.7812, 0.9765, 1.1718, 1.3671, 1.5624, ------

If i plot the range doppler plot by taking the range grid as the first one, the object is shown to be lets say at 9th sample, i.e. at 0.7407m

but i plot the range doppler plot by taking the range grid as the second one, the object is shown to be 1.5624 m (9th sample), which is about the double the distance. So which distance i should consider the object is at? This is how I got the doubt.

( just a suggestion: I think it can be resolved by knowing how does the mmWave SDK or postProc in mmWave Studio gets the range grid values. This may help you)

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Hi Palguna,

To plot the range spectrum, you could have your grid marking as Max range based on effective IF BW divided by FFT size. To test if this is accurate, you can capture data with a strong reflector at a known distance and observe a peak to that mapping and validate your findings.

Regards,
Kaushik

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

What do you mean by "grid marking as Max range"?

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Hi Palguna,

Apologies for the confusion. It was to be read as grid marking as (Max range based on effective IF BW divided by FFT size) i.e., this would be the range resolution that is effectively obtained in your spectrum.

Regards,

Kaushik

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

still don't understand it.

from mmwaveSensing-FMCW-offlineviewing_0.pdf (ti.com.cn), the max range is found as

, but you are saying max range as Bandwidth/rng_FFT_size

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Hi Palguna,

Not sure why he confusion here. The expression of Dmax and Dres you have pointed to are right. You can refer to the screenshot below to maybe clear your doubts:

Here, the rangeStep is what you would mark on each point on the x-axis of your 1D spectrum.

where max range = ((C * Fs) / (2 * S)) -> This is based on the bandwidth.

range step = (max range / FFT size) -> This is effectively (Max range/FFT size) which is what I mentioned earlier.

Hope this clears all confusion.

Regards,

Kaushik

+1 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

So, basically you are saying to take the value 0.1954 as the range resolution.

max range = 3*10^8*10*10^6/(2*29.982*10^12) = 50.03 meters

range resolution = 50.03/256 = 0.1954

As I see from the formula,

the actual bandwidth = slope * chirp time.

Here chirp time is not ADC sampling time, it is equal to number of ADC samples/sampling rate = 256/10^7 = 25.6uS

So using C/2*bandwidth gives = 0.1954m

thanks. could you tell where you got the above screenshot you posted

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

Could you tell from where did you take the screenshot

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Hi Palguna,

This information was obtained from an SDK used for 2nd generation devices and it can be found in the file called RF parser. Even the SDK for the AWR1843 should contain something similar to this.

Regards,

Kaushik

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

Thanks. I installed mmWave SDK 3.6.0.0. I search in the installation folder for RF parser. didn't find it. Could you tell where can i find the same by installing mmWave SDK 3.6.0.0. This helps in understanding other terms like how i understood this range resolution thing.

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Hi Palguna,

C:\ti\mmwave_sdk_03_06_00_00-LTS\packages\ti\demo\utils\mmwdemo_rfparser.c

Regards,

Kaushik

0 Palguna Kumar Reddy Gopireddy over 1 year ago in reply to Kaushik Gowda

Prodigy 141 points

Thanks

0 Kaushik Gowda over 1 year ago in reply to Palguna Kumar Reddy Gopireddy

TI__Mastermind 20605 points

Thanks for your confirmation. Will be closing the thead.

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AWR1843: How changing the "number of ADC samples" justifies, without changing the "ADC sampling time" and sampling frequency