Hi,
from datasheet:
why it have "Tmeas"? is there any explanation for that? thank you!
Best regards,
Yuan
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Hi,
from datasheet:
why it have "Tmeas"? is there any explanation for that? thank you!
Best regards,
Yuan
Hi, Yuan:
Please check the mmwave training on SNR calculation. You will find some explanation there.
Best,
Zigang
Hi Zigang,
Yes, I know this video, and this screenshot I pasted is from the video. after watching the video, I still cannot understand this formula, why here have "Tmeas" at numerator? is there any more explanation for that?
thank you!
Best regards,
Yuan
Hi, Yuan:
Without range FFT and Doppler FFT, the formula of SNR will be similar, just replace the measurement time as one ADC sample time.
But after range FFT, and Doppler FFT, there will be additional processing gain, SNR increases due to the coherent combination of signal. The processing gain of range FFT is the number of samples per chirp. And the processing gain of Doppler FFT is the number of loops per frame. Therefore, we get: T_adc * numAdcSample * numLoop = T_meas
Best,
Zigang
Hi Zigang,
The SNR before ADC can be get by: SNR= Power captured at RX /( Noise*F), where F is the noise factor from RX antenna to ADC. Noise equals kTB. As you mention, FFT will improve the SNR due to the processing gain. range FFT processing gain is the adc sample number, and velocity FFT processing gain is chirp number.
so the SNR should be Prx*Nsample*Nchirp/ (kTBF). Compared with the equation in the video, B and T_adc is missing. Could you help me to understand this?
The bandwidth BF in your formula can be computed as 1/T_adc.
Best,
Zigang
Zigang,
if the ADC sample number is less than FFT size, how do you choose the window size? let's say, we have 200 ADC samples and FFT size is 256. How do you choose the window size? 200 or 256? why?
You need to choose number of ADC samples. You can always fill zeros to use larger FFT size. But it does not increase number of samples that can be coherently combined to improve SNR.
Best,
Zigang