I have a quation about bandwidth, B in the thermal noise power, kTB in radar equation.
In FMCW, according to hforsten.com/third-version-of-homemade-6-ghz-fmcw-radar.html ,
"Noise bandwidth is clearly smaller than the sweep bandwidth
since the IF filter filters most of the RF noise out.",
(skip)
"Bandwidth of one FFT bin depends only of the lenght of the FFT.
With FMCW radar FFT length equals length of one sweep, ts
and bandwidth of one FFT bin is 1/ts."
So, in TI dev IWR1443, do the same way
B = 1/Tr
where Tr is Chirp Time (= sweep time).
Is this right?
Also, P.15 in TI document training.ti.com/.../Mmwave_webinar_Dec2017.pdf
SNR = Rx signal level * noise level
Rx signal level = Pt Gt (RCS) Gr lambda^2/ (4 PI)^3/R^4
noise level = Tf/{ kT (NF) }
Tf -> integration time
To be more exact,
SNR = Rx signal level * noise level
Rx signal level = Pt Gt (RCS) Gr lambda^2/ (4 PI)^3 /R^4 *N ( as integration effects)
noise level = 1 /{ kT (1/Tr) (NF) }
N -> Number of chirps
Is this understanding correct?