AWR1642: Clutter removal

Kei

Part Number: AWR1642

Hello,

I have a question about static clutter removal.
To remove static clutter, I suppose it has only to ignore zero th doppler data 'after' 2DFFT simply, but SDK subtracts zero th component, which is mean value of the input samples for 2D FFT, 'before' 2DFFT.
Could you let me know the reason why such way is used? Is there any technical reasons?

over 3 years ago

0 Kei over 3 years ago

Intellectual 740 points

Hello,

I would appreciate it if you would answer my question.
Please let me know if you need more information concerning this matter.

0 Aniket Mahajan over 3 years ago in reply to Kei

TI__Genius 15046 points

Hi Kei,

Sorry for the immense delay in my response.

Quite simply, the zeroth FFT bin is the DC component of the signal. If the mean of the signal is subtracted from it before taking the FFT, the DC component and therefore the zeroth FFT bin is zeroed out.

Regards,

Aayush

0 Kei over 3 years ago in reply to Aniket Mahajan

Intellectual 740 points

Hello Aayush,

Thank you for your reply.
I feel you didn’t fully understand what I meant, so please let me try again.

In order to remove static clutter (zeroth doppler objects), why does not SDK do the following procedure?
first: 1D-FFT
next: 2D-FFT
finally: delete zeroth doppler bin data

Regards,
Kei

+1 Aniket Mahajan over 3 years ago in reply to Kei

TI__Genius 15046 points

Hi Kei,

If the DC component isn't subtracted out, bin 0 will have a large value. When windowing, this will leak into the adjacent bins, possibly obscuring weak targets in them. This is why we prefer to remove the DC out altogether than just ignore bin 0.

Regards,

Aayush

0 Kei over 3 years ago in reply to Aniket Mahajan

Intellectual 740 points

Hi, Aayush,

Thank you for your answer.
I understand that clearly.

Regards,
Kei

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AWR1642: Clutter removal