UPGRADE YOUR BROWSER

We have detected your current browser version is not the latest one. Xilinx.com uses the latest web technologies to bring you the best online experience possible. Please upgrade to a Xilinx.com supported browser:Chrome, Firefox, Internet Explorer 11, Safari. Thank you!

cancel
Showing results for 
Search instead for 
Did you mean: 
Highlighted
Observer 062548775430
Observer
7,490 Views
Registered: ‎04-18-2008

Block floating point FFT output magnutude

Hi All!

 

I am working with block floating point FFT .

This method has more good accuracy in comparison with scaled FFT.

But I have problem to measure FFT output magnitudes for group of signals in the input. I wanna calculate approximately magnitudes of some  selected spectral components.

For example.

If one input signals will has high amplitude - exponenta will go to maximum and another spectral components magnitudes(i mean their mantisses) will have some level values.

If one input signals will has little amplitude - exponenta will go to low level and another spectral components magnitudes(i mean their mantisses) will have new level values.

Who have experience with block floating point FFT magnitudes?

 

Best Regards

 

Alexander

 

0 Kudos
1 Reply
Xilinx Employee
Xilinx Employee
7,481 Views
Registered: ‎08-01-2008

Re: Block floating point FFT output magnutude

You can use common exponent method. The block can be expressed in two components − a mantissa and a common exponent. The common exponent is stored as a separate data word. The value of the common exponent is determined by the data element in the block with the largest amplitude. In order to compute the value of the exponent, the number of leading bits has to be determined. If a given block of data consists entirely of small values, a large common exponent can be used to shift the small data values left and provide more dynamic range. On the other hand, if a data block contains large data values, then a small common exponent will be applied. Whatever the case may be, once the common exponent is computed, all data elements in the block are shifted up by that amount, in order to make optimal use of the available dynamic range

Thanks and Regards
Balkrishan
--------------------------------------------------------------------------------------------
Please mark the post as an answer "Accept as solution" in case it helped resolve your query.
Give kudos in case a post in case it guided to the solution.
0 Kudos