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Yes, but now supose im using 14 bits and i have 5 FIR filters, The filters output are 14 bits. If i sum the outputs of the filters overflow will occurs.
So how can i handle this?
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First, you need to work out if you need 14-bits of precision in your FIR filter output. If you know the input signal characteristics, and the FIR filter response, then you can calculate the RMS of the signal and the RMS of the noise, and from that you can determine the minimum number of bits you require. For now, lets assume you need 14-bits, but you should check.
When you add two B-bit values together, you get worst-case bit-growth of 1-bit, so for your 5 filters, you will get a bit growth of a little over 2-bits. If your input RMS is lower than its maximum possible value, then you probably only need to grow the result by 2 bits. If the noise in your signal is random, then its variance will add, so the variance in the sum will be 5 times the variance of the input (assuming each input has the same noise), so the noise power will increase by sqrt(5) = 2.24, i.e., a little over 1-bit, so your dynamic range might increase slightly (depending on whether the 'signal' is also noise-like or whether it is coherent).
I assume you have a 14-bit DAC, and that is why you have scaled your FIR filter outputs to 14-bits. However, your sum will be around 16-bits, so you will need to convergent round the wider sum to 14-bits so that you can drive the DAC.
Does that make sense?
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I want a 0 dBm for each FIR output seen in the DAC output.
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No you probably don't. You want the largest possible signal your DAC can output, within the constraints of the sampling bandwith, and the inverse sinc filter response (assuming you have such a filter).
The DAC output power into a 50-ohm load, or the power in dBm, will be whatever the board designer decided it will be. There's not much you can do about it, other than use what your board design has. You have not provided details on that, so I can't comment more.
Cheers,
Dave