the quantization noise power at FIR's output

Hi,
I know that there r two completely different types of noise in FIR. One is the type of noise that is present in the signal which
is filtered out by the stop-band of an FIR. I'll call this type of noise "SIGNAL NOISE" during this post. The other type of noise is the kind generated by the implementation of the filter itself, such as the quantization noise caused by truncating the long accumulated values back to a shorter length. I'll call this "IMPLEMENTATION NOISE".
Here, I focus on how to calculate the total quantization noise power at the ouput of a FIR.

The FIR is directly connected after an 8-bit ADC. With its coef quantized to 15 bits, this 16-tap fix-point FIR has 22-bit (8bit*15bit->22bits)multiplier and 26-bit (22bits+log2(16)) accumulator which could guarantee the multiply-accumulates performed on this FIR lose no one bit of accuracy.
N1 denotes the datawidth of the ADC which is the same as that of the FIR's input. So, The quantization error power due to the ADC is
Pe1= (1/12)(2Vref/(2^N1))^2
N2 denotes the datawidth of accumulator's result


Q1: If the accumulator's result is given to the following stage without any truncating, does Pe1 just equal to the total quantization noise power at the ouput considering there is no extra noise introduced ?

Q2:If the FIR uses the accumulator's upper 8-bit as its output, then
N3 denotes the datawidth of the FIR's output obtained from truncating accumulor's result
The quantization error power of this truncating is Pe3= (1/12)(2Vref/(2^N3))^2
Then, what is the total quantization noise power at the ouput, Pe1 or Pe3 or Pe1+Pe3?

Thanks！

At the accumulator output, the input quantization noise gets multiplied by the sum of the squared coefficients.
The truncation divides by 2^18 AND introduces a rounding error, that can be considered like like a further quantization.
So you got: Pe1*( sum(Ci^2)/2^36 + 1 ).

thanks, but could u explain further how to get this expression or give me some related material?
do u mean the noise introduced by truncating the accumulator is Pe1*sum(Ci^2)/2^36?
i dont know how does the 2^36 comes from.
And if I dont truncate the accumulator, what is the total quantization err pwer?

Thanks!

sum(Ci^2) is just a way to write "the sum of the squared coefficients".
Why? each time the FIR perfoms a multiplication the quantization noise associated with that sample is multiplied by the coefficient too - they are squared because we are computing noise power. Up to here you have ADC QN (Pe1) by sum(Ci^2).

Truncation from 26 to 8 bits, i.e. a left-shift of 18 bit, is equivalent to a division by 2^18 - but once again we are computing power, so (2^18)^2 = 2^36.

To this you must add the truncation noise. Here holds the same maths you did for the ADC, since you are againg dealing with 8 bits.

Oh, thanks so much!
so if i dont truncate the accumulator, the total quantization noise power is Pe1*sum(Ci^2).
but what should be the ci ? i mean the final fixed-point code of coef applied to the multiplier is a binary integer but it may represent various coef . For ex, the code 0111(2) may represent 0.111(2) or 01.11(2) or 011.1(2).

---------- Post added at 17:25 ---------- Previous post was at 17:23 ----------

Oh, thanks so much!
so if i dont truncate the accumulator, the total quantization noise power is Pe1*sum(Ci^2).
but what should be the ci ? i mean the final fixed-point code of coef applied to the multiplier is a binary integer but it may represent various coef . For ex, the code 0111(2) may represent 0.111(2) or 01.11(2) or 011.1(2).

Since you wanted to know the power BEFORE the truncation, 0111 represents just 0111 (I am following your example).

If you want to consider the point position, then you have coefficients of 15 bits AND a left-shift of 18 bits. In this case 0111 represents 0.0000111, because 15-18 = 3 zeroes added to the right of the point (I am not considering the sign).
It's just a matter of conventions if you want to consider the shift immediately (it is usual to); if you repeat the calculation in this case the 2^36 disappears - but the numerical result is the same.

I'm getting clearer but still a little confused.
Do u mean the the actual coef in the expression is Ci(the integer code)/2^18 due to the left-shift of 18 bits?
But it is amazing if I dont truncate the accumulator then there will be such a big err power Pe1*( sum(Ci^2) + 1 ) because the actual coef in the expression will be only Ci and this is a big number(a 15-bit integer).

---------- Post added at 15:15 ---------- Previous post was at 15:02 ----------

Oh , maybe it is my fault.
Without truncating , although the err power Pe1*( sum(Ci^2) + 1 ) is huge, the signal power at output Psin*sum(Ci^2) is also huge. So the SNR can also be a reasonable one.
It may be just a arithmetic game, right?
*Psin denotes the signal power at the input of FIR

Yes, if you think the coefs as Ci/2^18 then your 26 bit accumulator really is: 8 bit integer part POINT 18 bit fractional part.
Truncation just removes the fractional part (and introduces another quantization error).
It's just a matter of representation/conventions.

Oh , maybe it is my fault.
Without truncating , although the err power Pe1*( sum(Ci^2) + 1 ) is huge, the signal power at output Psin*sum(Ci^2) is also huge. So the SNR can also be a reasonable one.
It may be just a arithmetic game, right?
*Psin denotes the signal power at the input of FIR