rogeret
Member level 4
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!
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!
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