Increasing SQNR by Increasing Bits Per Sample

December 3, 2019

Increasing the number of bits per sample will decrease the quantization noise and, therefore, increase SQNR. Theorem 3 quantifies this qualitative relationship for a full-scale sine waveform input. Take note that the SQNR increases approximately 6dB for every additional bit.

Theorem 3

Let SQNR be the signal to quantization noise ratio for a full-scale sinusoid input.

Let b be the number of bits used by a converter. Then:

(1)   \begin{equation*} \text{SQNR}=\frac{3}{2}(2^b-1)^2 \text{ for }b\geq0 \end{equation*}

(2)   \begin{equation*} \text{SQNR}_{db}\triangleq10\text{log}_{10};\text{ SQNR}\approx[6.02b+1.76]\text{dB for }2^b>>1 \end{equation*}

Proof

(1) The power š‘ƒš‘  of a sine waveform Asin(2šœ‹š‘“cĀ  š‘” + šœ™) is A2/2

Proof: Let šœ = 1/š‘“cĀ  be one period of the sine waveform.

(3)   \begin{equation*} P_{s}\triangleq\frac{1}{\tau}\int_{0}^{\tau}[A\text{sin}(2\pi fct+\phi)]^2\text{d}t \end{equation*}

(4)   \begin{equation*} =\frac{\text{A}^2}{2}\int_{0}^{\tau}\text{sin}2(2\pi fct+\phi)\text{d}t \end{equation*}

(5)   \begin{equation*} =\frac{\text{A}^2}{2\tau}\int_{0}^{\tau}[1-\text{cos}(4\pi fct+2\phi)]\text{d}t\text{ by half-angle formula/squared identity} \end{equation*}

(6)   \begin{equation*} =\frac{\text{A}^2}{2\tau}\int_{0}^{\tau}1\text{d}t-\int_{0}^{\tau}\text{cos}(4\pi fct+2\phi)]\text{d}t\text{ by half-angle formula/squared identity} \end{equation*}

(7)   \begin{equation*} =\frac{\text{A}^2}{2} \end{equation*}

(2) For SQNR:

(8)   \begin{equation*} \text{SQNR}\triangleq\frac{\text{P}_{s}}{\text{P}_{n}}\text{ by definition of SQNR} \end{equation*}

(9)   \begin{equation*} =\frac{\text{A}^2/2}{\Delta^{2}12}\text{ by item (1) above and the proposition }\text{Var}v(n) \end{equation*}

(10)   \begin{equation*} =\frac{\Delta2}{12}\text{ from the lesson Quantization and Noise} \end{equation*}

(11)   \begin{equation*} =[(2^{\text{b}-1}-1/2)\Delta]^2/2\times\Delta^2/12\text{ from the lesson Quantization and Noise} \end{equation*}

(12)   \begin{equation*} =6(2^{\text{b}-1}-1/2)^2=6(1/2\times2^{\text{b}}-1/2)^2=\frac{6}{4}(2^{\text{b}}-1)^2=\frac{3}{2}(2^{\text{b}}-1)^2 \end{equation*}

(3) For SQNRdB:

(13)   \begin{equation*} \text{SQNR}_{dB}\triangleq10\log_{10}\text{SQNR}\text{ by the definition of }\text{SQNR}_{dB} \end{equation*}

(14)   \begin{equation*} =10\log_{10}[6(2^{\text{b}-1}-1/2)^2]\text{ by proof }(2) \end{equation*}

(15)   \begin{equation*} =10\log_{10}6+10\log_{10}[(2^{\text{b}-1}-1/2)^2]\text{ by property of logarithms} \end{equation*}

(16)   \begin{equation*} 10\log_{10}6+20\log_{10}(2^{\text{b}-1}-1/2)\text{ by property of logarithms} \end{equation*}

(17)   \begin{equation*} =10\log_{10}6+20\log_{10}[1/2(2^{\text{b}}-1)] \end{equation*}

(18)   \begin{equation*} =10\log_{10}6+20\log_{10}1/2+20\log_{10}(2^{\text{b}}-1)\text{ by property of logarithms} \end{equation*}

(19)   \begin{equation*} =10\log_{10}6+10\log_{10}(1/2)^2+20\log_{10}(2^{\text{b}}-1)\text{ by property of logarithms} \end{equation*}

(20)   \begin{equation*} =20\log_{10}(2^{\text{b}}-1)+10\log_{10}6/4\text{ by property of logarithms} \end{equation*}

(21)   \begin{equation*} \approx 20\log_{10}(2^{\text{b}})+10\log_{10}3/2\text{ for }2^{\text{b}}>>1 \end{equation*}

(22)   \begin{equation*} =20[\log_{10}2]\text{b}+10\log_{10}3/2\text{ by property of logarithms} \end{equation*}

(23)   \begin{equation*} \approx 6.02\text{b}+1.76 \end{equation*}

Over-Sampling with Quantization Uniformly Distributed White Noise

Theorem 3 is a theoretical ideal. In practice, an N bit converter has issues with inherent noise and doesn’t provide N bits of precision. Given this, system engineers want to know how many effective bits they can achieve. The b in Theorem 3 is equated in terms of SQNRdB to yield the following quantity called ENOB.

Let SINAD be the signal-to-noise-and-distortion ratio. Then:

(24)   \begin{equation*} \text{ENOB}\triangleq[\text{SNIAD}-1.76]/6.02 \end{equation*}