The 1 bit = 6 dB Rule of Thumb, Revisited.

Almost ten years ago I wrote an entry about the “1 bit = 6 dB” rule of thumb. This rule states that for each bit you add to a signal, you add 6 dB of signal to noise ratio.

The first derivation I gave then was focused on the noise, where the noise maximal amplitude was proportional to the amplitude represented by the last bit of the (encoded) signal. Let’s now derive it from the most significant bit of the signal to its least significant.

Let us suppose that the first bit represents half the maximum signal amplitude, A. Then the second one should represent a quarter, and the third an eighth, etc. For k bits, then, the maximum amplitude is

\displaystyle A_{max}=A\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots+\frac{1}{2^k}\right)=A\sum_{i=1}^k \frac{1}{2^i},

Or, alternatively,

\displaystyle A_{max}=A\sum_{i=1}^k \frac{1}{2^i}=A\left(1-\frac{1}{2^k}\right),

which means that the rest of the amplitude—noise—is

\displaystyle A\left(1-\left(1-\frac{1}{2^k}\right)\right)=A\frac{1}{2^k}.

Recall, the signal-to-noise ratio, the SNR, is given by the ratio of the powers of the signal and the noise, that is:

\displaystyle SNR=\frac{P_{signal}}{P_{noise}}=\left(\frac{A_{signal}}{A_{noise}}\right)^2.

Substituting A_{signal} and A_{noise} for their expressions parameterized by k we get that

\displaystyle SNR=\left(\frac{A_{signal}}{A_{noise}}\right)^2=\left(\frac{A\left(1-\frac{1}{2^k}\right)}{A\frac{1}{2^k}}\right)^2,

where the As cancel out. We can rewrite the ratio as

\displaystyle SNR=\left(2^k\left(1-\frac{1}{2^k}\right)\right)^2.

To get decibels out of the expression of the power, we write

\displaystyle dB(SNR)=10 \log_{10} SNR,

or, in our case,

\displaystyle dB(SNR)=10 \log_{10} \left(2^k\left(1-\frac{1}{2^k}\right)\right)^2.

Let us see what this expression yields.

\displaystyle dB(SNR)= 20 \log_{10} 2^k\left(1-\frac{1}{2^k}\right)

\displaystyle =20 \log_{10} 2^k + 20 \log_{10} \left(1-\frac{1}{2^k}\right)

\displaystyle = 20 k \log_{10} 2 + 20 \log_{10} \left(1-\frac{1}{2^k}\right).

Since 20 \log_{10} 2\approx 6.0206, and 20 \log_{10} (1-2^{-k}) goes rapidly to zero, we can write

\displaystyle dB(SNR;k)\approx k 20 \log_{10} 2 \approx 6.0206 k.

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So how fast does the last part, 20 log_10 (1-2^{-k}). go to zero? Intuitively, we should agree that as k grows, 1-2^{-k} gets closer and closer to one, and that, therefore, 20 log_10 (1-2^{-k}) goes to zero. But how fast? A quick plot reveals the following picture:

which indeed goes rapidly to zero. When we combine the two results, we see that the full expression and the approximation go quite agree as soon as k is somewhat large:

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So when a device tells you it has a SNR of 60 dB (like an 1980s cassette deck), what it really tells you is that it’s gives you about 10 bits resolution. CDs, with 16 bits, gives you approximately 96 dB; 24 bits bring you in the realm of 144 dB. Well, good luck finding a device that will indeed guarantee this kind of SNR.

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