Chapter 2, Problem 83GP

On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about $0.\text{28 }\mu \text{m}$ . A CD player's readout laser scans along the spiral's sequence of bits at a constant speed of about 1.2 m/s as the CD spins, (a) Determine the number Nof digital bits that a CD player reads every second, (b) The audio information is sent to each of the two loudspeakers 44,100 times per second. Each of these samplings requires 16 bits, and so you might expect the required bit rate for a CD player to be

$N_0=2\left(44,100\frac{\text{samplings}}{\text{s}}\right)\left(16\frac{\text{bits}}{\text{sampling}}\right)=1.4\times {10}^6\frac{\text{bits}}{\text{s}},$

where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that N₀is less than the number N of bits actually read per second by a CD player. The excess number of bits (= N - N_o)is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error- correction?