02:39
Physics for Scientists and Engineers with Modern Physics
(II) On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about 0.28$\mu \mathrm{m}$ . A CD player's readout laser scans along the spiral's sequence of bits at a constant speed of
about 1.2 $\mathrm{m} / \mathrm{s}$ as the CD spins. (a) Determine the number $N$
of 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 one would (at first glance) think the required
bit rate for a CD player is $N_{0}=2\left(44,100 \frac{\text { samplings }}{\text { second }}\right)\left(16 \frac{\text { bits }}{\text { sampling }}\right)=1.4 \times 10^{6} \frac{\text { bits }}{\text { second }}$$
$$\begin{array}{l}{\text { where the } 2 \text { is for the } 2 \text { loudspeakers (the } 2 \text { stereo channels). }} \\ {\text { Note that } N_{0} \text { is less than the number } N \text { of bits actually read }} \\ {\text { per second by a CD player. The excess number of bits }} \\ {\left(=N-N_{0}\right) \text { is needed for encoding and error-correction. }}\end{array}$
What percentage of the bits on a $\mathrm{CD}$ are dedicated to
encoding and error-correction?