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**Homework Statement**

On an audio compact disc, digital bits of information are encoded sequentially along a spiral path.

**Each bit occupies about 0.28 micrometers.**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 N of digital bits that a CD player reads every second.

b) The audio information is sent to each of the two loudspeakrs 44,100 times per second. Each of these samplings requires 16 bits and so one would think the required bit rate for a CD player is

**N**The excess number of bits(N-N

_{0}= 1.4*10^{6}bits/second._{0}) is needed for encoding and error-correction. What percentage of the bits on a CD are dedicated to encoding and error-correction?

**The attempt at a solution**

1.2 / (0.28 * 10^-6) = 4.3 * 10^6 bits per second.

((4.3-1.4)/ 4.3) * 100 = 67% for encoding and error correction.

Having seen the high percentage, I've thought that something may have gone wrong, so I ask for your help to make sure if I solved this correctly or not.