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Solved: On an audio compact disc (CD), digital bits of
Chapter 2, Problem 83GP(choose chapter or problem)
On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about \(0.28\ \mu m\). 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 spins.
Determine the number
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 you might expect the required bit rate for a CD player to be
\(N_{0}=2\left(44,100 \frac{\text { samplings }}{s}\right)\left(16 \frac{\text { bits }}{\text { sampling }}\right)=1.4 \times 10^{6} \frac{\text { bits }}{s}\),
where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that \(N_{0}\) is less than the number of bits actually read per second by a CD player. The excess number of bits \(\left(=N-N_{0}\right)\) is needed for encoding and error-correction. What percentage of the bits on a
are dedicated to encoding and error-correction?
Equation Transcription:
Text Transcription:
0.28 mu m
1.2 m/s
N_0=2(44,100 {samplings over s})(16 {bits over sampling})=1.4x10^6 {bits over s}
N_0
(=N-N_0)
Questions & Answers
QUESTION:
On an audio compact disc (CD), digital bits of information are encoded sequentially along a spiral path. Each bit occupies about \(0.28\ \mu m\). 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 spins.
Determine the number
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 you might expect the required bit rate for a CD player to be
\(N_{0}=2\left(44,100 \frac{\text { samplings }}{s}\right)\left(16 \frac{\text { bits }}{\text { sampling }}\right)=1.4 \times 10^{6} \frac{\text { bits }}{s}\),
where the 2 is for the 2 loudspeakers (the 2 stereo channels). Note that \(N_{0}\) is less than the number of bits actually read per second by a CD player. The excess number of bits \(\left(=N-N_{0}\right)\) is needed for encoding and error-correction. What percentage of the bits on a
are dedicated to encoding and error-correction?
Equation Transcription:
Text Transcription:
0.28 mu m
1.2 m/s
N_0=2(44,100 {samplings over s})(16 {bits over sampling})=1.4x10^6 {bits over s}
N_0
(=N-N_0)
ANSWER:
Step 1 of 3
Space occupied by the bits:
Readout speed of the laser:
Bit rate of the CD player:
