EECS 314 Winter 2007 HW 10 Problem 1
Student's name ___________________________
Discussion section # __________
(Last name, first name,
IN INK)
© 2007 Alexander Ganago
The big picture
As you know from lecture notes, music on CD is recorded with 16 or 20bit resolution,
at the sampling rate of ~ 44 kHz. As you know from reading the labels, a blank CD has
the capacity of 700 MB (megabytes; 1 Byte = 8 bits; 1 MB ~ 10
6
Bytes) and is rated for
80 minutes of music recording. In this Problem, you will “put 2 and 2 together” in order
to see whether all of the above numbers make sense.
Problem
Part 1 (15 points)
Assume that the music is recorded in two channels (stereo = left and right) that have the
same resolution of 16 bits and the same sampling rate of 44 kHz. Calculate the amount of
memory in MB required to record 1 minute of music.
Your answer: _______________________ MB per minute of music
Assume that a blank CD has the capacity of 700 MB. Calculate how many minutes of
music can be stored on the CD.
Your answer: T = _______________________ minutes on CD
If your answer above is less than 80 minutes, it means that a data compression algorithm
is used to record the data on music CDs. The compression can be easily quantified: for
example, if you estimate the needed memory as 1 MB and the actual record requires only
0.8 MB, then the compression ratio equals 1/.8 = 1.25, which is very modest compared to
that used in MP3 technology.
Discuss your answer: if you found T > 80 minutes, no compression is needed. If,
however, you found T < 80 minutes, compression is needed, and you are asked to
calculate the compression ratio.
Your answer on compression: _______________________________________________
Part 2 (10 points)
Repeat Part 1 for 20bit music recording.
Your answer: _______________________ MB per minute of music
Your answer: T = _______________________ minutes on CD
Your answer on compression: _______________________________________________
Show your work for both parts on a separate page.
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Problem #1 Solution
Each sample is recorded as a 16 bit binary number. Since both channels share the same
parameters, we can solve for one channel and double the result.
Starting with the sampling rate and the resolution, we can find the number of MB required for 1
minute of recording
min
56
.
10
2
min
1
60
10
1
8
1
1
16
sec
44000
6
MB
s
byte
MB
bits
byte
sample
bits
ond
sample
⋅
=
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
Now divide the capacity of the blank CD by the storage rate.
.
min
29
.
66
.
min
56
.
10
700
⋅
=
⋅
⋅
MB
MB
The answer is less than the 80 minutes which means that a data compression algorithm was used
to record the data. Next, find the compression ratio.
207
.
1
min
29
.
66
min
80
=
⋅
⋅
Repeat the same calculations for the 20bit music recording. More storage space is expected
since the resolution is improved.
min
12
.
12
2
min
1
60
10
1
8
1
1
20
sec
44000
6
MB
s
byte
MB
bits
byte
sample
bits
ond
sample
⋅
=
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⋅
⋅
⋅
⎟
⎠
⎞
⎜
⎝
⎛
⋅
.
min
76
.
57
.
min
12
.
12
700
⋅
=
⋅
⋅
MB
MB
39
.
1
min
76
.
57
min
80
=
⋅
⋅
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 Fall '07
 Ganago
 Binary numeral system, Logic gate, Alexander Ganago

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