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```					Digital Information Storage
Contents:
•Binary vs decimal
•CDs and DVDs
Binary:     Bin to Dec:
1101 = 1x23 + 1x22 + 0x21 + 1x20 = 13
0   0000
237 to binary:
1   0001
2   0010      Dec to Bin:    Place
3   0011      237-1x128      128     1 - msb
4   0100      109-1x64       64      1
5   0101      45 - 1x32      32      1
6   0110      13 - 0x16      16      0
7   0111
13 - 1x8       8       1
8   1000
9   1001      5 - 1x4        4       1
10   1010      1 - 0x2        2       0
1 - 1x1        1       1 - lsb
TOC
237 = 11101101
Whiteboards:
Binary Conversions
1|2|3|4

TOC
Convert 11001 to decimal:

11001 = 1x24 + 1x23 + 0x22 + 0x21 + 1x20 = 25

25                                              W
Convert 11011010 to decimal:

11011010 =
1x27 + 1x26 + 0x25 + 1x24 + 1x23 + 0x22 + 1x21 + 0x20 =
218

218                                                       W
Convert 175 to binary:
Dec to Bin:   Place
175-1x128     128     1 - msb
175:        47 -0x64      64      0
47 - 1x32     32      1
15 - 0x16     16      0
15 - 1x8      8       1
7 - 1x4       4       1
3 - 1x2       2       1
1 - 1x1       1       1 - lsb

175 = 10101111
10101111                                    W
Convert 198 to binary:
Dec to Bin:   Place
175-1x128     128     1 - msb
198:        70 - 1x64     64      1
6 - 0x32      32      0
6 - 0x16      16      0
6 - 0x8       8       0
6 - 1x4       4       1
2 - 1x2       2       1
0 - 0x1       1       0 - lsb

198 = 11000110
11000110                                    W
Analog storage:              Digital storage:
Signal varies continuously   1 vs 0 less subtle
Taping a record or CD        Error Correction

TOC

Sample rate/depth
Same basic approach that we took for dec to bin
soundcards…
TOC
10 samples

1.2

1

0.8

0.6

0.4

0.2

0
1   2   3   4   5     6      7   8   9   10   11
25 samples

1.2

1

0.8

0.6

0.4

0.2

0
1   2   3   4   5   6   7   8   9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
50 samples

1.2

1

0.8

0.6

0.4

0.2

0
1   3   5   7   9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
200 samples

1.2

1

0.8

0.6

0.4

0.2

0
1   11 21 31 41 51 61 71 81 91 101 111 121 131 141 151 161 171 181 191 201

CDs sample at 44,100 Hz, 16 bit depth
Information Storage on a CD

0    1    0       1

Pass around CD Drive

TOC
Pits change laser path length
(destructive vs constructive)
2d = 1/2λ

Example – if the pits are 150 nm deep, what wavelength
of laser light will create destructive interference relative
to this depth?
2d = 1/2λ
2(150 nm) = 1/2λ
λ = 600 nm
Rayleigh Criterion

 = r = 1.22
d     b

 = Angle of resolution (Rad)
r = min distance separating pits (m)
d = distance to CD from lens (m)
 = Wavelength of laser (m)
b = Diameter of CD Lens (m)

DVDs ≈ 640 nm
CDs ≈ 780 nm

Central maximum of one is over
minimum of the other
8 bits bytes are actually mapped to 14 bits on CDs
Whiteboards:
Pits and interference
1|2|3

TOC
If an optical drive uses light that is 340 nm,
what pit depth would create destructive
interference?

2d = 1/2λ = 1/2(340 nm)
d = 85 nm

85 nm                                            W
A CD has pits that are 125 nm deep. What
wavelength of laser light would generate
destructive interference relative to that depth?

2(125 nm) = 1/2λ
λ = 500 nm

500 nm                                             W
Example: DVDs have pits that are about
0.74μm apart. If the lens has a diameter of 4.2
mm, and the laser has a wavelength of 780 nm,
what must be the maximum distance from the
lens to the disc so that the reader can resolve
the pits?
 = r = 1.22
d     b

d = (0.74E-6)(4.2E-3)/(1.22*780E-9) = .003266…

3.3 mm                                            W
Other methods of data storage

•RAM/cache
•Flash memory/Core
•Magnetic
•CD/DVD

```
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