14.1.1 solve problems involving the conversion between binary numbers and decimal numbers. 14.1.2...

14.1.1 Solve problems involving the conversion between binary numbers and decimal numbers.

14.1.2 Describe different means of storage of information in both analog and digital forms.

14.1.3 Explain how interference of light is used to recover information stored on a CD.

14.1.4 Calculate the appropriate depth of a pit given the wavelength of the laser light.

14.1.5 Solve problems on CDs and DVDs related to storage capacity.

14.1.6 Discuss the advantage of storage of information in digital rather than analog form.

14.1.7 Discuss the implications for society of ever-increasing capability of data storage.

Topic 14: Digital technology 14.1 Analog and digital signals

Solve problems involving the conversion between binary numbers and decimal numbers.

In order to understand other numbering systems, you have to understand the one you use (the base 10 numbering system or decimal system), which gives place value to each of a set of 10 digits (0,1,2,3,4,5,6,7,8,9).

Place value is determined by powers of ten in our numbering system:


FYIIs the Roman numeral numbering system a place value system? No, it is not. Each symbol I, V, X, L, C, M, etc. represents a fixed quantity wherever it is placed.

105 104 103 102 101 100 Power of 10 place value

__ __ __ __ __ __ Place position


Each digit in the place position is multiplied by it place value, and the results are added up.




EXAMPLE:

The decimal number 2301 looks like this: And means this: 2103 + 3102 + 0101 + 1100

21000 + 3100 + 010 + 11

2000 + 300 + 0 + 1 = 2301.

2 3 0 1

FYIThe decimal system probably came about because of our ten fingers. Perhaps it is no coincidence that each finger is known as a “digit.”


Now that you understand how a place-value numbering system works let us consider the binary system, which only has two digits (0,1) and is termed base 2.

Instead of powers of 10, we use powers of 2:




EXAMPLE:

The binary number 1101 looks like this: And means this: 123 + 122 + 021 + 120

18 + 14 + 02 + 11

8 + 4 + 0 + 1 = 13 (dec).

1 1 0 1



PRACTICE: Convert 110101 (binary) to its decimal value.

SOLUTION:



Then from our place position grid we have: 125 + 124 + 123 + 122 + 021 + 120

132 + 116 + 08 + 14 + 02 + 11

32 + 16 + 0 + 4 + 0 + 1 = 53 (dec).

1 1 0 1 0 1

FYIWe write 110101 (bin) = 53 (dec).It is because electronic switching devices use circuits that are either on or off that binary is the numbering system of computers.



EXAMPLE: Concert 25 (dec) to binary.

Find the biggest power of 2 which is less than or equal to 25: 32 is too big.

16 works. Subtract 16 from 25 and see how much more we need: 25 - 16 = 9.

We need 9. 8 works. Subtract 8 from 9 and see how much more we need: 9 - 8 = 1.

4 and 2 are too big…The 1 works.Thus 25 (dec) = 011001 (bin) [ or 11001 (bin) ].

Power of 2 25 24 23 22 21 20

Decimal 32 16 8 4 2 1

Binary 0 1 1 0 0 1

PRACTICE:

Convert 39(dec) into its binary equivalent.

SOLUTION:

Thus 39(dec) = 100111(bin).



Power of 2 25 24 23 22 21 20

Decimal 32 16 8 4 2 1

Binary 1(7 left)

0

(7 left)

0

(7 left)

1

(3 left)

1

(1 left)

1

(0 left)

FYIWe call binary digits bits. O is a bit. So is 1.8 bits in a string are called a byte.

[ 0010 0111 byte form ]


The least significant bit (LSB) in a binary number is the rightmost one (the 20 place).

The most significant bit (MSB) in a binary number is the leftmost non-zero one.

Thus in the binary number 00100111 of the previous example, the most significant bit is in red and the least significant bit is in blue. FYIThe most significant bit changes least frequently while counting sequentially. It represents the largest power of 2 in its place value.The least significant bit represents the smallest power of 2 (20) and changes most frequently while counting sequentially.


BINARY: DECIMAL: 000 0 0 010 0 1 020 1 0 030 1 1 041 0 0 051 0 1 061 1 0 071 1 1

Describe different means of storage of information in both analog and digital forms.

Consider the following signal (perhaps the sound wave coming from a musical instrument):

If your data storage device stores an exact replica of this data, say using magnetic media, vinyl, or silver-nitrate film then we say that the data is analog.


Intensity

/ Wm-2

FYIWhat this means is that for every point on the actual source wave there is a corresponding point with a unique value on the stored data medium.

Time / s

cassette

VHS

film

vinyl


Another way to store the same information is to convert it to binary.

Then your data storage device stores an inexact copy of this data, say using magnetic media, CDs or DVDs, memory sticks, or CCDs, and we say that the data is digital.

FYIWhat this means is that for many (but not all) points on the actual source wave there is a corresponding binary value stored in the medium.


money card

s

CDs/DVDs HDs CCDssti

cks

floppies


Since data “streams” from the real world are usually continuous, converting real data like sound into digital form takes some doing.

It is easy to convert, for example, a sound wave into an analog electrical signal by recording the intensity as a voltage.

Then we add a fixed voltage to the wave to bring each point into the positive voltage region.


Intensity

/ Wm-2

Time / s

Voltage

/ V

Time / s

7

6

5

4

3

2

1

0


Suppose we sample at one-second intervals. We say that the sample rate is 1 Hz (1 sample per second).

A digital representation for the above wave is found by finding the voltage at the right of each interval, and truncating downward. We’ll demonstrate this method.


Time / s

Voltage

/ V

7

6

5

4

3

2

1

05 6 4 1 0 1 1 1 3 6 7 5 2 3 4 3 3 4


Now we convert the decimal values for the voltage into binary strings.We need three bits to represent the 0-7 (dec):

A playback of this sound would look like this: It would be stored digitally in the binary string


Time / s

Voltage

/ V

7

6

5

4

3

2

1

05 6 4 1 0 1 1 1 3 6 7 5 2 3 4 3 3 4101 110 100 001 000 001 001 001 011 110 111 101 010 011 100 011 011 100

101 110 100 001 000 001 001 001 011 110 111 101 010 011 100 011 011 100.

An increase in the sample rate will increase the fidelity (with a corresponding increase in data).


The following is a block diagram of the key components in converting analog signals to digital signals:

The clock sets the sample rate (Hz). The faster the clock rate, the more samples per second.

The analog-to-digital converter does what we did by hand on the previous slide (just like it says).

The digital signal is then stored in a medium.


101 110 100 001 000 001 001 000 011 110 111 101 010 011 100 011 011 100

Analog Signal

Clock

Analog to Digital Converter

(ADC)

Digital Signal

011 011


EXAMPLE: The human ear detects frequencies ranging from 20 Hz to 20 kHz. What should the sample rate of our clock be set at to record music?

SOLUTION:

According to the FYI the clock should be set at 40 kHz.

This is equivalent to 40000 samples per second.


FYIAs a rule of thumb the sample rate (clock frequency) should be twice the maximum frequency of the sound being recorded if fidelity is to be maintained.

Explain how interference of light is used to recover information stored on a CD.

Digital data is stored in various mediums, but always as zeros and ones.

On a compact disk (CD), which can store about 0.75 GB (3/4 of a gigabyte), the zeros and ones are pits and bumps.

The top waveform is the reference string of zeros and ones. Think of it as the clock. The bottom shape is the actual data on the CD. In three dimensions a line of bits on the CD would look something like this:


punch tape

c l o c k

1 0 1 1 1 0 1 0 0 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 1 1 1 1 0 1 1 1 1 0 1

7 bits

clockdata

CD bits

1001 100

CD pits and bumps

record grooves

Grooves spaced about 100 m apart.

CD rows spaced about 1.6 m apart.

This is 100 / 1.6 = 63 times farther.

BlueRay


Recall that if two coherent beams of light are shifted so the they differ in phase by half a wavelength, destructive interference will occur.

This is how a CD (or DVD) reader interprets the pits and bumps on a disk.

If the bumps are exactly ¼ of a wavelength in height light waves reflected from pit and bump will be exactly ½ wavelength off.


path difference = (n + ½) condition for destructive interferencen is an integer

pit

Bump (or land)

/4


FYIThe photogate detects the interference.


constructive interference

constructive

interference

destructive interference

reflected beams

reflected beams

reflected beams

Calculate the appropriate depth of a pit given the wavelength of the laser light.


EXAMPLE: A CD player uses a laser having a frequency of 3.851014 Hz to illuminate pits and bumps on a disk during readback.

Calculate the depth of a pit.

SOLUTION: First find the wavelength of the laser.

From c = f we have = c/f = 3.00108/ 3.851014

= 7.7910-7 m = 779 nm.

Then since the depth of a pit is /4 D = /4 = 779 nm/4 = 195 nm.195 nm

Solve problems on CDs and DVDs related to storage capacity.

CDs are a bit less than 6 cm in radius.Typical adjacent tracks are illustrated here:

Tracking is in a spiral. The laser is on an arm which can move radially.

The angular speed of the disk must vary accordingly so that the linear tracking speed v = r remains constant.

FYIThese dimensions will of course vary, depending on the frequency of the laser reader.


2.25 m

125 nm

0.5 m

0.97

m



PRACTICE: A CD has a radius R of 5.75 cm and a track separation s = 2.25 m.

(a) Find the number N of “data grooves” in R.

(b) Estimate the total “data groove” L length by assuming each groove has a circular length equal to a circle of the average radius of the CD.

SOLUTION:R = 0.0575 m, and s = 2.2510-6 m so that

(a) N = r/s = 0.0575 / 2.2510-6 = 25600 grooves.

(b) ravg = R/2 = 0.0575/2 = 0.0288 m.

L = NCavg = 25600[2(0.0288)] = 4630 m.

2.25 m



PRACTICE: A CD has a radius R of 5.75 cm and a track separation s = 2.25 m.

(c) If the smallest pit (or bump) is p = 0.97 m, how many bits of data can be stored along the total length of the data spiral?

(d) If 1 byte B is 8 bits, how many bytes can be stored on the CD?

SOLUTION:

(c) bits = L/p = 4630 / 0.9710-6 = 7.77109 bits.

(d) (4.77109 bits)(1 B/8 bits) = 596 MB.

2.25 m

0.97

m

Discuss the advantage of storage of information in digital rather than analog form.

Analog signals can very accurately represent the source signal. During recording and playback, however, noise (a form of interference that distorts the signal) can add distortion to the signal. (Think of Xeroxed copies of copies…)

Furthermore, analog storage requires sensitive physical media such as vinyl grooves, magnetic tape, and photographic film.

All of these media degrade under normal usage, and environmental fluctuations such as temperature, humidity, or even magnetic fields.


Discuss the advantage of storage of information in digital rather than analog form.

Digital signals can very accurately represent the source signal. Furthermore, during recording and playback, noise is not a factor since digital storage requires only two discrete states which are recognizable as 0s and 1s even if they are in a “fuzzy” state.

Finally, because the data is stored in digital format, it can be easily fed into computer programs and be manipulated. Thus you can do Google searches and PhotoShop your digital pictures.


FYIDigital information also takes up far less space than analog media.

Discuss the implications for society of ever-increasing capability of data storage.

One implication is that due to the convenience of digital data (both in its storage and retrieval), more information is both kept and accessed by people world-wide. It should be obvious how this can be a two-edged sword.

Another implication is that due to the ability of anyone to post and manipulate data, although there is very much more information available to us then ever before, it is harder to evaluate this data respecting its validity.

And how about that plagiarism?




Power of 2 25 24 23 22 21 20

Decimal 32 16 8 4 2 1

Binary 0 0 1

4 left

1

0 left

12 left

12 left

0 0

Calculate the appropriate depth of a pit given the wavelength of the laser light.


To create the condition for destructive interference at a pit/land interface, the pit must be /4 in depth.

Thus is 4 times the pit depth.

4(150 nm) = 600 nm.



The LSB is the rightmost digit in a binary number, holding the place of the lowest power of 2 (20 = 1 dec).

= 023 + 122 + 021 + 020 = 4 (dec).

= 023 + 122 + 021 + 120 = 5 (dec).



There are 20 characters including spaces and punctuation.

Each character is represented by 8 bits.

Thus the phrase is represented by 20 8 = 160 bits.



Analog is paper or microfilm and is hard to store (and retrieve).

Could be bulky or temperature sensitive.

Reproduction is more susceptible to noise (consider copies of copies on Xerox).



Paper uses trees, microfilm uses chemicals.

Waste disposal issues. Analog requires larger quantities of media than digital.

14.1.1 solve problems involving the conversion between binary numbers and decimal numbers. 14.1.2...

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