ECE 101 An Introduction to Information

Technology

Analog to Digital Conversion

Information Path

InformationDisplay

Information Processor

& Transmitter

InformationReceiver and

Processor

Source ofInformation

DigitalSensor

TransmissionMedium

Sinusoidal functions

• Key in all of EE!• f(t) = A sin(t+), where A is the

amplitude, =2f, f = frequency = 1/T, T = period, = phase, = circumference / diameter of a circle = 3.14

• f(t) repeats itself when the argument (t+) increases by 2

• A pure tone has a single frequency

Sampling time waveforms

• Ts= Sampling Period (seconds/sample)

• fs= Sampling Rate = 1/ Ts (Hertz or Cycles per second)

Sampling images

• Images must be sampled in 2 dimensions• Use square grid Ts units per side (length per

sample) (perhaps Ls units is more descriptive)

• fs= Sampling Rate = 1/ Ts (samples per length)

• 3 dimensions > movies

Arbitrary Signals as Sinusoids• Any analog signal can be constructed by

using sinusoidal components with different frequencies and amplitudes

• Spectrum provides the relative amplitudes of the frequency components in a waveform

• Power of a frequency component = ½ the square of its amplitude

• Harmonics occur at multiples of a fundamental frequency

• Frequency range = bandwidth

Time (sec)

F(t) = sin (2t)

Time (sec)

F(t) = sin (2t) -1/2 sin (4t) + 1/3 sin (6t)

Time (sec)

f(t) = sin (2t) -½ sin (4t) + 1/3 sin (6t) – ¼ sin (8t) + 1/5 sin (10t)

Nyquist Sampling Criterion

• Must identify the highest frequency component in a waveform, fmax.

• In order to ensure that no information is lost in the sampling process, we must sample the signal at a frequency, fs>2 fmax.

• If fs<2 fmax, then aliasing may occur with one or more false frequencies appearing.

Aliasing Error

• If sampling frequency is less than two times the maximum frequency, then a new alias frequency may appear.

• If a single (hence max.) frequency, fo, exists, fmax = fo and the sampling fs < 2fo, then the alias frequency, fa = |fs- fo | = |fo- fs |

Cos 4tf0 = 2 Hzfs = 16 Hz

fs = 2.5 Hz

fs =1.5 Hz

Examples

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Number Representations

Base 10764 =764(10)=7x102 + 6x101 + 4x100, where in

the base 10, digits 0,1,2,…7,8,9 are permissible (NOT 10). Note 100 = 1

In 764 the 7 is the most significant digit (MSD) and 4 is the least significant digit (LSD)

Can use any Base n. Since digital work largely deals with two signals we select n=2

Information in Bits

• Binary digits (bit) form the basis for the information technology language.

• Computers have codes of them for numbers, sound, images, anything else represented by a computer.

• They use 1’s and 0’s only, hence base 2• 4-bit word 24 = 16 different messages• n-bit word 2n different messages

Binary Number Representations Base 2 1(10)=001(2)= or 0x22 + 0x21 + 1x20, where 20=1 2(10)=010(2)= or 0x22 + 1x21 + 0x20

3(10)=011(2)= or 0x22 + 1x21 + 1x20

4(10)=100(2)= or 1x22 + 0x21 + 0x20

10(10)=1010(2)= or 1x23 + 0x22 + 1x21 + 0x20

29(10)= 11101(2)= or 1x24+1x23+1x22+0x21+1x20

or 16 + 8 + 4 + 0 + 1 In 110010(2) the MSB=1 and LSD =0

Methods for Finding the Binary Form of a Decimal Number

#1- Repeatedly divide the decimal number by 2 and retain the remainder as the LSB.

• Find 29(10) • 29/2 = 14 rem 1, LSB = 1• 14/2 = 7 rem 0, next bit = 0• 7/2 = 3 rem 1, next bit = 1• 3/2 = 1 rem 1, next bit = 1• 1/2 = 0 rem 1, MSB = 1• So, 29(10) = 11101

Methods for Finding the Binary Form of a Decimal Number

#2- Find the largest power of 2 less than the number. That becomes the MSD. Subtract these numbers and repeat the process.

• Find 29(10) • 24 = 16, (MSB), 29-16 = 13,• 23 = 8, 13-8=5,• 22 = 4, 5-4=1,• 20 = 1, LSB• Therefore 29(10) = 11101

Binary Numbers

• 8 bits = 1 byte– 1 byte can represent 28 = 256 different

messages• 4 bits = a nibble (less frequently used)• 1 kilobyte = 210 = 1,024 bytes = 8,192 bits• 1 Megabyte = 220 = 1,048,576 bytes • 1 Gigabyte = 230 = 1,073,741,824 bytes• 1 Terabyte = 240 = 1,099,511,627,776 bytes

Bits and Bytes• Bits

– Often used for data rate or speed of information flow

– 56 kilobit per second modem (56kbps)– A T-1 Communication line is 1.544 Megabits

per second (1.544 Mbps)• Bytes

– Often used for storage or capacity (computer memories are organized in terms of 8 bit words

– 256 Megabyte (MB) of RAM– 40 Gigabyte (GB) Hard disk

Quantizer Concept

• To properly represent an Analog signal we need to depict discrete sample levels or “quantize” the signal.

• Key is the step size to generate a stair step pattern of values

• Each step then takes on a binary number value

Quantizer Design

• A quantizer producing b-bits has a staircase with the number of steps equal to Nsteps=2b

• The first step has the value of Vmin. The staircase has 2b –1 steps remaining each of a size

• The maximum value is Vmax=Vmin+(2b –1) • 2 types of errors:

– step size Δ being too large and that is related to the number of bits in the quantizer

– inadequate quantizer range limits that causes clipping

Quantizer Range• Assume that the limits of Vmax and Vmin are

not known such as in an audio system.• Measure the audio signal strength with a

meter that indicates the root-mean-square (rms) voltage value.

• The rms voltage value of a signal produces the same power as a battery with a constant voltage of the same value.

• Adjust the quantizer until Vmax =4 Xrms and Vmin = -4 Xrms and the step size is =8 Xrms/ (2b –1)

Signal-to-Noise Ratio

• An important measure of the performance of a system is evaluation of the signal-to-noise ratio (S/N or SNR) that divides the signal power by the noise power.

• Signal power is s2 = Xrms

2

• Noise power level n2 = 2/12

• SNRdB = 10 log10 s2/ n

2

Review of Logarithms• Why logarithms: simplify multiplying &

dividing and use in both signal to noise ratios and information theory

• Decimal system in powers of 10:– 100 = 1– 101 = 10– 102 = 100– 103 = 1000– 104 = 10000

• The exponent (=number of zeroes) is the logarithm

Logarithms Between 1 and 10

• If the log101 = 0 and log1010 = 1; what is the log10 of a number between 1 and 10?

• log103 = x or 10x = 3 (recall logs are exponents)

• Result: 100.4771 = 3; therefore, x = 0.4771 or• log103 = 0.4771

Logarithms

• Log A x B = log A + log B• Recall exponents add in multiplication

– 1000 x 10000 = 103 x 104 = 107 = 10,000,000– 297 x 4735 = 1,406,295– 102.4728 x 103.6753 = 106.1481 =1,406,294.998

Decibels – Application of Logs• Decibel (dB) – 1/10 of a Bel (named for

Alexander Graham Bell)– Logarithmic expression of the ratio of 2 signals

• PowerdB = 10 log P2 / P1

Voltage or CurrentdB = 20 log (V2 / V1)dB = 20 log (I2 / I1)

Signal-to-Noise Ratio

• An important measure of the performance of a system is evaluation of the signal-to-noise ratio (S/N or SNR) that divides the signal power by the noise power.

• Signal power is s2 = Xrms

2

• Noise power level n2 = 2/12

• SNRdB = 10 log10 s2/ n

2

Base n • Binary – 2• Octal – 8• Decimal – 10• Hexadecimal – 16• When dealing with collection of bits like binary

words representing text characters using the ASCII (American Standard Code for Information Interchange) code – it is inconvenient to deal with each individual bit

• So may use octal (3 bit) words or hexadecimal (4 bit) words

Octal - Base 8

• Base 8, using first 8 numerals: 0, 1, 2, 3, 4, 5, 6, 7

• Because 8 is a power of 2, can use octal numbers to represent a group of 3 bits

• Example:• 1001112 = 1+2+4+32 = 3910

• 478 = 4x8 + 7 = 3910

Hexadecimal - Base 16 • Base 16, using first 16 numerals including 6

letters: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

• Because 16 is a power of 2, can use octal numbers to represent a group of 4 bits

• Example:• 001111102 = 21 + 22 + 23 + 24 + 25 =

• 2 + 4 + 8 + 16 + 32 = 6210

• 3E16 = 3x16 + 14 = 6210