ece 101 an introduction to information technology analog to digital conversion
DESCRIPTION
Sinusoidal functions Key in all of EE! f(t) = A sin( t+ ), where A is the amplitude, =2 f, 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 frequencyTRANSCRIPT
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ECE 101 An Introduction to Information
Technology
Analog to Digital Conversion
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Information Path
InformationDisplay
Information Processor
& Transmitter
InformationReceiver and
Processor
Source ofInformation
DigitalSensor
TransmissionMedium
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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
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Sampling time waveforms
• Ts= Sampling Period (seconds/sample)
• fs= Sampling Rate = 1/ Ts (Hertz or Cycles per second)
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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
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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
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Time (sec)
F(t) = sin (2t)
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Time (sec)
F(t) = sin (2t) -1/2 sin (4t) + 1/3 sin (6t)
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Time (sec)
f(t) = sin (2t) -½ sin (4t) + 1/3 sin (6t) – ¼ sin (8t) + 1/5 sin (10t)
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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.
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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 |
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Cos 4tf0 = 2 Hzfs = 16 Hz
fs = 2.5 Hz
fs =1.5 Hz
Examples
3.12 and
3.13
from Kuc
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Severe Undersampling
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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)
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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
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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
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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
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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