Transcript Lecture 3
ECE 101 An Introduction to Information Technology Analog to Digital Conversion Information Path Source of Information Information Display Digital Sensor Information Processor & Transmitter Transmission Medium Information Receiver and Processor 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 4t f0 = 2 Hz fs = 16 Hz Examples 3.12 and fs = 2.5 Hz 3.13 from Kuc fs =1.5 Hz Severe Undersampling If f s 2f o, then an aliasing occurs, and an alias frequency results, f a f s f o f s/ 2 But, if f s is very small this may not be the case, hence we must repeat the process to get the alias frequency. As before let f o 2 Hz but let f s 0.8 Hz (Ts 1.25 sec.) then f a f s f o 0.8 2 1.2 f s/ 2 0.4 Hz (problem) so f a f s f a 0.8 1.2 0.4 f s/ 2 (so, ok, barely) Number Representations Base 10 764 =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-tonoise ratio (S/N or SNR) that divides the signal power by the noise power. • Signal power is s2 = Xrms2 • Noise power level n2 = 2/12 • SNRdB = 10 log10 s2/ n2 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 • Power dB = 10 log P2 / P1 Voltage or Current dB = 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-tonoise ratio (S/N or SNR) that divides the signal power by the noise power. • Signal power is s2 = Xrms2 • Noise power level n2 = 2/12 • SNRdB = 10 log10 s2/ n2 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