1. ECE 101 An Introduction to Information Technology Analog to Digital Conversion

2. Information Path Information Display Information Processor & Transmitter Information Receiver and Processor Source of Information Digital Sensor Transmission Medium

3. 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 frequency

4. Sampling time waveforms T s = Sampling Period (seconds/sample) f s = Sampling Rate = 1/ T s (Hertz or Cycles per second)

5. Sampling images Images must be sampled in 2 dimensions Use square grid T s units per side (length per sample) (perhaps L s units is more descriptive) f s = Sampling Rate = 1/ T s (samples per length) 3 dimensions > movies

7. 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

8. Time (sec) F(t) = sin (2  t)

9. Time (sec) F(t) = sin (2  t) -1/2 sin (4  t) + 1/3 sin (6  t)

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

11. Nyquist Sampling Criterion Must identify the highest frequency component in a waveform, f max. In order to ensure that no information is lost in the sampling process, we must sample the signal at a frequency, f s >2 f max. If f s <2 f max, then aliasing may occur with one or more false frequencies appearing.

12. 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, f o, exists, f max = f o and the sampling f s < 2f o, then the alias frequency, f a = |f s - f o | = |f o - f s |

13. Cos 4  t f 0 = 2 Hz f s = 16 Hz f s = 2.5 Hz f s =1.5 Hz Examples 3.12 and 3.13 from Kuc

14. Severe Undersampling

15. Number Representations  Base 10  764 =764 (10) =7x10 2 + 6x10 1 + 4x10 0, where in the base 10, digits 0,1,2,…7,8,9 are permissible (NOT 10). Note 10 0 = 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

16. 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 2 4 = 16 different messages n-bit word 2 n different messages

17. Binary Number Representations  Base 2  1 (10) =001 (2) = or 0x2 2 + 0x2 1 + 1x2 0, where 2 0 =1  2 (10) =010 (2) = or 0x2 2 + 1x2 1 + 0x2 0  3 (10) =011 (2) = or 0x2 2 + 1x2 1 + 1x2 0  4 (10) =100 (2) = or 1x2 2 + 0x2 1 + 0x2 0  10 (10) =1010 (2) = or 1x2 3 + 0x2 2 + 1x2 1 + 0x2 0  29 (10) = 11101 (2) = or 1x2 4 +1x2 3 +1x2 2 +0x2 1 +1x2 0  or 16 + 8 + 4 + 0 + 1  In 110010 (2) the MSB=1 and LSD =0

18. 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

19. 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) 2 4 = 16, (MSB), 29-16 = 13, 2 3 = 8, 13-8=5, 2 2 = 4, 5-4=1, 2 0 = 1, LSB Therefore 29 (10) = 11101

20. Binary Numbers 8 bits = 1 byte –1 byte can represent 2 8 = 256 different messages 4 bits = a nibble (less frequently used) 1 kilobyte = 2 10 = 1,024 bytes = 8,192 bits 1 Megabyte = 2 20 = 1,048,576 bytes 1 Gigabyte = 2 30 = 1,073,741,824 bytes 1 Terabyte = 2 40 = 1,099,511,627,776 bytes

21. 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

22. 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

23. Quantizer Design A quantizer producing b-bits has a staircase with the number of steps equal to N steps =2 b The first step has the value of V min. The staircase has 2 b –1 steps remaining each of a size  The maximum value is V max =V min +(2 b –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

24. Quantizer Range Assume that the limits of V max and V min 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 V max =4 X rms and V min = -4 X rms and the step size is  =8 X rms / (2 b –1)

25. 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  s 2 = X rms 2 Noise power level  n 2 =  2 /12 SNR dB = 10 log 10  s 2 /  n 2

26. 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: –10 0 = 1 –10 1 = 10 –10 2 = 100 –10 3 = 1000 –10 4 = 10000 The exponent (=number of zeroes) is the logarithm

27. Logarithms Between 1 and 10 If the log 10 1 = 0 and log 10 10 = 1; what is the log 10 of a number between 1 and 10? log 10 3 = x or 10 x = 3 (recall logs are exponents) Result: 10 0.4771 = 3; therefore, x = 0.4771 or log 10 3 = 0.4771

28. Logarithms Log A x B = log A + log B Recall exponents add in multiplication –1000 x 10000 = 10 3 x 10 4 = 10 7 = 10,000,000 –297 x 4735 = 1,406,295 –10 2.4728 x 10 3.6753 = 10 6.1481 =1,406,294.998

29. 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 P 2 / P 1 Voltage or Current  dB = 20 log (V 2 / V 1 )  dB = 20 log (I 2 / I 1 )

30. 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  s 2 = X rms 2 Noise power level  n 2 =  2 /12 SNR dB = 10 log 10  s 2 /  n 2

31. 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

32. 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: 100111 2 = 1+2+4+32 = 39 10 47 8 = 4x8 + 7 = 39 10

33. 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: 00111110 2 = 2 1 + 2 2 + 2 3 + 2 4 + 2 5 = 2 + 4 + 8 + 16 + 32 = 62 10 3E 16 = 3x16 + 14 = 62 10

34. Logarithms – Base “a” then a=2 In information theory we need logs to the base 2, not 10 Conversion of bases in general: –log a x = (log 10 x)/ (log 10 a) –If a = 2, then use log 10 2 =.301 –log 2 x = 0.332 (log 10 x); OR –Recall log a N = x or a x = N –So log 2 N = x or 2 x = N

35. Logarithms – Base 2 2 0 = 1; log 2 1 = 0 2 1 = 2; log 2 2 = 1 2 2 = 4; log 2 4 = 2 2 3 = 8; log 2 8 = 3 2 4 = 16; log 2 16 = 4 2 5 = 32; log 2 32 = 5