Understanding Digital Communications

1. Understanding Digital Communications Lighthouse Chapter SCTE March 19th, 2008 Doug Pieri Senior Sales Engineer Douglas.pieri@arrisi.com 303-410-0034

2. 2 SCTE 03/19/08 Agenda Analog to Digital A to D and D to A Sampling Rate Nyquist Theorem Aliasing Bit resolution Effects on SNR Total Data Rate Digital Transmission QPSK and QAM Constellation Data Rate vs Bandwidth Effects of Noise “Quality” Measurements BER and MER Noise vs Data Rate

3. Analog to Digital

4. 4 SCTE 03/19/08 It’s an Analog World…. Despite what you might have heard, we communicate in a analog world. Voice Our voice is a time varying signal that changes in both amplitude and frequency The average voice contains components from roughly 300Hz to 3KHz The average human hearing ranges from 20Hz to 20KHz Visual Our vision is also dependant on a time varying signal that changes in both amplitude and frequency Visible portion of the optical spectrum is 400 to 700nm However, we use Analog and Digital transmission systems to more efficiently use the rest of the spectrum shown above

5. 5 SCTE 03/19/08 Analog and Digital Transport Analog transport systems (similar to Voice and Visual) use signals that vary over time in Amplitude and/or Frequency. The signal to be transported is typically used to modulate (change amplitude and/or frequency) a carrier frequency Digital communications by comparison are typically transmitted at a fixed frequency and are comprised of a string of On and Off states usually referred to as One’s and Zero’s The signal to be transported must be mathematically manipulated and converted into a string of On/Off states The required manipulation usually means the digital transmission system uses more complex (and therefore more costly) electronics then their analog counterpart. So why use Digital Transport?

6. 6 SCTE 03/19/08 The Advantage of DigitalTransport All transmission systems add impairments. Analog System Non-linear distortions occur during modulation Noise accumulates as the signal propagates Received signal has been degraded by both noise and distortion which can not be removed Digital System Digitizing Error (Noise) occurs during A to D conversion Noise accumulates as the signal propagates, however… As long as the receiver can accurately distinguish between the One and Zero state, the signal can be reconstructed and performance will be limited only by Digitizing Error (Noise)

7. 7 SCTE 03/19/08 A to D Conversion Two important factors in the A to D conversion Sampling Rate Bit Resolution Sampling Rate Interval at which samples of the analog signal are taken Referred to as Samples per Second or Sampling Frequency Nyquist Sampling Theorem governs the minimum sampling rate Minimum sampling frequency must be at least twice the highest frequency of the signal to be digitized Example: Return band from 5-42MHz must be sampled at 84MHz or greater What is the sampling rate of an Audio CD? Hint: Human hearing is roughly 20Hz to 20kHz Bit Resolution Number of allowable amplitude for each sample taken Each bit can represent “1” or “0” only, but multiple bits can be strung together as “words” of “n” number of bits Number of levels per bit can be calculated as 2^n, where “n” is the number of bits. Example: 8 bits leads to 2^8 = 256 levels

8. 8 SCTE 03/19/08 One-bit Resolution With one bit you can only represent two levels, namely levels 0 and 1. In a one bit system, resolution is quite poor and the reconstructed digital signal does not resemble the original analog signal. Frequency of the original signal can be determined, but any amplitude “information” carried on that signal will likely be lost.

9. 9 SCTE 03/19/08 Quantization Noise The difference between the analog signal and its digitized equivalent is referred to as quantization error or quantization “noise” because it is equivalent to noise when the signal is re-created. The quantization error creates a theoretical limit on Signal-to-Noise (SNR) for the re-created signal at a given bit-resolution.

10. 10 SCTE 03/19/08 Bit-Resolution and the effect on SNR With each increase in bit-resolution, the number of levels that can be represented also increases As will be seen, more levels allows for a more accurate representation of the original signal which equates to better signal-to-noise on the reconstructed signal The maximum theoretical SNR of the reconstructed is shown in the table above for different bit-resolutions The actual achievable SNR at any given bit-resolution is typically several dB lower than the theoretical levels.

11. 11 SCTE 03/19/08 Two-bit Resolution As was shown in the table, a two-bit system can represent 4 levels, namely states 00, 01, 10 and 11. In a two bit system, resolution is still quite poor, but significantly better than a one bit system.

12. 12 SCTE 03/19/08 Four-bit Resolution A four-bit system allows 16 different resolution levels from 0000 through 1111. Again, we see that quantization error is substantially reduced with each increase in bit-resolution

13. 13 SCTE 03/19/08 Eight-bit Resolution An eight bit system allows for 256 different resolution levels 00000000 through 11111111. At 8-bit resolution, the digital representation of the signal is quite good. The minimum recommendation for digital return systems is 8-bit resolution. Most newer systems are 10-bit or better.

14. 14 SCTE 03/19/08 Visual Representation of “Noise” Visually, we can see how each increase in bit-resolution reduces the quantization “noise” on the digitized signal

15. 15 SCTE 03/19/08 Sampling Rate Don’t forget, sampling rate is also important… Nyquist Sampling Theorem governs the minimum sampling rate Minimum sampling frequency must be at least twice the highest frequency of the signal to be digitized Nyquist Theorem causes some practical limitations A 6MHz baseband signal requires a sampling frequency of 12MHz minimum A 42MHz return band requires 84MHz minimum To digitize the entire forward band, we would need to sample at 1.1GHz (550MHz system) to 2.0GHz (1GHz system). No chip-sets currently available…

16. 16 SCTE 03/19/08 Sampling-Rate When sampled at a sufficient sampling rate, the reconstructed digital signal will accurately represent the original signal. In this example the original signal is comprised of two frequencies A lower frequency signal that defines the “envelope” A higher frequency that is roughly 10 times the lower frequency At sufficient sampling rates, both are well represented in the digitized signal

17. 17 SCTE 03/19/08 Sampling-Rate Too Low Sampling at too low a frequency can cause loss of signal information during reconstruction. The example above shows sampling at roughly 1/5th the required frequency. The lower frequency is maintained, but the higher frequency is lost…..or is it???

18. 18 SCTE 03/19/08 Digital Aliasing Digital Aliasing is a phenomena that causes frequencies higher than the Nyquist frequency to be mapped to a frequency below the Nyquist frequency. The “ladder” shown above can be used to determine what higher frequencies will be mapped to given a known sampling frequency (Example: 100MHz). In this example, 75, 125, 175, 225, and 275MHz would all get mapped to 25MHz. Also 87.5, 112.5, 187.5, 212.5 and 287.5MHz would map to 12.5MHz. If this was a digital return system (5-42MHz) with leakage from the forward path (50MHz+), the forward could potentially be mapped to frequencies within the return band.

19. 19 SCTE 03/19/08 Total Data Rate The total data rate for any given digitized signal can be calculated as follows. Determine the minimum sampling rate. As discussed, this is always at least 2X the highest frequency of the signal to be digitized. EXAMPLE Typical Return band is 5-42MHz Minimum Sampling frequency is 84MHz (2*42MHz) For simple math, we will use 100MHz or 100 Million samples/second Determine the bit resolution. As discussed, this will be largely dependant on the SNR required EXAMPLE 8-bit and 10-bit are the most typical for return band digitization For simple math we will use 10-bit resolution or 10 bits/sample Multiply bit resolution and sampling rate EXAMPLE 100 Million samples/second * 10 bits per sample = 1,000,000,000 bits/second Approximately 1 Gb/s required to digitize the return band How about a 550MHz forward band requiring 52dB SNR? 1.1 Giga samples/second * 10 bits per sample = 11.0Gb/s!!!

20. 20 SCTE 03/19/08 100MHz Sampling at 10 bits/sample Another way to look at it…. In the return, we need roughly 100MHz sampling and 10bits per sample At 100MHz sampling with 10 bits per sample, 10 bits of data are generated every 10nS

21. Digital Transmission

22. 22 SCTE 03/19/08 Analog, Digital … or Both!?! A signal my undergo many “A to D” or “D to A” conversions during transmission Legacy CATV plants are analog transmission systems Digital information needs to be converted to a quasi-analog format in order to work in legacy systems QPSK and QAM are techniques of transmitting data across analog systems Both QPSK and QAM can be considered digital information on an analog carrier

23. 23 SCTE 03/19/08 Quadrature Phase Shift Keying Quadrature Phase Shift Keying (QPSK) is a technique for transmitting digital information across an analog transport system QPSK works by transmitting two “carriers” of the same frequency such that they are at a 90o phase relative to one another “I” is the reference or “In-Phase” carrier and “Q” is the 90o shifted or “Quadrature” carrier At fixed time intervals, a “symbol” is created by the levels of the two carriers relative to one another. In the case of QPSK, each axis represents one bit and therefore transmits 2-bits per symbol (remember, each bit supports two levels) QPSK is sometimes referred to as QAM-4

24. 24 SCTE 03/19/08 Quadrature Amplitude Modulation Quadrature Amplitude Modulation (QAM) is similar to QPSK except that each axis is allowed to have more than two levels In the case of QAM-16 (as shown above), each axis represents two bits and therefore transmits 4-bits per symbol At any given symbol rate, QAM-16 is twice as efficient as QPSK The graph to the right is usually referred to as the “constellation” Many pieces of test equipment have the capability of displaying the constellation. As we will discuss later, it can be useful for troubleshooting

25. 25 SCTE 03/19/08 QAM-16 and QPSK Another way to look at it… QPSK (QAM-4) supports 2 bits/symbol QAM-16 supports 4 bits/symbol QAM-64 (not shown) will support 6 bits/symbol The symbol rate is essentially the bandwidth available for transmission What this means is that for any given available bandwidth, each increase in bits/symbol increases the data rate that can be transmitted

26. 26 SCTE 03/19/08 Data Rate and Bandwidth The above chart shows several modulation formats and the typical bandwidth (Symbol Rate) The Data Rate for any modulation rate can be calculated as follows: Start with the bandwidth (not including guardbands) The bandwidth is essentially in units of symbols/second Example: Bandwidth of a QAM-256 carrier is 5M symbols/second Multiply by the number of bits/symbol Example: QAM-256 transmits at 8 bits/symbol Example: 8 bits/symbol * 5M symbols/second = 40M bits/second So, if QAM-256 is more efficient than the lower modulation rates, why not use it everywhere???

27. 27 SCTE 03/19/08 The Effects of Noise In an ideal world, every symbol would be transmitted exactly in the center of the appropriate box. Unfortunately, noise and other system impairments cause the symbol to drift away from the center Noise on the “I” carrier cause the symbol to drift up and down Noise on the “Q” carrier cause the symbol to drift left and right Phase and distortion issues can also have an impact as we will discuss

28. 28 SCTE 03/19/08 The Effects of Noise At the receiver, as long as the symbol has not drifted outside the appropriate box, the transmitted data can be recovered error free If, however, the data has drifted beyond the appropriate box, the data recovered will be have errors One method of measuring the “Quality” of the data transmission system is to use bit-error rate

29. 29 SCTE 03/19/08 Bit Error Rate Bit Error Rate (BER) is a measurement of the bits received incorrectly. Specifically, BER is the number of received errors divided by the number of transmitted bits Usually, BER is expressed in a 10-x format Example 1: One error in 1 million transmitted bits = 1/1000000 = 10-6 Example 2: One error in 1 billion transmitted bits = 1/1000000000 = 10-9 Example 3: Four errors in 1 billion transmitted bits = 4/1000000000 = 4*10-9

30. 30 SCTE 03/19/08 Modulation Error Rate Another method of measuring the quality of the received signal is Modulation Error Rate Modulation Error Rate (MER) is a measurement of the average deviation from the center of the symbol region Specifically, MER is a ratio of the average symbol deviation to the average symbol magnitude expressed in dB Higher MER means better signal quality (similar to CNR)

31. 31 SCTE 03/19/08 MER and BER Small changes in MER correspond to big changes in BER However, MER is frequently the measurement of choice for technicians Accurate BER measurements typically require a “closed loop” test Closed Loop essentially means that the test set needs to know what data pattern was sent in order to interpret if errors are present This means that true BER measurements are an invasive test in that the actual data must be removed and replaced with “test data” MER by comparison is non-invasive in that it only looks at how far the average symbol is away from the “ideal” location

32. 32 SCTE 03/19/08 Noise Effects on Different Data Rates One of the other things you may have noticed is that there are different curves for the different QAM formats The question was posed earlier why not just use QAM-256 everywhere since it is more bandwidth efficient than the lower modulation rates. This is the main reason why. Each step up from QPSK (QAM-4) to QAM-16, 64, 256 and so on required approximately 6dB better MER (CNR, SNR) in order to maintain the same Bit-error rate (BER)

33. 33 SCTE 03/19/08 Noise Effects on Different Data Rates The approximate 6dB increases in MER can be better understood if we refer back to the constellation. As the modulation rate increases, the symbol regions decrease and are 1/4th the “area” of the prior region. On a power scale, this is 10*log(1/4) = -6dB This means total noise power must be 6dB lower, which means... NPR, CNR, and SNR must increase by 6dB for each increase in modulation rate if the same BER is to be maintained.

34. 34 SCTE 03/19/08 System Noise

35. 35 SCTE 03/19/08 Gain Compression

36. 36 SCTE 03/19/08 Phase Noise

37. 37 SCTE 03/19/08 Coherent Interference

38. 38 SCTE 03/19/08 Summary Analog to Digital A to D and D to A Sampling Rate Nyquist Theorem Aliasing Bit resolution Effects on SNR Total Data Rate Digital Transmission QPSK and QAM Constellation Data Rate vs Bandwidth Effects of Noise “Quality” Measurements BER and MER Noise vs Data Rate

39. 39 SCTE 03/19/08