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TV & SAT TV

Some Theory

Data transmission via cable TV networks
Cable network is a system receiving and distributing RF signals mainly inside apartment buildings. The borderline between a community antenna system and a cable TV network can be defined in many different ways, e.g. in Poland a cable network is defined as RF installation located in more than one building and having more than 250 outlets.
Cable TV was primarily planned to be signal installation allowing distribution of large (above 60) number of programs to large or very large group of subscribers. Currently, thanks to common use of HFC (Hybrid Fiber Coaxial) networks, it is possible to create networks serving tens of thousands of subscribers. In the beginning there were used the same channels as for terrestrial television only. To keep up with customers' demands for new channels, there have been utilized frequencies between the bands of terrestrial TV - so called cable channels - marked S.
Transmission capabilities of cable TV
In Poland there is used 8 MHz raster in D/K standard, with PAL color system. SECAM system is also allowed for original SECAM broadcasts (e.g. some satellite French or Russian channels). For stereo programs there is used digital Nicam Stereo system, but many cable networks also distribute some programs in A2 analog stereo.
The total capacity of such system is 99 channels, but the channels 1-5 are currently not used for distribution of TV programs. The rest - 94 channels - take continuous frequency range from 110 MHz to 862 MHz.
Older networks usually used only a subset of the available channels, which illustrates the table below:
Table 1. Frequency bands typically used in antenna community systems and cable TV networks in Poland

Range

Frequency band [MHz]

Channel's symbols

Old networks

New networks

FM (CCIR)

87.5 - 108.0

FM

Always

Always

Lower special band

110 - 174

S01 -S08

Often

Always

VHF III

174 - 230

K06 -K12

Always

Always

Upper special band

230 - 302

S09 - S17

Often

Always

Hyper-band

302 - 470

S18 -S38

Seldom

Almost always

UHF IV (lower UHF)

470 - 606

K21 - K37

Always

Always

UHF V (upper UHF)

606 - 862

K38 - K69

Often

Almost always


Detailed TV channel frequency list can be found here.
Data transmission to the subscriber.
There are usually used only 60 channels for television transmission. On the assumption that, for the reason of possible distortions, there are left free the channels of terrestrial transmitters (usually 8) and 4 channels for modulators (of e.g. VCRs), it is still 22 channels available for data transmission. In practice, the number is further decreased because some channels have to be skipped due to distortions caused by other transmitters etc., so the real number is around 10.
These channels can be used for digital data broadcasting to subscribers (forward direction). Due to high quality of cable transmission (high S/N ratio, even including some specific distortions in cable networks), especially in the direction to subscribers, it is possible to use complex multi-level modulations. Such modulations ensure fast transmission within low bandwidth, i.e. high efficiency.
Typical examples are the 16QAM and 64QAM modulations. In practice, they are only used for broadcast channels, because they require relatively high signal to noise ratio. The advantage of this kind of modulation is high channel capacity, equal 4 b/Hz/s for 16QAM, and 6 b/Hz/s for 64QAM.
Reverse transmission - the return channel.
Obviously the users must have possibility of reverse transmission to the head station. Because of use of distribution amplifiers, the only possibility is frequency division, i.e. the forward transmission is performed in the range of television channels and the reverse transmission - in 5-65 MHz range. On account of the specificity of this kind of transmission, it is required to use interference-resistant modulations.
For this purpose there are usually used BPSK and QPSK modulations. Their basic advantages are high resistance to distortions and simplicity of the modulators and demodulators. These are the simplest phase modulations, with binary phase-shift keying and quadrature phase-shift keying adequately. The channel capacity is equal 1 b/Hz/s for BPSK and 2 b/Hz/s for QPSK.
Band selection for the return channel.
We have mentioned before that for the reverse transmission in cable networks there have been chosen frequencies lying below forward band, i.e. the 5-65 MHz range. It's worth trying to understand why.
There were two possible variants, either using the band lying below the lowest TV channel, or above the highest one. The frequencies above 862 MHz are less vulnerable to external interferences, as the range is a subject to regulations and the transmitters have limited output power.
However, distribution of so high frequency signals in cable TV networks encounters various problems, related to increase of cable attenuation and decrease of shielding effectiveness. In addition, the higher frequency, the bigger trouble with making filters with steep edges of frequency characteristics.
By contrast, the band below 65 MHz is the most commonly used frequency band, thus the environment is full of interferences. It is interfered by CB transceivers, household devices, car ignition systems, lighting controllers, computers etc. However, the basic advantage of this band for cable applications is low attenuation of the cables and possibility of making efficient filters. Besides that, it is easier to build active devices working in lower frequency bands.
In the very beginning, the upper frequency of the return channel was 30 MHz, to avoid any possibility of interference with the lowest TV channel beginning at 47 MHz. Later, since the lower channels were not used any more, the band has been widened.
The bands of forward and return channels

Return channel

Bandwidth

Forward channels

Remarks

5-30 MHz

25 MHz

47-862 MHz

47 MHz - the lowest frequency of CH2 (raster B)

5-45 MHz

40 MHz

55-862 MHz

 

5-55 MHz

50 MHz

65-862 MHz

65.5 MHz - the lowest frequency of FM-OIRT band

5-65 MHz

60 MHz

85-862 MHz

87.5 MHz - the lowest frequency of FM band (CCIR)

Return channel bandwidth varies from 25 to 60 MHz, however we should remember that a part of it is useless due to too high interference levels.
Throughput of the return channel.
Now we will try to estimate the throughput of the return channel. Transmission speed depends on the available bandwidth and spectral efficiency of the modulation used.
Rb=B*n
Where:
Rb - transmission speed in bps (bits per second)
B - bandwidth in Hz
n - spectral efficiency in bps/Hz, showing the number of bits that can be coded by one change of the carrier; n describes capacity of the modulation used (it is limited by the ratio of total signal power over the bandwidth and total noise power over the bandwidth).
Transmission speed is proportional to available bandwidth and channel capacity.
The more complex modulation, the higher n, reaching even 10 for 1024QAM modulation.
modulationn (real value)η (theoretical)
4QAM1.72
2FSK0.81
BPSK0.81
QPSK1.92
8PSK2.63
16PSK2.94
BPSK=2PSK, QPSK=4PSK, 4QAM=4PSK, 4QAM=4PSK
It seems that it is better to use modulations with larger
modulationQPSK4QAM16QAM64QAM
BER

C/N [dB]

10-39.610.317.022.9
10-613.513.820.626.7
10-915.515.722.628.7
10-1217.116.923.330.1
Influence of C/N parameter on choice of modulation.
It is not a problem to achieve signal to noise ratio better than 40dB in forward channel, so multi-level modulations are usually chosen e.g. 16QAM or 64QAM, that have better transmission characteristics than PSK. It's worth to remember that M-QAM (quadrature AM) modulations are identical to M-APK (amplitude phase keying).
The situation in the return channel is much worse - the signal to noise ratio is always smaller, and changeable within the band. In this case, there have to be chosen modulations featuring high resistance to distortions - usually QPSK, sometimes BPSK.
At this stage we can already determine the throughput of the return channel. We assume that the bandwidth of a single channel is 4 MHz (typical value) and we use QPSK modulation.
Rb=B*n=4MHz*2b/s/Hz
Rb=4MHz*2b/s/Hz=8Mbps
For the return channel we can use several such channels, e.g. five, so the total available throughput may reach 40Mbps.
Similarly, we can determine the throughput of the forward channel. Let's assume that single channel bandwidth is 6 MHz (allowance for some margin) and we use 64QAM modulation.
Rb=B*n=6MHz*6b/s/Hz
Rb=6MHz*6b/s/Hz=36Mbps
Using several channels e.g. six, we get maximum throughput of 216 Mbps - a really high-speed data transmission network. Additionally, thanks to using several channels, we get increased reliability of the system. Different modulation types in the return and forward channels cause differences in transmission speed, however they match well with the traffic flow - the subscribers normally receive much more information than they send - the transmission asymmetry is not significant for a typical user.
Bandwidth of the return channel.
The total bandwidth of the return channel depends on the required channel capacity and the kind of modulation (which is connected with adequate equipment for the head station and that used by the subscribers). As we mentioned before, more complicated modulations are out of question in this channel, so the natural tendency for utilization of the whole available band.
The source of problems is interference, both external and generated within the network.


There can be defined parameter describing availability of return channel. It says what percent of return (reverse) channel width can be used for data transmission. We should be conscious that it can vary in different parts of network. In practice, it turns out that availability of reverse channel is only a fraction of its width. It is necessary to measure availability of the channel, or at least to estimate the ranges that are not available due to disturbing signals. This measurement should be performed with a spectrum analyzer capable of operation in return channel range.
Example spectrum of the return channel
The knowledge of spectral distribution of interfering signals and their levels, as well as of available signal to noise ratio, allows us to choose adequate sub-ranges ensuring proper data transmission, with required BER level.
Practically the strongest interferences occur on bands used by short-wave radio communication, on 27 MHz band and about 50 MHz, on the lowest frequencies, and on intermediate frequencies of radio and television receivers. Their arrangement is disordered which favors use of considerably narrow "elementary" channels. The narrow channels can be fitted between the interfering spectral lines.
Unfortunately the price we pay for it is complication and increase of the number of cable modems in head station. If we want to achieve the same bandwidth, we have to compensate narrow elementary bands by increased number of the the transmission channels. It requires additional modems in head station and increases the total cost of equipment.
There are used different bandwidths:

Company

Name

Bandwidth [MHz]

Throughput [Mb/s]

Modulation type

NetGame

NeMo

2.5 /2.6

1.8/5.12

QPSK

Cisco System

MC11

0.2 - 3.2

5/10

QPSK/16QAM

Nortel Networks

LANcity

6

10
 

Phasecom

SpeedDemon

1.66

2.5

QPSK

Bandwidth of the broadcast channel
Bandwidth of the broadcast channel cannot be wider than width of television channel. In D/K system it is equal to 8 MHz. The most common bandwidth in cable modems is equal to 6 MHz. One of reasons is compatibility with NTSC standard used in the US and other countries, with 6 MHz channel bandwidth.
Possibility of using wide broadcast channel results from better transmission parameters in comparison with return channel. Broadcast channel is usually located in UHF band, where the number and amplitudes of outer interferences are smaller. In addition, in this band there is significantly lower level of internal interference.
Large C/N ratio, at least 43 dB, allows to use multi-level modulations ensuring high transmission speeds. It is possible to achieve speeds of 10-40 Mbps for one broadcast channel.

Company

nazwa

Bandwidth [MHz]

Throughput [Mbps]

Modulation type

NetGame

NeMo

6

10

QPSK

Cisco System

MC11

6

27/40

64QAM/256QAM

Nortel Networks

LANcity

6

10

QPSK

Phasecom

SpeedDemon

6

31

64QAM

Interference caused by strong signals from cable modems.
Cable modems, both those working in head stations and those installed at subscribers' places have high output levels. They typically are 120 dBuV. It is usually not a problem for broadcast channels (other signals distributed in the network have comparable levels), but the locally injected return signal may interfere some TV channels in local televisions.
Risk of interferences caused by cable modem
Because of limited separation between outputs of typical splitter (usually 25dB), at the input of TV set or stereo there is strong signal from cable modem. Assuming that modem's output level is 120dBuV, on the input of the receiver there is signal from 5-65 MHz band at 95 dBuV. Such high signal may cause strong intermodulation effects. This problem can be solved by the use of:
  • filters blocking return channel band,
  • splitters / multimedia outlets with increased separation between R, TV and D (data transmission) outputs.
Interferences generated on passive elements of the network.
Another problem is non-linearity of passive elements. Primarily it concerns nodes of the network, which in unidirectional network operate at levels not exceeding 100 dBuV. However, in bidirectional networks the levels in return channel significantly exceed 100 dBuV.
Some passive elements are built with the use of ferrite cores that may become nonlinear when the levels are too high. Nonlinear characteristics are the reason of nonlinear distortions and generation of products that may interfere useful band. The countermeasure is to reduce these effects by using specially designed taps not susceptible to high signal levels.
Other source of nonlinear distortions are bad connections (non-ohmic contacts), caused by oxidized surfaces, wet cables and connectors. They make parasitic diodes that deform the signals and generate nonlinear distortions. To eliminate the unwanted effects, the designer should choose renown components, and the installer should perform the work according to best standards, including protection against water and moisture.
The cables used should be gel-filled to avoid penetration of water/moisture in case of damaged outer sheath. Other components employed in the network should be designed for operation in systems with return channel (high signal levels).
Filters, diplexers and outlets.
Use of filters, diplexers and multimedia outlets allows to avoid overdriving of televisions by strong output signal of cable modem (reverse channel).
High-pass filters that strongly attenuate the band of reverse channel, should be used at receiver's input, which is connected to cable network together with the cable modem. Additionally, it is needed to group subscribers - those who have modem should be connected to the branch that is directly connected to building network, and the rest - should be connected to building network through high-pass filter.
Grouping of subscribers
The rejection band is usually 5-65 MHz (the upper value depends on chosen reverse channel band), and the pass band is 87-862 MHz. Minimum attenuation at cut-off frequency should be > 40dB, pass-band ripple less than 1-2 dB.
Return loss in pass band (input and output) should be 16.5 +/- 1.5 dB/octave. It's worth to notice that there are other, more expensive filters available, ensuring also matching in rejection band.
For subscribers that don't use cable modem, instead of high-pass filters connected in front of the subscriber's outlet, there can be used outlets with built-in filters. It both simplifies the system and protects the network against interferences (coming from the televisions).
Outlets with high-pass filter in TV path and band-pass filter in FM path.
There are outlets in two versions: with splitter or with tap; sometimes they are without a band pass filter in FM path - in this case both outputs (sockets) pass the whole 87-862 MHz band.
For subscribers that have modem there are used "multimedia" or "data transmission" outlets. They have three outputs (sockets): two for unidirectional transmission (R/FM and TV), and one for bidirectional transmission (D) - to connect the cable modem.
Basic parameters include isolation between outputs R or TV and D, and attenuation of interferences generated by TV tuner. The data transmission path should have low attenuation, in both directions.
It should be noticed that use of a splitter in FM/TV path causes that attenuation of TV and R outputs is the same and both sockets provide the whole 87-862 MHz band. Application of 87-108 MHz pass-band filter improves conditions of FM reception. Sometimes the R (FM) output is connected via a tap, with 8-10 dB attenuation.
Internal diagrams of multimedia outlets with splitter and tap.
In the case of multimedia outlets, modem is connected through a tap with 10 dB attenuation, which improves isolation between D input/output and TV/FM outputs.
Telkom-Telmor company manufactures interesting multimedia outlet with improved attenuation parameters. Advanced multimedia outlet GMF-351 has additional filters in data transmission line, thanks to which the insertion loss (D) has been reduced to 1 dB and the isolation between D and R outputs, or between D and TV outputs, is by 10 dB higher than in the typical case with one high-pass filter.
It is very good solution for subscribers using reverse channel, because it practically eliminates the problem of distortions caused by modem and those generated by receivers.
Internal diagram of GMF-351 outlet
In the case when modem is installed in other place than TV set, division of the signal should not be made byTV set, but in other available place. Standard splitters are not suitable for this purpose - the TV set or stereo would be overdriven by strong return signal from the modem. The path leading to the television has to contain high-pass filter (it may be located in the outlet). The signal can be divided by typical splitterss only after such a filter. In the case of splitting the signal at the input (see the diagram below), the splitter has to be prepared for levels about 120 dB, and the R/TV outlet has to be equipped with high-pass filter.
Diagram of home installation with television and modem placed in different rooms - the R/TV outlet has to be equipped with high-pass filter
Typical solution is the use of multimedia distributors (diplexers or crossovers). The simplest ones are only equipped with tap for cable modem, and high-pass filter in the TV output. The more advanced distributors contain several filters, which improves isolation but is the reason of higher costs.
Multimedia distributors with tap and with splitter
The main advantage of these distributors is integration, in one device, of the tap/splitter and set of filters blocking penetration of unwanted signals from the reverse channel and those generated in the TV tuners. Similarly to the outlet of Telkom-Telmor company, the attenuation in dataa line is asymmetric, 1 dB from the modem and 10 dB to the modem.
There are also available variants with splitters instead of taps, that are characterized by the same attenuation on broadcast line (between R/TV and D paths, ca. 4dB) and lower isolation between these outputs.

The sources of internal distortions.
Broadcast channel.
Resistance of cable network to external distortions must be connected with elimination of internal distortions generated in the network. Due to structure of cable networks, penetration of interferences to the broadcast band (87.5-862 MHz) is significantly minimized because of directional characteristics of passive devices. The major source of distortions are cascades of amplifiers, between the head station and the destination point.
The sources of distortions in broadcast channel
Return channel.
There is a bigger problem with the return channel, where the distortions from different parts of the network "accumulate" in the signal. Interference and noise generated by amplifiers and other active devices sum up in the node, decreasing parameters of the signal combined from the signals of individual modems and received by devices in the head station.
The sources of distortions in return channel
The main sources of distortions are TV, radio devices, and computers connected to the cable network. Each of these devices generates some unwanted signals (e.g. receiving equipment - the intermediate frequencies) usually in the range of reverse channel. The level of distortions injected to the network by a TV set can reach 50 dBuV. It should be noticed that some distortions can be generated even when the device is in stand-by mode.
Additionally, there is a lot of interferences in the city environment - both generated in the home (household appliances), or coming from outside. Some of them can penetrate the network through the connected devices (i.e. TV sets, stereo, computers).
It is the reason why return channel should be provided only to the points where it is really needed. The rest of home network should be separated by high-pass filter/s. In the place of using cable modem all receivers have to be connected via diplexers or multimedia outlets.
The conclusion is that the network has to be as tight as possible, especially for the unwanted signals in the return channel band. The network has to be composed of suitable equipment (e.g. passive elements linear with high levels), and carefully tested/measured. There should not be used pass-through configurations.
Level of network traffic generated by subscribers.
The level of the traffic generated by subscribers enforces adequate means to ensure good quality of service. Let's try to estimate traffic capacity of data transmission network based on cable TV network. To simplify the issue, we assume one broadcast and one return channel and the following parameters:
  • - broadcast channel with 6 MHz bandwidth, 64 QAM modulation, transmission speed 31 Mbps,
  • reverse channel with 1.66 MHz bandwidth, QPSK modulation, transmission speed 2.5 Mbps.
As we mentioned before, for most subscribers the asymmetry of transmission speed in different directions isn't a problem - they usually download much larger amounts of data than upload to the network.
Now we have to determinate what transmission speed we want to reserve for each subscriber. On account of that no subscriber would accept lower transmission speed than that offered by telephone / ISDN modem, we have to assume that the minimal transmission speed can't drop below 64 kbps for a single subscriber.
The possible number of data channels can be calculated from the formula:
N=Rb/P
  • N - number of channels,
  • Rb - transmission speed in b/s,
  • P - transmission speed for single subscriber in b/s
  • Nd=31*10E6/64*10E3=484 (number of broadcast channels - we assumed 64kbps for one channel)
  • Nz=2.5*10E6/6.4*10E3=390 (number of reverse channels)
We have assumed, on account of traffic asymmetry, that reverse traffic to broadcast traffic ratio is 1:10 (6.4 kbps).
Presented values inform that maximal number of simultaneously working subscribers can't exceed 400. If the number of working subscribers began to increase, the only way to service the increased traffic is to decrease transmission speed for single user.
Obviously, not all subscribers work simultaneously, so the maximum number of subscribers can be greater. It also depends on the traffic generated by the active subscribers. Some approximation is given by the ratio of active subscribers to the total number of subscribers.
A measure of an average traffic generated in the network is Erlang (average traffic intensity). 1E means that user generates continuous traffic, e.g. 0.1E - the user user generates traffic only in 10% of network using time.
In the case of telephony it is assumed that subscriber generates traffic with 0.1E intensity, in the case of data transmission: 0.06 - 0.1E. Estimated number of subscribers connected to one node using one channel can be counted with the use of the following patterns:
nz=Nz/a=390/0.06=6500 (nz - number of serviceable subscribers - reverse channel, a - average traffic intensity)
nd=Nd/a=484/0.06=8066 (nd - number of serviceable subscribers - broadcast channel, a - average traffic intensity)
It can be assumed that in our case the number of connected subscribers cannot exceed 8100. Obviously, 64kbps isn't what subscribers really want; cable networks usually offer much higher speeds. Due to proportion, it is easy to estimate that 4050 subscribers can be serviced at 128 kbps, 2025 at 256 kbps, 1012 at 512 kbps, 506 at 1Mbps, and so on.