Of CATV time spread a net

   Current, cable television enterprise is developing in high speed, CATV net will from the tradition one-way onefold (transmit TV signal only) , to two-way and seesaw pattern muti_function way develops (INTERNET, VOd,
Data switch, broadcast of digital TV, number) , network structure becomes optical fibre cable to mix a net by the onefold cable network in the past (HFC) , the core that its develop is become geminate by one-way net to transmission net. Since be two-way,transmit a net, with respect to existence one answers arraign to inscribe reversely, the noise that transmits a passageway reversely namely in the most outstanding issue in two-way CATV system accumulates a problem, namely alleged " funnel effect " .
One, the generation of noise
   What normally retrorse passageway transmits is data signal, transmission main technique index is to be piled up by accident rate, and HFC cable allocates a net is comprised by branch alloter, in the environment that transmission passageway is shared to make its are in a complex electromagnetic wave by a large number of users, the interference of shortwave broadcast, electric home appliances
2, restrain a measure that passes passageway noise
   1, the screen capacity that increases coaxial cable to allocate a network, prevent interference to be invaded from cable, alloter of the terminal box of the user, branch all should be chosen good screen function and segregation spend more expensive product. Additional, coaxial cable network should have good ground connection system, ground connection resistor should be less than 8 ohm.
   2, elaborate design cable allocates a net, outside decreasing, disturb invade. Designing us to be familiar with quite before to what transmit to cable television system, but to the design of retrorse passageway we still may compare silent to be born, retrorse passageway talks to design the issue that wants a consideration below. A, signal transmits a system to the difference with retrorse transmission loss.
   Arrive from user terminal unit between front can be nodded as time, temperature and ground and differ somewhat, it is the edge is answered for the most part among them travel way port of each branch, alloter arrives the cable loss between amplifier, because branch, alloter is to be to transmit offer suitable n, answering the case that send to fall, the n of signal each not identical, and the n that transmitting a passageway reversely is existing bigger difference, a spur track that allocates amplifier with now is exemple try to explain, see a picture one:
Be being designed to the passageway is to press top job frequency to choose each branch implement, make signal is in each branch implement output port n is consistent, such making that in raising a mouth to reach user home, each have an appropriate n, however, time pass method loss to will appear bigger difference, the data that expresses a kind is a its very good specification.

Express number of 1 branch port

T1 T2 T3 T4 T1 arrives T4
Total loss of 750MH branch port 27db 25db 26db 27db 0db
Total loss of 40MHZ branch port 27db 21.2db 16.8db 9.3db 17.7db

   Example specification is above on the Hui Chuanpin rate of 40MHz, to different port time pass method loss difference to be able to amount to 17.7dB, so big difference has very big input range with respect to requirement front receiver (very big dynamic range) but the system has the restriction that carries comparing of a confusion of voices, should answer so sending signal is unallowed when n is less, be like the example above, the first branch implement answer the n that pass to set for 60dB, so the 4th branch answers the signal n that pass to have 42dB only, when answering the signal that send by the 4th branch apparently, carrying a confusion of voices to compare do not accord with index, general we ask the terminal unit of the client side is like machine top box or CABLE MODEM place to send time when passing signal to reach front must close, so how to accomplish this? Namely how each branch implement Where is the loss normalization of port? A kind of simple and practical method uses equalizer namely. Because equalizer is to be designed,enclothe whole to with retrorse frequency band (5-750MHZ) , such equalizer can amend a loss difference that passes a passageway at the same time even also the response to the passageway. For example the equalizer of a 6dB will be in attenuation 6dB in 5MHZ, and be in in 40MHZ attenuation 4.6dB, , attenuation 0dB is in in 750MHZ, this equalizer inserts caustic to be 1dB, be in loss to be 1dB in 750MHZ so, 5.6dB is in 40MHZ place, analyse with the example in front again, in each branch implement port place inserts equalizer: If the graph is shown 2 times, time pass 40MHZ to manage each branch implement if express,port is worth to the attenuation of amplifier 2:

Express number of 2 branch port

T1 T2 T3 T4 T1 arrives T4
Total loss of 750MH branch port 27db 26db 27db 28db 1db
Total loss of 40MHZ branch port 27db 26.8db 27db 28.8db 1.8db

By the watch 2 in knowable, equalizer reduced the loss difference of retrorse passageway greatly, accomplish branch implement the normalization method loss of port, another advantage was to reduce the come or flow together that enters cable system to get together noise. Joining balanced (set interference n to be 20dBmv) , the interference intensity that mouth of the first branch enters is - 7dBmv, the interference that mouth of the last branch enters is 10.7dBmv; Use balanced later, the n of all port became - 7dBmv left and right sides, such making of whole system carry comparing of a confusion of voices to be able to improve.
   B, every HZ secures power law: Of full load time pass a system to be comprised by a lot of channel, the bandwidth that each channel place occupies and modulation means (QPSK, SQAM) wait to differ possibly, how to add up to comfortable n for every channel allocation so? Because answer the laser of the catapult that send light to input n to have strict restriction, exceed its when n when n of the biggest enable input, laser can appear to cut wave phenomenon and make signal serious and lack fidelity. Above all we should analyse laser after all to be able to increase how old radio frequency power, the bandwidth that takes up general power by every channel next (total bandwidth is 35MHZ) more or less will allocate each channel. The cent recipe method that can use a kind of every HZ to secure power here: All allot general power to whole Hui Chuanpin to be taken from the Bu Changping of every HZ above all (the decision of place of highest input n that the numerical value with chief condition can accept by laser, if laser is highest,input n is 45dbmV so long average power is every HZ pace - 30DBMV. Algorithm is as follows: Set every HZ pace to grow average power to be X, x= laser machine is highest input n - HZ of 10lg total bandwidth, x=45dbmV-10lg[35*10E6]=-30.44dbmV, anthology - 30dbmV) the bandwidth that has with each channel next allots power. For example a certain channel has bandwidth to be 1MHZ, the power value that it assigns is - 30DBMV+10lg (10E6) =30DBMV, another channel place occupies bandwidth to be 10KHZ, the power value that then its assign is - 30DBMV+10lg (10E4) =10DBMV. From here we can see, narrow channel gathers together noise is little power is little also at the same time, wider channel susceptive gathers together noise is big but signal power is strong also at the same time, assured to have between each channel so more balanced carry comparing of a confusion of voices.
   3, node breaks up: Already had carried in front, the user that cable network place leads is more, gather together noise effect is bigger, the method that for this we can use much node opening light will break up cable network place to lead an user. General a smooth node brings the range that the user can assure 500 times to answer those who pass a passageway to carry a confusion of voices to set than be in inside.

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