Here are some concepts and definitions that may help you to understand or solve the problem:
- Check that the input level has an S/N ratio between 6 and 7 dB.
- As you can see in the TDT specifications, the baud rate is between 20 and 30 Mbaud, this means a real Group Bit Rate (GBR) of twice this value: between 40 and 60 Mbit/sec at the input of the TDT.
- Make sure that you know the correct values of the transmission parameters, in other words, the QPSK baud rate or Mbit/sec (GBR) and the transmission FEC.
- The term known as useful rate (RU) is calculated as follows:
- RU = GBR x FEC x Reed Solomon ratio in real terms,
- However, as the Reed Solomon correction is added again in the QAM modulation, the following formula is sufficient:
RU = GBR x FEC
- At the QAM modulator, we have to filter the data stream by using special raised cosine shaped filters with a 15% roll off in order not to affect the output pulses.
- The bandwidth at the output is slightly higher than the theoretical one:
Theoretical QAM symbol rate = (RU) x (1/Reed Solomon ratio) x (1/bits per symbol)
- Due to the raised cosine shaped filter, the bandwidth is larger:
1,15 times larger (filter roll-off) than the theoretical one.
- In our specifications we indicate the theoretical maximum QAM baud rate: 6,9 Mbaud, for the above mentioned conditions.
- When this value is higher, the QAM modulator will not be able to work.
Example 1:
- QPSK input rate =27500 Mbaud (GRB = 55 mb/s)
- FEC = 3/4 (typical Astra transmissions),
The RU should be:
RU= 55000000 x (3/4) x (188/204) = 38014 mb/s
If we do not take the Reed Solomon ratio into account, the calculations would be as follows:
RU = 55000000 x (3/4) = 41250 mb/s
Also, each symbol has 6 bits (2^{6} = 64) so:
64 QAM symbol rate = RU x (204/188) x (1/6) = 38014 x (204/188) x (1/6) = 6.875 Mbaud
If we do not take the Reed Solomon value into account, as mentioned above, the result would be the same:
64 QAM symbol rate = 41250 x (1/6) = 6,875 Mbaud which is less than the 6,9 Mbaud specified.
The system will work because the QAM modulator can handle this amount of data (real output channel bandwidth is 7.9 MHz = 6,87 x 1,15).
Example 2:
- Symbol rate = 26,85 Mbaud
- FEC = 7/8
then:
GBR = 26,87 x 2 = 53,7 mbit/sec.
RU = 53,7 x (7/8) = 46,9875 mbit/sec
64 QAM symbol rate = 46,9875 mbit/sec x (1/6) = 7,83 Mbaud
This means that it is out of the QAM capability.
The QAM modulator will be de-engaged and it will not work, which is indicated by the LED off. (The real output channel bandwidth is 9 MHz)
Example 3:
The European bouquet transponder in the Asiasat 2 has the following specifications:
- Symbol rate = 28,125 Mbaud
- FEC = 3/4
then:
GBR = 28,125 x 2 = 56,25 mbit/sec.
RU = 56,25 x (3/4) = 42,1875 mbit/sec
64 QAM symbol rate = 42,1875 mbit/sec x (1/6) = 7,03 Mbaud
so the real bandwidth is:
7,03 x 1,15 = 8.08 MHz
which is greater than 8 MHz, the standard VHF and UHF channel bandwidth.
The Televes TDT cannot work with this transponder.
In conclusion:
- Check the input signal
- Take note of the baud rate and FEC of the QPSK satellite transmission.
- Compute the baud rate
- Baud rate < 6.9 Mbaud. the system should work
- Baud rate > 6.9 Mbaud. the system will not work, as it is out of its range.