We have two digital modulations, both are the same, the only different is the symbol rate. Usually low symbol rate modulation have less BER than high symbol rate modulation. But my question is if in the case of a bad carrier stability, high symbol rate modulation could have less BER than low symbol rate modulation. I think answer is YES, but i am not sure.

# Symbol rate versus BER question

Started by ●November 8, 2009

Reply by ●November 8, 20092009-11-08

JAlbertoDJ wrote:> We have two digital modulations, both are the same, the only different is > the symbol rate. > > Usually low symbol rate modulation have less BER than high symbol rate > modulation. But my question is if in the case of a bad carrier stability, > high symbol rate modulation could have less BER than low symbol rate > modulation. I think answer is YES, but i am not sure.The answer is NO. High bitrate: 0101010101010101.... Low bitrate: 0000111100001111.... Think about it. VLV

Reply by ●November 8, 20092009-11-08

> >The answer is NO. > >High bitrate: 0101010101010101.... >Low bitrate: 0000111100001111.... > >Think about it. > >VLV >But, for example, suposse the case of a BFSK at 1 baud. F0= 1000 Hz F1= 2000 Hz Frecuency carrier error maximum: +-10 Hz. Then, signal received during interval of 1 bit, for example could be: F0= 1006 Hz F1= 1997 Hz At 1 baud signal received is out of the range of the macht filter.

Reply by ●November 8, 20092009-11-08

Reply by ●November 8, 20092009-11-08

JAlbertoDJ wrote:>>The answer is NO. >> >>High bitrate: 0101010101010101.... >>Low bitrate: 0000111100001111.... >> >>Think about it. > > But, for example, suposse the case of a BFSK at 1 baud. > > F0= 1000 Hz > F1= 2000 Hz > > Frecuency carrier error maximum: +-10 Hz. > > Then, signal received during interval of 1 bit, for example could be: > > F0= 1006 Hz > F1= 1997 Hz > > At 1 baud signal received is out of the range of the macht filter.That only tels that your receiver is not optimal in case of unstable carrier. Modify the receiver accordingly. VLV

Reply by ●November 8, 20092009-11-08

>That only tels that your receiver is not optimal in case of unstable >carrier. Modify the receiver accordingly. > >VLV > >For example, with Fast FHSS?

Reply by ●November 8, 20092009-11-08

JAlbertoDJ wrote:>> The answer is NO. >> >> High bitrate: 0101010101010101.... >> Low bitrate: 0000111100001111.... >> >> Think about it. >> >> VLV >> > > But, for example, suposse the case of a BFSK at 1 baud. > > F0= 1000 Hz > F1= 2000 Hz > > Frecuency carrier error maximum: +-10 Hz.You make things up. How can you get that kind of frequency error? cheap crystals are good to a part in 10e-7. Nobody with any sense would assign two frequencies in simple harmonic relation.> Then, signal received during interval of 1 bit, for example could be: > > F0= 1006 Hz > F1= 1997 Hz > > At 1 baud signal received is out of the range of the macht filter.The longer the symbol lasts, the more time there is to integrate. Integration reduces noise. Jerry -- Drink deep or taste not the Perian spring. A little knowledge is a dangerous thing. Alexander Pope. �����������������������������������������������������������������������

Reply by ●November 9, 20092009-11-09

JAlbertoDJ wrote:>>That only tels that your receiver is not optimal in case of unstable >>carrier. Modify the receiver accordingly. >> >>VLV >> >> > > > For example, with Fast FHSS?Idiot

Reply by ●November 9, 20092009-11-09

>JAlbertoDJ wrote: >>> The answer is NO. >>> >>> High bitrate: 0101010101010101.... >>> Low bitrate: 0000111100001111.... >>> >>> Think about it. >>> >>> VLV >>> >> >> But, for example, suposse the case of a BFSK at 1 baud. >> >> F0= 1000 Hz >> F1= 2000 Hz >> >> Frecuency carrier error maximum: +-10 Hz. > >You make things up. How can you get that kind of frequency error? cheap >crystals are good to a part in 10e-7. Nobody with any sense would assign>two frequencies in simple harmonic relation.In the case of an example, somebody with a sense could assign two frequencies in simple harmonic relation.> >> Then, signal received during interval of 1 bit, for example could be: >> >> F0= 1006 Hz >> F1= 1997 Hz >> >> At 1 baud signal received is out of the range of the macht filter. > >The longer the symbol lasts, the more time there is to integrate. >Integration reduces noise.Integration reduces noise, but if frecuency is not estable you can be integrating only noise.

Reply by ●November 9, 20092009-11-09