A Bell 202 modem sends audio signals to represent digital data.
This signalling method can be called AFSK (audio frequency shift keying.)
Binary ones (marks) are represented by 1200Hz (+/- 10Hz) audio.
Binary zeroes (spaces) are represented by 2200Hz(+/- 10Hz) audio.

Is the above correct and, assuming the modem is running at 1200 baud, does that mean that one bit is 833.33 microseconds long, and thus one bit can contain only one cycle of 1200Hz audio or 1.833 cycles of 2200Hz audio?

Thanks.

 

Too hard or too easy to bother with?

 

Hello,

I found this in the wiki:
https://en.wikipedia.org/wiki/Bell_202_modem

Bertus

 

Yes, the change between the two bits is "filled in" to keep the phase correct.
AFAIK the output is basicaly coming from a sine table fed into a DAC. At the bit border the sample frequency of the table going to the DAC switches such that it produces 1200 or 2200 Hz sine output.

see this for example https://www.google.com/search?q=har...v4Q9QEwAnoECAMQDA#imgrc=ZJ_r-LfsOGcB7M:&vet=1

 

Hello,

I found this in the wiki:
https://en.wikipedia.org/wiki/Bell_202_modem

Bertus

Thanks for the reply.

 

Yes, the change between the two bits is "filled in" to keep the phase correct.
AFAIK the output is basicaly coming from a sine table fed into a DAC. At the bit border the sample frequency of the table going to the DAC switches such that it produces 1200 or 2200 Hz sine output.

see this for example https://www.google.com/search?q=hart+waveform&client=firefox-b&sxsrf=ACYBGNRGooaw3M5SqlgRHk5nqQjJj3uGlg:1569357138851&tbm=isch&source=iu&ictx=1&fir=ZJ_r-LfsOGcB7M%3A%2C8-ZwQjU7-xxMgM%2C_&vet=1&usg=AI4_-kQK4sBtCoJ-cQkfEOQryVTeBg_hbw&sa=X&ved=2ahUKEwiStKX2purkAhVQxhoKHVatDv4Q9QEwAnoECAMQDA#imgrc=ZJ_r-LfsOGcB7M:&vet=1

Thanks for the reply. That link led me to a very nice App Note from Analog Devices.