## Re: ACT200L and Ultimate TV

 Re: ACT200L and Ultimate TV From: Karl Bongers - 2003-05-12 05:13:46 ```Yes, lirc_sir is locked at 38khz freq on output signals. lirc_serial can do any frequency on output. Below is some info on how it could be changed to output at 56khz. On Wed, May 07, 2003 at 08:57:10PM -0400, dbakker@... wrote: > I have been succesfull and programming the receiving end of the ACT200L= =20 > and LIRC. The problem I have is transmitting codes to the ultimate TV=20 > receiver. Iv read that the driver is locked at a certain freq.. Does=20 > anyone how to change the driver to transmit at a diff freq and what freq= =20 > an UltimateTV receiver is looking for?? >=20=20 LIRC SIR mode analysis: Baudrate selection and modulation: Current driver sends a '[' (5bH) out at 7N1, 115200 baud. It also samples data at a rate of the baudrate. Each byte received represents a pulse. Each byte not received represents a space. +12v _____ _____ _____ | | | | | | 0 0v | | | | | -12v |-----|-----| |-----|-----| |-----|-----| 1 start 1H 2H 4H 8H 10H 20H 40H stop Above example: '['(5b) going out serial port at 115200,7,N,1. At 115200, each bit is 1/115.2k =3D 8.68us With a desire modulation frequency of 38khz, we want 26.3us total period, or 13.15us for half the cycle(a bit). _____ | | | | | |_____| |<-26.3us ->| 38khz 8.68us*3bits =3D 26.04us, so 3 bits is roughly 38khz time period. What if we want 57.6khz? 57.6khz is 17.361us period. So we want a bit time of 8.6805 _____ | | | | | |_____| |<17.361us> | 57.6khz Can we derive this from a 115200 bit time of 8.68us? 8.68us * 2 =3D 17.36 So two bit times at 115200k baud can be used to represent the 57.6khz frequency nicely. We just need to work around the fixed start and stop bits, maybe adjust the number of data bits to make it fit. +12v _____ _____ _____ _____ _____ | | | | | | | | | | 0 0v =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20= =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20 | | | | | | -12v |-----| |-----| |-----| |-----| |-----| 1 start 1H 2H 4H 8H 10H 20H 40H 80H stop Tx byte=3D55H Heres one way with 8 databits(8N1), another way with 6 databits(6N1): +12v _____ _____ _____ _____=20=20=20=20=20=20=20 | | | | | | | | 0 0v =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20= =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20 | | | | | | -12v |-----| |-----| |-----| |----- 1 start 1H 2H 4H 8H 10H 20H stop Tx byte=3D15H Receive sample rates @115200K: 8N1(10 bits) : 86.8us 7N1(9 bits) : 78.1us 6N1(8 bits) : 69.4us The 6 data bits would be nicer as you would get a finer grain sample rate. ```

 ACT200L and Ultimate TV From: - 2003-05-08 00:57:17 Attachments: Message as HTML ```I have been succesfull and programming the receiving end of the ACT200L and LIRC. The problem I have is transmitting codes to the ultimate TV receiver. Iv read that the driver is locked at a certain freq.. Does anyone how to change the driver to transmit at a diff freq and what freq an UltimateTV receiver is looking for?? ```
 Re: ACT200L and Ultimate TV From: Karl Bongers - 2003-05-12 05:13:46 ```Yes, lirc_sir is locked at 38khz freq on output signals. lirc_serial can do any frequency on output. Below is some info on how it could be changed to output at 56khz. On Wed, May 07, 2003 at 08:57:10PM -0400, dbakker@... wrote: > I have been succesfull and programming the receiving end of the ACT200L= =20 > and LIRC. The problem I have is transmitting codes to the ultimate TV=20 > receiver. Iv read that the driver is locked at a certain freq.. Does=20 > anyone how to change the driver to transmit at a diff freq and what freq= =20 > an UltimateTV receiver is looking for?? >=20=20 LIRC SIR mode analysis: Baudrate selection and modulation: Current driver sends a '[' (5bH) out at 7N1, 115200 baud. It also samples data at a rate of the baudrate. Each byte received represents a pulse. Each byte not received represents a space. +12v _____ _____ _____ | | | | | | 0 0v | | | | | -12v |-----|-----| |-----|-----| |-----|-----| 1 start 1H 2H 4H 8H 10H 20H 40H stop Above example: '['(5b) going out serial port at 115200,7,N,1. At 115200, each bit is 1/115.2k =3D 8.68us With a desire modulation frequency of 38khz, we want 26.3us total period, or 13.15us for half the cycle(a bit). _____ | | | | | |_____| |<-26.3us ->| 38khz 8.68us*3bits =3D 26.04us, so 3 bits is roughly 38khz time period. What if we want 57.6khz? 57.6khz is 17.361us period. So we want a bit time of 8.6805 _____ | | | | | |_____| |<17.361us> | 57.6khz Can we derive this from a 115200 bit time of 8.68us? 8.68us * 2 =3D 17.36 So two bit times at 115200k baud can be used to represent the 57.6khz frequency nicely. We just need to work around the fixed start and stop bits, maybe adjust the number of data bits to make it fit. +12v _____ _____ _____ _____ _____ | | | | | | | | | | 0 0v =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20= =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20 | | | | | | -12v |-----| |-----| |-----| |-----| |-----| 1 start 1H 2H 4H 8H 10H 20H 40H 80H stop Tx byte=3D55H Heres one way with 8 databits(8N1), another way with 6 databits(6N1): +12v _____ _____ _____ _____=20=20=20=20=20=20=20 | | | | | | | | 0 0v =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20=20= =20=20=20=20=20=20=20=20=20=20=20=20=20=20=20 | | | | | | -12v |-----| |-----| |-----| |----- 1 start 1H 2H 4H 8H 10H 20H stop Tx byte=3D15H Receive sample rates @115200K: 8N1(10 bits) : 86.8us 7N1(9 bits) : 78.1us 6N1(8 bits) : 69.4us The 6 data bits would be nicer as you would get a finer grain sample rate. ```