Unfortunately, when testing my base station I found that it was nowhere near as sensitive as the development board – the base station was getting CRC errors whilst the development board was still receiving the signal fine. \u00a0After a small bit of investigation I found out why. \u00a0The problem is all down to the fact that I want to use the CC1125 at the lowest baud rate (300 bps), and consequently a very narrow receive bandwidth (3.8 kHz).<\/p>\n
<\/p>\n
Before I get to the problem, we need to understand some theory. \u00a0CatTrack uses a modulation scheme called Frequency Shift Keying (FSK). \u00a0I am using a carrier frequency of 868 MHz and a frequency deviation of 1 kHz. \u00a0This deviation is the value recommended by TI for a baud rate of 300 bps. \u00a0This means that in order to transmit a ‘0’, the transmitter transmits a signal at (868 MHz –<\/strong><\/span> 1 kHz) for (1 \/ 300) seconds, so 867.999 MHz for 3.3 milliseconds. \u00a0In order to transmit a ‘1’, the transmitter switches to (868 MHz +<\/strong><\/span> 1 kHz), so 868.001 MHz and sits there for 3.3 milliseconds. \u00a0Bit by bit our entire message is transmitted. \u00a0Makes sense right?<\/p>\n Carson’s bandwidth rule tells us the occupied bandwidth of the transmitted signal:<\/p>\n Where:<\/p>\n An important thing to note here (and the source of much confusion in various forums!) is that So in our case, the bandwidth occupied by the transmitted signal is:<\/p>\n Now, in the receiver we have a receive bandwidth of 3.8 kHz. \u00a0The receive bandwidth needs to be set to a value such that the transmitted signal fits comfortably inside it.<\/p>\n In theory this sounds perfect – our receive bandwidth is 3.8 kHz and our transmitted signal only occupies 2.3 kHz, everything should work just fine!<\/p>\n However… this maths assumes that the transmitter transmits its signal centred\u00a0exactly<\/em>\u00a0on 868 MHz, and that the receiver places its 3.8 kHz receive window\u00a0exactly<\/em> around 868 MHz. \u00a0The reality is somewhat different.<\/p>\n We have frequency references inside our transmitter and receiver from which the transmit and receive frequencies are derived. \u00a0These have an inherent accuracy, expressed in\u00a0parts per million (ppm)<\/em>. \u00a0A 1 MHz reference oscillator with an accuracy of \u00b110 ppm could in reality output any value from 0.99999 MHz to 1.00001 MHz. \u00a0This doesn’t look too bad, but by the time this signal is multiplied up to our transmit frequency, using a reference oscillator with an accuracy of \u00b110 ppm means that instead of transmitting on\u00a0exactly<\/em> 868 MHz, we could be transmitting at 868.008680 MHz – as much as 8.7 kHz off the correct frequency! \u00a0Remembering that our receive window is only 3.8 kHz, transmitting 8.7 kHz away from the correct frequency is a big problem – our receiver will never even see our transmitted signal!<\/p>\n![\"CBR=2({\Delta}f+f_m)\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-835df9d93de34b237bbc5fb648073079_l3.png\" "\\"Rendered")
<\/p>\n\n
![\"CBR\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-f2d37e797a04c8f739bedbf3887cad4f_l3.png\" "\\"Rendered")
is the bandwidth occupied by the signal<\/li>\n![\"{\Delta}f\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-3710e689137417be96dc09da58177865_l3.png\" "\\"Rendered")
is the frequency deviation<\/li>\n![\"f_m\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-3f8595b11fdcb51a5c71a3f70b5397b9_l3.png\" "\\"Rendered")
is the modulating bit rate<\/li>\n<\/ul>\n is equal to the ![\"\frac{baud rate}{2}\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-e8c082215b4d0f5317a213896961474b_l3.png\" "\\"Rendered")
in our case. \u00a0Our highest baud rate will be when sending an alternating 0-1-0 sequence. \u00a0As we saw above, this means we transmit for 3.3 milliseconds on one frequency, then 3.3 milliseconds on another. \u00a0Thus the period is 6.6 milliseconds, therefore the bit rate is ![\"\frac{1}{6.6 ms}=150\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-e16a9cf9507c2bdc7242872ac502c8b4_l3.png\" "\\"Rendered")
.<\/p>\n![\"2(1000+150)=2.3 kHz\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-f8af7ed12c7ce8162dea6e1b0f9fad0b_l3.png\" "\\"Rendered")
<\/p>\n
![\"(868000000 + 1150) + ((868000000 + 1150)(2\times10^{-6})) = 868.002886 MHz\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-f3ada9ae37162ef53f97d91c792bd360_l3.png\" "\\"Rendered")
<\/p>\n![\"(868000000 - 1150) + ((868000000 - 1150)(2\times10^{-6})) = 868.000586 MHz\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-15a138a7c05bf2a48c44222ecb9a02db_l3.png\" "\\"Rendered")
<\/p>\n![\"(868000000 - 1900) - ((868000000 - 1900)(2\times10^{-6})) = 867.996364 MHz\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-9e6dc81e56a907a15b02ddeced4e5312_l3.png\" "\\"Rendered")
<\/p>\n![\"(868000000 + 1900) - ((868000000 + 1900)(2\times10^{-6})) = 868.000163 MHz\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/ql-cache\/quicklatex.com-0072835ed24923b9cca959de01042326_l3.png\" "\\"Rendered")
<\/p>\n![\"\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/uploads\/2018\/04\/cattracktx.png\")
<\/p>\n![\"\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/uploads\/2018\/04\/frequency-offset-registers.png\")
<\/p>\n![\"\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/uploads\/2018\/04\/widened-rx-bandwidth.png\")
<\/p>\n![\"\"](\"https:\/\/www.burningimage.net\/cattrack\/wp-content\/uploads\/2018\/04\/receive-bandwidth-matching-transmit-bandwidth.png\"){kind=spacer}
<\/p>\n