RF Questions (Channel Spacing, OOK Modulation, 434MHz...) - Page 1
Forum section: electronics / projects

author: David
author_rank: Frequent Contributor
timestamp: October 04, 2013, 11:11:34 am
content: Hi all,I am trying to improve on my RF knowledge and so I'm considering a simple project with a 434MHz transciever.My question is regarding channel spacing and occupied bandwidth.According to ETSI EN 300 220-1 (Page 28), if you use a frequency between 434,040 MHz and 434,790 MHz then you have no duty cycle restrictions (and up to 10mW E.R.P) as long as the channel spacing is smaller or equal to 25kHz.I am planning to use a simple OOK modulation scheme. Is my understanding correct that my channel bandwidth cannot be larger than 25kHz so to avoid interference with adjacent channels? What if I were to only use a single channel? How do I determine the bandwidth required for my needs? Is this dependent upon the bit rate of the modulated signal?I have many more questions but any help with the above would be great!Cheers,Dave

author: w2aew
author_rank: Super Contributor
timestamp: October 04, 2013, 10:26:51 pm
content: The bandwidth occupied by OOK will depend on two things: the bit rate, and perhaps more importantly, the filtering of the baseband data signal at the modulator. Fast RF rise and fall times of the OOK pulses will dramatically increase the occupied BW.

author: w2aew
author_rank: Super Contributor
timestamp: October 05, 2013, 12:23:22 am
content: About the best you can do is BW equal to the NRZ bit rate. If the modulation is filtered to the point where an alternating 1010... pattern looks like a sine wave, then the spectrum will look a bit like an AM signal that is 100% modulated with a sine at half the bit rate, thus will occupy BW equal to bit rate. If you don't filter this much, occupied BW will be >> bit rate.

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: October 05, 2013, 06:54:55 am
content: Quote from: David on October 04, 2013, 11:11:34 amYes. It is probably in the standard somewhere, how much your signal needs to be attenuated at +/- 12.5 KHz from the carrier frequency. Note that this requirement also means your carrier frequency has to be very stable, so it doesn't drift across multiple channels.Quote from: David on October 04, 2013, 11:11:34 amNot sure what you mean. You are only using one channel at a time, period. Attempting to use multiple channels simultaneously in order to increase bandwidth would probably count as an attempt at circumventing the bandwidth restrictions.Quote from: David on October 04, 2013, 11:11:34 amTo expand on the reply given by W2AEW: If you modulate a carrier with a low(er) frequency sinewave, the occupied bandwidth is 2x modulating frequency. How strong the sidebands are (IE. if they are powerful enough to need to be taken into account) depends on the relative amplitudes of the carrier and modulating signal.An unfiltered square wave, like a bit stream from a MCU, contains - in theory - infinitely many mixed sine waves, with the lowest frequency being the bit rate clock. So, as stated by W2AEW, if your MCU signal could be filtered down to a sine wave at the bit rate clock frequency (it cannot), the maximum allowed bit rate would be 25 KHz (allowed bandwidth) / 2 = 12.5 KHz.In practice it needs to be a fair bit lower than this, as you do need the higher modulating frequencies in order to convey the on/off information. Just how much lower depends on the details of the transmitter, and the capabilities of the receiver. I doubt any generalized 'canned' answer is possible here. You need a detailed analysis for each specific circuit.

author: ElectroIrradiator
author_rank: Frequent Contributor
timestamp: October 05, 2013, 07:23:43 am
content: Here is a crude example to start with. (What follows won't stand up to a detailed analysis, so use with caution.)If your transmitter has a simple OOK modulator, with no filtering of the keying signal, then the spectrum of the modulating signal is that of a hard square wave. This means the harmonics of the keying rate have known maximum amplitudes:An= 2 / (n * Pi), for n odd.The spectrum of a square wave only contains the odd harmonic frequencies of the fundamental. This does change, once you add your modulation information, but let us ignore that for a moment.Imagine your signal needs to be down 30 dB, 1 part in 1000, at +/-12.5 KHz. (I have no idea what the actual limit may be, I just pulled that figure out of the thin air.)What is the highest bit clock you can use, if you don't filter the modulating signal at all?1/1000 > 2 / (n * Pi), n odd <=>n > 2000 / Pi, n odd <=>n > 636.62..., n odd =>n = 637, minimum.So as a crude estimate your maximum bit rate clock for a maximum signal 30 dB down at +/- 12.5 KHz will be 12500 / 637 = 19.6 Hz...Which is why you mostly want to low pass filter the modulating signal before applying it to the transmitter. An off the shelf module with a fixed, digital 'keying input' is thus likely to be trouble, unless it contains its own filtering.

author: SArepairman
author_rank: Frequent Contributor
timestamp: October 05, 2013, 04:34:33 pm
content: how well policed are such laws?

author: G7PSK
author_rank: Super Contributor
timestamp: October 05, 2013, 09:23:26 pm
content: Quote from: SArepairman on October 05, 2013, 04:34:33 pmProbably not at all, that is until you step on someones toes. I have seen somewhere that some of the 433 and 434 is used by the police in the UK and even worse for when it comes to transgressing is the MOD and CAA (ministry of defence and civil aviation) not a good idea to get the wrong side of these boys.

author: David
author_rank: Frequent Contributor
timestamp: October 06, 2013, 06:56:16 pm
content: Thank you for all the responses! I will sit down and digest these when I get a chance. No doubt this will produce many more questions!