Tuesday 31 December 2013

Can a Kenwood Radio be used to Communicate with a Motorola Radio?

To answer your question (that is, after all, why I’m here): It depends entirely on which models you are planning to use. For example, if you had two PMR446 variants that were both on the same band, they ought to work fine (even if one was Kenwood and the other was Motorola).

If two radios are the same basic type and set to the same channel, then I don’t personally see why they wouldn’t work. However, if they aren’t of the same type, then they probably won’t work, it’s that simple.

Two-way radio technology is both simpler than you’d think it would be and more complicated than it first appears (if that makes sense), so its always best to make sure you have access to good kit that is easy to use. Personally, (if it is at all possible for you) I’d suggest spending out a bit and getting a new set, I’ll explain why below…

Your question does incur a dangerous element, so I feel I’d better warn you. Make sure that you understand, beyond a shadow of a doubt, that your license covers the frequency you are using. For most of the frequencies you’d likely to be using in order to connect the two devices, a special license is required. Transmitting without a license is a serious crime and you could spend up to two years in prison.

As ‘Robert J’ from ‘Yahoo! Answers’, suggests, the reason it’s considered to be so serious an offence is because you can actually endanger lives by interrupting radio transmissions from ambulances or police cars. I hate to sound like a square here, but in this instance, the rules exist for a reason.

Anyway, you really need to do your homework on this one and it may actually turn out to be easier and cheaper for you to just buy a new set of radios from either Kenwood or Motorola. If you were running a business, I’d go as far as to actually recommend you take this step.

However, if you’re only using the radios for personal projects, then provided you check them out and stick to the rules, you shouldn’t have a problem.

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