Thayne said:
Nope, I (still)don't get it. I assume you are not talking about injection as far as the local oscillator freqs where you end up with the IF's.
All of the DCS and CTCSS modulation only serves to open the squelch when it matches, right??
Anyway, I appreciate your efforts to explain it; I will try to read up on it.
Just getting that kenwood programmed so I could talk to those camp employees was what I needed, so I am happy.
OK. Let's try it this way... when a CDCSS waveform is modulated, there are 1s and 0s - the 1s swing the carrier up and the 0z swing the carrier down in frequency. If, in the receiving process, the carrier is inverted, your signal is turned upside down and the 0s would represent a high swing and the 1s a low swing. This ends up making the digital waveform upside down from what it was. A 0=1, and a 1=0. Hence, inverting the code re-inverts the waveform and you end up with an upright waveform.
On voice or CTCSS, it doesn't matter if the signal is inverted or not since it's all sine waves. An upside down since wave sounds the same. But, with digital signals, reversing the 1s and 0s is a totally different waveform from what you started with.
This:
___I''''''I______I''''''I___
becomes this:
''''''I___I''''''''''''I___I''''''
If it's the injection that has you confused, all you really need to understand is that different injecton will make the signal a mirror image if itself - inverting it. Using an inverted code doubles the inversion and makes it 'look' right to the decoder. But, let's try this explaination: A receiver can either have high or low injection relative to the receiving frequency. If it's low injection, the difference of the carrier will be the same at the IF - highs are higher and lows are lower. BUT, if it uses high injection, the difference between the LO (Local Oscillator) and the carrier will be greater when the signal goes low, and less difference when the signals goes high. Remember, you are looking at the signal from the 'other wide' of the frequency. So, the IF shifts will be inverted from the true carrier shifts.
With the different injection, think of it as a bouncing ball, only one that is bouncing digitally - either high or low at any given point. If you watch the ball from the bottom, the higher it is, the farther the distance from you. All well and good since close=0 and far=1. But, if you watch the ball from above, the higher it is, the closer it is to you, and close=0 and far=1 digitally. So, on peaks of height, you are seeing the ball closer (a '0') when in fact it is in reality a '1'.
Did either of those examples help at all?
Joe M.
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