It's not harmonics of 1 MHz that you need: it's harmonics of 500 kHz: your regenerative detector tunes over a 500 kHz range.
I'll explain that in depth . . . .
Example:
To receive signals in the freq. range 7 to 8 MHz with your regenerative detector tuning 2.5 to 3.0 MHz, for the input range of 7.5 to 7.0 MHz, the local osc. will need to be 10.0 MHz (subtractive mixing): 10.0 minus 7.5 = 2.5; 10.0 minus 7.0 = 3.0.
For an input range of 8.0 to 7.5 MHz, the local osc. will need to be 10.5 MHz (subtractive mixing): 10.5 minus 8.0 = 2.5; 10.5 minus 7.5 = 3.0.
Therefore the local osc. needs to be 10.0 MHz and then 10.5 MHz . . . and nothing in between.
Now extend that concept over the entire signal input range: 5.0 to 15.0 MHz.
On the basis that the oscillator is on the high side of the signal, we have [osc. freq.] minus [RF freq.] = [regen. tuning freq.]
So, 8.0 MHz minus 5.0 MHz= 3.0 MHz; 8.0 MHz minus 5.5 MHz = 2.5 MHz: regen. tunes 3.0 to 2.5 MHz
And: 8.5 MHz minus 5.5 MHz = 3.0 MHz; 8.5 MHz minus 6.0 MHz = 2.5 MHz: regen. tunes from 3.0 to 2.5 MHz
and so on, up to:
17.0 MHz minus 14.0 MHz = 3.0 MHz; 17.0 MHz minus 14.5 MHz = 2.5 MHz.
17.5 MHz minus 14.5 MHz = 3.0 MHz; 17.5 minus 15.0 MHz = 2.5 MHz
This covers the entire R.F. input range from 5.0 to 15.0 MHz.
Thus, since the R.F. stage tunes from 5.0 MHz to 15.0 MHz, the local osc. needs to be in 500 kHz steps from 8.0 MHz to 17.5 MHz.
Now if the regen. detector tuned over a 1 MHz bandwidth, then the local osc. would be in 1 MHz steps . . . in essence, what happens in the Racal RA-17 using a 1 MHz crystal osc. and a harmonic generator: the "kHz tuning" covers a continuous 1 MHz bandwidth; the "MHz tuning", in effect, selecting the appropriate harmonic from the 1 MHz crystal, and thus acting as a 'harmonic selector switch'.
Finally, it is obvious that if only harmonics of 1 MHz were required, rather than harmonics of 500 kHz, the total number of osc. freqs. required would be halved. I suspect that this, in turn, would lead to an easier method of producing the required local osc. freqs.
Over to you, Lawrence!
Al.
I'll explain that in depth . . . . Example:
To receive signals in the freq. range 7 to 8 MHz with your regenerative detector tuning 2.5 to 3.0 MHz, for the input range of 7.5 to 7.0 MHz, the local osc. will need to be 10.0 MHz (subtractive mixing): 10.0 minus 7.5 = 2.5; 10.0 minus 7.0 = 3.0.
For an input range of 8.0 to 7.5 MHz, the local osc. will need to be 10.5 MHz (subtractive mixing): 10.5 minus 8.0 = 2.5; 10.5 minus 7.5 = 3.0.
Therefore the local osc. needs to be 10.0 MHz and then 10.5 MHz . . . and nothing in between.
Now extend that concept over the entire signal input range: 5.0 to 15.0 MHz.
On the basis that the oscillator is on the high side of the signal, we have [osc. freq.] minus [RF freq.] = [regen. tuning freq.]
So, 8.0 MHz minus 5.0 MHz= 3.0 MHz; 8.0 MHz minus 5.5 MHz = 2.5 MHz: regen. tunes 3.0 to 2.5 MHz
And: 8.5 MHz minus 5.5 MHz = 3.0 MHz; 8.5 MHz minus 6.0 MHz = 2.5 MHz: regen. tunes from 3.0 to 2.5 MHz
and so on, up to:
17.0 MHz minus 14.0 MHz = 3.0 MHz; 17.0 MHz minus 14.5 MHz = 2.5 MHz.
17.5 MHz minus 14.5 MHz = 3.0 MHz; 17.5 minus 15.0 MHz = 2.5 MHz
This covers the entire R.F. input range from 5.0 to 15.0 MHz.
Thus, since the R.F. stage tunes from 5.0 MHz to 15.0 MHz, the local osc. needs to be in 500 kHz steps from 8.0 MHz to 17.5 MHz.
Now if the regen. detector tuned over a 1 MHz bandwidth, then the local osc. would be in 1 MHz steps . . . in essence, what happens in the Racal RA-17 using a 1 MHz crystal osc. and a harmonic generator: the "kHz tuning" covers a continuous 1 MHz bandwidth; the "MHz tuning", in effect, selecting the appropriate harmonic from the 1 MHz crystal, and thus acting as a 'harmonic selector switch'.
Finally, it is obvious that if only harmonics of 1 MHz were required, rather than harmonics of 500 kHz, the total number of osc. freqs. required would be halved. I suspect that this, in turn, would lead to an easier method of producing the required local osc. freqs.
Over to you, Lawrence!

Al.








