?

Not really.

I would guess the confusion arises because there are two different inversions happening. The first is what you experienced with the DDS -- the frequency span is the same but TUNING is inverted. That is mostly tradition as we ordinarily count going up in frequency as we turn the dial clockwise. In the case of the BITX40, if the VFO tunes from 4.7 MHz to 5 MHz, then the RF tunes from 7.3 MHz to 7.0 MHz. So to get a "proper" tuning we actually have to reverse or "invert" the VFO signal. That is what you have actually described in your rig.

The second type is much more subtle. If one thinks of sidebands, the USB extends lower audio frequencies from the carrier outward UP in frequency, and the LSB does the reverse, i.e., the higher audio frequencies are DOWN in frequency. Merely shifting the USB to the LSB area will invert the audio; now the upper audio frequencies are nearer the carrier and the lower ones further away. That is not the way we hear things.

How then does one decode USB on LSB side (or vice versa)? Clearly some other mechanism has to happen which also inverts the effect of the LSB into one of an USB (ibid). Then the phases will be correct and the original audio will reappear. This is not a simple problem.

Most of the traditional 9 MHz IF and 5 MHz VFO rigs (the 20/60m rigs) did not have this problem. Generally if the filter were a lattice filter, they used both a USB and LSB crystal and adjusted the BFO frequency to either sideband; That is, they use dual BFO frequencies. They also used (as does Farhan in his 20/40 rig) dual bandpass filters to accommodate each of the different sidebands. This practice continues; some BITX rigs use the same mechanism even though they do not use lattice filters (they generally use Cohn filters).

With the advent of the sharp bandpass crystal filter rigs (such as the BITX) this did not work any more and there is ONE BFO frequency. Note that the modulating frequency on transmit is the IF. The received modulated frequency is the RF. The VFO is passive and merely adds or subtracts enough to produce either an RF output or IF input. So where can sideband inversion of the second type take place? In the mixer. Note that the IF is the actual modulated RF signal on transmit.

? "The resulting upper and lower sidebands each contain both of the modulating frequency components. The upper-sideband components consist of the sum of each modulating frequency plus the carrier frequency, and no inversion takes place. The lower-sideband components consist of the carrier frequency minus each of the modulating frequencies, and they become inverted. Inversion occurs in any frequency-translating process when the mixing frequency is higher than the signal frequency and the difference products are selected in the output circuit. This principle can be used for sideband switching in both transmitters and receivers, since by this means an upper-sideband signal is converted to a lower-sideband signal or vice versa."

Source:SSB - QRZ Israel HAM radio portal.htm

Since our mixing frequency is the VFO? (5 MHz), and the output is the RF (7 MHz), no inversion of the second kind takes place. But there is another product (VFO + IF) where it does because the IF is greater than the VFO (-IF + VFO). This product is suppressed by the final BPF. Consider that BPF. It is critical. It keeps out non-40m frequencies from the receiver, and also suppresses non-40m signals from the transmitter, including out-going inversion products.

I know this is confusing. I wonder all the time if I have gotten it right. The math is somewhat complex. Hope this helps anyway.


john
AD5YE


---In BITX20@..., <chase8043@...> wrote :

Hi John

I had a chance to try some testing with sideband inversion.

BFO = 11.9986 MHZ and VFO = 4.8486 MHZ this gives me a starting frequency of 7.15 MHZ(RF = BFO - VFO). Set the encoder to increase frequency in CW direction. But this to increase VFO or decreasing RF. So I switched it so RF goes up in CW direction.

I usually listen in on a net at 7.163 MHZ. To tune in the voice I had to increase RF. So I find that LSB is on the wrong side of the (non-existent) carrier.

Is this the sideband inversion?

73

Ken VA3ABN

On Thu, Dec 8, 2016 at 11:56 PM, iam74@... [BITX20] <BITX20@...> wrote:

?

Think of it this way: VFO - IF = RF. Does this result in sideband inversion? (hint: -IF + VFO. It it positive or negative?) Why don't you try it and see what happens. If you get nothing but garble, then the answer is "no...it does not.

If the results are clear signals, then you must decide if the LSB is being inverted, or is it truly USB. Remember that the ordinary condition by convention is that the LSB on 20m is garble or non-existent. Therefore you must hear USB inverted to LSB or true USB for intelligibility. Be aware of AM; test your results through several different QSOs.

Actually, it doesn't matter which it is -- as long as communication takes place.

john
AD5YE

---In BITX20@..., wrote :

Opps. Not?18.998600 - 4.998600 = 14.0. Should be 25.998600 - 11.988600 = 14 or is this the wrong sideband?

73 Ken

?

There is an excellent and easy to remember rule of thumb on sideband inversion.? Explained here:

On Thursday, December 15, 2016 9:52 PM, "Ken Chase chase8043@... [BITX20]" <BITX20@...> wrote:

?

Hi John

I had a chance to try some testing with sideband inversion.

BFO = 11.9986 MHZ and VFO = 4.8486 MHZ this gives me a starting frequency of 7.15 MHZ(RF = BFO - VFO). Set the encoder to increase frequency in CW direction. But this to increase VFO or decreasing RF. So I switched it so RF goes up in CW direction.

I usually listen in on a net at 7.163 MHZ. To tune in the voice I had to increase RF. So I find that LSB is on the wrong side of the (non-existent) carrier.

Is this the sideband inversion?

73

Ken VA3ABN

On Thu, Dec 8, 2016 at 11:56 PM, iam74@... [BITX20] <BITX20@...> wrote:

?

Think of it this way: VFO - IF = RF. Does this result in sideband inversion? (hint: -IF + VFO. It it positive or negative?) Why don't you try it and see what happens. If you get nothing but garble, then the answer is "no...it does not.

If the results are clear signals, then you must decide if the LSB is being inverted, or is it truly USB. Remember that the ordinary condition by convention is that the LSB on 20m is garble or non-existent. Therefore you must hear USB inverted to LSB or true USB for intelligibility. Be aware of AM; test your results through several different QSOs.

Actually, it doesn't matter which it is -- as long as communication takes place.

john
AD5YE

---In BITX20@..., wrote :

Opps. Not?18.998600 - 4.998600 = 14.0. Should be 25.998600 - 11.988600 = 14 or is this the wrong sideband?

73 Ken

?

So does this mean the Bitx40 board that just arrived is now out of date? What has been updated, is there an update for existing boards or do I buy another Bitx40?

Bruce

VK2RT