Sorry for the incomplete email.?
Pushed the wrong button on the keyboard while still compiling this draft before any corrections..
OK.Let's see if I can make it easy.
Lets say that we have a 12 MHz carrier and we want to convert it to transmit to the 28 MHz band.
We could either mix the 12 MHz carrier with a 16 MHz signal (12 MHz + 16 MHz = 28 MHz) or a 40 MHz (40 MHz - 12 MHz = 28 MHz).
In the case of mixing the 12 MHz carrier with the 40 MHz signal we will also get 52 MHz which we do not require and which we will filter out.
Let's say that now we have a 12 MHz filter which passes the lower side-band and we want to mix the filter output signal to transmit on the 28 MHz band.
Let's say that the carrier frequency is exactly 12 MHz and we modulate the carrier with a 1 kHz audio tone.
In a normal AM transmitter, we end up with the carrier and two side bands, one which is 12 MHz + 1 kHz = 12,001,000 MHz which is the Upper Side Band (USB) and the other 12 Mhz - 1 kHz = 11.999,000 MHz which is the lower side band?
Now let's take the same example as above but the signals are passed through a filter which passes the lower side-band (LSB) (11,999,000 MHz in the example) and filters out the Upper Side-band (USB) (12,001,000 MHz).
The carrier is also attenuated somewhat by the filter as it is placed down the slope of the filter curve to cut the lower voice frequencies as we are not after generating a hi-fi signal and are not really required for intelligibility apart from other considerations while the carrier is further heavily attenuated in the balanced modulator. Theoretically, if the balanced modulator was perfectly balanced and everything screened, the carrier attenuation will be infinite, but practically it is attenuated to a very very low level as to be considered as being suppressed.
Now let's say that the carrier was not suppressed but the lower side-band was passed by the filter and we mix them with a 40 MHx oscillator. (Remember that the upper side-band was suppressed by the filter and theoretically is not present, leaving us with the 12 mHz carrier and the 11.999,000 MHz signals)
The output of the mixer will contain the 40 MHz oscillator, the 12 MHz carrier and the 11.999.000 MHz lower side-band including other mixer products which for the present purpose we shall ignore.
By using a balanced mixer as used in the 12 MHz side-band generator the 40 MHz signal can be cancelled (attenuated to a very low level) and we will have an output signal from the mixer of 40 MHz + 12 MHz = 52 MHz and 40 MHz - 12 MHz = 28.000,000 MHz and also 40 MHz + 11.999,000 MHz = 51.999,000 MHz and 40 MHz - 11.999,000 = 28.001,000 MHz.
However, since we used a balanced modulator at 12 MHz, the 12 MHz carrier was suppressed and only the lower side band was allowed to pass through, which simplifies the matter as when we mix the lower side band from the filter with the 40 MHz oscillator will will only get??40 MHz + 11.999,000 MHz = 51.999,000 MHz and 40 MHz - 11.999,000 = 28.001,000 MHz.
At the output of the mixer we put a filter to select our wanted 28 MHz signal and strip the unwanted 52 MHz signal.
This clearly shows how the lower side-band from the filter (11.999,000 MHz) in relation to the original carrier frequency of 12 MHz has been turned into an upper side-band by the mixer when we used an oscillator frequency for mixing higher then the output frequency required.
Now if we take the lower side band which is 11.999.000 Mhz and mix it with a 40 MHz oscillator, we can also get on the 50 MHz band, but in this case it is easy to see that we will still end up with a side-band transmission if you work out the simple mathematics.
Now let's say that instead of mixing our crystal filter output with 40 MHz, we mix it with 16 MHz to get on the 28 MHz band (16 MHz + 12 MHz = 28 MHz).
But since we have suppressed the carrier and we only have an 11.999,000 MHz signal, 16 MHz + 11,999,000 = 27.999,000 signal which in this case will be outside the 28 MHz band but serves to illustrate our purpose.
16 MHz - 11.999,000 Mhz will also give us 4,001.000 MHz which we can easily filter out.
If our side-band filter passed the upper side-band, the opposite will be true
As you can see it is simple mathematics but may be difficult to understand without some numbers.
Hope this explained the difficulty.
Regards
Lawrece
Michael Shreeve
Michael Monteith