Syllabus Sections:-

Mixer and Local Oscillator

4m.1 Understand the function of a mixer, the generation of intermediate frequencies (IF) and other mixer products.

Understand that for given RF and IF. frequencies, there is a choice of two local oscillator (LO) frequencies. Understand the reasons for the choice and calculate the frequencies.

If you have a mixer with an incoming RF signal and a local oscillator they mix together and give you an IF output.

Each section of a radio receiver must be well designed and constructed. The oscillator in particular should be free of distortion and unwanted modulation, as poor construction leads to unwanted mixer products reaching the audio stages.

You can have what is called oscillator high or oscillator low. Let's look at that in more detail.

If you have an RF signal coming in of 1MHz and an IF out of 0.5MHz to obtain the IF you could have a local oscillator into the mixer of either 1.5MHz of 0.5Mhz.

Traditionally we use oscillator high 1.5MHz and this is associated with Image problems discussed below.

Calculation of Frequencies

We know that mixing two frequencies produces :-

  1. the RF frequency,

  2. The Local oscillator frequency,

  3. the SUM of the RF frequency and the Local oscillator frequencies and

  4. the difference of the RF frequency and the Local oscillator frequencies

If the RF frequency F1, the Local oscillator frequency is F2, and we want the output IF frequency = F.

Thus if F1 = 1MHz and F2 = 1.5MHz the frequencies produced from above will be  :-

  1. = 1MHz

  2. = 1.5Mhz

  3. = 2.5MHz

  4. = 0.5Mhz

And if F1 = 1MHz and F2 = 0.5MHz the frequencies produced from above will be :-

  1. = 1MHz

  2. = 0.5Mhz

  3. = 1.5MHz

  4. = 0.5Mhz

So for two different local oscillator frequencies the same IF frequency can be produced.

So how does one decide whether to use the high or the low frequency. If a low IF is chosen then the filtering is more difficult as it is closer to the wanted frequency so it calls for a high Q tunable filter.

So on balance you might then say that you would chose the high frequency. This would be fine if the signals on the bands were spread well apart but they are not.

The result is the double superhet the high frequency is used first followed by the low frequency.

4m.2 Understand the origin of the second channel or image frequency and calculate the frequency from given parameters.

The origin of the second channel or image frequency is the attempt of a receiver to receiver two RF signals apparently on the same frequency at the same time. The unwanted signal is known as 'second-channel' (or 'image') and is an interference to the wanted signal to the receiver.

This second channel interference only occurs when there is the unwanted signal on the "second channel".

So if you were transmitting in the 1.8MHz band and someone was tuning to the medium wave band and the signal they wanted was twice the IF below your transmission they could hear your transmission as the second channel.

Look at the diagram above and you will see the relationship between the wanted Broadcast band signal on the receiver and the unwanted amateur Band transmission as the second image.

So the existence of second channel interference depends on the response of the signal input circuit of the receiver's mixer to a frequency which is separated from the resonant frequency of the input circuit by twice the IF. Changing the IF would reduce the incidence of image interference in this case.

So if the wanted and the unwanted signals are an equally distance away from the LO (local oscillator) they will both appear at the audio output.

When you look in a mirror, your image in the mirror appears to be as far away from you as you are to the mirror thus apparent mirror image.

So let us look at another example:-

If we have a 28MHz band receiver and an intermediate frequency of 8MHz where would the image frequency be ?

So first you must find the Local oscillator frequency ....

28 - LO = 8 so LO = 28 - 8 = 20MHZ

or LO - 28 = 8 so LO could be 28 + 8 = 36 MHz

With the LO at 20MHz and the frequency wanted at 28MHz the image would be at 12MHz

with the LO at 36MHz and the wanted frequency at 28MHz the image would be at 44MHz

another way of looking at this is :-

Frequency + (2 x IF) = image frequency

thus 28Mhz + (2 x 8) = 44MHz

and

Frequency - (2 x IF) = image frequency

thus 28Mhz - (2 x 8) = 12 MHz


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