Syllabus Sections:-

Mixer and Local Oscillator

3I3  26  Understand the advantages and disadvantages of high and low intermediate frequencies and the rationale for the double superhet and triple superhet.

High IF gives good image rejection and poor selectivity.

Low IF gives good selectivity and poor image rejection.

You have to decided which to use. For instance if you had very low IF frequency, on a receiver on 28MHz, you may get good selectivity but you may suffer IMAGE that is receiving signals twice removed from the frequency you are interested in. On the other had if you had High IF you would get good image but poor selectivity.

HIGH IF followed by LOW IF  in the double superhet

In receiver we tend to have what is called a double superhet. We convert the signal frequency down to a high IF to give good image response and then further convert it to a second LOW IF to give good selectivity.  You might be able to remember this order by remember a usual greeting "Hi low everybody"!

So in the diagram above the IF1 is the high IF and IF2 is the low frequency.

Q Factor

The Q factor - if you look at a parallel tuned circuit and it has a high Q then the response curve will be very sharp. If the Q is poor we get a low Q circuit where the response is flat and broad.

The triple Superhetrodyne has an additional IF stage further ensuring that the correct modulated frequency reaches the audio amplifier..


  ------------------------------------------------------------------------------------------------------

3I3  26 continued Understand that for given RF and IF. frequencies, there is a choice of two local oscillator (LO) 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        the original frequency

  2. = 1.5Mhz     the LO frequency

  3. = 2.5MHz  the SUM frequency

  4. = 0.5Mhz  the difference frequency

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

  1. = 1MHz       the original frequency

  2. = 0.5Mhz    the LO frequency

  3. = 1.5MHz    the SUM frequency

  4. = 0.5Mhz   the difference frequency

So for two different local oscillator frequencies the same IF frequency can be produced frequency in item 4 in each case.


3I3  26 continued Understand the reasons for the choice and calculate the frequencies.

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 -- remember a usual greeting "Hi low everybody"!.

3I3  26 continued 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 receive 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 of the wanted signal to the receiver.

So whilst not on the calculation sheet you are provided with in the exam you have to know that:-

Second Channel ( of Image Frequency ) = Your wanted Frequency + 2( the IF )

But with all mixing you get addition and subtraction so

Second Channel ( of Image Frequency ) = Your wanted Frequency - 2( the IF )

Your wanted frequency can be either higher or lower than the LO frequency by the value of the IF (intermediate frequency).

This second channel, or 'image frequency', interference only occurs when there is an unwanted signal on the "second channel" or image 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 0.84MHz on the receiver and the unwanted Amateur Band transmission 1.84MHz 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 ?

The Image Frequency = Frequency + (2 x IF) or put it another way Frequency + (2 x IF) = Image Frequency

or

The Image Frequency = Frequency - (2 x IF) or put it another way Frequency - (2 x IF) = Image Frequency

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

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

In an exam question you could be asked what is the local oscillator frequency when given the wanted frequency and the unwanted frequency (image frequency) .

To work that out you need to subtract the lower frequency from the higher  and divide the result by 2 as the Local oscillator sits in the middle of the wanted frequency and the unwanted frequency as shown in the diagram above.

But you could be asked to find the IF then you would do the calculation above and then subtract the Local oscillator frequency from either the wanted of the unwanted frequency.







brats copyright logo