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..
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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 :-
-
the RF frequency,
-
The Local oscillator frequency,
-
the SUM of the RF frequency and the Local oscillator
frequencies and
-
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 :-
-
=
1MHz the
original frequency
-
=
1.5Mhz the LO frequency
-
=
2.5MHz the SUM frequency
-
=
0.5Mhz the difference frequency
And if
F1 = 1MHz and F2 = 0.5MHz the frequencies produced from
above will be :-
-
=
1MHz the
original frequency
-
=
0.5Mhz the LO frequency
-
=
1.5MHz the SUM frequency
-
=
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.
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