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Message-ID: <7D653C9C42F5D411A27C00508BF8803D01A9F1B0@mail.dstl.gov.uk>
From: "Talbot Andrew" <ACTALBOT@mail.dstl.gov.uk>
To: rsgb_lf_group@blacksheep.org
Cc: "Tottingham Brian J" <BJTOTTINGHAM@dstl.gov.uk>, 
 Linda_Holtby@ntlworld.com
Date: Mon, 28 Apr 2003 09:24:53 +0100
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Subject: LF: RE:  A conumdrum for the weekend - Image Cancelling Mixer.
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We thrashed out this solution on Friday afternoon, it explains why the maths
appears to fail but the concept doesn't.  Judging by relies on Friday
evening and Saturday, several readers of the reflecter were getting there,
but a lot of red-herrings were also appearing. 

The ambiguity arises here.  In the second configuration (repeated below),
the one that appears to fail in the way I described it, the output from the
top mixer is SIN(A+B) - SIN(A-B)

But we know that SIN(-X) = -SIN(X)since the Sine function has odd order
symmetry, so the mixer output can justifiably be written as :
SIN(A+B) + SIN(B-A) 
Then after the phase shift we have COS(A+B) + COS(B-A)

Now, we also know that COS(-X) = COS(X) The function has Even symmetry, so
it is now perfectly valid to write the output as :
COS(A+B) + COS(A-B)

Which, by addition or subtraction with the other path, cancels one of the
sidebands while reinforcing the other as expected.  For once it is the maths
that causes the problem, not the concept.

There were quite a few comments from readers along the lines of "broadbancd
phase shifters" to cope with both sideband frequencies.  But this wasn't
really an issue.  In some cases it is the input frequency sideband that is
being cancelled - which needs a bit of mental agility to think though these
two designs backwards, noting the two input terms that give the same output
term in any mixer.

Where image cancelling mixers are used in practice these days is in digital
receivers where the IF is at a very low frequency, like baseband to DC.  The
90 degree phase shift is then just an inherent part of the DSP process,
although diagrams showing how the systems work often show a physical phase
shift at this point.  In real analogue hardware, such as the radar receiver
block diagram that generated this problem, the phase shifters are usually
placed in the most convenient position - usually where the relative
bandwidth is at its narrowest such as the RF and LO ports on a
downconverting superhet receiver, with the IF output being summed. BUT on
the matching transmitter just reversing the signal flow, the non-phase
shifted channel now becomes one of the inputs, mixed with the LO to form the
RF output signal.  So the two configurations in the condrum become receiver
and transmitter respectively in a classic single conversion design.

In the radar receiver however, a different situation arises.  As it has to
cover a wide tuneable Rf input range of 6 - 18GHz a broadband phase shifter
here is impractical.  The tuneable must LO has to have one, and this is just
a case of designing the LO with a bank of suitable phase shifters.  The IF
at a VHF frequency is a few tens of MHz wide and a 90 degree phase shift
network at these freqeuncies is straightforward.

Andy G4JNT

=========================================================
NOW,  if we move the second 90 degree phase shift to the output of the
mixers (the IF) rather than the LO, as shown below, intuition and common
sense tells us it should still work since the mixing process is fully
reversible and actual direction of signal flow is irrelevant.  Furthermore,
many real designs of SSB exciters and receivers prove this really does work
in practice.

        ---90---X----90----|
Signal -|       |         +/- IF
        ------------X------|
                |   |
                | - |
                  |
              Local Osc.

But here is the conumdrum :

Keeping the same terminology of Signal = SIN(A), and LO = SIN(B), the inputs
to the top mixer become  
COS(A) . SIN(B)
and the output given by the product rule :
COS(A).SIN(B)     =  SIN(A+B) - SIN(A-B)

After the output 90 degree phase shift, the SIN terms beocome COS so we
have, in the top output leg :
COS(A+B) - COS(A-B)

The output from the bottom mixer is, as before :
SIN(A).SIN(B)   =  COS(A-B) - COS(A+B)

Which is the SAME as in the top leg, and will either cancel or reinforce
both sidebands.  
So it doesn't appear to work at all !
Moving the output 90 degree shift to the lower leg still fails to cancel one
sideband only.

So where is the conundrum?    Both forms of image cancelling mixer do indeed
work, and the trig identities can be taken from any reference.
==================================================
PS.
I do have one rather weak explanation, but it doesn't give that warm cozy
feeling expected when theory falls into place!
Andy


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