Just at make my contribution to the confusion :
assume in the 2nd setup that we swap signal and LO. It shouldn't make no
difference as it is just a matter of interpretation what you call signal
and LO.
Thus ,
Keeping the same terminology of Signal = SIN(A), and LO = SIN(B)
becomes Signal = SIN(B) and LO = SIN(A)
At the upper leg : SIN(A) * COS(B) = SIN(A+B) + SIN(A-B) (ignoring the
divide by 2)
after the 90 degrees phase shift at the output = COS(A+B) + COS(A-B)
At the lower leg : SIN(A)*SIN(B) = COS(A-B) - COS(A+B)
Adding : COS(A+B) + COS(A-B) + COS(A-B) - COS(A+B) = 2*COS(A-B)
Subtracting : COS(A+B) + COS(A-B) - COS(A-B) + COS(A+B) = 2*COS(A+B)
So that way theory seems OK.
73, Rik ON7YD
At 11:52 25/04/2003 +0100, you wrote:
The following little theoretical problem has occured several times during my
working life, and has cropped up once again, when another engineer here
asked me to explain how a radar receiver front end, as shown in the block
diagram in a manufacturers data sheet, could possibly work.
A well-known supplier of packaged mixers once had the same conumdrum to
solve when producing a customised mixer for us. Concern about it resulted in
them developing a special 90 degree network for operating over 0.3 to 5 MHz,
when all that was really needed was a couple of simple designs that operated
at HF/VHF. If common sense had ruled, the mixer would have been produced
much quicker and at less cost.
Andy G4JNT
=========================================================
Image Cancelling Mixer connundrum.
In the classic image cancelling mixer, suppression of one mixer sideband is
obtained by combining two 90 degree shifted input signals with two 90 degree
shifted versions of a local oscillator. The outputs from the two channels
are then added or subtracted depending on which sideband is desired.
---90---X---------|
Signal -| | +/- IF
------------X-----|
| |
90 |
| - |
|
Local Osc.
If the signal is represented by SIN(A) and the Local Oscillator by SIN(B)
then the signal input to the top mixer is now COS(A) as it has passed
through a 90 degree phase shift, and its respective LO is COS(B)
The mixers multiply the two signals, and from standard trignometric
identities taken from any mathematical reference, the products of two sines
can be expressed as a sum and difference equation
So the outputs of the mixers (ignoring the factors of 2 in the trig
identities) are :
Top COS(A).COS(B) = COS(A-B) + COS(A+B)
Bottom SIN(A).SIN(B) = COS(A-B) - COS(A+B)
So the cancellation / reinforcement at the output is obvious.
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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