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Date: Fri, 25 Apr 2003 16:38:57 +0200
To: rsgb_lf_group@blacksheep.org
From: "Rik Strobbe" <rik.strobbe@fys.kuleuven.ac.be>
In-reply-to: <7D653C9C42F5D411A27C00508BF8803D01A9F1AC@mail.dstl.gov.uk>
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Subject: Re: LF: A conumdrum for the weekend - Image Can	celling Mixer.
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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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