Message-ID: <3EAE44BE.5010704@virgin.net>
Date: Tue, 29 Apr 2003 10:24:14 +0100
From: "Stewart Bryant" <stewart.bryant@virgin.net>
User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US; rv:1.0.2) Gecko/20030208 Netscape/7.02
MIME-Version: 1.0
To: rsgb_lf_group@blacksheep.org
References: <7D653C9C42F5D411A27C00508BF8803D01A9F1B0@mail.dstl.gov.uk>
Subject: Re: LF: RE:  A conumdrum for the weekend - Image Cancelling Mixer.
Content-Type: text/plain; charset=us-ascii; format=flowed
Content-Transfer-Encoding: 8bit
Sender: <owner-rsgb_lf_group@blacksheep.org>
Precedence: bulk
Reply-To: rsgb_lf_group@blacksheep.org

Andy

Since there can only be one solution, there must be an error
in your original transformation i.e.

ie SIN(A-B + 90) = - COS(A-B) and not +COS(A-B) as
originally asserted.

Stewart

Talbot Andrew wrote:
> 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
> 
> 
> "This e-mail is intended for the recipient only.  If you are not the
> intended recipient you must not use, disclose, distribute, copy, print,
> or rely upon this e-mail. If an addressing or transmission error has
> misdirected this e-mail, please notify the author by replying to this e-mail."
> 
> "Recipients should note that all e-mail traffic on MOD systems is
> subject to monitoring and auditing."
> 
>