Message-ID: <3EA9774C.6000500@virgin.net>
Date: Fri, 25 Apr 2003 18:58:36 +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
Cc: "ACT \(E-mail\)" <g4jnt@thersgb.net>, 
 "Tottingham Brian J" <BJTOTTINGHAM@dstl.gov.uk>,
  Linda_Holtby@ntlworld.com
References: <7D653C9C42F5D411A27C00508BF8803D01A9F1AC@mail.dstl.gov.uk>
Subject: Re: LF: A conumdrum for the weekend - Image Can	celling 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

I have a problem with the output 90deg phase shifter. I can see how you
can shift the LO and the signal by 90deg, but the output from the
top mixer has sin(A+B) and sin(A-B) signals present. Unless you have
a frequency sensitive phase shifter, it is only going to shift one
component by 90deg and the other by some strange ammount.

Is the phase shifter frequency sensitive?

Stewart G3YSX

Talbot Andrew 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
> 
> 
> "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."
> 
>