Yes that is correct according to the sheet of trig identities on the wall above
my desk.
But as I wrote it, the top mixer input is not SIN(A) and COS(B), as the LO
signal does not have the 90 degree shift in it.
The inputs to the top mixer are .......
COS(A).SIN(B) = SIN((A+B) - SIN(A-B) which is much the same as you give, but
with A and B reversed and using the knowledge that SIN(A-B) = -SIN(B-A)
Andy
-----Original Message-----
From: Alan Melia [SMTP:[email protected]]
Sent: 2003/04/25 15:29
To: LF-Group
Subject: LF: something for the weekend ??
Hi Andy, I am afraid my Terman "Radio Engineers Handbook" 1943 Page 19
gives
Sin(A).Cos(B)= Sin(A-B)+Sin(A+B) ....if we forget about the 0.5 factor.
but it agrees on the others. This gives you Cos(A-B) + Cos(A+B) out of the
top after the 90deg. phase shift and Cos(A-B) - Cos(A+B) out of the bottom
....again cancelling one sideband.
I was toying with the idea that Sin(A+pi/2)=Cos(A) but Cos(A+pi/2)= -Sin(A)
but that seems irrelevant.
Cheers de Alan G3NYK
[email protected]
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