mrv@drune.UUCP (VagherMR) (11/02/85)
I've had some difficulty recently in trying to resolve an apparent
difference between some simple mathematical analysis and practical
results in terms of some basic spectral analysis. I decided to
seek the help of specialists on the net before looking at another
useless textbook on the subject. I'm sure the answer must be simple
but anyway here it is:
If there exists a baseband signal x(t)=Kcos(Wmt) where Wm is the
angular frequency and this signal is input to a multiplier or
a balanced mixer where the multiplying signal or local oscillator
is y(t)=Xcos(Wlt) at frequency Wl the output is
s(t)=KXcos(Wmt)cos(Wlt). This leads to an equation for the
output of s(t)=(KX/2)cos(Wl-Wm)t + (KX/2)cos(Wl+Wm)t . Taking
the Fourier transform of this gives a "delta" function in the
frequency domain of area KX/4 at frequencies Wl-Wm , Wl+Wm and
also at the negative of these frequencies (which accounts for
the extra 1/2 factor. However it should be perfectly valid
to work with positive frequencies and real functions of time
only. The output of this would be a DSB-SC signal.
Calculating the power of these spectral components is where I
have my dilemna. The expected value or mean square power of
the baseband component is K*K/2 in a one ohm system and
the power of each component of s(t) is K^2*X^2/8 or the total
power in s(t) equals K^2*X^2/4. Assuming K=X=1 the power in the
baseband signal would be 10log(1/2) or -3dBW. The power in each
of the components of s(t) would be -9dBW. This leads to a loss
of 6dB from the baseband to one of the components of s(t).
However, in working with real double balanced diode mixers the
loss from the baseband input to the output is from 5-7dB.
Actually this loss is specified as the conversion loss from
RF to IF but this is equivalent. The manufacturer's literature
briefly states that 3dB of the loss is due to theoretical
considerations and that the rest is due to thermal and mismatch losses.
Can anyone see where my mistake is in calculating 6dB of loss?
stephany.WBST@Xerox.ARPA (11/05/85)
re: problem with power in a balanced modulator I think your problem is that you assumed that the diode characteristics fit a pure square law . Actual, ideal diodes contain higher terms, third, fourth, etc. power. An ideal diode is I=Io(1-exp(eE/kT)), which must be expanded in a power series. In the balanced configuration all odd terms cancel so you are left with the square, fourth power and higher even powers. Your math only took into account the first square term. I did not work it out but I believe that is the problem. Joe N2XS