I'm trying to understand an analysis of a oscilloscope probe
compensating circuit (two parallel RC circuits in series).

The formula I have is:
Vout / Vin (f) = R2/(R1+R2) * sqrt ( 1 + w^2*T^2 ) / sqrt ( 1 +
w^2*Te^2 )
the phase is tan^-1(wT) - tan^-1(wTe) ...
How is that phase found?
I usually use inverse tangent of imaginary of real but in this case I
have (1 + jwT) / (1 + jwTe)
What is the real part? What is the imaginary part?

I could not really tell what you were trying to say. Nevertheless, I
think I can explain how these probes work. This assumes that inductance
(including strays) is not significant.
The resistor divider enables slow input signals to be divided
accurately. To take care of high frequencies, a capacitive divider with
the same ratio as the resistive divider is used. The biggest capacitance
item usually is the input capacitance of the scope. To this is added the
capacitance of the cable from the probe to the scope input. To get the
capacitance ratio correct, a variable capacitance is usually placed in
the probe. This is adjusted to the correct ratio while observing a
square waveform on the scope. This division ration increases the input
resistance and decreases the parallel capcitance at the probe compared
to that at the input connector. This increased input impedance is
achieved by reducing the signal level at the scope input connector.
The signal can be distorted by using a probe by its capacitance shunting
out high frequencies. However, distorted as it may me, a good probe will
present a smaller copy of the waveform to the scope as it exist in the
circuit. This distortion or detuning if you wish, is why you try to keep
probe capacitance small. Thus, a 100:1 probe will present less
capacitance to the probed point than a 10:1 probe. The cost again, is a
reduced signal at the scope.
You can go through the detailed circuit analysis, but it will not give
you much greater understanding.
Bill

Is this correct?
Then if I remember correctly in polar notation you divide the
magnitude by the magnitude and the angle by the angle to get
the answer.
Is this correct?
Thanks Professor.
peace
dawg

It's been awhile but I seem to recall that the angles use the same rules as
exponents, IE: subtract for division, add for multiplication, multiply for
raising to a power etc ....
bg

Thanks Don, and everyone else.
Considering 1/1+jy made things "click" for me. For some reason the
idea of converting the numerator to polar, converting the denominator
to polar, and then dividing the two of them didn't cross my mind. Now
things are making sense.
Thanks again.

Polytechforum.com is a website by engineers for engineers. It is not affiliated with any of manufacturers or vendors discussed here.
All logos and trade names are the property of their respective owners.