Phase angle of RC circuit

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?
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In article

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
--
Private Profit; Public Poop

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Thanks Bill. The probes were just an example being used while learning the circuit analysis though.
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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
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Wecan do it wrote in message

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
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( > Magnitude of A/B) @ phase) =1

------------ bg has it right. the angle in this case would be a-b Remember that the complex number usage is based on De^jd =D (cos(d) +jsin(d)
--

Don Kelly snipped-for-privacy@shawcross.ca
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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.