Input and output impedance in RC circuits

Question

As far as I learned, the rule of thumb for the ratio of output impdance to load(input impedance)is 1 to 10. Thevenin resistance(output resistance) Rth should be very small compred to the load resistance Rload. Rth/Rload must be small in order to avoid “loading the circuit”. A maximum ratio of 1/10 is often used as a design rule of thumb.

I was trying to understand how this would result in RC circuits. I came across to the following link which is part of a text book where it explains rule of thumb in RC case (see page 19):

http://sites.fas.harvard.edu/~phys123/classnotes/class2_notes_headerfile_rev_july14.pdf

Basically the text says:

Since impedance of an RC circuit is:

Z = Xc // R = [R * (-j/(w * C))] / [R + (-j/(w * C))]

we have a frequency dependent impedance here and we should apply the worst case scenario.

To make my question clear I add the following circuit:

In above circuit filter A is driving filter B. Considering the circuit, here is what I understand from his arguments on input and output impedance:

1-) We want ideally Zout (in filter A) to be as low as possible. That means in worst case Zout will be high which happens when frequency is very low. And he approximates filter A to R here.

2-) We want ideally Zin to be as high as possible. That means in worst case Zin will be very low which happens when frequency is very high. This approximates Filter B to R. I can see that from the Z = Xc // R relation. I can picture this in my mind when I think with "phasor representation". If the frequency goes too high, Xc goes to zero and Z becomes R. This looks more straightforward.

My question is about the first argument above. I will repeat again:

Lets consider Filter A again in above circuit. We want ideally Zout to be as low as possible because we don't want loading. That means in worst case Zout will be too high which happens when frequency is very low.

But when I think with phasor representation. If the frequency is set to too low, Xc goes to infinity and Zout doesn't become R.

Where am I wrong here?

Edit:

Here is how I derive mathematically so far(let me know if agree):

lim[f-->0]: Zout = ?

lim[f-->0]: Xc --> infinity

lim[f-->0]: (-j * R * Xc) / (R - j*Xc) = (-j * R * infinity) / (R - j * infinity)

|Zout| = (R*infinity) / sqrt(R^2 + infinity^2)

|Zout| = (R*infinity) / infinity

|Zout| = R

Answer 1

We can't achieve the 'ideal'. But if Zin is 10x Zout of the previous stage then the loading, attenuation, and interaction between stages will be minimal.

If the frequency is set to too low, Xc goes to infinity and Zout doesn't become R

Zout becomes R plus the source impedance. Provided we stick to the 1:10 rule (Rb = 10 x Ra) we are OK. The only problem is that each stage multiplies the impedance by 10, so 3 stages is about the limit for a practical circuit.

However, this only applies if you need low attenuation and minimal interaction between stages. If you don't mind a bit of loss and roll-off in the passband then a lower ratio may be acceptable.

Answer 2

Z = Xc // R = [R * (-j/(w * C))] / [R + (-j/(w * C))]

divide numerator and denominator by 'w'

Z = [R * (-j/(w * C))/w] / [R + (-j/(w * C))/w]

Z = [R * (-j/(C))] / [R/w + (-j/(C))]

at very high freq, 'w' -> infinity, R/w term vanishes

Z = [R * (-j/(C))] / [(-j/(C))]

Z = R (worst case at high frequencies)

At very low freq, the source impedance is R//Xc and Xc is very large (tends to infinity) the result

Zout = R*Xc/(R+Xc)

Divide numerator and denominator by Xc,

Zout = R/(R/Xc+1)

R/Xc term can be dropped as Xc is very large,

Zout = R (low freq worst case)