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While reading The Art of Electronics (3rd Ed.), I was stumped by a particular assertion in section 1.7.1C, "Blocking capacitor":

For instance, every stereo audio amplifier has all its inputs capacitively coupled, because it doesn't know what dc level its input signals might be riding on. In such a coupling application you always pick R and C so that all frequencies of interest ... are passed without loss ... That determines the product RC ... You've got the product, but you still have to choose individual values for R and C. You do this by noticing that the input signal sees a load equal to R at signal frequencies ... so you choose R to be a reasonable load, i.e., not so small that it's hard to drive, and not so large that the circuit is prone to signal pickup from other circuits in the box.

What do the authors mean with a small resistance being "hard to drive"? Also, why does a high-value resistance affect signal pickup?

I can only guess, but my hypothesis is that signal pickup may pop up because the cutoff frequency is lowered when R is increased, and lower frequencies pass through. A low resistance may become hard to drive because of wasted power?

The schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ Small resistance are "hard to drive" because the current is needed to "drive them" into desired voltage level. And high-value resistance are "prone" to pick a noise because high-value resistance times "small" current is equal to large voltage. And this voltage will be amplified by a amplifier. \$\endgroup\$ – G36 Dec 26 '16 at 17:43
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You should read that sentence in the context of what was explained in the introductory chapters about signal sources and voltage dividers.

As can be shown the signal source will see the amplifier input as a simple resistance (R) in the passband. Once you set the bandwidth (i.e. the RC product) you are left with one degree of freedom, i.e. choosing the R value.

The R value is chosen as a trade-off between a very large (ideally infinite) resistance (good to avoid attenuation, but bad for noise pickup) and a very low (ideally zero) resistance (no noise pickup, but no signal transfer either!).

To be more explicit, in the passband the circuit is like this:

schematic

simulate this circuit – Schematic created using CircuitLab

$$ V_{in} = V_s \cdot \frac{R}{R_s + R} $$

So, if R is much bigger than Rs, then Vin will be almost equal to Vs (no attenuation). But making R bigger will increase the input impedance of the amplifier, thus increasing the chance for noise pickup (noise sources can couple capacitively to the input terminal, and they appear as high impedance sources, so we want their (noise) signal to be attenuated, hence low R).

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