But, how much?
I started off thinking about how the load is connected in a simple resister divider virtual ground (SRDVG) CMoy.
I did not include the input circuitry to keep things clean. The values of the resistances are too large to have any significant current flow.
Then using small-signal analysis model, power supply rails are converted to ground:
It is clear that the two 4.7kΩ resistors in parallel is too high a resistance compared to the load. Most of the voltage from the amplifier will appear across the 4.7kΩ resistors (effectively 2.35kΩ)
So a capacitor is placed in parallel to the 4.7kΩ resistors to reduce the impedance to AC signals.
The capacitor appears as a short-circuit to AC, so no AC current flows across the 4.7kΩ resistors, so they can be removed in the AC model.
Wait...
The circuit on the right is the "capacitor in series with speaker load omg" which has somehow become a taboo. (As in, anything with output capacitor is immediately labeled as "suck".) Actually, the left circuit is worse than the right, because the finite impedance of the capacitor results in crosstalk when the other channel is connected to the same capacitor.
I still need to do my initial objective - to see by how much this is bad.
One thing good about using capacitors, (assuming ideal) is that for a fixed load the performance does not depend on volume, while active virtual grounds can handle a certain amount of current, above which unexpected things happen.
On the other hand, the circuit's performance becomes frequency-dependent, requiring huge capacitor values to avoid high impedance at lower frequencies and noticeable bass roll-off.
With a 33Ω load, in order to have a cutoff frequency of 20Hz, 241µF is needed. There's the second channel too, so 482µF, lets round it down to 470µF.
Actually, this isn't as bad as I thought. With higher-impedance loads, the capacitance required would be even less.
The impedance of a 235µF capacitor @ 20Hz is -j33.86Ω. Well, it should be around there, since |Vtotal| = √2*Re(Vtotal). Or in English, just know that in an RC high-pass-filter or low-pass-filter, the cutoff frequency is the frequency at which the capacitor's reactance, or 1/2πfC = the resistor's resistance R.
But, compare this with the 1-digit Ω or less impedance that active devices provide, 33.86Ω is a whole lot. Well... "whole lot" enough to reduce the voltage across the speaker load by √2 yea?
Lets see what happens at 100Hz -
Capacitor reactance = 6.77Ω
Resulting attenuation = -0.18dB
Phase shift = -11.6º
Oh yea there's still phase shift to worry about... how much phase shift is bad is debatable, I'll just leave the numbers here.
In comparison, lets try an active virtual ground with 10mA for each channel. Less attenuation issue, less crosstalk issue, less phase shift, but 0.33V max with a 33Ω load. Maximum allowable voltage increases as load impedance increases though. If a buffer capable of 100mA (for each channel) is used, then max voltage is 3.3V with 33Ω and 15V with 150Ω. That's more than ever needed.
As you can see, both capacitor-coupled output and active virtual ground methods have their own pluses and minuses. If cost and complexity permits I'd go for high-current active virtual ground, but very high current solutions can be expensive to have sometimes.
Back to the topic in the post title, how is the simple resistor divider virtual ground (SRDVG)(inclusive of capacitor C1) solution? A big fat no as I see it. While most circuits have their pros and cons vs others, the SRDVG is simply outclassed by the typical method of using DC-blocking capacitors. Unless you really have no space or budget for two additional 470uF capacitors (lets overkill a bit, shall we?)
Moral of story: Just because it doesn't seem to be "in series with the output" doesn't mean it isn't.
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