The problem arises from adding the \$7.5 \Omega\$ at the output. If you calculate the Thevenin Equivalent you'll get the following circuit, where the \$5 \Omega\$ resistor is the parallel equivalent of \$7.5 \Omega\$ and \$15.5 \Omega\$.
simulate this circuit – Schematic created using CircuitLab
You can see that the output voltage has an insertion loss from the factor \$\frac {7.5} {7.5+15.5}\$ and it is where the 9.6 dB comes from.
The \$5 \Omega\$ equivalent resistor explains why the frequency cutoff changes to 2100 Hz.
Your amplifier has a fixed output impedance and finite voltage swing. To get the most power out of it, the load impedance has to match the output impedance. Two speakers in parallel have half the impedance of one speaker. This is apparently too low for your amp to drive properly.
Probably your "wallwart" supplies are collapsing under the heavy drain of two speakers in parallel. The lower supply voltage makes the 1.5 V or so deadband in your output a larger fraction of the overall, significantly increasing distortion. You don't say what kind of opamps you are using, other than they are not rail to rail. The supply voltage may be collapsing to the point where there is little active region left between the deadband in the output and the output range of the final opamp.
In addition, parts of your circuit don't make sense and could rather easily be replaced by a better design:
- You have symmetric ± supplies. That's good. So why why why are you level shifting the input away from ground-centered?
Upon closer inspection, it seems you are level shifting to compensate for the input signal being centered around 2.5 V. This is just plain silly.
You can't hear DC. Even "HiFi" audio only goes down to 20 Hz. The obvious way to deal with input DC offsets is to AC couple the signal. Get rid of all the nonsense to the left of the positive input of the second opamp. Replace it with a 1 µF cap in series followed by a 10 kΩ resistor to ground.
- You seem to be willing to live with the deadband in the dual emitter follower output stage, but at least include it in the feedback loop so that the opamp can try to compensate for it. All this requires is to connect the 10 kΩ (Argh, no component designators) feedback resistor to the output of the whole amp instead of the output of the opamp.
Here is your basic circuit with the obvious points mentioned above fixed:
Note that this is both simpler and will work better.
There are ways to significantly reduce the final stage deadband. Two diodes is a very common approach.
What I usually do is use a couple more transistors in the output stage to give it a gain of 2. The previous stage then only has to drive to ± half the supply range. That gets around needing a rail to rail opamp, which generally aren't available at the ±12 V or more you want to run them at.
Added in response to scope traces
You have even more problems than you realize.
Your circuit is oscillating under load, almost certainly by feeding back thru the power supplies. I should have explicitly mentioned this, but that's what C3 and C4 in my circuit are intended to prevent. Try the circuit I posted. It uses mostly the same parts but should perform better.
You can also see evidence of the output stage deadband on the scope trace. Again, including the output stage in the feedback look will help with this, although it won't fix it.
I now see the opamp is a LM324. That's not a good choice for audio. I'd use a TL07x with ±12 V supplies at least. That will probably mean beefier output transistors, possibly with heat sinks.
Once you get this working, I can show you how to get more voltage swing and less deadband from the output stage, but one thing at a time. That would be for a new question anyway.
Best Answer
Plot the phases for the three voltages and you'll find that they're out of phase at the crossover points:
The problem here probably is the phase of the midrange driver. Focus on that. You should also take into account the actual phase of each of the three drivers (which are just 8 Ohm resistive loads in your simulation).
Here is the result assuming a simple phase reversal (180-deg) in the midrange driver:
And here you have the total power delivered to the 3 loads, as suggested by The Photon, where you can see the approx -6dB power down at the crossover frequencies:
Also note that you may be looking after two things that are quite different from each other:
If you want to predict the frequency response that you would measure with a microphone, you need to factor in the response of the drivers (you just can't add them together), their physical positioning, etc.
If you want to predict the frequency response of the overall load seen by the amplifier's output, then you need to replace the 8 Ohm loads by the equivalent impedance of each driver.
1 and 2 can be very different things to simulate and measure, and will be highly dependent on the characteristics of each driver.