cutoff frequencyfiltergain
I'm doing some work on filters, but all the references I can find use a 'perfect' gain characteristic, with the trace starting at 0dB. Is the measurement for corner/cutoff frequency always taken at exactly -3dB? What if the maximum gain isn’t 0dB?
For example, see the below figure, where the maximum gain is -3dB. Surely taking the corner frequency at -3dB in this case would not be useful- instead would we take it at -6dB, which is -3dB from the actual maximum gain?
Thanks for your help.
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.
The problem with this prototype is in the absence of bypass capacitance on power rail. If a model for power lead of 30cm in length (400nH inductance) is added to LTspice simulations, the output does match observations - -3dB fall-off at 15 MHz. 
Best Answer
It's not 3dB absolute, it's 3dB down from the peak, or some sort of nominal attenuation. So in your case, where the passband is -3dB, 3dB down is at -6dB.
Note that some filters (e.g. Chebychev) have significant passband ripple; if this exceeds 3dB then the "3dB down" figure loses meaning. In that case, or just if it's what matters to the system designer, a different definition of bandwidth may be chosen.