Electronic – LC low pass filter: intuitive understanding

circuit analysiscircuit-designfilterfrequency responseresonance

The figure below is an LC low pass filter and its bode plot.
As we can see the ratio Vout/Vin is larger than 1 around the resonant frequency.
Is there an intuitive explanation why the ratio is larger then 1 here?

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Best Answer

Imagine you have a car steering wheel connected to a heavy flywheel by a flexible, stretchy rubber band. If you turn the steering wheel slowly the flywheel starts moving and, if you are turning at a constant rate, the flywheel catches up in speed and all the energy stored in the rubber band becomes transferred to the flywheel rotation.

An inductor is the rubber band and a capacitor is the flywheel. Constant speed on the steering wheel is a dc voltage.

Now, should you rotate the steering wheel back and forth at the "right" rate, the flywheel will also start to oscillate and, in the absence of friction losses (aka resistors), that flywheel oscillation will build in amplitude and continue to build until eventually something gives out. This would be called destructive resonance and happens in electric circuits too.

The math for the electrical case and the mechanical case is virtually the same.

Should you have attached a motor instead of the steering wheel and applied a step change from zero speed to so many rpm then the flywheel would accelerate until it reaches motor rpm then continue to accelerate until all the stored energy in the rubber band was extracted. Given very low losses, the flywheel would attain a speed of double the motor speed whereupon, it starts slowing down and dumping energy back into the rubber band. At some point in time later, the flywheel will decelerate to zero speed and the process will start again.

This cycle time represents the resonant frequency of the flywheel and rubber band. In mechanical terms it relates to mass and stiffness.