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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1\$\begingroup\$ It's all about the voltage across the inductor. The thing about inductors is that the current can't change instantaneously - something has to give otherwise the equations won't add up - so the voltage goes to a value to make everything work. This is why you can make big sparks with inductors \$\endgroup\$– WillMar 13, 2017 at 18:27
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\$\begingroup\$ This answer makes some sense. However, why this only happens with certain frequencies? \$\endgroup\$– emnhaMar 13, 2017 at 18:57
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\$\begingroup\$ You can think about the sine wave at various frequencies being delayed/advanced by capacitance/inductance and these effects creating constructive/destructive interference as the frequency changes. I think @biggidvs has a point on this, the phasor maths might well be the most intuitive way to understand what is happening. \$\endgroup\$– WillMar 13, 2017 at 19:37
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\$\begingroup\$ The oscillation mechanism implies endless energy swinging between L and C in a non-dissipative way if there is no ohmic loss (damping) as in the example. If you excite the circuit by bringing energy while the oscillating energy peaks - meaning exciting the network right at the resonance frequency -when i(t) is max in the inductor or v(t) is max across the cap, the energy in the circuit increases cycle by cycle and output amplitude grows. In theory, as you cancel the denominator at \$s=\omega_0\$, the magnitude of the TF should be infinite but given stray components in L and C, it is finite. \$\endgroup\$– Verbal KintJul 2, 2017 at 17:42
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4 Answers
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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.
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Consider your child on a swing.
Each time you give them a push you make them swing. If you add energy in resonance to the natural swing time, they will swing higher and higher.
It is the same with this circuit, at resonant frequency voltage/current "swings" between the inductor and the capacitor. If you apply a signal at the same frequency... it swings higher.
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It has to do with the math behind the filter. Simple answer is that the peak is when the imaginary portions of the impedance/admittance cancel each other out. This can cause a very large spike in voltage (hince the dotted lines and why there is no real max to it). Adding a resistor helps to dampen this down.
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\$\begingroup\$ the maths describes what happens, things don't happen because of maths.... \$\endgroup\$– WillMar 13, 2017 at 18:32
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\$\begingroup\$ I said it was a simple answer. You can read more here: en.wikipedia.org/wiki/Electrical_resonance \$\endgroup\$– biggi_Mar 13, 2017 at 18:35
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\$\begingroup\$ @biggidvs: I understand the math and why there is the peak from the transfer function. However, I am looking for an intuitive understanding of this. \$\endgroup\$– emnhaMar 13, 2017 at 18:53
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Just like a pendulum which stores energy in TWO modes: kinetic/motion and potential/height, so does this LC circuit store energy in TWO modes: magnetic fields (inductors) and electric fields (capacitor).
The pendulum has pivot friction and air friction to absorb energy.
The pure LC has no losses. Here is your circuit, with Q=1,000 (0.1% loss per radian)
and here is the frequency response
answered Mar 14, 2017 at 4:16