passive-networkssystemresponse
Ive got a question about identifying the overdamped, underdamped, and critically damped waveforms of an RLC circuit.
From my (very basic understanding), underdamped decays while oscillating. Critically damped decays the fastest without oscillating, and the overdamped decays without oscillating (but critically damped decreases faster)
In the question below, I feel like the the graph the key shows as overdamped is the critically damped response since it decays the fastest. Which is correct?
What you have is a series RLC circuit. One of the most common introductory circuit in EE. For example, look under Series RLC Circuit in this wikipedia page:
en.wikipedia.org/wiki/RLC_circuit
These are equations copied from the same page for the under-damped case. Given R, C along with measured \$\omega_d\$, L can be solved from these equations. (You don't even need to measure the attenuation factor \$\alpha\$.)
If the RLC do want to oscillate, the node where D1 cathode is connected will go negative. Therefore, depending on how D1 is driven, it may conduct during the negative half cycle and kill the oscillation.
There are quite a few subtle differences between band-pass and low/high filters but, for a simple LCR band-pass filter, damped and un-damped resonant frequencies are the same numerically and formulaically, \$\dfrac{1}{2\pi\sqrt{LC}}\$.
This is also the (un-damped) natural resonant frequency for high/low pass filters.
When damping is added, the natural frequency stays the same but it can (eg for a low pass filter) rotate anticlockwise in the pole zero plane and this leaves (in the jw axis) what is known as the damped resonant frequency, \$\omega_d = \omega_n\sqrt{1 - \zeta^2}\$. See lower part of 1st set of pictures below (it's basically Pythagoras and right-angle triangles).
And for low pass filters, there is the frequency at which peaking occurs and this is slightly different to damped resonant frequency, \$\omega_p = \omega_n\sqrt{1 - 2\zeta^2}\$.
The amplitude that this peaks at is \$\dfrac{1}{2\zeta\sqrt{1-\zeta^2}}\$
The proof of these three different frequencies for a low-pass filter is relatively straightforward but a little long winded. Below is an extract of a design paper for a 2nd order low pass filter where it is shown that a varying Q-factor moved the "peaking" frequency away from the natural frequency (100 Hz in this example) but, as always Q is the value of the peak at the natural resonant frequency: -
Maybe a close up view will be more exciting: -
Best Answer
Only graph 1 can be critically damped since there is an undershoot, but no subsequent overshoot.
Assuming the capacitor has an initial condition, then the voltage across the three components in parallel is:
\$v=-\frac{1}{C}\large \int \small I_C \: dt \: =L\large\frac{dI_L}{dt}\small \:=R\left(I_C-I_L\right) \$
Solving the equations simultaneously for \$\small I_L\$ gives:
\$\large \frac{d^2I_L}{dt^2}\small +\large\frac{1}{RC}\frac{dI_L}{dt}\small +\large\frac{1}{LC}\small I_L\$=0
Let \$\small LC=1\$, i.e. natural frequency = 1 rad/sec, for simplicity, but without losing generality.
Then, for critically damped (equal roots) the solution is of the form:
\$\small I_L=(A+Bt)e^{-t}\$, where \$\small A\$ and \$\small B\$ are constants.
The voltage across the inductor (and R and C) is \$v=\small L\large \frac{dI_L}{dt}\$, giving:
\$v=\small L(B-A-Bt)e^{-t}\$
Which has the characteristic shape of Graph 1