What are the resonant frequencies for this RLC circuit?

Question

What are the resonant frequencies in the following circuit?

Full disclosure: this is question 34 on page 655 of Pearson's Electronics Fundamentals 8th Ed.

I recognise the parallel section comprising \$C, R_{W_1}\$ and \$L_2\$ and compute a resonance at 2599Hz using \$f_r = \frac{1}{2\pi\sqrt{LC}}\$, and find \$V_{out} = 9.8\$V (given \$V_1 = 10V\$) at this frequency because the parallel section is equivalent to a single \$41671\Omega\$ resistor.

However, the question implies that there is more than one resonant frequency... what are the others and how are they deduced?

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

One way to determine the resonant frequencies is via simulation. You can apply a 1V source at various frequencies (e.g. start at 100Hz and increment). Then plot the current provided by the source vs. frequency (admittance plot). You can apply a 1A source as well, then plot the voltage at terminals of source vs. frequency (impedance plot).

I did both for your circuit in ATPDraw (gui for ATP) and they are plotted below (swept from 100Hz to 50kHz in 100Hz steps).

First one is impedance plot, since we are injecting 1A the vertical axis (which is labeled V) is equivalent to Z. Peak is around 2.9kHz (horizontal axis is frequency).

You can narrow it down once you have roughed out where the peaks are. Below is the plot of current vs. frequency when source is 1V. This is admittance plot.

UPDATE: Based on Tony Stewart's tutoring my plot below is Vout/Vin instead of input impedance/admittance like i show above. This one is logarithmic on horizontal axis.

Answer 2

The fun is in determining the transfer function (TF) symbolically otherwise you don't know what elements contribute to the resonant frequencies. You can use the brute-force analysis to determine this TF - good luck with that - or use the fast analytical circuits techniques or FACTs described in the book I published. By determining the time constants of this circuit by inspection (no algebra at all) and assembling them at the end, you determine the TF of this 3rd-order network.

The various time constants are here:

and the final transfer function is here:

So there are three zeroes and three poles. The more difficult exercise is to factor this beast in a second-order polynomial multiplied by a low-frequency pole and zero. Once this is done, you can determine analytically the resonant frequencies you want:

This is an approximated expression and you can see a dc attenuation of 48.2 dB and a notch appearing at 6 kHz.

Addendum

To show you the principle behind the small sketches, I have added below the steps I used to determine the denominator poles and the high-frequency gains. For the poles, you reduce the excitation \$V_{in}\$ to 0 V (replace the symbol by a short circuit) and you temporarily disconnect the considered energy-storing element. You then "look" through its connecting terminals to see the resistance \$R\$. Then, depending if it is a capacitor or an inductor, you have \$\tau=RC\$ or \$\tau=\frac{L}{R}\$. With passive circuits like in this example, you can inspect the circuit and simply find the resistance by reading the arrangement: no algebra and easy corrections if necessary.

For the numerator, you can either go through a null double injection (NDI) or determine high-frequency gains \$H\$. These high-frequency gains allow me to re-use the natural time constants (the \$RC\$ and \$\frac{L}{R}\$ values) to determine the numerator. Simply put, it is an excellent means to "see" if any energy-storing element contributes a zero: place the considered element in its high-frequency state (a short for a capacitor or an open circuit for an inductor) and check if there is a non-zero gain \$H\$ linking \$V_{in}\$ to \$V_{out}\$. If there is a gain, meaning that the stimulus in this mode can propagate and create a response, the tested element contributes zero. Without doing anything on this circuit, you can immediately count 3 zeros there.

The other difficult thing is to factor the raw polynomial into meaningful expressions where poles, zeroes and resonant frequencies appear. It is not always easy as factorization depends on the time constants and how they compare with each other. For instance, it seems possible to define three separated poles considering three distinct positions. The plots show a fairly good approximation with the full-blown expression:

I recommend to read this document written by Prof. Erickson from Colorado Edu which explains in details how to do it.

Answer 3

Any closed path, with series inductor(s) and series capacitor(s), will resonate.

In a simulation you may not see the resonance IF the Q is too low.

Note "closed path" is required, so the bimodal energy ( exchanging between electric field and magnetic field energy storage) is part of the circuit behavior.

To see these modes, make the Qs very high.

Change those 4 ohm resistors to 0.004 ohms.

And that resistor from the source also dampens --- make it 10x higher (if needed)

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resonant LOOPS are

• L1+C, though this loop is very damped by "R"

• L2+C

Answer 4

This only takes a few minutes to see where your and @relayman357 's error are.

Andy was correct about the series parallel effects.

The top L mainly controls the series pole but is low Q. The bottom L controls both pole and zero and the zero appears deep but -3dB is about the same for Pole and zero.

So the peak gain is at 3.1kHz and notch at 6.0 kHz but this not what you want.

The problem mainly is you did not specify or ask for what you should have wanted. i.e. a design spec.

e.g. Design Spec.

• fbp = 3kHz BW = tbd ?
• fbs = 6kHz BW = tbd @ -3dB and
  • BW = tbd at -xx tbd dB

Simulated Proof of Concept (without specs)