# LC circuit analysis using perfect inductors and capacitors

#### dcbingaman

Joined Jun 30, 2021
720
Granted you would never see this in the real world. Maybe if we use superconducting wires for the inductors? Does anyone see any mistakes in the math? The results seem rather interesting. See attached.

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#### ZCochran98

Joined Jul 24, 2018
214
The math appears to be correct. What you are seeing is a pole at $$\omega = 0$$ and $$\omega = \pm1$$, which causes the circuit (effectively a 3rd-order/4th-order filter [3 finite poles, 4 zeros]) to act as an open circuit, so you wouldn't be able to drive power through it (infinite power dissipation). In the real world, as you know, there would be a resistive element, which would allow some current, but not much. How good your filter would be at stopping that flow is the quality factor (which also controls how narrow that point is).

In regards to your question about superconductors: if superconductors are operating in DC, they act kind of as you expect, but not as they are commonly thought of. To clarify: superconductors are not perfect conductors; they have a finite conductivity (called the "normal conductivity"), which dictates the maximum current they can carry (they also have parameters like the critical current and critical magnetic field that, once passed, force the superconductor to stop acting as a superconductor and instead transition to a normal conductor). In YBCO (a "high-temperature" superconductor - 90 Kelvin, roughly), the normal conductance can be up to 1.5-2 MS/m, which is slightly better than steel (1.2 MS/m - not a good conductor), but not as good as titanium (2.4 MS/m), and is nowhere near as conductive as gold, copper, or silver (44, 58, 62 MS/m, respectively).

In AC, superconductors get even weirder due to how they work. Effectively, in AC, they obtain what appears to be a surface resistance that increases cubically (in normal metals, surface resistance increases with the square root of frequency, so superconductors' resistance grows much more rapidly!). The superconductor also gets a tiny inductance, though its value is independent of frequency (it's called the "kinetic inductance"), and is mostly geometry-based (like a normal wire), though with some added temperature dependency. So superconductors in AC are not loss-free, sadly, and can start having really bizarre effects.

Basically, to perfectly realize this and see this behavior in the real world, we would need a perfect conductor, rather than a superconductor. Alas, those do not exist except in simulation. Superconductors are lossless at DC (to a limited degree), but not at AC.

#### dcbingaman

Joined Jun 30, 2021
720
The math appears to be correct. What you are seeing is a pole at $$\omega = 0$$ and $$\omega = \pm1$$, which causes the circuit (effectively a 3rd-order/4th-order filter [3 finite poles, 4 zeros]) to act as an open circuit, so you wouldn't be able to drive power through it (infinite power dissipation). In the real world, as you know, there would be a resistive element, which would allow some current, but not much. How good your filter would be at stopping that flow is the quality factor (which also controls how narrow that point is).

In regards to your question about superconductors: if superconductors are operating in DC, they act kind of as you expect, but not as they are commonly thought of. To clarify: superconductors are not perfect conductors; they have a finite conductivity (called the "normal conductivity"), which dictates the maximum current they can carry (they also have parameters like the critical current and critical magnetic field that, once passed, force the superconductor to stop acting as a superconductor and instead transition to a normal conductor). In YBCO (a "high-temperature" superconductor - 90 Kelvin, roughly), the normal conductance can be up to 1.5-2 MS/m, which is slightly better than steel (1.2 MS/m - not a good conductor), but not as good as titanium (2.4 MS/m), and is nowhere near as conductive as gold, copper, or silver (44, 58, 62 MS/m, respectively).

In AC, superconductors get even weirder due to how they work. Effectively, in AC, they obtain what appears to be a surface resistance that increases cubically (in normal metals, surface resistance increases with the square root of frequency, so superconductors' resistance grows much more rapidly!). The superconductor also gets a tiny inductance, though its value is independent of frequency (it's called the "kinetic inductance"), and is mostly geometry-based (like a normal wire), though with some added temperature dependency. So superconductors in AC are not loss-free, sadly, and can start having really bizarre effects.

Basically, to perfectly realize this and see this behavior in the real world, we would need a perfect conductor, rather than a superconductor. Alas, those do not exist except in simulation. Superconductors are lossless at DC (to a limited degree), but not at AC.

You seem really knowledgeable in this area. Thanks for the excellent information on superconductors. It appears there is a lot to know about them and their limitations. Are you a physicist?

#### dcbingaman

Joined Jun 30, 2021
720
The math appears to be correct. What you are seeing is a pole at $$\omega = 0$$ and $$\omega = \pm1$$, which causes the circuit (effectively a 3rd-order/4th-order filter [3 finite poles, 4 zeros]) to act as an open circuit, so you wouldn't be able to drive power through it (infinite power dissipation). In the real world, as you know, there would be a resistive element, which would allow some current, but not much. How good your filter would be at stopping that flow is the quality factor (which also controls how narrow that point is).

In regards to your question about superconductors: if superconductors are operating in DC, they act kind of as you expect, but not as they are commonly thought of. To clarify: superconductors are not perfect conductors; they have a finite conductivity (called the "normal conductivity"), which dictates the maximum current they can carry (they also have parameters like the critical current and critical magnetic field that, once passed, force the superconductor to stop acting as a superconductor and instead transition to a normal conductor). In YBCO (a "high-temperature" superconductor - 90 Kelvin, roughly), the normal conductance can be up to 1.5-2 MS/m, which is slightly better than steel (1.2 MS/m - not a good conductor), but not as good as titanium (2.4 MS/m), and is nowhere near as conductive as gold, copper, or silver (44, 58, 62 MS/m, respectively).

In AC, superconductors get even weirder due to how they work. Effectively, in AC, they obtain what appears to be a surface resistance that increases cubically (in normal metals, surface resistance increases with the square root of frequency, so superconductors' resistance grows much more rapidly!). The superconductor also gets a tiny inductance, though its value is independent of frequency (it's called the "kinetic inductance"), and is mostly geometry-based (like a normal wire), though with some added temperature dependency. So superconductors in AC are not loss-free, sadly, and can start having really bizarre effects.

Basically, to perfectly realize this and see this behavior in the real world, we would need a perfect conductor, rather than a superconductor. Alas, those do not exist except in simulation. Superconductors are lossless at DC (to a limited degree), but not at AC.

From what I understand to determine the 'poles' in the filter you take the denominator set it equal to zero and solve for all w that gives zero and for the zero's you take the numerator set it equal to zero and apply the same principle. I can see the poles you mention as all three 0,1,-1 for w gives zero in the denominator. I am not sure how you got the zero's? Obviously I am wrong about that as the numerator is a constant at 1 and does not have w in it. What is the proper way to determine the poles and zeros in a LC circuit?

#### crutschow

Joined Mar 14, 2008
30,763
Below is the LTspice simulation Bode Plot amplitude from 0.1Hz to 0.5Hz:

#### dcbingaman

Joined Jun 30, 2021
720
Below is the LTspice simulation Bode Plot amplitude from 0.1Hz to 0.5Hz:

View attachment 246280
What is interesting about your data and the one I did in excel. There are two poles that happen at:
w=1-(1/e)
and
w=2-(1/e)

The infamous e showing up again.

Last edited:

#### ZCochran98

Joined Jul 24, 2018
214
You seem really knowledgeable in this area. Thanks for the excellent information on superconductors. It appears there is a lot to know about them and their limitations. Are you a physicist?
I am both a physicist and electrical engineer. My Master's thesis was in EE, but focused on a (slightly impractical and unfeasible) superconductor/GNRFET hybrid system for generating extremely high frequency sinusoids for VCO-style applications, so I did a lot of research on the limitations of superconductors relatively recently (within the past year).

#### Papabravo

Joined Feb 24, 2006
19,262
From what I understand to determine the 'poles' in the filter you take the denominator set it equal to zero and solve for all w that gives zero and for the zero's you take the numerator set it equal to zero and apply the same principle. I can see the poles you mention as all three 0,1,-1 for w gives zero in the denominator. I am not sure how you got the zero's? Obviously I am wrong about that as the numerator is a constant at 1 and does not have w in it. What is the proper way to determine the poles and zeros in a LC circuit?
When the numerator contains a constant, it means that the transfer function has no zeros, only poles.

#### crutschow

Joined Mar 14, 2008
30,763
What is interesting about your data and the one I did in excel. There are two poles that happen at:
w=1-(1/e)
and
w=2-(1/e)

The infamous e showing up again.View attachment 246284
Yes, I didn't simulate to a low enough frequency (modified below):

#### ZCochran98

Joined Jul 24, 2018
214
From what I understand to determine the 'poles' in the filter you take the denominator set it equal to zero and solve for all w that gives zero and for the zero's you take the numerator set it equal to zero and apply the same principle. I can see the poles you mention as all three 0,1,-1 for w gives zero in the denominator. I am not sure how you got the zero's? Obviously I am wrong about that as the numerator is a constant at 1 and does not have w in it. What is the proper way to determine the poles and zeros in a LC circuit?
To determine poles/zeros, you generally need to put everything over a common denominator. In the case of your system, your total impedance is:
$j\left(\omega-\frac{1}{\omega}+\frac{1}{\frac{1}{\omega}-\omega}\right)$
But this can be turned into:
$j\frac{\omega^2\left(1-\omega^2\right)-\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Which can be simplified to:
$j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Now, the actual transfer function is going to be a little messier, so the final pole/zero location may be different. $$H=V_{out}/V_{in}$$, so, with our voltage divider here, that becomes (with the impedance of the capacitor and inductor in parallel being called $$Z_{tank}$$):
$H = \frac{V_{out}}{V_{in}} = \frac{Z_{tank}}{Z_T} = \frac{j\frac{\omega}{1-\omega^2}}{j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}} = \frac{\omega^2}{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2} = \frac{\omega^2}{\omega^2-\left(1-\omega^2\right)^2}$
So the transfer function itself is actually a fourth-order pole and a second-order zero. If we expand the denominator we get:
$H = -\frac{\omega^2}{\omega^4-3\omega^2+1}$
If we let $$\Omega \equiv \omega^2$$, then the bottom becomes an easy-to-solve quadratic (we just gotta remember that we'll have both positive and negative solutions for $$\omega$$. The pole equation becomes $$\Omega^2-3\Omega+1$$, which has solutions:
$\Omega = \frac{3\pm\sqrt{5}}{2}$
Solving then for $$\omega$$ gives $$\omega = \pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$. This can actually be simplified: $$\frac{3\pm\sqrt{5}}{2}\ = \left(\frac{\sqrt{5}\pm1}{2}\right)^2$$. This latter ratio should look familiar: it's the golden ratio, $$\phi$$! So we can say $$\omega=\pm\phi$$ and $$\omega=\pm\frac{1}{\phi}=\pm\left(\phi-1\right)$$, which corresponds to poles at roughly 258 mHz and 98.4 mHz, which is what @crutschow's simulation shows.
Your observation of seeing peaks at 1-1/e and 2-1/e in Excel are actually approximates of the golden ratio: 1-1/e = 0.6321, while $$\phi^{-1}=\phi-1= 0.6180$$. Similarly, 2-1/e = 1 + (1+1/e) = 1.6321, and $$\phi = 1+\phi^{-1} = 1.6180$$, so that's kind of neat.

#### dcbingaman

Joined Jun 30, 2021
720
To determine poles/zeros, you generally need to put everything over a common denominator. In the case of your system, your total impedance is:
$j\left(\omega-\frac{1}{\omega}+\frac{1}{\frac{1}{\omega}-\omega}\right)$
But this can be turned into:
$j\frac{\omega^2\left(1-\omega^2\right)-\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Which can be simplified to:
$j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Now, the actual transfer function is going to be a little messier, so the final pole/zero location may be different. $$H=V_{out}/V_{in}$$, so, with our voltage divider here, that becomes (with the impedance of the capacitor and inductor in parallel being called $$Z_{tank}$$):
$H = \frac{V_{out}}{V_{in}} = \frac{Z_{tank}}{Z_T} = \frac{j\frac{\omega}{1-\omega^2}}{j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}} = \frac{\omega^2}{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2} = \frac{\omega^2}{\omega^2-\left(1-\omega^2\right)^2}$
So the transfer function itself is actually a fourth-order pole and a second-order zero. If we expand the denominator we get:
$H = -\frac{\omega^2}{\omega^4-3\omega^2+1}$
If we let $$\Omega \equiv \omega^2$$, then the bottom becomes an easy-to-solve quadratic (we just gotta remember that we'll have both positive and negative solutions for $$\omega$$. The pole equation becomes $$\Omega^2-3\Omega+1$$, which has solutions:
$\Omega = \frac{3\pm\sqrt{5}}{2}$
Solving then for $$\omega$$ gives $$\omega = \pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$. This can actually be simplified: $$\frac{3\pm\sqrt{5}}{2}\ = \left(\frac{\sqrt{5}\pm1}{2}\right)^2$$. This latter ratio should look familiar: it's the golden ratio, $$\phi$$! So we can say $$\omega=\pm\phi$$ and $$\omega=\pm\frac{1}{\phi}=\pm\left(\phi-1\right)$$, which corresponds to poles at roughly 258 mHz and 98.4 mHz, which is what @crutschow's simulation shows.
Your observation of seeing peaks at 1-1/e and 2-1/e in Excel are actually approximates of the golden ratio: 1-1/e = 0.6321, while $$\phi^{-1}=\phi-1= 0.6180$$. Similarly, 2-1/e = 1 + (1+1/e) = 1.6321, and $$\phi = 1+\phi^{-1} = 1.6180$$, so that's kind of neat.
Nice thanks! In other words my mistake is not reducing the numerator and denominator to simply polynomials?

#### ZCochran98

Joined Jul 24, 2018
214
Nice thanks! In other words my mistake is not reducing the numerator and denominator to simply polynomials?
Basically. To do transfer funciton analysis, you need to make it so everything is over a common denominator, which is generally a LOT of tedious math, but you get "nice"-looking polynomials generally by the end that give you all the info you'll need.

#### dcbingaman

Joined Jun 30, 2021
720
To determine poles/zeros, you generally need to put everything over a common denominator. In the case of your system, your total impedance is:
$j\left(\omega-\frac{1}{\omega}+\frac{1}{\frac{1}{\omega}-\omega}\right)$
But this can be turned into:
$j\frac{\omega^2\left(1-\omega^2\right)-\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Which can be simplified to:
$j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}$
Now, the actual transfer function is going to be a little messier, so the final pole/zero location may be different. $$H=V_{out}/V_{in}$$, so, with our voltage divider here, that becomes (with the impedance of the capacitor and inductor in parallel being called $$Z_{tank}$$):
$H = \frac{V_{out}}{V_{in}} = \frac{Z_{tank}}{Z_T} = \frac{j\frac{\omega}{1-\omega^2}}{j\frac{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2}{\omega\left(1-\omega^2\right)}} = \frac{\omega^2}{\left(\omega^2-1\right)\left(1-\omega^2\right)+\omega^2} = \frac{\omega^2}{\omega^2-\left(1-\omega^2\right)^2}$
So the transfer function itself is actually a fourth-order pole and a second-order zero. If we expand the denominator we get:
$H = -\frac{\omega^2}{\omega^4-3\omega^2+1}$
If we let $$\Omega \equiv \omega^2$$, then the bottom becomes an easy-to-solve quadratic (we just gotta remember that we'll have both positive and negative solutions for $$\omega$$. The pole equation becomes $$\Omega^2-3\Omega+1$$, which has solutions:
$\Omega = \frac{3\pm\sqrt{5}}{2}$
Solving then for $$\omega$$ gives $$\omega = \pm\sqrt{\frac{3\pm\sqrt{5}}{2}}$$. This can actually be simplified: $$\frac{3\pm\sqrt{5}}{2}\ = \left(\frac{\sqrt{5}\pm1}{2}\right)^2$$. This latter ratio should look familiar: it's the golden ratio, $$\phi$$! So we can say $$\omega=\pm\phi$$ and $$\omega=\pm\frac{1}{\phi}=\pm\left(\phi-1\right)$$, which corresponds to poles at roughly 258 mHz and 98.4 mHz, which is what @crutschow's simulation shows.
Your observation of seeing peaks at 1-1/e and 2-1/e in Excel are actually approximates of the golden ratio: 1-1/e = 0.6321, while $$\phi^{-1}=\phi-1= 0.6180$$. Similarly, 2-1/e = 1 + (1+1/e) = 1.6321, and $$\phi = 1+\phi^{-1} = 1.6180$$, so that's kind of neat.
I updated my document per your changes. I had to work through all the algebra and displayed all the steps, but it sure was a pain to verify it. How you ever knew that the solutions had a square that equaled the solutions is beyond me. But I did back check it in the math by carrying it out, and sure enough you are 100% correct. I was not familiar with the golden ratio until now. Looked it up on Wikipedia as well. Here is my updated document:

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#### ZCochran98

Joined Jul 24, 2018
214
Figuring out the bit about the golden ratio was partially out of familiarity with the ratio, a flash of intuition, and the observation that when I checked what your estimate of 2-1/e and 1-1/e were, I noticed it looked close-ish to the golden ratio. So when I got to the simplification step, I, out of curiosity, decided to see if $$\frac{3+\sqrt{5}}{2} = \phi^2$$, and was pleased to see that it was. I wouldn't say I was clever enough to be able to derive it in the first place. But I suppose a way we could derive it from scratch is to say:
$\frac{\sqrt{5}\pm3}{2} = (a\pm b)^2 \rightarrow a^2\pm 2ab+b^2$
If I assume, for simplicity of the math, that the sum of the square stuff gives a rational number and the 2ab gives some fraction of a root, then I can check by a (mildly-annoying) set of equations:
$a^2+b^2 = \frac{3}{2}, \pm2ab=\pm\frac{\sqrt{5}}{2}$
$4ab = \sqrt{5} \rightarrow b = \frac{\sqrt{5}}{4a} \rightarrow b^2 = \frac{5}{16a^2}$
$a^2+\frac{5}{16a^2} = \frac{3}{2}\rightarrow 16a^4+5=24a^2\rightarrow 16a^4-24a^2+5 = 0$
$A\equiv a^2 \rightarrow 16A^2 - 24A + 5 = 0 \rightarrow A = \frac{24\pm\sqrt{576-320}}{32} = \frac{3\pm2}{4}\rightarrow a=\pm\frac{\sqrt{3\pm2}}{2}$
So $$a = \frac{1}{2}$$ or $$\frac{\sqrt{5}}{2}$$ (we'll ignore the negative version for now). Solving for $$b$$ will get you:
$b = \frac{\sqrt{5}}{4\cdot \frac{1}{2}} = \frac{\sqrt{5}}{2}, b=\frac{\sqrt{5}}{4\cdot\frac{\sqrt{5}}{2}}=\frac{1}{2}$
Basically, whichever $$a$$ you use, $$b$$ will be the opposite one. Combining the two, we end up with:
$\frac{\sqrt{5}\pm3}{2}=(a\pm b)^2= \left(\frac{\sqrt{5}}{2}\pm\frac{1}{2}\right)^2$
Using the negative version of $$a$$ doesn't change the answer (you would end up with $$(-a\mp b)^2 = (-(a\pm b))^2 = (a\pm b)^2$$), and nor does swapping $$a$$ and $$b$$, as $$(a\pm b)^2 = (b\pm a)^2$$.

So, that's how I would have derived that surprising relationship had I not initially seen your numerical result and made an educated guess based on it.

Tl;dr: initially it was a lucky guess based on intuition, but actually deriving/proving it is a little longer, requiring some arithmetic gymnastics.