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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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?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 3rdorder/4thorder 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 "hightemperature" superconductor  90 Kelvin, roughly), the normal conductance can be up to 1.52 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 geometrybased (like a normal wire), though with some added temperature dependency. So superconductors in AC are not lossfree, 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.
Hope this answers your question!
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?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 3rdorder/4thorder 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 "hightemperature" superconductor  90 Kelvin, roughly), the normal conductance can be up to 1.52 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 geometrybased (like a normal wire), though with some added temperature dependency. So superconductors in AC are not lossfree, 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.
Hope this answers your question!
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 VCOstyle applications, so I did a lot of research on the limitations of superconductors relatively recently (within the past year).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?
When the numerator contains a constant, it means that the transfer function has no zeros, only poles.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?
Yes, I didn't simulate to a low enough frequency (modified below):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
To determine poles/zeros, you generally need to put everything over a common denominator. In the case of your system, your total impedance is: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?
Nice thanks! In other words my mistake is not reducing the numerator and denominator to simply polynomials?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:
\[ j\frac{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2}{\omega\left(1\omega^2\right)} \]
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 \(Z_{tank}\)):
\[ H = \frac{V_{out}}{V_{in}} = \frac{Z_{tank}}{Z_T} = \frac{j\frac{\omega}{1\omega^2}}{j\frac{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2}{\omega\left(1\omega^2\right)}} = \frac{\omega^2}{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2} = \frac{\omega^2}{\omega^2\left(1\omega^2\right)^2} \]
So the transfer function itself is actually a fourthorder pole and a secondorder zero. If we expand the denominator we get:
\[ H = \frac{\omega^2}{\omega^43\omega^2+1} \]
If we let \(\Omega \equiv \omega^2\), then the bottom becomes an easytosolve quadratic (we just gotta remember that we'll have both positive and negative solutions for \(\omega\). The pole equation becomes \(\Omega^23\Omega+1\), 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 \(\omega=\pm\frac{1}{\phi}=\pm\left(\phi1\right)\), 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 11/e and 21/e in Excel are actually approximates of the golden ratio: 11/e = 0.6321, while \(\phi^{1}=\phi1= 0.6180\). Similarly, 21/e = 1 + (1+1/e) = 1.6321, and \(\phi = 1+\phi^{1} = 1.6180\), so that's kind of neat.
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.Nice thanks! In other words my mistake is not reducing the numerator and denominator to simply polynomials?
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: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:
\[ j\frac{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2}{\omega\left(1\omega^2\right)} \]
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 \(Z_{tank}\)):
\[ H = \frac{V_{out}}{V_{in}} = \frac{Z_{tank}}{Z_T} = \frac{j\frac{\omega}{1\omega^2}}{j\frac{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2}{\omega\left(1\omega^2\right)}} = \frac{\omega^2}{\left(\omega^21\right)\left(1\omega^2\right)+\omega^2} = \frac{\omega^2}{\omega^2\left(1\omega^2\right)^2} \]
So the transfer function itself is actually a fourthorder pole and a secondorder zero. If we expand the denominator we get:
\[ H = \frac{\omega^2}{\omega^43\omega^2+1} \]
If we let \(\Omega \equiv \omega^2\), then the bottom becomes an easytosolve quadratic (we just gotta remember that we'll have both positive and negative solutions for \(\omega\). The pole equation becomes \(\Omega^23\Omega+1\), 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 \(\omega=\pm\frac{1}{\phi}=\pm\left(\phi1\right)\), 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 11/e and 21/e in Excel are actually approximates of the golden ratio: 11/e = 0.6321, while \(\phi^{1}=\phi1= 0.6180\). Similarly, 21/e = 1 + (1+1/e) = 1.6321, and \(\phi = 1+\phi^{1} = 1.6180\), so that's kind of neat.
by Jake Hertz
by Jake Hertz
by Kate Smith