We’re particularly and intentionally going to do with out the Smith chart right this moment and take a look at some transmission line properties from a really fundamental perspective.
Contemplate a transmission line, a coaxial cable only for the sake of the quick dialogue, of arbitrary size and divided right into a cascade of one-foot lengths that are additional divided into 2400 sectors of 0.005 inches every (Determine 1). (Notice: 2400*0.005 inches = 12 inches = 1 foot) These numbers are arbitrary, however for the sake of this dialogue, they are going to be helpful.
Determine 1 A hypothetical one foot of transmission line divided into 2400 sectors.
Now, let’s think about two simplified circuit fashions of this transmission line which is related to a load (Determine 2).
Determine 2 Two circuit fashions of the coaxial transmission line proven in Determine 1.
We have now two circuit fashions as a result of we’ve a selection on the far left to both finish the mannequin with the R2 C2 mixture or with the L1 R1 mixture. Notice that these two schematics solely present 5 sectors. The fashions may actually go off to the left eternally and eternally and…You get the concept.
The load is taken as a collection mixture of a resistance known as “R” in collection with a reactance known as “L” as if that reactance had been an inductor. Bodily, that reactance would in all probability be inductive with the numerical worth of L as some optimistic quantity, however mathematically, L is also a unfavorable quantity. That isn’t mathematically prohibited.
Within the following evaluation in Determine 3, primarily based on the higher schematic of Determine 2, we see an algebraic derivation of impedance Z1 within the type of a brand new collection mixture of a brand new worth of R and a brand new worth of L. As soon as we’ve these two new values of R and L for Z1, we are able to derive them for Z2 in the identical trend, then derive them for Z3, and so forth., and carry on going if we had been of a thoughts to take action. This could go on eternally.
Determine 3 Algebraic derivation of impedance Z1.
In brief, we are able to derive the enter impedance that will get offered on the left-hand finish of the road versus the road’s size for the higher schematic of Determine 2. It occurs to end up that utilizing the decrease schematic doesn’t actually have an effect on the top outcome by a lot, only a slight addition to the brand new R and the brand new L, so we are able to purposely ignore that schematic.
A typical 50 Ω coaxial cable, RG-58 sort for instance, is understood to have 30 pF of capacitance per foot between the middle conductor and the defend braid. The worth of C2 within the above fashions is then 30 pF / 2400 = 0.0125 pF per sector. The attribute impedance, name that Zo which we’ve already stated is 50Ω, yields an equation for L1 as L1 = Zo² * C1 = .03125 nH per sector.
These arbitrarily chosen 0.005-inch sectors are sufficiently small that we are able to calculate the Z1, Z2, Z3, and so forth., impedances with an inexpensive diploma of confidence. Within the following illustrations proven in Determine 4, a frequency was arbitrarily taken as 2.4 GHz, the free area wavelength could be very almost 300 / 2400 meters = 0.125 meters = 4.92 inches, for which our sector size of 0.005 inches is just 0.001016 of 1 free area wavelength, which is a lot adequate.
Beginning with a load R of 100 Ω, a load L of 0 Ω of reactance, and an idealized lossless line mannequin with R1 and R2 each at zero, we are able to draw a couple of plots of impedance versus cable size.
Determine 4 Impedance versus size of a lossless line.
Ranging from the load of R = 100 and L = 0, we comply with a clockwise round trajectory across the impedance chart. Notice that when the size equals one-quarter-wavelength as within the shaded sketch, the offered impedance is 25 + j0 which is the basic impedance transformation for a “quarter wave stub”.
We take additional word that for 0.06958 ft = 0.83496 inches = 0.0212 meters = ¼ wavelength at 2.4 GHz, the place at no cost area, one wavelength could be very almost 300 / 2400 meters = 0.125 meters. If we take the ratio of the bodily lengths of (4 * 0.0212) / 0.125, we get the cable’s velocity issue as 0.678, which could be very almost the recognized velocity issue of RG-58 sort cable.
As we go to higher lengths, we comply with the round trajectory over and again and again, eternally, which is to date so good, however now we take a look at a lossy cable.
We have now two loss elements in our mannequin: R1 and R2. The R2 was included within the derivation, however for the reason that conductors are proper up towards, and in touch with, the dielectric materials between them, I’ve chosen to depart R2 numerically as zero. The worth of R1 nevertheless, is one other matter.
We are able to let R1 be some non-zero worth and run our repetitive impedance calculations once more. Once we achieve this, one thing completely different occurs as seen in Determine 5.
Determine 5 Impedance versus size for a really lossy line.
Letting R1 rise from 0 Ω to 0.1 Ω, we see the previously round trajectory of impedance as a operate of size develop into a spiral. Whereas the 0.1 Ω worth is absolutely too excessive when in comparison with a real-world cable, it provides a graphical outcome that demonstrates the idea.
Notice that the spiral converges on a resistance part that could be very almost the attribute impedance of the cable, but additionally word on this sketch, that the convergence level’s reactive part doesn’t go to zero.
As proven in Determine 6, I let the non-zero R1 develop into a lot smaller, happening to 0.001 Ω, and let the size be a lot higher, going as much as fifty ft as I’ve chosen, the spiral converges inward far more slowly, and the convergence level will get a lot nearer to 50 + j0.
Determine 6 Impedance versus size for much less and fewer lossy traces.
From a sensible standpoint, this evaluation and diagramming confirms that the offered impedance of a lossy, very lengthy cable converges fairly carefully on the worth of that cable’s attribute impedance.
A lot ado about one thing (With apologies to the Bard.) however with it we come to a easy and well-known conclusion.
John Dunn is an electronics advisor, and a graduate of The Polytechnic Institute of Brooklyn (BSEE) and of New York College (MSEE).
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