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Say you have this simple RF circuit : enter image description here

I "understand" that \$Z_0\$ must match \$Z_s\$ and \$Z_L\$ for maximum power transfer to the load. However, it has never been clear to me how to treat \$Z_0\$ in a circuit analysis. Is the voltage at the \$(Z_o,Z_L)\$ node $$V_{(Z_0,Z_l)}{=^?}\frac{V_in\cdot Z_L}{Z_s+Z_0+Z_L}$$

For example, in a \$50\Omega\$ system, this would make the voltage at \$V_{(Z_0,Z_l)}= \frac{V_{in}}{3}\$. I don't believe it is (and I can't build this right now to prove it to my self and I'm not familiar with transmission lines on Spice), but then I can't explain what \$Z_0\$ physically is --ie, it is an impedance, but not really!. If it were, then the load would have to be matched instead to \$Z_s+Z_0\$.

I'd appreciate it if someone could set me straight on this. I believe the answer is that if \$Z_0 = Z_s=Z_L\$, then it is acts as a lossless connection between \$Z_s\$ and \$Z_L\$, but I don't know that I can really explain this. I someone can help me understand this (preferably with math behind it) or refer me to additional literature on the topic, I would greatly appreciate it. I already reviewed transmission lines and impedance matching on Wikipedia and other sources, and if the explanation for this "specific" aspect of \$Z_0\$ was there, then I did not understand it or make the connection. So please, don't just refer me back to those websites without some additional clarification or explanation of a particular section or sections where this is discussed.

Thanks for the help.

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    \$\begingroup\$ Are you trying to operating point, DC, AC, or transient analysis in SPICE? \$\endgroup\$
    – The Photon
    Jun 27 at 18:12
  • \$\begingroup\$ Does this answer your question: How is xΩ impedance cable defined? \$\endgroup\$
    – The Photon
    Jun 27 at 18:15
  • \$\begingroup\$ Yes, it's an actual impedance in that it really does impede current flow and varies with frequency. But that doesn't mean it's easily or even statically measurable, nor does it mean the equipment to directly measure it is common. Perhaps the reason you can't explain what it physically is is that it is not lumped; it is distributed. You can't just point to it without just pointing to the wire as a whole. A bit like "is capacitance a real thing?" before anyone has built a discrete capacitor. \$\endgroup\$
    – DKNguyen
    Jun 27 at 18:56
  • \$\begingroup\$ For maximum power transfer to the load you need Zs to be zero and, it can be without fear of reflection. Current must begin travelling down a wire before it gets to sense what's at the far end hence, it's a real (but transient) impedance. \$\endgroup\$
    – Andy aka
    Jun 27 at 18:57
  • \$\begingroup\$ @Andy aka --Zs to be 0? can you explain? I thought it needs to be =ZL and to avoid reflections at the interface with the cable, it has to be = Z0 as well..... \$\endgroup\$
    – jrive
    Jun 27 at 19:06
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Is the voltage at the \$(Z_o,Z_L)\$ node $$V_{(Z_0,Z_l)}{=^?}\frac{V_in\cdot Z_L}{Z_s+Z_0+Z_L}$$

No, this equation doesn't apply, because the characteristic impedance of the transmission line is not an impedance connected between the two ends of the line.

As the other answers have said, and some previous Q&A's have explained, the characteristic impedance of the transmission line is the ratio of voltage to current that allows a wave to propagate along the line without reflections.

To work out the voltage at the load, in addition to the characteristic impedance of the line you also need to know its electrical length at the frequency of the signal you are applying.

From there you need to determine the four-port representation of the line in a convenient format, such as S-parameters, Z-parameters, or Y-parameters. Then you can construct a system of equations describing the system and solve for the voltage across the load.

Because there is a phase delay between the signal at the source end of the line and the load end of the line, there's no equation as simple as the one you proposed to find the load voltage. In some special cases (for example, either the source or load impedance is closely matched to the line impedance, or the length of the line is exactly one quarter wavelength, etc.) you might be able to find a simple equation, but in the general case it will be somewhat more complicated.

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  • \$\begingroup\$ why is the electrical length of the transmission line important? Isn't the characteristic impedance of a 50ohm coax 50ohm regardless of length and frequency? \$\endgroup\$
    – jrive
    Jun 28 at 0:35
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    \$\begingroup\$ The characteristic impedance of the line is independent of the length. But the effect in the circuit also depends on the length of the line. In any introductory electromagnetics textbook this material will be covered over 2 or 3 chapters. \$\endgroup\$
    – The Photon
    Jun 28 at 0:55
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The Z0 of the transmission line is only an impedance in the sense that it's a ratio between voltage and current.

A transmission line can support a wave in each direction. For that wave, the ratio of its voltage to its current is Z0. Always.

It may sometimes seem that this ratio is broken for a transmission line. For instance, connect a battery to an open circuit transmission line, and you appear to have finite voltage with no current. However, what you actually have is the superposition of a wave in each direction, with the voltages adding and the currents subtracting to give you an instantaneous voltage with zero current everywhere.

If the source happens to launch a wave with the voltage and current already in the line's impedance ratio, and if the sink can absorb a wave also with that ratio, then the transmission line is perfectly matched, and the source will transmit power to the load.

If on the other hand the load is mismatched, then it will reflect some of the signal

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A cable can be modelled with KVL using 0 Ohms or some small DC resistance for all frequencies from DC to about 1/10th of a wavelength of the cable length and also at 1/2 wavelength and all multiples of that frequency.

But at wavelengths in between those examples the transmission line acts as a conjugate impedance transformer. So at 1/4 wavelength with an open circuit you have a short circuit to the say 75 or 50 Ohm source. This means you have a high Q notch filter at that length. Then at 1/2 wave it can pass thru. But if the end resistance does not match then a reflection coefficient or Return Loss determines the attenuation.

  • the Zo has several effects depending on the wavelength of the signal and for KVL low frequency purposes, it is 0 and lossless matched impedance in the stead-state average it is 0 Ohms loss.

So the characteristic impedance only has an effect on cables ,longer than 1/10th of wavelength in time, time domain reflectometer and also 4 , spectral impedance & frequency response domains known as scattering or “s parameters”.

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