Impedance matching.

[Jimbz]

Hello everybody, small question about impedance matching,

As you know if you match an impedance source with the the same value load, you get maximum power transfer in high frequency applications, the source and load value is usually 50 ohms.

My question is more about the source, measuring equipment like network analyzers and the transmission lines.

We have:
1) Measuring equipment with 50 ohm impedance (more active and some reactive???).
2) Transmission cables with 50 ohm impedance (some active and more reactive, frequency dependant).
3) Load with 50 ohm impedance (full active).

I would like to know how does the measuring equipment match its source impedance to lload impedance.

Correct me if I make a mistake here, but is the process following: the load is a constant 50 ohms active, the cable is a constant some ohms active and rest reactive depending on the frequency. We need to sum up the cables and measuring instrument impedance to match our load ?? The only thing that can therefore change is the measuring equipment output (source), depending on the frequency, it will internally shift its active resistance, to make sure the that the source and cable sum remain 50 ohms ? Is this how its done or have I made a gigantic mistake somewhere ?

Thank you for reading

[Paul Price]

You've made a gigantic mistake somewhere.

[c4757p]

You've made a gigantic mistake somewhere.

Helpful.

Characteristic impedance isn't purely reactive, per se. It's the flow of energy into the cable as the signal physically travels down its length. The cable reactance helps to determine it, as the reactances of the cable tell how much energy is stored in the cable per unit voltage or current, but it does not end up being frequency-dependent. Ideally, a 50 ohm cable has a fixed 50 ohm characteristic impedance. (We all know what "ideal" means, I'm sure - there is some variation. Not much, though, over the proper operating bandwidth of the cable.)

Because it's all about propagation velocity, the same signal will never simultaneously encounter both cable and load, so you don't sum the impedances. A given time-point in the signal is either moving down the cable, or has popped out the end, never both. The signal transitions from one to the other. That's why you match them. This allows the energy wave moving down the cable to come out the end and dissipate into the load, rather than hitting a discontinuity where some reflects back.

[macboy]

...
Because it's all about propagation velocity, the same signal will never simultaneously encounter both cable and load, so you don't sum the impedances. A given time-point in the signal is either moving down the cable, or has popped out the end, never both. The signal transitions from one to the other. That's why you match them. This allows the energy wave moving down the cable to come out the end and dissipate into the load, rather than hitting a discontinuity where some reflects back.

Very nice description. 

[T3sl4co1l]

As for the way equipment achieves that match... usually it's with a big honking resistor.  What could be simpler?

RF equipment, where power efficiency is more important than strict impedance matching, typically does not have a dynamic impedance match.  That is, if you transmit power at the output port, expect to see some reflecting back.

For a possible analogy, consider molten metallic copper:
http://commons.wikimedia.org/wiki/File:Copper_just_above_its_melting_point.jpg#mediaviewer/File:Copper_just_above_its_melting_point.jpg
even though it is glowing (incandescent), it doesn't cease to be shiny and pink.  Under dim conditions, the self-light is too much to see that reflection, but under bright conditions, it's still obvious that it is shiny.  So, too, you can have electronic equipment which is outputting power, yet still reflective.

So, there is an important distinction between a pure resistive match, the dynamic impedance, and the "maximum power" impedance.

The most important, most pedestrian case, is an audio power amplifier: the amplifier is designed to achieve maximum power output into a 4 or 8 ohm load, but has a dynamic impedance of milliohms.  Matching to the milliohm value would indeed maximize your power output for a given output voltage (say, a millivolt), but the amplifier is limited to a few amperes before it ceases to be a linear source, so you'd only get milliwatts.

Amplifiers can also be designed for very high output impedances, in which case they resemble current sources; but again, too high a load resistance and, in this case, the source will run out of voltage before delivering much power.

Tim