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I was wondering why many generators ect. have 50 ohm output resistance.

Before you answer here is what I know:

Many transmission lines are 50 ohm, and terminated with a 50 ohms load to get maximum power transferred to the load and minimize reflection. So for the load it is obvious why 50 ohm is needed, but why for the source?

I would get more power transferred to the transmission line (or load), if the source impedance was say 10 ohms.

Is it to cancel any reflections hitting the source from the load?

Or does it just mean that the source has been matched to 50 ohms load?

EDIT: After searching google for hours, I found this link which explains my question: http://users.tpg.com.au/users/ldbutler/OutputLoadZ.htm

Conclusion: It usually means the source has been matched for 50 ohms. Sometimes in precision signal generators (not RF power amplifiers), the source has 50 ohms input impedance, this is to cancel re-reflections hitting the source from the load. But for RF power amplifiers and etc., the output has been matched for 50 ohms, since half the power would be lost inside the equipment, if the output impedance was actually 50 ohms.

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2 Answers 2

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You are apparently talking about signal generators, not power generators. The signal can't usually be used immediately at the signal generator output. That means it has to be somehow transmitted to where it is needed. Since the shape of the signal is important, the manufacturer of the signal generator has to assume you will want to use a carefully ballanced transmission line for getting the signal from the generator to the load. As you say, lower impedance is better, so they assume you are using 50 Ω coax, which is the lowest impedance commonly available high frequency cable. Matching the signal generator output to the transmission line impedance allows for the least loss of signal integrity and absorbs reflections at the signal generator end due to any mismatches at the other end.

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  • \$\begingroup\$ Whoever downvoted this, please explain what you think is wrong. I read what I wrote again, and still think it is correct. \$\endgroup\$
    – Olin Lathrop
    Apr 7, 2012 at 15:16
  • \$\begingroup\$ not, me but since your at zero perhaps it was a corrected misclick. \$\endgroup\$
    – kenny
    Apr 7, 2012 at 15:39
  • \$\begingroup\$ Sorry for the downvote, but your answer is a bit confusing. Check following comments. \$\endgroup\$
    – Juancho
    Apr 7, 2012 at 16:19
  • \$\begingroup\$ 50 Ohm is not the mimnimum impedance for t-lines. It is the most standard for RF and instrumentation. It is midpoint between minimum losses (75 Ohm, used by CATV) and max power transfer (30 Ohm), and thus became a standard. \$\endgroup\$
    – Juancho
    Apr 7, 2012 at 16:20
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    \$\begingroup\$ Signal integrity would mainly be relevant for (high speed) digital signals, not for modulated (RF) signals. For RF, you are dealing with a modulated sinusoidal carrier, and superposition of sinusoidals (due to reflections) is still a sinusoidal. In the case of RF you are mainly concerned with maximum power transfer and matched impedances throughout in order to have null reflection coefficients. ((you used the term balanced, where matched is more appropriate; balanced refers to bal/unbal lines)) \$\endgroup\$
    – Juancho
    Apr 7, 2012 at 16:24
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Maximum power is transferred when the source impedance = the load impedance.

So for a 50 ohm load you want a 50 ohm source and a 50 transmission line.

If you had a 75 ohm load you would want a 75 ohm transmission line and a 75 ohm source.

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  • \$\begingroup\$ Yes true, but that only applies when the source resistance is constant, we can get maximum power from the source by making the load equal to the source. But if I got a 50 ohms load, and use a 10 ohm source (with same power capability), I would get more power transferred to the load. \$\endgroup\$
    – JakobJ
    Apr 7, 2012 at 13:52
  • \$\begingroup\$ @JakobJ - how do you figure that a 10 ohm will deliver more power to a 50 ohm load than a 50 ohm source? Remember we are talking about power delivered to the load, not power wasted as heat in the transmission line due to high SWR losses. \$\endgroup\$
    – JonnyBoats
    Apr 7, 2012 at 13:55
  • \$\begingroup\$ Hmm. I maybe wrong here, but if a 50 ohm transmission line is terminated with a 50 ohm load, looking into the transmission line would seem just like a 50 ohm load. Now if I got a ideal voltage source with a 50 ohms resistance, I would loose half the power in the 50 ohm source resistor? If I got say 1 ohms source resistance, and same voltage source, I would get more power transferred to the 50 ohms load? \$\endgroup\$
    – JakobJ
    Apr 7, 2012 at 14:09
  • \$\begingroup\$ You should not confuse maximum power transfer to the load with maximum efficiency (i.e. not losing half the power at the source resistance). Max power transfer is achieved when source and load impedances are complex conjugates (or equal if they are real-valued). Max efficiency is when your source impedance is smallest. Check en.wikipedia.org/wiki/… \$\endgroup\$
    – Juancho
    Apr 7, 2012 at 16:11
  • \$\begingroup\$ @Juancho - Quote from the link you send: "The theorem states how to choose (so as to maximize power transfer) the load resistance, once the source resistance is given, not the opposite. It does not say how to choose the source resistance, once the load resistance is given. Given a certain load resistance, the source resistance that maximizes power transfer is always zero, regardless of the value of the load resistance." - Again I could be misunderstanding something, but this states my point exactly. If we can lower source resistance when constant load (50ohm), this we deliver more power.? \$\endgroup\$
    – JakobJ
    Apr 7, 2012 at 18:04

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