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in this question i tried to find the solution like this but i could'nt get the success

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What's mistake i am doing can anyone help?

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    \$\begingroup\$ Transmission lines are measured in propagation delays, not in meters. \$\endgroup\$ – Ale..chenski Aug 29 '17 at 21:07
  • \$\begingroup\$ 0.615 is \$\beta\ell\$. What about \$\ell\$? \$\endgroup\$ – Vladimir Cravero Aug 29 '17 at 21:08
  • \$\begingroup\$ @VladimirCravero L will come as 0.97 \$\endgroup\$ – Rohit Aug 29 '17 at 21:14
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    \$\begingroup\$ Is it realistic that Zin would be 50 ohms with a 100-ohm load? \$\endgroup\$ – davidmneedham Aug 29 '17 at 21:15
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    \$\begingroup\$ So the actual question is "how to minimize VSWR in an intentionally mismatched system by tuning the length of the line"? \$\endgroup\$ – rackandboneman Aug 30 '17 at 2:32
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The issue here is that you can't fully match your 50 ohm internal impedance to the 100 ohm load with just some length of a transmission line. You need a stub, or a cap, or an inductor too, in conjunction with the transmission line, in order to get the 50 ohm Zin you want.

This may not be quite visible with the Zin equation, but if you were to see this on the Smith chart, you'd soon realize that a transmission line by itself won't get you the 50 ohm you're looking for.

I don't know if you're familiar with the Smith chart, but take a look at the following:

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The point "2.0" represents on that diagram your 100 ohm load (All impedances are normalized with respect to a 50 ohm system then 100 ohm/50 ohm = 2). By adding some length of a transmission line between your load and your generator, you rotate in a circle, you're transforming your impedance (that would be Zin in your problem). The center of the circle is the point marked as "1.0"

Now take a look at the following chart:

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You see that by just rotating in a circle (which is adding length of a TL between the load and the source) I never get it to exactly cross through the 1.0 point, which is the center and the 50 ohm Zin you're looking for. The different points you see on the circle are Zin for different lengths of a TL. The all represent a complex number with real and imaginary non-zero parts.

You need something else to get the 50 ohm full match (maybe a cap or inductor in parallel together with the TL). There are techiniques to accomplish this, but that's beyond the scope of what you've seen so far.

Hopefully that explains a bit why you won't get a 50 ohm match out of the TL alone. So technically speaking, you could still find which om of those lengths given as possible answers results in the best case scenario for the power, but if you're trying to get it according to the maximum power theorem (Zinternal=Zin), you're out of luck with just the TL to fully match the system.

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It's a trick question. The VSWR, mismatch and power transfer are the same for any length of line.

A 100 Ohm load has an VSWR to a 50 Ohm source, of 2:1. After 1/4 a wavelength of line, the 100 Ohms has been transformed to 25 Ohms. Still 2:1 VSWR to the 50 Ohm source. After 1/2 wavelength, 5 m, it's back to 100 ohms. The same for all multiples of a quarter wave, and for any other line length, just the impedance will be somewhat capacitive or inductive.

Maybe the examiner meant to specify a 100 Ohm line and a 50 Ohm load, then there would be fewer correct answers.

Now to answer your actual question:

You are solving the wrong equation. For perfect maximum power transfer \$Z_L=Z_S^*\$, (you forgot the conjugate). Divide both sides by 50, multiply out the denominator. It's clear the equation has no solution. This is because No length of line will achieve perfect matching. Your starting assumption is wrong.

You are actually looking for the maximum power transfer that occurs over length, so you need to:

  • write an expression for power transfer, which would be the magnitude of \$\Gamma={(Z_L-Z_S)}/{(Z_L+Z_S)}\$ where \$Z_L\$ is the transmission line equation you used above, and \$Z_S\$ is the source impedance, 50.
  • or convert \$\Gamma\$ to VSWR, which is almost the same thing
  • simplify if you dare
  • differentiate over \$l\$, or even just over your \$x\$

You should find that \$|\Gamma|\$ does not depend on line length, if you're lucky you might remove the \$tan(\beta l)\$ entirely in the simplification.

Hence all answers have the same power transfer.

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  • \$\begingroup\$ "After 1/2 a wavelength of line, the 100 Ohms has been transformed to 25 Ohms." Would that not be what would happen at 1/4? \$\endgroup\$ – rackandboneman Aug 30 '17 at 16:06
  • \$\begingroup\$ rackandboneman yes, 1/4 wave. Corrected. @user287001 no, it really is that simple. Break the problem anywhere convenient, and use the Smith chart to analyse the right hand half. This is what the left hand half sees. I break it between the source and the line. So the source, looking into the line, sees an impedance, and hence a VSWR. That's it, it works for any line impedance, no reflections to worry about. \$\endgroup\$ – tomnexus Aug 30 '17 at 17:47
  • \$\begingroup\$ @tomnexus ok, I removed the lousy comment. \$\endgroup\$ – user287001 Aug 30 '17 at 18:35
  • \$\begingroup\$ +1 Since the transmitter end is terminated with a matching resistor, there will be no secondary reflection back into the line, so it doesn't matter how long the line is, I totally agree. The irritation comes that the stupid example sheet doesn't have an option "any of the above". Poor students, their instructor should be fired. \$\endgroup\$ – Ale..chenski Aug 31 '17 at 0:41
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The lengths given are, at velocity factor 1, full wavelength, half, quarter, and eight.

Since we are mismatched, we likely want to save what can be saved by voltage matching - so we want the load (including transmission line!) to present the highest possible impedance it can.

This is where going for the 5m line (half wavelength) will come in handy: this always makes the load impedance appear 1:1 at its input, so we are looking into 100 Ohms. That is probably your solution.

The 10m will behave just the same, but be a bit more lossy.

The quarter-wave line is useless here - even as it can do impedance transformation, a load higher than its specific impedance will always end up transformed to an impedance BELOW it - which seems to be the worse match here.

All that I could find about eigth-wavelength lines (1.25m) suggests that you would get a capacitive behaviour in that setup, which is probably dependent on the absolute capacitance of the line - since that is not given, it is probably not meant to be the answer.

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The input of the line is matched and the output is not. There is single reflection. The reflection coefficient at the end of the line does not depend on the length of the line. The reflected wave sinks to the source due the perfect matching and loss-free propagation. => The power transmission does not depend at all on the line length in this case. The missing of secondary reflections quarantees that all energy that is absorbed by the load from the coming wave is taken at the first meeting of the wave and the load.

Your mistake was to assume that max. power means =matching. You started demanding the impossible. The impossiblity should have been obvious at last when the same X cannot make both the real and imaginary parts of Z right at the same time.

By only changing the line length you cannot get full impedance match as proved informally here, more formally in other answers with Smith's chart and also by the dead end of your own trial. No othe methods were allowed, so accept that max. power which is available by changing the line length. In this case all lengths are as good.

The hard way i.e. how to get the same with equations: Assume an input source voltage. Make a formula for the power that is absorbed by the combination of the line and the load. All of it is taken by the load because the line dissipates nothing. As well you can calculate the load power starting from the incident voltage wave and reflection coefficient. Finally the line length drops out of the formula if you can manipulate properly the complex expression.

OFFTOPIC: Another answer suggested that there may be is an error in the original problem. It was suggested that maybe the terminating resistances are both 50 Ohm and the line impedance =100 Ohm. In that case the line length affects. Max power transfer would be the same as with fully matched 50 Ohm system, if the line length were a multiple of half wavelength.

ADDENDUM: The comments fight. I did a simulation. My line is ideal, 50 Ohm and its propagation delay is 0,1 microseconds. I did a frequency sweep from 100 kHz to 20 MHz. That means length/wavelength ratios from 0,1 to 2 are tested. The plot shows constant attenuation from the voltage source to the load. The attenuation is about 3,5 dB as one can easily calculate from the resistances 50 and 100 Ohm by using the voltage division formula.

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  • \$\begingroup\$ "The reflection coefficient does not depend on the length of the line" - maybe your coefficient doesn't depend on it, but what the sink actually sees in a mismatched system like this depends on it a lot - remember lecher line experiments? \$\endgroup\$ – rackandboneman Aug 30 '17 at 2:37
  • \$\begingroup\$ @rackandboneman the impedance of the combination line + mismatched load depends a lot of the length of the line. True. But in this case the power transmission doesn't. The reflected wave stays as powerful. \$\endgroup\$ – user287001 Aug 30 '17 at 7:56
  • \$\begingroup\$ So if I put the 100 Ohm load at an antinode vs at a node, power dissipation in that load would vary. If the power dissipation in the load varies, so does the reflected power, unless anything not dissipated to the load is radiated. Where am I missing out according to your model? \$\endgroup\$ – rackandboneman Aug 30 '17 at 12:10
  • \$\begingroup\$ @rackandboneman if you place something elsewhere than the questioner had stated, it's totally different problem.. There's one 100 ohm resistor at the end of the line. Keep it! There's a 50 ohm resistor at the beginning. Keep it, too. Do not add more! You can vary the length of the line, but the maximum amplitude point of the standing wave (=antinode) stays at the end of the line as long as the load is a resistor and bigger than the line impedance (=50 ohms). \$\endgroup\$ – user287001 Aug 30 '17 at 12:35
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    \$\begingroup\$ Yes, but to be clear, it's the magnitude of the reflection coefficient of 100 ohms, and a 50 Ohm line, to 50 Ohms source impedance, that is the same for any line length. The phase changes. The power delivered is the same for all choices. \$\endgroup\$ – tomnexus Aug 30 '17 at 17:56

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