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My question has to do with the rule of thumb we generally use when dealing with transmission lines.

We say that if the length of the line is 10% or less of the wavelength, we could neglect the effects of the transmission lines—and that makes sense if we look at from the perspective of the time delay it takes the wave to travel along a short vs a long line.

But when looking at it from the perspective of the input impedance equation, the 10% rule of thumb doesn't always hold. For example,

enter image description here

$$ Z_{in}=Z_o\dfrac{Z_L+jZ_o\tan{(\frac{2\pi}{\lambda}L})}{Z_o+jZ_L\tan{(\frac{2\pi}{\lambda}L})}$$

For some values of \$Z_o\$, and \$Z_L\$, (with \$L=0.1\lambda\$), you don't necessarily get an input impedance close to the value of the load one (which I think would mean we could ignore that the TL is even there).

Take for example, \$Z_o=50\$, and \$Z_L=300\$, with \$L=0.1\lambda\$, then \$Z_{in}=23+j64\$. The line does transform the impedance seen by the source even though it is "short" as per 10% the rule of thumb.

Also, even if the line effects were negligible somehow, the reflection coefficient would still be nonzero since it's defined by:

$$\Gamma =\dfrac{Z_L-Z_o}{Z_L+Z_o}$$

What would be the effect of having a nonzero \$\Gamma\$ even when the line effects are considered negligible? (Hopefully this makes sense!)

What am I missing here?

Thanks

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    \$\begingroup\$ In power transmission lines, 'short' is defined in the order of 60km, which is 1% of the wavelength at 50Hz. \$\endgroup\$ – Chu Sep 9 '18 at 9:02
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    \$\begingroup\$ The 10% rule applies, as far as I am aware, to digital transmission of data so that reflections don't cause data corruptions. \$\endgroup\$ – Andy aka Sep 9 '18 at 9:10
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    \$\begingroup\$ Examining the (hardcopy!) Smith Chart, it's apparent that the large mismatch between line and load impedances is the problem. It appears the 10% rule can't deal with a factor of 6. \$\endgroup\$ – Chu Sep 9 '18 at 9:48
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Also, even if the line effects were negligible somehow, the reflection coefficient would still be nonzero since it's defined by: $$\Gamma =\dfrac{Z_L-Z_o}{Z_L+Z_o}$$ What would be the effect of having a nonzero Γ even when the line effects are considered negligible?

This is exactly what you should expect. The reflection coefficient does not go to zero when the line length goes to 0.

If you have a generator with impedance \$Z_0\$ driving a load with impedance \$Z_L\$, with no line in between, the voltage across the load will not be equal to the generator's nominal voltage (the voltage it would drive on a matched load), indicating the presence of both forward and backward waves.

For some values of \$Z_o\$, and \$Z_L\$, (with \$L=0.1\lambda\$), you don't necessarily get an input impedance close to the value of the load one (which I think would mean we could ignore that the TL is even there).

Take for example, \$Z_o=50\$, and \$Z_L=300\$, with \$L=0.1\lambda\$, then \$Z_{in}=23+j64\$.

This equivalent input impedance doesn't look very close to 300 ohms.

But consider, if we drive a 300-ohm load directly from a 1 V 50-ohm generator, the voltage across the load will be 0.86 V.

If we drive the composite load (300 ohms at the end of a 0.1-wavelength line) with a 1 V 50-ohm generator, the voltage at the output of the generator will be about \$0.61-0.34j\$, which has a magnitude of 0.70.

When you're used to working with a computer that does calculations to 10 digits, 0.70 doesn't seem particularly close to 0.85 (it's about a 20% error). But if you were working with a slide rule like early rf engineers did, this error might not be the biggest one in a complex calculation.

Practically, you might only know your \$Z_0\$ to \$\pm\$10% and your load impedance (at a particular frequency in your operating band) to \$\pm\$10%, and you might not know what reactive parasitics are associated with the load, so trying to calculate the load effect more accurately than this would not be sensible anyway.

Of course if your application requires greater accuracy, you are free to adopt your own rule of thumb, such as only ignoring transmission line effects when the line length is less than 1/20 or 1/50 of the wavelength.

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As the rise time equates to the fundamental BW of a step pulse f=0.35/T in most delay lines , filters and transmission line ( but not always e.g. ladder filter delay lines) 10%f is less than, 35% f above so the spectrum is not significantly altered.

There exist lower level harmonics in a step pulse above this ratio , but not always significant, except when a Nyquist antialiasing filter is needed on high resolution ADC’s

Similar arguments can be made for this “Rule of Thumb” for RF for Ref. Coeff. as it impacts fairly small acceptable transmission loss. An exception might be high power Tx where 0.1 dB unnecessary loss is significant. Another is Group Delay distortion in modem cables such that the bandwidth must be split up into many sub-bands each with a flat equalized group delay so that an aggregate bit rate can exceed the cable wavelength by a large amount trading off SNR for BW.

When the frequency that corresponds to <10% wavelength the staircase effects on step pulse echoes or phase shift and impedance transformation effects have an acceptable tolerance for most application except for the echo ripple becomes negligible.

So your understanding or assumption that it always applies is false.

It is a function of design spec tolerances for parameters such as Return Loss.

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Thanks everyone for the comments and answers. I think I was able to gather enough insight to answer my own question with what @The Photon and @Chu said.

I ran a few tests to find how the impedance would change with length ("short" lengths < 0.1\$\lambda\$ or 10% of the wavelength). I had \$L=0.1\lambda,0.05\lambda, 0.07\lambda, 0.1\lambda\$. With that I computed the different input impedances for a system with \$Z_o=50\Omega, Z_g=50\Omega\$ and a sweep of load impedances from 10 to 1000 Ohms.

enter image description here

The magnitude of the input impedance vs the actual load impedance looks like this:

enter image description here

As it shows, for the different load impedances, the magnitude of the input impedance comes the closest to the load at 1% of the wavelength (which makes sense since the tangent term in the input impedance equation approaches zero). This jives with @Chu's comment.

What I found interesting was the magnitude of the voltage at the output of the generator (as mentioned by @The Photon). For a 1V generator, with impedance \$Z_g=50\Omega\$, if we could ignore the TL, the voltage at the output of the generator should be:

$$V_o = V_G\dfrac{Z_L}{Z_G+Z_L} $$

Obviously since the TL changes the impedance with length, this changes the voltage.

This a plot of the magnitude of voltage at the output of the generator for different TL lengths. The legend indicates what the voltage is (a fixed number) when ignoring the presence of the TL, as in the voltage divider equation previously shown.

enter image description here

For values of the TL length < 0.1\$\lambda\$, the error between the voltage considering the TL vs ignoring it is <20%. After 0.1\$\lambda\$, the error in the magnitude of the voltage grows rapidly. The green plot shows a perfect match, that is why there is no error in voltage and the input impedance equals the load's.

This what the error plot looks like (deviation from the output voltage when TL is ignored).

enter image description here

Maybe the max 20% error at 10% of the wavelength is OK for voltage measurement (think RS485 for example). For RS485, just +/-200mV on the diff signal means a digital 1 or 0. So if you are actually getting 80% of a 3.3 VDD rail for a high, that should be plenty to get a clean 1 or a 0 (one signal on the diff pair will be high and the other low). This agrees with @Andy Aka's comment, for voltages (1s, 0s) this works.

A similar argument can be made about the power (I did some calculations), I just didn't want to turn this into a longer answer.

Thanks everyone for the help!

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