Smith Chart...

[ms660]

Could someone confirm if I've got this approx. in the right ball park, need Zin result for 50 Ohm Zo coax of 0.6 of a wavelength long, load = 25 Ohms R, my printer's out of action at the moment so can't print off a chart, had to manage with an online chart on a small computer screen ! but not sure if I'm right, I make it approx. 33+J24ohms?

Lawrence.

[russell_w_b]

Lawrence,

From my spreadsheet I make it 33.74 + j24.1

[ms660]

Cheers for checking that, yes, 0.6 is electrical wavelength (I should have dropped the word coax)

Lawrence.

[G8HQP Dave]

You could do a plausibility check by remembering that a halfwave does nothing to impedance, so 0.6 wavelengths is the same as 0.1 wavelengths. 0.1 is smallish, so the cable will act (to a first approximation) as either a series inductor or a shunt capacitor. The 25R load is smaller than the cable impedance, so the cable will act as an inductor. You should expect to see something in the region of 25R plus some inductance, but raised somewhat in impedance because 0.1 is almost halfway to 0.25 (which would give 100R).

I was taught how to estimate the result of a calculation in A-level physics - necessary when using a slide rule! Still useful when using a calculator or a computer.

[ms660]

Thanks, supposing the same wavelength (0.1) and same Zo but the load is changed to 100R, what will be the reactance term be at the input?

The reason I ask is because I'm a bit confused on this one (the 100R scenario)

Lawrence.

[kalee20]

That's a great example of a sanity check, Dave!

[ms660]

Originally Posted by ms660

Thanks, supposing the same wavelength (0.1) and same Zo but the load is changed to 100R, what will be the reactance term be at the input?

The reason I ask is because I'm a bit confused on this one (the 100R scenario)

Lawrence.

Hopefully goof solved....forgot to extend the line down into the capacitive reactance curves Got approx. 50-J34 ohms?

Lawrence.

[russell_w_b]

Indeed. 49 - j35 near enough.

[ms660]

Cheers again

Waiting for printer ink

Lawrence.

[G8HQP Dave]

100R on 0.1 wavelengths of a 50R line would be expected to give a capacitive reactance in parallel, then the whole thing reduced in magnitude a bit because you are halfway to a 0.25 wave line, then when you turn it into series form the individual parts (resistive and reactive) get smaller than the parallel form. Harder to estimate than the first example!

[G0HZU_JMR]

You can also cross check your answers using a simple analysis and a few simple equations. Both cases are for a VSWR of 2. If you work out the (mag)reflection coefficient from the VSWR it is 0.3333 in both cases as (2-1)/(2+1) = 0.3333

0.1 wavelength = 72 degrees clockwise rotation for the reflection coefficient (or on a smith chart). So for the 25R case (where the initial reflection coefficient is 0.3333 / 180deg) the reflection coefficient after 0.1 wavelength should be 0.333/(180-72)deg = 0.3333 / 108deg

The 100R case is easy to solve because 25R and 100R are both cases of 2:1 VSWR and will always be 180deg apart on a 2:1 VSWR circle no matter what the length of the 50R tline. So the reflection coefficient for the 100R case will be 0.3333 / (108-180) = 0.3333 / -72deg

Of course you could get the same answer starting from 100R with a reflection coefficient of 0.3333 /0deg and rotating clockwise to get -72degrees. You can then convert these reflection coefficients into rectangular or polar impedance format as required using the classic equations (eg Zc = Zo * ((1+r)/(1-r)) ) or you can use an online converter to check your answers

Of course, if you used a real transmission line like RG58 coax then you would get a slightly different answer for 0.6 wavelength compared to 0.1 wavelength because of the loss in the coax.

[russell_w_b]

You can do all that, or you can write an excel spreadsheet (other spreadsheets are available) using the transmission line equation and a few trig functions (don't attempt to manipulate complex numbers on Excel: it's a pain). I wrote mine when I was in the mood and I pull it down off the computer when I just want to know what's at the other end, rather than ratch for a pencil and a chart and a ruler and a compass.

[G0HZU_JMR]

Yes, the most powerful method is to solve it with a computer Another computer based method is to use a linear simulator program that allows the user to tweak everything in real time, eg the load, the line length and the frequency. The results can be very intuitive to watch as they move around on a computer generated smith chart. Another solution would be to use the smith chart program by Fritz Dellsperger although this has fairly grainy resolution. It is a brilliant tool for simple impedance matching and is very powerful.

But I did wonder if this was a homework question on the smith chart so I gave a method to solve this using a numerical equivalent of the old school 'compass' method used with a paper copy of the smith chart.

[ms660]

I cross checked using the reflection coefficient r=Zl-Zo/Zl+Zo then measuring that off on the radial scale and transferring that to the chart on the 0.1 wavelength line.

Lawrence.

[Radio Wrangler]

There is a rather useful Smith hart calculator for the PC called LLSmith (The writer was Lance Lascari, hence the LL)

It's available free from his website: www.rfdude.com

Run the programme, click on setup to enter the frequency, and 25Ohms as the starting impedance, and Zo as 50 Ohms. Close setup. On the bottom toolbar click series line and it'll give you 90 degrees of 50 Ohm line as a default. Your scroll wheel on your mouse will vary the length, or you can click on the line in the list at the left of your screen and type in values. You can see whats happening on the smith chart, but to the bottom of the list area the result is also given numerically.

This is a great piece of software.It's age is showing, but It works well. I have it on W7 at work, and my macbook runs it on XP on a virtualised machine so it's pretty robust.

To do it in your head is easy. 25 Ohms is a 2:1 VSWR on the axis, to the left of centre. 0.6 of a wavelength is a phase shift of 0.6 * 2 * 360 degrees = 432 degrees, so it goes once round and leaves 72 degrees remainder, which is pretty much the way Jeremy did it.

david

David