Understanding Instantaneous Current Equation

[PerArdua]

I'm on the edge of buying a copy of Joel Dunsmore's Handbook of Microwave Component Measurements, wanting to understand more about the underlying math behind microwave measurements.

I've worked through what's available on the preview by Amazon, but I'm unsure how a couple of equations are derived - where have I gone wrong?

Using the following figure 1.1:



For equation 1.10, we assume (assuming a very long transmission line and the line impedance at the reference point V₁ is the same as the source), and told the instantaneous current going into the transmission line is:



But by my workings



On this train of thought, it seems eq 1.12 isn't derived using V_FI_F, instead it comes from only using equation 1.11 - so a bit of a logical inconsistency, but it doesn't help me figure out what I_F should be, as using V_S/(2Z₀) gives:



Using the result of 1.10 produces a result that's different to this, so I can't work backwards either - is it the case that there's been a duplication error with 1.11?

[MarkT]

The equation 1.10 from the book isn't dimensionally consistent - looks like a lack of technical proofreading...

[radiogeek381]

Gonna try this with math notation… here goes.

But first… I haven’t seen the book, but at the point where you are in the discussion, the author is introducing forward and reflected power.  You probably get this, but just in case there are other readers out there…. There are a lot of different ways to wrap your head around this.  It all depends on how you’re wired. Here’s one of the ways.

Assume that the voltage everywhere was 0 at the start. When a voltage of any sort is introduced from Vs down the line it will travel down the transmission line as a change from whatever the voltage was, to whatever it is right now — think of a wave advancing down the line. The *forward* voltage and *forward* current along the line are related strictly to the line impedance and the voltage and current introduced at the start. It has to be this way, because otherwise the source end of the line would know what was at the load end of the line. That can’t happen because information can’t travel faster than light. So the source won’t know what is at the far end until one round trip. 

But for now, we’re only concerned with the forward trip.

On the forward trip, the source only knows the impedance looking in to the line — Z0.  So Vf (V1 at the beginning of time) is the result of a voltage divider between Zs and Z0. hence

\$V_f = \frac{V_s Z_0}{Z_0 + Z_s}\$

But what is the current?  It must be the same through Zs as it is into the transmission line.  So

\$I_f = V_f / Z_0\$

So now we’ve got the forward current and the forward voltage

\$P_f = V_f I_f = V_f V_f / Z_0 = \frac{V_s Z_0}{Z_0 + Z_s} \frac{V_s Z0}{Z_0 + Z_s}  \frac{1}{Z_0}\$

Cancelling out a Z0 and consolidating

\$P_f = \frac{V_s^2 Z_0}{(Z_0 + Z_s)^2}\$

Where the units (volts squared over impedance) makes me feel good about the units .

Now lets set Z0 to be Z_s. We get

\$P_f = \frac{V_s^2 Z0} {(2 Z_0)^2} = \frac{V_s^2}{4 Z_0}\$

Which feels pretty good.

Soon after all this, the author should introduce the backward wave Vr and Ir. I’m not sure how he’ll introduce it, but it will follow some of the same lines. In all the developments, it becomes important to remember superposition and Kirchoff’s rules, especially the current-into-a-node rule. And I’ve found it helpful to work through all this assuming a step function at Vs. This helps reasoning about it from intuition.  Once you are convinced of that case, application to sinusoids is a little matter of mathemagic using Mssr. LaPlace.

Apologies if there was a post prior to this covering the same thing — I had to step away in the middle of this.

And, in case I screwed up the math… Here’s all that stuff de-latexized.

Vf = (Vs Z0) / (Z0 + Zs)

If = Vf / Z0

Pf = Vf If
= Vf Vf / Z0
= (Vs Z0)/(Z0 + Zs) * (Vs Z0)/(Z0 + Zs)  / Z0

Pf = (Vs^2 Z0)  /  (Z0+Z0)^2

if Z0 = Zs

Pf = (Vs^2 Z0) / (2 Z0)^2
= Vs^2 / (4 Z0^2)

[wd5jfr]

Thanks for your post but I've never encountered your symbol nomenclature using: $ \frac }{
$Vf = \frac{V_s Z_0}{Z_0 + Z_s}$
Please explain.
Thanks
Hank

[RoGeorge]

Thanks for your post but I've never encountered your symbol nomenclature using: $ \frac }{
$Vf = \frac{V_s Z_0}{Z_0 + Z_s}$

Those are TeX/LaTeX/mathjax syntax for formulas, except a backslash symbol is missing.  If it were there, the forum will display them as math formulas, i.e.:

 \$ Vf = \frac{V_s Z_0}{Z_0 + Z_s} \$

For mathjax posted on EEVblog, the $ character has to be preceded with a backslash, or else it will just show as text.  If you forget which simbol was it, you can always press the omega (ohms) symbol at the end of the smiley icons un the posting editor. and replace the backslash omega with your equation

 

It is recommended to have the editor in text mode (can be changed from the 'Toggle View' icon in the posting editor).
More info about posting formulas here:  https://www.eevblog.com/forum/beginners/how-does-posting-math-notation-work/

[radiogeek381]

Ok… I think I’ve fixed it.  Hope it helps.  (I edited the original reply.)

[wd5jfr]

Hey r..k381
I have never used that text editor. Good job correcting, thanks.  I followed your clear logic all the way through and wondered if you have the time and initiative to use the same logic to derive the return loss formula?
In both editions formula 1.12, the Vs should be squared.
Hank

[PerArdua]

Thank you radiogeek, your explanation makes perfect sense!