Well, I am finally ready to give public answers to Maurizio's quiz.
>1a) An ideal 50 Ohm SWR meter, with no line lenght, is placed directly
>between a 100 Ohm pure resisistive load and a 50 Ohm source. What does the
>meter read ?
Answer> The meter reads an SWR of 2:1
>1b) Are there standing waves between source and load ?
Answer> Yes. The impedance mismatch creates a standing wave condition.
>1c) How much power is on the load ?
Answer> 88.9% of the transmitted power is delivered to the load. For a
100 Watt signal this is 88.9 Watts.
> 1d) Where is the lost power gone ?
Answer> The lost power is absorbed by the 50 Ohm source. The power
reflection took place at the 50/100 Ohm interface. Therefore, to the
reflected waves which have never left a 50 Ohm environment, the 50 Ohm
source appears as a 50 Ohm load.
>
>2a) The generator frequency is X and a quarter wave line of 70.71067812 Ohm
>is placed between its output and the load. Is it the effective matcher loss
>equal to the typical line loss at frequency X or is it different ?
Answer> The only loss in the transmission line would be its
characterisitic insertion loss.
>2b) Why ?
Answer> The loss in a transmission line is always its characteristic loss.
>What's the power on the load this time ?
The power on the load this time is 100% or 100 Watts (excluding the
insertion loss of the coax). The reason is that the 70.71067812 Ohm line
is a quarter wave transformer. A 1/4 line with impedance of
Square_root(Zsource*Zload) will transform Zload into an impedance of
Zsource at a quarter wave length. 70.71067812 Ohms is the square root of
the product of 50 Ohms and 100 Ohms.
Since the 50 Ohm source sees a 50 Ohm load, no power is absorbed in the
source. Sure, there is a standing wave pattern on the 70.71067812 coax,
but it is within the coax. Since the match only occurs at a line length
of 1/4 wavelength, this is inherently a narrow band match.
>
>3a) A quarter wave shorted stub is made of the same line and connected to the
>generator output of the previous
>example, what does it happen with impedances, losses and SWR ?
Answer> Nothing. A 1/4 wave shorted stub has an infinite impedance.
Therefore it appears not to be there.
>3b) What is the VSWR on the short end of the stub ?
Answer> Infinity
>3c) And at 1/8 wavelenght from the short ?
Answer> Also infinity
>3d) What power is dissipated by the shorted stub ?
Answer> None
>
>4a) The source is 50 Ohm resistive, the load is 100 Ohm resistive, the
>generator frequency is X. Two quarter wawe lines of 70.71067812 Ohm are
>connected back to back with a nominal 50 Ohm SWR/impedance analyzer in the
>connection point between the lines.What does the analyzer reads in term of
>SWR and impedances (in the form of real and imaginary part) ?
Answer> VSWR between the coax sections is 1:1. Imepdance is 50+j0 Ohms.
>4b) What would it read on the source side ?
Answer > Since the VSWR meter has no line length, taking it out of the
circuit is the same as connecting the two pieces together. You then have
a 1/2 wavelength of line. A half wavelength repeats the impedance at the
far end.
Therefore, the impedance is 100 Ohms at the source.
>4c) And on the load side ?
If you put the 50 Ohm SWR/impedance meter at the output of the coax,
directly on the load, the load would still read 100 Ohms. Since the
output of the coax at that point is 50 Ohms (50 Ohm source transformed by
a 1/2 wavelength of coax), the VSWR meter would read 2:1.
Regardless though of the length of coax, the impedance of the load
looking directly at the load would ALWAYS read 100 Ohms.
73,
Jon
KE9NA
--------------------------------------------------------------------------
The Second Amendment is NOT about duck hunting!
Jon Ogden
jono@enteract.com
www.qsl.net/ke9na
"A life lived in fear is a life half lived."
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