A SWR conundrum

[northern35s]

First off the question isn't one I came up with, the credit for this goes to Steve Hunt, G3TXQ.

Considering we've been talking about SWR and transmission lines a fair bit, I thought it might help some of us get a better grip of the subject at hand with this little teaser, I'll post the first part of the teaser now and see where you get with it op:


If you had 8 coax patch leads, each a quarter-wave long, some with a characteristic impedance of 52 Ohms and some of 48 Ohms, and you connected them all in series with a perfect 50 Ohm termination on the end, what's the best SWR you could see at the input and what's the worst SWR?

 

[hookedon6]

... what's the best SWR you could see at the input ...

at the "input' of what? the first jumper? the load?
I'm not sure I understand the question correctly,.....
is it gonna be a "gotcha" type answer? doesn't make any difference, I'll play along

here is one (possible, depending on the exact configuration) answer: (for 52 ohm to 48 ohm at the load)
my keyboard won't make the "magnitude of reflection" symbol, but here is the formula"

Z0 -Z1/Z0 +Z1 = mag of reflection

52- 48/52 + 48 = 4/100 = .04

VSWR = 1 + mag of reflection/ 1 - mag of reflection

1 + .04/1 - .04 = 1.04/.96 = 1.08:1 VSWR



this is for one 52 ohm section feeding one 48 ohm section,.. if you want different configurations, then just put the correct numbers in the formula and plug and chug.

without doing the math, I would guess the worst VSWR would be from 52 to 48 to 52 to 48 to 52 to 48 to 52 to 48 ohm sections and the best would be 48 to 48 to 48 to 48 to 52 to 52 to 52 to 52


YMMV

 

[hookedon6]

...without doing the math, I would guess the best VSWR would be 48 to 48 to 48 to 48 to 52 to 52 to 52 to 52


YMMV

now that I have thought about it, and I havn't done the math , but I'm thinking that it wont make any diference between 48 to 48 to 48 to 48 to 52 to 52 to 52 to 52 or vice versa,.....

either way, it is the same 2 ohm mis-match into a 50 ohm load

 

[RADIOOMAN]

How many of each and in what configuration?
I have done the math on four 48 ohms and four 52 ohms alternating one and then the other.
What I came up with is 1.8212:1 in one case and 1.8241:1 in the other.

 

[W5LZ]

From just the numbers you have used it seems fairly safe to say that the importance of SWR isn't all that much. By using the strictly 'SWR' numbers, the end result isn't as definite as when you use the full numerical value of that 'SWR', the 'R+/-J'. What would the result then be, with that "+/-J" factored in? Try it, see what happens. It won't be what you expect...
- 'Doc

(No, I won't do it, YOU do it. See for your self.)

 

[RADIOOMAN]

From just the numbers you have used it seems fairly safe to say that the importance of SWR isn't all that much. By using the strictly 'SWR' numbers, the end result isn't as definite as when you use the full numerical value of that 'SWR', the 'R+/-J'. What would the result then be, with that "+/-J" factored in? Try it, see what happens. It won't be what you expect...
- 'Doc

(No, I won't do it, YOU do it. See for your self.)

We'll let the computer do that for us.
I did my calculations ignoring the miniscule inherent +j residing in some coaxial cables.

 

[RADIOOMAN]

From just the numbers you have used it seems fairly safe to say that the importance of SWR isn't all that much. By using the strictly 'SWR' numbers, the end result isn't as definite as when you use the full numerical value of that 'SWR', the 'R+/-J'. What would the result then be, with that "+/-J" factored in? Try it, see what happens. It won't be what you expect...
- 'Doc

(No, I won't do it, YOU do it. See for your self.)

How can you possibly know what I expect it to be?!
That is making a very bold assumption on your part!

 

[northern35s]

Ok chaps, thanks for giving it a go, I've had a sleep so here's the worst case scenario:

Load > 52 > 48 > 52 > 48 > 52 > 48 > 52 > 48 > source

What VSWR would you see at the source, and if you were to arrange these lines, for best case, what would you see, so two questions:

Worst case VSWR

Best case VSWR


I'm out of town overnight so I won't be giving the answer until tomorrow evening (UK time of course )

 

[RADIOOMAN]

Ok chaps, thanks for giving it a go, I've had a sleep so here's the worst case scenario:

Load > 52 > 48 > 52 > 48 > 52 > 48 > 52 > 48 > source

What VSWR would you see at the source, and if you were to arrange these lines, for best case, what would you see, so two questions:

Worst case VSWR

Best case VSWR


I'm out of town overnight so I won't be giving the answer until tomorrow evening (UK time of course )

Answer for scenario shown on this page is 1.897:1 and the answer for the configuration that gives the best swr is load > 52 > 52 > 48 > 48 > 52 > 52 >48 > 48 = 1.0:1.
The above answer are assuming the the tx line from the signal source to the swr meter is 50 ohm.
If not 50 ohm but 52 ohm then my answers will ajust by 4 %.

 

[hookedon6]

... here's the worst case scenario:

Load > 52 > 48 > 52 > 48 > 52 > 48 > 52 > 48 > source )

did i win the kewpie doll?

 

[W5LZ]

RADIOOMAN,
It's a bold statement, yep, but it's also a very general statement and you know how those are. While you are considering those (+J) thingys, you might also consider the (-J) thngys, they won't be the same in all cases. And since both of them can vary in whatever 'device' is being included in the calculation, the resulting answer will also not be the same. The more of those 'devices' there are the larger the difference can be, in both directions.
Will those (+/-j)s make a big difference in every situation? Of course not, sometimes you get 'lucky'. It's a conundrum...
- 'Doc

 

[hookedon6]

...the answer for the configuration that gives the best swr is load > 52 > 52 > 48 > 48 > 52 > 52 >48 > 48 = 1.0:1.

1.0:1,.........show the math, I don't think so
any way you configure it, there will be at least a 2 ohm (4%) mismatch into a 50 ohm load, so the VSWR can not be 1.0:1



actually the BEST VSWR (1.08:1) will be 48-48-48-48-52-52-52-52, or exactly the inverse.

4-48's in series to 4-52's in series are exactly the same as one 48 to one 52.


VSWR = line impedance/load impedance @ the load

 

[RADIOOMAN]

1.0:1,.........show the math, I don't think so
any way you configure it, there will be at least a 2 ohm (4%) mismatch into a 50 ohm load, so the VSWR can not be 1.0:1



actually the BEST VSWR (1.08:1) will be 48-48-48-48-52-52-52-52, or exactly the inverse.

4-48's in series to 4-52's in series are exactly the same as one 48 to one 52.


VSWR = line impedance/load impedance @ the load

Here is the formula I used:
Transformed load value = Zo*Zo/Zl

 

[hookedon6]

Here is the formula I used:
Transformed load value = Zo*Zo/Zl

ok you showed a formula,.... now show the actual math,.... like I did in my first post.


or are you just using some software to compute an answer? perhaps a Smith Chart?

when you said this :"Load > 52 > 48 > 52 > 48 > 52 > 48 > 52 > 48 > source" and didn't realize that you had it backwards, I started to suspect

bottom line,.... no matter how you configure the coax sections, there WILL be an impedance MISMATCH, and a VSWR of 1.0:1 is impossible.

 

[hookedon6]

there are numerous software programs that will "compute" a number for you,.......
for "some" reason, most of them will only allow one 1/4 wl section to be used, I wonder why.


try this one, just insert your numbers and plug and chug on a calculator

Zin = Z0[Zload*cosh(x) + Z0*sinh(x)]/[Z0*cosh(x) + Zload*sinh(x)]

Given Zin as a known, solving for the unknown Zload yields:

Zload = Z0[Z0*sinh(x) - Zin*cosh(x)]/[Zin*sinh(x) - Z0*cosh(x)]