100 feet of RG-58A/U SWR effects shown visually.

[The DB]

Below are many pictures showing what 100' of RG58A/U coax does to your apparent SWR reading.

This shows the effects of coax on SWR to those who don't understand the math behind it, or are simply to lazy to do it themselves.

The SWR graphs have two lines, the top line is from plugging the terminator directly into the analyzer, the bottom line is from adding 100 feet of RG58A/U coax in between the terminator and the analyzer and rescanning.

If you don't know what a terminator is, think of it as a dummy load with a different impedance.

The middle line is the center frequency of the CB band here in the US.

The first graph was to demonstrate the effects of a 1.5:1 SWR antenna. The bottom line shows 1.2:1 SWR. So if you use 100 feet of RG-58A/U and see 1.2:1 SWR what you are actually getting is 1.5:1 SWR.

The second graph demonstrates 2:1 SWR.

The third graph demonstrates 3:1 SWR.

The fourth graph was meant to demonstrate a 5:1 SWR, however they did not have a resistor of the correct size so I chose the closest one available. This one thus shows What 4.75:1 SWR would look like.

Obviously if you have different lengths of this type of coax the results will vary. Less length and the effects will be lessened, more length and the effects will be greater. Different types of coax, such as RG-8X will have different results as well.



The DB

 

[hookedon6]

huh?,............. is the next thing you are gonna say is that coax CHANGES the real VSWR?

coax only passes RF. put a 52 ohm load on the end of your RG 58 and see what the VSWR really is.
when you change the impedance of the "terminator" and/or the resonance freq, you start talking apples and oranges.

you posted :
Obviously if you have different lengths of this type of coax the results will vary. Less length and the effects will be lessened, more length and the effects will be greater."
the above statement is absolutely incorrrect,.... added coax in (say) 1/2 wl multiples will have NO effect on true (or apparent) VSWR.

the true VSWR is a function of the LOAD impedance only

 

[W5LZ]

It would also be helpful if you knew the actual impedance of that RG-58 cable, not what's printed on the jacket. I think that would also account for a bit of the discrepancy in the graph's readings.
- 'Doc

 

[Captain Kilowatt]

I think the intent of the OP is to show that if the coax is not terminated in it's characteristic impedance then the SWR as indicated at the transmitter is NOT the true SWR as presented by the load. As far as the length of cable affecting the outcome, it is NOT a case of less cable = less of an effect and more cable is more of an effect. It is a case of the cable affecting the reading regardless of length unless it is an exact multiple of half waves long. The closer one gets to an odd multiple of 1/4 waves long the greater the impedance transformation and thus the greater the discrepancy in the indicated SWR versus the real SWR.

 

[bob85]

DB is not saying coax length effects vswr under the conditions he is testing,
he's showing that what you read with the vswr meter placed some distance from the load through a length of coax with loss effects the reading NOT the real vswr,

vswr measures lower the further away from the load you place the meter due to loss in the coax but real vswr does not change,
what he is showing is true so long as there are no common mode currents on the coax outer shield,

if you do have common mode currents then the common mode impedance of the outer shield is seen in parallel with the load impedance, then and only then does coax length effect the real vswr because common mode impedance changes with the length of the coax outer shield,

if moving your meter a few feet along the coax effects vswr readings significantly you have common mode currents on the outer shield or a faulty meter or very lossy coax,

the only thing that effects vswr is the load impedance in relation to the coax characteristic impedance,

its all explained in walt maxwells reflections and i trust walt cebik and the other arrl guys knew what they was talking about,

some words of wisdom from reflections,

 

[The DB]

Removed by The DB.

The DB

 

[northern35s]

Well summed up Bob

One thing to remember is that loss affects the signal in both directions, the signal and it's reflection, it's also important to note that there isn't just one forward and one reflected wave, the reflected wave bounces backwards and forwads a number of times eventually dissipating in the coax, with each pass it loses a little more, the longer the coax, the greater the loss

 

[RADIOOMAN]

When the coax is terminated with a load that is the same as its characteristic impedance then there is no power reflected for the swr meter to measure; Not much very interesting about that.
It is when the coax is NOT terminated with a load that is the same as its characteristic impedance is when things start to become interesting.
Before I go on with this, something needs to be stated because this is where the problem is; There is 'real coax' (transmission line) that has loss and then there is 'theoretical coax' (transmission line) that has no loss.
What we are talking about in this thread is 'real coax' and since it has loss then the swr at the load end, if coax is terminated in a value that is NOT the same as its characteristic impedance, is going to be a value that is more than the swr value that is measured at the measured end.
Since SWR is a ratio of the forward power to the reflected power then it stands to reason that if we use real coax then some of the power that is sent to the load is going to be lost in the coax and then what gets reflected back is also going to suffer from that same loss in the coax and is going to be less than what it started out as from the mismatched end.
So with less power reaching the load and then less power being reflected back to the swr meter, the swr meter see less reflected power and then displays a lower value than what is at the mismatched end.
This is what causes the swr value to be less than the actual value is at the mismatched end of the coax.
Even if you use 1/2 wave lengths (velocity factor corrected length) the swr is going to be less on the measured end than it really is on the mismatched end.
Imagine that you replace your old coax with new coax and then discover that the swr goes higher with the new coax.
Most people would be suspect of the coax causing the problem; And they would be wrong.
It is the antenna where the problem is; The old coax was lossy due to age and was causing the swr to look better than it really was.
Think about this theoretical situation: You have a theoretical lossless coax that is infinitely long, and you have no load on the end of the coax; (complete mismatch, all of the power is reflected and returned to the other end) What do you think the SWR will measure?

 

[hookedon6]

...There is 'real coax' (transmission line) that has loss and then there is 'theoretical coax' (transmission line) that has no loss.
What we are talking about in this thread is 'real coax' and since it has loss then the swr at the load end,... is going to be a value that is more than the swr value that is measured at the measured end.?

what that really means is,........... the meter is not in the correct place

you are saying is that "line loss" creates VSWR.op: we all know that is not correct,

VSWR is the RATIO of power/voltage/current/ect. reflected from and measured at the LOAD.

 

[RADIOOMAN]

what that really means is,........... the meter is not in the correct place

you are saying is that "line loss" creates VSWR.op: we all know that is not correct,

VSWR is the RATIO of power/voltage/current/ect. reflected from and measured at the LOAD.

No, no and no.

 

[W5LZ]

There's a problem with that explanation. The returning current will be lesser because of any resistive losses but the ratio (SWR) will stay the same or have a very, very slight change, not really noticeable. The same 'standing wave' is still there and is being constantly 'refreshed' with 'new' current because the transmitter is still transmitting. When that transmitter quits transmitting then the standing wave dies so no SWR. There isn't a gradual lessening of that standing wave (actually there is but the rise/fall time isn't going to be measurable except by calculation or some -very- expensive equipment). For all practical purposes it's instantaneous.
- 'Doc

 

[RADIOOMAN]

Well summed up Bob

One thing to remember is that loss affects the signal in both directions, the signal and it's reflection, it's also important to note that there isn't just one forward and one reflected wave, the reflected wave bounces backwards and forwads a number of times eventually dissipating in the coax, with each pass it loses a little more, the longer the coax, the greater the loss

Yep, loss in both directions and the longer the coax the greater the loss.

 

[The DB]

Think about this theoretical situation: You have a theoretical lossless coax that is infinitely long, and you have no load on the end of the coax; (complete mismatch, all of the power is reflected and returned to the other end) What do you think the SWR will measure?

SWR would appear to be 1:1 as in a theoretical infinite length of "perfect" no loss coax there would be no far end, and the signal would continue to travel down that coax ad infinitum, thus causing no reflected signal to measure.

If there is a far end as you describe it would in theory show an SWR that approaches infinity, assuming you waited long enough for the signal to get to the end of the near infinitely long "perfect" no loss coax and make the return trip to make the measurement. Until that signal makes its way back to the measuring device it will appear to be a perfect SWR match.

Am I right?


The DB

 

[hookedon6]

#1 ....if you do have common mode currents then the common mode impedance of the outer shield is seen in parallel with the load impedance, then and only then does coax length effect the real vswr....


#2 ...the only thing that effects vswr is the load impedance in relation to the coax characteristic impedance,...

well, which one is it?
my money is on #2

 

[RADIOOMAN]

SWR would appear to be 1:1 as in a theoretical infinite length of "perfect" no loss coax there would be no far end, and the signal would continue to travel down that coax ad infinitum, thus causing no reflected signal to measure.

If there is a far end as you describe it would in theory show an SWR that approaches infinity, assuming you waited long enough for the signal to get to the end of the near infinitely long "perfect" no loss coax and make the return trip to make the measurement. Until that signal makes its way back to the measuring device it will appear to be a perfect SWR match.

Am I right?


The DB

Absolutely correct.
If the forward never makes it the end to be reflected (because the end is infinitely too far away) then there will be no reflected to be measured.
And the swr meter will indicate 1:1 because in spite of the fact that there is forward power to be measured there is no reflected power to be measured.
I just put this in the post 'cause I thought some might find it interesting and is the basis for how a swr meter measures swr; And that is the measure of the ratio of the forward to the reflected.