VSWR specified for an RF output?

[GV]

I thought I had some grasp on the concept of VSWR but I don't understand what it means for an RF output to have a VSWR specification on it's own (i.e. without knowing what is connected to it).

I have a Keysight/Agilent N9344C handheld spectrum analyzer with a tracking generator output. The datasheet has the following specification for the tracking generator:
VSWR    < 2.0:1    (Nominal)
(see bottom of page 7 in the data sheet: https://literature.cdn.keysight.com/litweb/pdf/5990-7193EN.pdf)

What does this mean exactly?

My first guess would be that it means that the tracking gen. output must be connected to a load with a VSWR of 2.0:1 or less. If I am correct, does that mean that the output can handle -9.5 dBm of reflected power if I have the tracking gen. output set to 0dBm but it can only handle -19.5 dBm of reflected power if I have the tracking gen. output set to -10dBm? (this is assuming that a VSWR of 2:1 means that the reflected power is 9.5 dB lower than the forward power)

[Tony_G]

It's about enabling you to calculate mismatch uncertainty. Check out the example here:

http://www.microwavejournal.com/articles/6166-calculating-mismatch-uncertainty

TonyG

[rfengg]

Beg to differ though I may be wrong.......the tracking generator can handle any reflected power from what its sending out.
I think the VSWR of the  tracking generator specified , is a measure of the source impedance at the output port across  its bandwidth of 5 MHz to 7 GHz.
In other words , from 5 MHz to 7 GHz, you can expect the worst case source impedance of the tracking generator to , for e.g., vary anything between 25 and 100 ohms which is a VSWR of 2.
Any other views on this?

[Tony_G]

I assume that you're disagreeing with the OP. For the OP, when you connect the tracking gen to the DUT there will be some mismatch uncertainty that will impact the 'known' value of power the DUT sees. There will also be mismatch on the connection from the DUT to the SA that impacts the 'known' value that the SA sees. In the end you should be able to say something to the effect of 'I measured X plus or minus Y dB'.

TonyG

[TheSteve]

Beg to differ though I may be wrong.......the tracking generator can handle any reflected power from what its sending out.
I think its the VSWR of the  tracking generator specified , is a measure of the source impedance at the output port across  its bandwidth of 5 MHz to 7 GHz.
In other words , from 5 MHz to 7 GHz, you can expect the worst case source impedance of the tracking generator to , for e.g., vary anything between 25 and 100 ohms which is a VSWR of 2.
Any other views on this?

This is what it means. It is also why you see people place low value attenuators on the output of the tracking gen and/or input of the SA before running a normalization. It should improve the 50 ohm matching between the filter etc under test and the TG and SA at the sacrifice of some dynamic range.

[Tony_G]

Great point Steve. You'll often see 10dB attenuators in calibration setups for exactly this reason - here is a Minicircuits app note on this that comments on the dynamic range reduction (maybe coincidently if references a program created by a Steve from Agilent):

https://www.minicircuits.com/app/AN70-001.pdf

TonyG

[tkamiya]

What does it mean to have UP TO and [NOMINAL] though? 

Nominally less than 2.0 but if it's 2.1, don't call it a defect?

[GV]

Thank you all for your input. Let me see if I got this right:

I intend to connect the tracking generator output (with a VSWR of 2.0:1) to an amplifier (Mini-Circuits TVA-11-422A+: https://ww2.minicircuits.com/pdfs/TVA-11-422A+.pdf) which has an input VSWR of 1.7:1.

Calculating the respective reflection coefficients I get:
γ_TG = (1-2.0)/(1+2.0) = -0.333
γ_AMP= (1-1.7)/(1+1.7) = -0.259

which then gives me uncertainty range:

20*log(1-γ_TG*γ_AMP) = -0.785 dB
20*log(1+γ_TG*γ_AMP) =  0.720 dB

The formulas are taken from (the aforementioned) Steve's article on page 124 of EDN's February 1st, 2001 issue: https://m.eet.com/media/1145806/22087-20101di.pdf

Assume I set my tracking generator output to 0 dBm. Do the above numbers mean that the amplifier would receive a signal somewhere between -0.785 dBm and 0.720 dBm?

[radioactive]

You're on the right track, but you shouldn't get gain from a mismatch or even from a perfect match.  Best you can do is 0dB loss.  As others in this thread pointed out, for VSWR of 2:1,  your impedance will be somewhere between 25 and 100 ohms.  (50/2 to 50*2).  Similar for the 1.7:1,  would be between 29.4 ohms and 85 ohms.  So, worst case would be between 29.4 and 100 ohms for the mismatch.

Calculate the reflection coefficients for that.   For cases where ZL is a real number,        ρ=abs((ZL-Z0)/(ZL+Z0))   (see this page: https://www.microwaves101.com/encyclopedias/voltage-standing-wave-ratio-vswr)

p= abs( (29.4-100)/(29.4+100)) = 0.5455950540958269

Then the mismatch loss is -10*log10( 1-p^2)  = -1.53dB

So,  you could expect your amp (load) to see from 0 to -1.53 dBm if you output 0 dBm from your source.

Hopefully I did that right.

[Tony_G]

You're on the right track, but you shouldn't get gain from a mismatch or even from a perfect match.  Best you can do is 0dB loss. 

I think there is a key point here that GV should see is that the calculation is for uncertainty not a specific statement of power. For example, if you put a power sensor on the tracking generator and do the calulation for uncertainty (let's ignore the sensor system uncertainties and just consider the SWR reflection, basically the sensor perfectly sees the non-reflected power) the the reading on the sensor is +/-X from the actual power being delivered.

So as you tweak the TG to get your 0dBm reading the true power there will be 0dBm +/-XdB.

Personally I always try to abide by the laws of thermodynamics (I hear the fines can be a bitch) so the overall system, assuming that it is passive, will have a net loss - When you get to the end of the system, as radioactive said, you should be whatever you started with minus some power however your final measurement will still have some +/- variation.

Given the original situation of input to an amplifier, you can consider the output, attenuators, cable and input to be passive and calculate what the power is, +/-, that the active part of the amplifier sees.

TonyG