Thanks Frank; it seems reasonable to me that it, as a loss would be negative. However, if I were in sales, a loss seems like something that a customer wouldn't want to pay for.?
Stuart K6YAZ
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-----Original Message-----
From: Frank S <ka2fwc@...>
To: nanovna-users <
[email protected]>
Sent: Sun, Aug 4, 2019 12:17 pm
Subject: Re: [nanovna-users] NanoVNA Under The Covers
Not to keep this thread going (long), I have been working in the RF
field for way over 30 years. When ever we talked about loss. it is
ASSUMED (you know what that means) that it is negative but I never said
neg return loss, I would just state the loss (if it was gain, it would
be positive) but I don't know of anyone (that I have worked with) that
said "I had a positive gain of 40db. I have said I had negative gain
(meaning the AMP is bad? and I get less power out then I? put in.
My point is with a toy like this, use it for what it is (and cost) and
drive on.
Just my $0.02
Frank
On 8/4/2019 3:02 PM, Stuart Landau via Groups.Io wrote:
? ?
? ? ? ? - r? ? ? ? {\displaystyle RL(\mathrm {dB} )=10\log _{10}{P_{\mathrm {i} } \over P_{\mathrm {r} }}I
? ?
? ? -
I suppose we could start a fight over this, but here is what Wikipedia has to say about the sign of return loss. There is no claim made that this is "the last word" on this subject. I learned to use a negative sign a long time ago.
Sign[edit]
Properly, loss quantities, when expressed in decibels, should be positive numbers.[note 1] However, return loss has historically been expressed as a negative number, and this convention is still widely found in the literature.[1]The correct definition of return loss is the difference in dB between the incident power sent towards the Device Under Test (DUT) and the power reflected, resulting in a positive sign:
? ? ? ? -? ? R L (? d B? ) = 10? log? 10? ?? ? P? i? ? P? r? ? ? ? {\displaystyle RL(\mathrm {dB} )=10\log _{10}{P_{\mathrm {i} } \over P_{\mathrm {r} }}}
However taking the ratio of reflected to incident power results in a negative sign for return loss;
? ? ? ? -? ? R? L ¡ä? (? d B? ) = 10? log? 10? ?? ? P? r? ? P? i? ? ? ? {\displaystyle RL'(\mathrm {dB} )=10\log _{10}{P_{\mathrm {r} } \over P_{\mathrm {i} }}}
? ? ? ? - where RL'(dB) is the negative of RL(dB).
Return loss with a positive sign is identical to the magnitude of ¦£ when expressed in decibels but of opposite sign. That is, return loss with a negative sign is more properly called reflection coefficient.[1] The S-parameter S11 from two-port network theory is frequently also called return loss,[2] but is actually equal to ¦£.Caution is required when discussing increasing or decreasing return loss since these terms strictly have the opposite meaning when return loss is defined as a negative quantity.
Stuart Landau K6YAZLos Angeles, USA
-----Original Message-----
From: Dr. David Kirkby from Kirkby Microwave Ltd <drkirkby@...>
To: nanovna-users <[email protected]>
Sent: Sun, Aug 4, 2019 8:29 am
Subject: Re: [nanovna-users] NanoVNA Under The Covers
On Sun, 4 Aug 2019 at 08:46, tuckvk3cca <tuckvk3cca@...> wrote:
Thanks Gary. I contacted my supplier, EBay and ask if I could get another
set of cal8bration loads.? No luck. The 50 ohm was of very high quality,
return loss -70dB compared to -40dB from another source I had, up to 1GHz.
Anyone knows where to get such loads? Sent from my Samsung Galaxy
smartphone.
Return loss of loads should not stated as a *positive* number as it¡¯s a
loss. A return loss with a negative number could only be obtained from a
negative resistance, such as a Gunn diode amplifier.
A return loss of 70 dB can only found in fairyland! I measured 44 dB, but
stated that the measurement would be subject to large errors as the return
loss of the loads in my 85052B calibration kit were specified as 48 dB.
Really, 70 dB can only be found in fairyland. Another VNA might indicate 70
dB, but the result is meaningless.
Dave G8WRB
