Hallo, ich habe das Problem das mein NanoVNA als USB Gerät erkannt wird und ich habe kein zugewiesenen COM Port. Somit kann ich mit der App auch nicht drauf zugreifen. Welchen Treiber muss ich für Win

I have a spreadsheet I wrote to calculate and design low noise amplifiers and power amplifiers with microwave FET's. It uses s-parameters. Some of the equations are huge. They involve adding and subtr

But either format represents a complex number. Maybe "polar" and "rectangular" would be better. 73, Maynard W6PAP

No, it isn't. It's just a complex number. A complex number does not equate to an impedance. How do you explain the fact that an s21 of zero indicates no power out of the device for any power in? And a

Hi Andy It's my misuse of "Cartesian", I really should have said "complex" rather than Cartesian. Jeff

Hi Jeff, If we have a polar format of R / theta conversion to Cartesian format can be done via: R*(cos(theta) + j*sin(theta)) So your polar value of 0.5 angle 90deg in Cartesian form is: 0.5*(cos(90)

Forgot to add If you take a network that is merely a 150 ohm resistor then the Reflection Coefficient and S11 will change if you view it in 75 ohm system rather than a 50 ohm one. You won't have a con

Hi Andy I didn't miss out anything really, I was not trying to explain how you transformed from Polar to Cartesian forms (plotting on a Smith Chart is an easy way), just highlighting that S parameters

Take S11 for example and a purely resistive case; to achieve a 3:1 VSWR in a 50 ohm system you can use a 150ohm resistor, but 225ohms is a 75 ohm system. Yes, the Reflection Coefficient will be 0.5 at

Jeff, You've missed out a couple of steps. S11 in polar form as 0.5 angle 90deg is 0 + j0.5 in Cartesian format. You chose an easy one to work out. Other easy ones might be a short circuit: 1 angle 18

I think this is where we differ. A VSWR of 3:1 will have identical S parameters whether it be in a 50- or 75-ohm system. Half the power being reflected is what matters in a system and whatever the con

Sorry, I don't understand your point. If you take 0.5 at 90deg in a 75 ohm system the VSWR remains at 3:1 but the impedance will be 45+j60 ohms. The system impedance must be stated (or assumed to be 5

Again, it's a ratio, not an impedance! David -- SatSignal Software - Quality software for you Web: https://www.satsignal.eu Email: davidtaylor@... Twitter: @gm8arv

... and if that system is 75-ohm rather than 50-ohm? The S11 is a ratio expressed as a complex number, not an impedance itself. David -- SatSignal Software - Quality software for you Web: https://www.

It is a measure of the impedance and phase shift through the network presented to the incident wave. Regards Jeff

Further to my last and to make the point clearer: let's take an example of an S11 expressed in polar form of 0.5 at 90 degrees (reflection co-efficient of 0.5 at 90 degrees). That can also be expresse

I don't see how S21 (for example) is an impedance. David -- SatSignal Software - Quality software for you Web: https://www.satsignal.eu Email: davidtaylor@... Twitter: @gm8arv

Yes, but that complex number is an impedance!!! It can be plotted directly on Smith Chart as an impedance, just as Y parameters are admittances and can be plotted as such. Regards Jeff

Are these filters you are trying to measure active or passive? If they are passive, you wouldn't need to power up the radio. You have s11 and s21 backwards. S11 is the reflection coefficient of the in

Sorry to burst your bubble Jeff, but the S parameters are not impedances, and like any complex number it can be represented in polar or Cartesian format. Read the first few pages here, you’ll see that