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I'm a bit confused on output impedance of function generator and coaxial cable impedance.

Consider a function generator with output impedance of \$50 \Omega\$. Suppose that I connect to it a BNC \$50 \Omega\$ coaxial cable and then i connect it to my circuit.

I can think of this configuration as in picture

schematic

simulate this circuit – Schematic created using CircuitLab

My question is: how big is \$R_{int}\$? \$50 \Omega \$ or \$100 \Omega\$? And what does contributes to \$R_{int}\$? The function generator output impedance or the coaxial cable impedance or both?

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  • \$\begingroup\$ Output current will vary as the loaded) Generator output voltage varies; this deltaIout causes variation in the transconductance of the pushpull ClassB bipolar output devices; changes in the transconductance change the Rout of the generator, leading to distortion in the output voltage. \$\endgroup\$ – analogsystemsrf Feb 19 '17 at 22:58
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There are two ways to represent this, one which corresponds to an ideal model, and one which is realistic.

The idea model:

You have an ideal voltage source (with zero output impedance) with a \$50\Omega\$ resistor in series (\$R_{int}\$), and so the total output impedance is 50Ω. In your schematic above, the voltage source is ideal, and so \$R_{int}\$ would be \$50\Omega\$.

The realistic model:

The voltage source is not ideal and has some output impedance, and then the series resistor is selected so that the total output impedance is \$50\Omega\$. For instance, let's say there's an op-amp in there (quite likely). The op-amp has some output impedance, \$R_o < 50 \Omega\$, because it is not an ideal voltage source, and then the series resistor \$R_{er}\$ (internal to the function generator, not internal to the op-amp) is selected so that \$R_o + R_{ser} = 50 \Omega\$, and thus the total output impedance is \$50\Omega\$.

If you are modelling circuits in a simulator like LTSpice, you can do this two ways: you can add a \$50\Omega\$ resistor to an ideal voltage source, or you can set the series resistance of the voltage source to \$50\Omega\$. When designing a real circuit, more care is necessary and you must carefully read the datasheet to determine the output impedance of your driver.

schematic

simulate this circuit – Schematic created using CircuitLab

To sum up:

...And what does contributes to \$R_{int}\$? The function generator output impedance or the coaxial cable impedance or both?

The internal impedance of the voltage driver and a resistor inside the function generator contribute to \$R_{int}\$. The cable impedance does not contribute to this. The function generator's output impedance will be \$50\Omega\$.

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  • \$\begingroup\$ Thanks for the answer, but then, if I have the BNC cable connected to the output (and then to circuit, which means that current flows) there will be a voltage drop along the BNC anyway (due to a \$50 \Omega \$ resistance) right? \$\endgroup\$ – Sørën Feb 19 '17 at 18:10
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    \$\begingroup\$ @Sørën No, because the cable does not have a 50Ω resistance (measure it with a multimeter). At DC, it has a very low resistance. Only at higher frequencies does the impedance come into play. This is a very involved subject, and I would suggest you read up transmission line theory and the nature of electromagnetic radiation if you wish to better understand this. \$\endgroup\$ – uint128_t Feb 19 '17 at 18:16

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