High output impedance from voltage source to a low pass filter?

I'm a first year EEE undergrad student. One of my labs was about low-pass filters (simple RC ones). When connecting the voltage source to the low pass filter circuit, the instructions (of the lab exercise) said to set the function generator to High Output Impedance.

simulate this circuit – Schematic created using CircuitLab

• Mar 2, 2019 at 18:37
• This may help: youtu.be/fOJ1aExSkhw
– Big6
Mar 2, 2019 at 18:44
• It means that you have to switch off the 50 ohm ouptut resistor. Mar 2, 2019 at 18:58

Typically function generators have a low output impedance amplifier in series with a 50 ohm resistor to the output BNC connector.

If you set the output to 50 ohms it will produce double the set amplitude at the amplifier output in anticipation of an external 50 ohm cable and termination dividing that voltage equally. If you apply only a very light load you will see that doubled voltage at the terminals.

If you set high impedance, the generator assumes very light loading and produces the set amplitude at the internal amplifier output node.

Your loading approaches 100 ohms at high frequencies so it is neither high nor low in comparison to 50 ohms, so to measure your filter response you should measure both the input and the output amplitudes.

• Agreed - it does not change the output of the signal generator at all, just the voltage that is produced for a given setting. Mar 2, 2019 at 21:28

Your low-pass filter circuit is incomplete, missing the proper Thevenin equivalent of the function generator:

simulate this circuit – Schematic created using CircuitLab

Many function generators default to having Rs=50 ohms. Some audio applications expect an internal resistance of 600 ohms, and can switch from one Rs to another Rs. A good function generator takes care to appear as an accurate Thevenin equivalent.

• The function generators I have used do not change the output impedance, they just alter the calibration of the unit to reflect that the load is high impedance. Mar 2, 2019 at 21:30

simulate this circuit – Schematic created using CircuitLab

simulate this circuit – Schematic created using CircuitLab

This is a buffer before the LPF. It is an impedance transformer, it has very big impedace at the input and very low impedance at the output. In such way the source impedance has no effect on the RC circuit.

What you asked is to switch the source to Hi Z. Well the function generator usually has 50 ohm impedance in series which can be also bypassed on some models. What you get is a low impedance output, not high. Also having a source with high impedance makes no sense, probably it's a typo, they meant switch the source to low impedance.

The transfer function of buffered LPF is $$\H(s)=\dfrac{1}{1+sRC}\$$. If you interconnect three same circuits you get: $$\H(s)=\dfrac{1}{(1+sRC)^3}\$$ But if you ommit the buffer, then the transfer function is different, since they are not isolated and they interact.

IMO the scope of the exercise is to get a transfer function of this LPF, so you have to feed it with low impedance.