Trying to quantify attenuation with Function Generator

[ErikG]

Hello all,

I am trying to verify if me theoretical predictions for a low pass filter match the real world attenuation.
My setup is a function generator, outputting a sine wave at 1Vpp, which with no load is really 2Vpp.
The generator is connected to the input of my low pass filter. Just a simple RC circuit with a 68 Ohm resistor and a 1.3uF cap to ground.
The load on the output is matched to the 50 ohm series resistance in the function generator.
When the LPF is connected to the function generator, the output gets loaded down a little bit. In this case its about .3v so the output is 1.7v

My question is this:
Should i use my unloaded voltage to calculate attenuation or the loaded voltage to figure out the attenuation in the circuit?
IE 20*log(1.7Vpp/output of LPF)
or the unloaded output of the function generator
IE 20*log(2Vpp/output of LPF)

Thanks for your help.
-ErikG

 

[BradtheRad]

All resistances seen by the capacitor play a part in the time constant.

Suppose the function generator (or preceding stage) were to have a different output resistance. Then your LP filter will display a different rolloff curve.

As for measuring the attenuation, it is relative to which you consider the previous state:

(a) whether the 68 ohm load was already in place, then you added the capacitor, or
(b) whether you added both the capacitor and 68 ohm load together.

Notice that 'A' is a case of attaching a capacitor at the node of a resistor divider. This could make for a more visible indication of the capacitor's effect.

 

[ErikG]

Thanks. That makes sense about the LPF having a different roll off curve. It is something i modeled in LTspice.

The current state is (B). The resistor and capacitor was added together, as was the 50 ohm load. Here is my schematic:


I hope this helps
-Erik

 

[D.A.(Tony)Stewart]

The output will be rated voltage (1v) when the load matches the source impedance.

 

[chuckey]

In general filters are measured at the system impedance, which is usually 50 or 75 or 600 ohms, though they can be any value you choose. The generator output impedance must equal the system impedance must equal the load impedance. Now the voltage across the load will be .5 of the generators voltage and can be taken as zero loss (0 dB). Now you insert your filter between the generator and load, leave the generator level control where it was, swing its frequency and measure the output voltage with respect to the 0dB level.
Now you can decide on your system impedance, you can shunt the generators output terminals to reduce its output impedance, or put a series resistor in to increase it to match the load resistor. set your 0dB level across the load. insert the filter and away you go.
Frank

 

[ErikG]

Thanks Chuckey, your explanation helped immensely.
Everything makes sense now.
Thank you all for your help.

Have a wonderful weekend,
-ErikG