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I observed a strange fact while using a simple RC high pass filter. Here is the circuit I used. enter image description here On \$CH1\$ I measured \$V_{out}\$, while on \$CH2\$ I observed the value of \$V_{in}\$.

I kept the function generator at \$V_0=20V\$ pk-pk (sinusoidal), and I increased the frequency from \$10 Hz\$ to \$10^4 Hz\$ in steps.

The strange thing is that voltage on \$CH2\$ (which is \$V_{in}\$) decreased while the frequency was increased! That is, at low frequency \$V_{in} =V_0=20 V\$ (pk-pk), but then it decreased to \$4.5V\$ at \$10^4 Hz\$!

Nevertheless the filter seemed to work ok because if I plot the ratios \$\frac{Vout}{Vin}\$ (whith \$V_{in}\$ measured with oscilloscpe, hence decreasing) vs frequency (on log scale) I get the right curve, corresponding to my cutoff frequency \$f=\frac{1}{2\pi RC}\$.

Obviously doing a simulation on Multisim or other simulators I get that \$V_{in}= V_{out}\$ (besides little variations).

Then I tried to build another filter with a different capacitor and I did not observe this strange behaviour, which makes me think that this was probably caused by the capacitor I used.

What I would like to know is if there can be any explanation for this behaviour of the filter wich is due to the characteristics of the capacitor used.

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  • \$\begingroup\$ that's a very low impedance filter you have there. What's the output impedance of the generator? Was the 'other filter' built with a different value capacitor? \$\endgroup\$
    – Neil_UK
    Feb 9 '17 at 17:43
  • \$\begingroup\$ It was a normal \$50 \Omega\$ output impedance. I tried to put a different capacitor (0.1 \$\mu F \$) (the function generator was exactly the same) and I observed that \$V_{in}\$ was approximately constant increasing the frequency \$\endgroup\$
    – Sørën
    Feb 9 '17 at 17:52
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As frequency rises, the impedance of the capacitor drops and, at a significantly higher frequency than the natural 3 dB point (2605 Hz) of the filter, you could make an argument that the capacitor becomes, in effect, a short circuit hence, the output of your signal generator will be loaded by the 13 ohm resistor.

If the signal generator's output impedance is 50 ohm, the 13 ohm resistor loading the output would reduce the amplitude down to 20% of the unloaded value. This of course adds another level of complexity to things but, providing you plot the ratio of Vout to Vin it won't make a difference.

Reducing the 4.7 uF to 100 nF means the cut-off point when Xc = 13 ohms is much higher at 122 kHz but you would see the same effect as you approached this higher frequency.

Ideally, with a 50 ohm impedance signal generator, to keep the input voltage largely constant across frequency, you should use a much higher value for the load resistor i.e. something like 1 kohm. If this were repeated with the 100 nF capacitor, the 3 dB point would be nearly the same as the original at 1591 Hz. The output would drop slightly at higher frequencies and tend to reduce to a value of about 95% of its low frequency value.

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  • \$\begingroup\$ Thanks a lot for this answer! Since I do not know exactly how output impedance work in my function generator: isn't the function generator supposed to provide the voltage difference choosen "beyond" its output impedance? In other words, if I set voltage to 20Vpp isn't there a voltage of 20Vpp at the terminals of the function generator? Or should I really subtract \$|Z_{out} I|\$ if I want the voltage actually delivered to the load? (if it is useful I used a AimTTi TG2000 and I selected 20Vpp on the menu). \$\endgroup\$
    – Sørën
    Feb 11 '17 at 23:19
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    \$\begingroup\$ A function generator is a perfect voltage source in series with its specified output impedance. There can be no magic or it would not be compatible with driving a transmission line because it would not be able to absorb reflections. So, if you load the output with a resistor equivalent to its output impedance, you get half the voltage. \$\endgroup\$
    – Andy aka
    Feb 12 '17 at 0:29
  • \$\begingroup\$ Doing again the calculations this really seems to explain the data I got! So I must consider a \$50 \Omega\$ resistance in series! If I may ask one last thing, when I did the measurements I connected to the function generator a BNC coaxial cable with characteristic impedance of \$50 \Omega\$. Must the cable impedance be added to the output impedance of the function generator (\$50 \Omega \$ too)?(that would give a total of \$100 \Omega\$, which would be strange) Or is the impedance of cable already taken into account in the output impedance of function generator? \$\endgroup\$
    – Sørën
    Feb 19 '17 at 18:30
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    \$\begingroup\$ The 50 ohm of the coax tells you what impedance the coax has to be terminated in the avoid reflections and standing waves. At the frequencies you are considering, there are negligible effects so, ignore it. \$\endgroup\$
    – Andy aka
    Feb 19 '17 at 20:13

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