Low-pass filter

Yesterday in the lab we were supposed to calculate the step response of a standart simple RC low-pass filter that looks exactly like the network above.

However, when we try to plot the step function and the step response, it looks like the following graph: Graph

From what I have been told, the plots shouldn't look like that. The capacitator doesn't even charge until it is at 5V. The resistor is 300kΩ and capacitator is 100nF. I think the only possible explanation would be that the two elements aren't proportional to one another, which throws off the ability of the capacitator to fully charge. I don't think it could be measurement error, since it would have to be the same mistake for each period, and I don't think that is likely.

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    \$\begingroup\$ How did you capture the two signals? Did you use a DSO? If yes: the simplest explanation is that you used different offsets and scales for both channels (the relevant buttons of the DSO are called "Position" and "Scale" or "Var"). \$\endgroup\$ – Curd Jan 1 '17 at 22:07

Clearly the output of the filter is being loaded by something, and that something is not just a simple resistor to ground. From the information given, we can calculate the Thevenin equivalent of what is connected to the output of the filter.

By looking only at the settled DC levels, we can ignore the capacitor. This means we have two cases where both the input and output are known. At 0 V in, you get 1 V out. At 5 V in you get 2.5 V out.

Since a change of 5 V on the input results in 1.5 on the output, the gain is 0.3. Since the gain comes only from the resistance of the Thevenin source, we can think of this as a voltage divider. This divider has a top resistor of 300 kΩ and a gain of .3. Working the voltage divider formula backward, this means the bottom resistor is 130 kΩ.

Now we know the resistance of the Thevenin source is 130 kΩ:

We can use any known input and output to find the voltage. I'll use the case of 0 V in producing 1 V out. Again working thru the voltage divider math, we get that V1 must be 1.43 V.

So the Thevenin equivalent connected to the output of the filter is 130 kΩ and 1.43 V. However, it is unlikely you actually have a voltage source and resistor connected to the output. It is more likely that there are two resistors, one to 5 V and the other to ground.

In other words, we'll assume this is what you really have:

From the Thevenin equivalent, we know that the R2/R3 voltage divider produces 1.43 V when unloaded, and the R2//R3 = 130 kΩ. After solving the simultaneous equations, we find that R2 = 455 kΩ and R3 = 182 kΩ.

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