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I've been struggling to understand how filter capacitors specifically low pass filters achieve their output signal waveform. If I take the example from https://www.electronics-tutorials.ws/rc/rc_3.html:

enter image description here

The output waveform for the right side looks like a saw-tooth/triangualr type waveform. I'm confused why it doesn't look something like this though:

enter image description here

My confusion stems from the thought that measuring across the right side there its in parallel with the capacitor and if there is a high signal across the nodes going to the capacitor then how come we wouldn't see that on the measure points to the right of the capacitor?

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    \$\begingroup\$ Neither drawing is correct, though as your link eventually admits in the fine print the first one might be approximate (though they neglect to mention out of scale) for pulse durations substantially below the RC time constant. The thing you need to realize is that when a resistor limits current, it takes time for a capacitor to change voltage, and because the current through a resistor depends on the voltage change across it, the behavior is self-referential and therefor exponetional. To get an actual triangle, you need a voltage-independent current source, not a resistor. \$\endgroup\$ – Chris Stratton Aug 9 '18 at 0:37
  • \$\begingroup\$ So I get that they are not exactly correct since it wouldn't be perfectly triangular, what I still don't understand even after reading your comment is why I dont see anything that resembles a square on the Vout side of things \$\endgroup\$ – csteifel Aug 9 '18 at 1:06
  • \$\begingroup\$ Don't expect a square wave on the output. Think about what happens to voltage in the time after you throw a switch putting a limited current into a capacitor. Then throw the switch the other way taking a limited current out. That's basically what your square wave does, alternately charging it up and draining it down - both operations which take time. \$\endgroup\$ – Chris Stratton Aug 9 '18 at 1:09
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    \$\begingroup\$ In this question - link - the same source figure is used, and I've detailed in the answer how to obtain the output signal via bode analysis. The website you've linked in the question actually does a fine job explaining how the output waveform is obtained by combining step responses. I can't understand where your doubt lies. \$\endgroup\$ – Vicente Cunha Aug 9 '18 at 4:39
  • \$\begingroup\$ Remember how mobile batteries charge up linear then slower. Batteries are like massive caps and voltage slope depends on current. I=CdV/dt \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Aug 9 '18 at 5:30
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Both square and triangular waveforms are possible, as the asymptotic shapes in the limits of very high or very low frequencies into the RC filter. At middling frequencies, you get a an exponential shape.

enter image description here

The legend is the number of RC time constants that a single high or low input period, a half cycle, lasts.

You can see that for 0.2, the output amplitude is small. The voltage across the resistor doesn't change much during the high or low time, so the current is more or less constant. The output waveform approximates to a triangle, and will approximate more closely as the number of RCs per half cycle falls.

For 10 RCs per half cycle, the output is almost square. For 18, it's more nearly square again.

The 'classic' RC output we tend to draw is that for 3 RCs per half cycle. Clearly neither square nor triangular.

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Well, we can calculate that mathematically. We know that your circuit is a voltage divider (using the s-domain):

$$\text{V}_{\space\text{o}}\left(\text{s}\right)=\frac{\frac{1}{\text{s}\text{C}}}{\frac{1}{\text{s}\text{C}}+\text{R}}\cdot\text{V}_{\space\text{i}}\left(\text{s}\right)=\frac{1}{1+\text{s}\text{C}\text{R}}\cdot\text{V}_{\space\text{i}}\left(\text{s}\right)\tag1$$

A positive square wave with lower value \$0\$ V and higher value \$\hat{\text{u}}\$, we get for \$\text{V}_{\space\text{i}}\left(\text{s}\right)\$:

$$\text{V}_{\space\text{i}}\left(\text{s}\right)=\frac{1}{1-\exp\left(-\text{T}\cdot\text{s}\right)}\cdot\int_0^\text{T}\text{V}_{\space\text{i}}\left(t\right)\cdot e^{-\text{s}t}\space\text{d}t\tag2$$

Where \$\text{T}=\frac{1}{\text{f}}\$ (the frequency of the input signal).

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