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I am trying to intuitively understand the response of the RC circuit for a triangular input. My understanding is as follows:

Let the ramp be mt.u(t)

Initially, the voltage drops mainly across the resistor and generates a ramp current. This current passes through the capacitor and then produces a parabolic response. In steady state, the capacitor gets charged to mt and introduces a constant current of mC. So, there is a constant delay of mRC between input and output.

For a triangular waveform, I understood it behaves the same as that of ramp input. What I don't understand is how the output waveform behaves at the peaks of the triangular wave. The output almost looks like a sine wave.

enter image description here

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It's an illusion. If you calculate the response of a 1st order lowpass with a ramp input and then differentiate its response, you'll find these:

$$\begin{align} H(s)&=\dfrac{1}{s\tau+1} \\ X(s)&=\dfrac{1}{s^2} \\ y(t)&=\mathcal{L}^{-1}\left(H(s)X(s)\right)=\tau\mathrm{e}^{-t/\tau}-\tau+t=t-\tau(1-\mathrm{e}^{-t/\tau}) \tag{1} \\ \dfrac{\mathrm{d}y(t)}{\mathrm{d}t}&=1-\mathrm{e}^{-t/\tau} \tag{2} \end{align}$$

You'll probably recognize (2) as the step response, which means (1) is the integrated step response. No wonder, since a ramp is an integrated Heaviside. So what you see is a "smoothed out" version of the input, which is a pulsed exponential (infinitely differentiable over the interval). And it's smooth at the peaks because the derivative of (1) at zero is zero.

The output almost looks like a sine wave.

The illusion continues: a symmetrical triangular waveform has Fourier components at:

$$\sum_{k=0}^\infty{(-1)^k\dfrac{\sin(2\pi(2k+1)t)}{(2k+1)^2}} \tag{2}$$

and you can see that the denominator is squared, so the harmonics vanish pretty quickly. By comparison, a square wave has the denominator without a power (and no \$(-1)^k\$ term). So the difference will be exactly (1) vs (2).

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