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


1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.