The output voltage of an RC integrator circuit is proportional to the integral of the input voltage.

This means that a triangular wave should transform into a quadratic curve (since the integral of a linear function is a parabola).

Why does it convert a triangular wave into a sine wave then? What am I missing here?

## Best Answer

So, a square wave has harmonics at all odd multiples of the fundamental, with strength diminishing as the number of the harmonic:

$$x_s(t) \simeq \sum_{n = 0}^\infty \frac{\cos 2 \pi (2n + 1) t}{2n + 1}$$

A triangle wave is just the square wave, integrated, with the appropriate constant added to make things tidy:

$$x_\Delta(t) \simeq \sum_{n = 0}^\infty \frac{\sin 2 \pi (2n + 1) t}{(2n + 1)^2}$$

A parabolic wave is the same thing, again:

$$x_p(t) \simeq \sum_{n = 0}^\infty -\frac{\cos 2 \pi (2n + 1) t}{(2n + 1)^3}$$

Because of that cubic in the denominator, the difference between a sine wave and this "pseudo-sine" wave is very small. And it turns out that it's pretty hard to see the deviation on a graph anyway.

If you plot the quadratic pseudo-sine wave on a graph, superimposed on a

realsine wave, you'll see the (slight) difference.