I made this integrator circuit by following this diagram. Now, according to the wikipedia, an integrator circuit is an active low pass filter which you can see in the picture below:

So, the cutoff frequency is

$$f_{cutoff} = 159.1 \ \text{Hz}$$ and the unity gain frequency, $$f_{0dB} = 1.59 \ \text{kHz}$$

However, this calculation is only consistent with any sinusoidal waves. The unity gain is different for a square wave.

For a square wave I found that it was:

$$f_{0dB} = 111.5 \ \text{kHz}$$

My questions is, am I misinterpreting the equations or is there something I am missing? If not then why am I getting different cut-off and 0 dB frequencies for a different wave.

enter image description here enter image description here enter image description here enter image description here

  • 7
    \$\begingroup\$ Look up "Fourier transform" and/or "Laplace transform". Also, stop using a 741, they were obsolete before I was born. \$\endgroup\$
    – Hearth
    Sep 21, 2023 at 19:43
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    \$\begingroup\$ The reason I pointed you in that direction is because you seem to be acting like a square wave has only one frequency, when it has infinitely many frequency components. \$\endgroup\$
    – Hearth
    Sep 21, 2023 at 19:47
  • 7
    \$\begingroup\$ Look up "Fourier transform". \$\endgroup\$
    – Hearth
    Sep 21, 2023 at 19:51
  • 6
    \$\begingroup\$ @Ghost All those sine waves of all those different frequencies are independently affected by the integrator. \$\endgroup\$
    – Hearth
    Sep 21, 2023 at 19:56
  • 5
    \$\begingroup\$ What makes you think it doesn't follow the "quintessential" equations? Plot the frequency response. The X-axis is frequency. Singular. The response for a single frequency. A waveform containing a single frequency. (Hint: A sine wave.) Now, your square wave contains many frequencies as @Hearth et. al. have been trying to tell you. Each of those frequencies will be attenuated by a different amount exactly according to the "quintessential" equations. \$\endgroup\$
    – John D
    Sep 21, 2023 at 20:11

2 Answers 2


My questions is, am I misinterpreting the equations or is there something I am missing?

this calculation is only true for any sinusoidal waves

Yes, the equations are only for sinusoids. For other wave shapes obtain the Fourier harmonic sinusoids with their attendant amplitudes. Apply the formulae to each harmonic and sum the results to obtain the amplitude of the frequency of the square wave. Notice that the wave shape has changed. You will be summing phasors.

The wave shape remains the same for sinusoids only.

You will find one or both of the equations are approximations for Rf>>Ri.

  • \$\begingroup\$ I have edited my post and uploaded the actual picture where I found the unity gain for the square wave. Also, it no longer remained a triangular wave. \$\endgroup\$
    – Ghost
    Sep 22, 2023 at 14:51

I am only going to focus on the case where you show a scope image. From it I will find the following for my comparative simulation purposes:

  • \$1\:\text{V}_\text{PK}\$ square wave, centered on \$0\:\text{V}\$, running at \$2.1\:\text{kHz}\$ and 50% duty cycle, and a rise and fall time of \$745\:\text{ns}\$, as shown there.
  • Component values as shown.

A square wave contains all of the positive odd integer harmonics of the primary frequency. As \$C=100\:\text{nF}\$ then \$X_C\$ at the primary frequency is about \$758\:\Omega\$. But given that the parallel \$10\:\text{k}\Omega\$ is at right angles, it has almost no impact at all. The feedback is essentially all capacitive. So this is a simple integrator at this frequency (and at all the odd harmonics above it.) The output will average towards zero, given time.

So this just means that the input resistor determines the current sourced during half the square wave cycle and the current sunk during the other half. In this case, that's \$I=\pm 1\:\text{mA}\$. (I'm ignoring the rise/fall times here.) The capacitor integrates this for half a cycle, or for about \$\Delta t=\frac{0.5}{2100\:\text{Hz}}\approx 238.1\:\mu\text{s}\$, and then de-integrates for the next half a cycle. The capacitor equation says: \$\Delta V=\frac{I}{C}\cdot\Delta t\approx\frac{1\:\text{mA}}{100\:\text{nF}}\cdot 238.1\:\mu\text{s}\approx 2.381\:\text{V}_\text{PP}\$.

Let's look at LTspice's take on this using a decent (expensive) rail-to-rail opamp:

enter image description here

Note that it reports a difference of \$\approx 2.366\:\text{V}\$. This is pretty close to prediction.

Now, if you were to reduce the frequency enough, say an order of magnitude below the cross-over frequency, then the opamp output would look a lot like another square wave with rounded edges but with the magnitude determined by your DC gain figure which here is 10 (well, -10, really.) So the square wave would have tops and bottoms at \$\pm 10\:\text{V}\$ in this case.

In between, you get something in-between-looking.

  • \$\begingroup\$ Thank you for your very elaborated answer. However, it actually went over my head. My question was why am I getting a different unity gain frequency for the square wave which is 111.5 kHz? And, it does not follow the cut-off and unity gain frequency. \$\endgroup\$
    – Ghost
    Sep 22, 2023 at 15:57
  • \$\begingroup\$ @Ghost It's okay. Not every answer connects well. The other answer you have is also correct and another good way to go. Someone may yet add another. The basic idea in what I'm saying is that everything important has already happened within an order of magnitude of the cross-over frequency and, in between, it's an amalgam. Here, at 1/10th and below, the capacitor doesn't matter and all you have is the remaining DC gain. So out looks like in with gain. At 10X and above, the feedback resistor doesn't matter and its an integrator and not enough time for DC gain, so tiny peak-to-peak triangle. \$\endgroup\$ Sep 22, 2023 at 17:43
  • \$\begingroup\$ @Ghost In between, the very RC edge on each side of the square wave at 1/10th and below starts to gradually dominate the edges, expanding their relative widths, while shrinking the width of the DC gain tops and bottoms, until there's no longer any time for the DC gain of 10 tops and bottoms and the edges take over, completely. There's still the RC shape, but now the top-to-bottom height shrinks down because there's no time to reach them, now. Eventually, all you have left is the rapidly rising/falling part of edges, which look almost straight. And here, the triangle has re-appeared. \$\endgroup\$ Sep 22, 2023 at 17:47
  • \$\begingroup\$ @Ghost You will get all the same thing using RusselH's explanation. But to get there, you will have to look at the many sums, collectively, while treating everything at the input as a sum of odd-X harmonic but also perfect sine waves, which are at the input and at the output but with differing magnitudes at the input and therefore also different at the output, before re-summing them back together. It all just works. \$\endgroup\$ Sep 22, 2023 at 17:49

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