# Summing Integrator cut off

I would like to know how to calculate cut off frequency of this summing integrator circuit. I fail to find anywhere on the internet how to calculate cut off frequency in this configuration. Then how do I calculate unity-gain frequency this circuit?

• The unity gain, you can get it from Andy's answer, it's 0dB intersection point 1/Ti = 1/RC. – Marko Buršič Jul 20 at 18:43
• Do i just sum them up? (1/R1C)+ (1/R2C) – Sundark12 Jul 20 at 18:54
• Of course not, these are two separate inputs, with separate integration constants, therefore the unity gain is different for each input. – Marko Buršič Jul 20 at 18:55
• Vo(t) shud be -1/R1C.... The output is the negative sum of two integrals – Tony Stewart Sunnyskyguy EE75 Jul 20 at 18:56
• By the way the Vout equation is wrong. It's $V_o=- \dfrac{1}{R_1 C} \int{v_1 dt} - \dfrac{1}{R_2 C} \int{v_2 dt}$ – Marko Buršič Jul 20 at 19:00

Both passive and active integrators are -6dB / octave and do not have an upper limit defined by -3dB.

More likely you might consider the passive integrator has a lower limit frequency perhaps defined by -6dB or more depending on your error tolerance.

The true integrator needs an initial condition or a conditional reset or slow leakage to some desired level.

The link above shown below gives you some interactive tools to see the difference. I would like to know how to calculate cut off frequency of this summing integrator circuit.

Integrators ideally don't have a cut-off frequency: - Then how do I calculate unity-gain frequency this circuit?

For each input, calculate $$\\tau\$$ where $$\\tau=C\cdot R\$$ then, take the reciprocal. That answer will be in radians per second so, to get it in hertz, divide by $$\2\pi\$$. With different input resistor values there are different unity-gain frequencies for each input. Picture from this slide-player

• Couldn't a frequency, when the output amplitude falls below -3dB be called as cut-off – Marko Buršič Jul 20 at 18:27
• @MarkoBuršič an ideal integrator has infinite gain at DC so, what frequency would be 3 dB down on infinity? Of course for a practical op-amp integrator this can be estimated based on knowledge of the open-loop transfer function but, that isn't disclosed in the question. – Andy aka Jul 20 at 18:31
• Frankly, I was never thinking about integrator's cut-off frequency, but if there would be a such frequency, I would choose the -3dB with respect to the input signal amplitude, just thinking, of course without any phase shift it would be difficult to call it a cut-off. – Marko Buršič Jul 20 at 18:40

A real integrator circuit (using real opamps) is in fact a first order lowpass with a very low 3dB-cut-off frequency wo (caused by the finite open-loop gain of the opamp).

However, as far as the integrator function is concerned, this frequency wo could be seen as a kind of "start frequency" for the begin of the integrating property. But note that real integration means "90 deg phase shift". And this is the case (real amplifiers) for one single frequency w(90)=SQRT(wp/T) only. (wp:opamps open-loop pole frequency, T: Integrator time constant).

But in practice there is a broader frequency region which is used for integrating purposes with a phase shift between app 89.5...90.5 deg.

This allowable deviation from the ideal phase shift could be used for defining a corresponding frequency range and two "cut-off frequencies" which describe the practical range of integration. But these "cut-off" frequencies have, of course, nothing to do with any 3dB-values.