First, let's clarify that you mean active (op-amp) circuits. IMO the integrator is always stable (741 is a godd example of "an integrator inside an op-amp":) while the differentiator needs a correction (a resistor in series to the input capacitor and a capacitor in parallel to the resistor) to be stable. A simple explanation is that these elements limit the overall gain at high frequencies... and the circuit is actually a band-pass filter...

Dear Cyril

Integrator and differentiator circuits that have op-amp is non linear circuit because of the presence of active element and we can't apply BIBO stability analysis on non linear circuit as it is only applicable to LTI system. Hence we can't judge the stability of such active circuit by BIBO analysis.

In my question, integrator and differentiator, that i have mention is simply a RC circuit

My doubt is that " Is it really unstable according to BIBO stability criterion " . If NO, then can we generate a ramp signal by this passive integtrator, while input is a step signal?

I wonder how a passive circuit can be unstable?

To generate a ramp signal by a passive C integtrator, the input signal has to be a step current signal.

passive circuits will not necessarily be stable under all stability criteria. For instance, a resonant series LC circuit will have unbounded voltage output for a bounded voltage input, but will be stable in the sense of Lyapunov, and given bounded energy input will have bounded energy output.

In control theory, a continuous linear time-invariant system is exponentially stable if and only if the system has eigenvalues (the poles of input-to-output systems) with strictly negative real parts. (i.e., in the left half of the complex plane). Systems that are not LTI are exponentially stable if their convergence is bounded by exponential decay. For example if output is a sinusoidal signal then it starts oscillate and system become marginal stable, but if we this output is mutiplied by function exp(-t) then this starts to decay the oscilations and system achieve a constant steady state value.

An exponentially stable LTI system is one that will not "blow up" (i.e., give an unbounded output) when given a finite input or non-zero initial condition. Moreover, if the system is given a fixed, finite input (i.e., a step), then any resulting oscillations in the output will decay at an exponential rate, and the output will tend asymptotically to a new final, steady-state value. If the system is instead given a Dirac delta impulse as input, then induced oscillations will die away and the system will return to its previous value. If oscillations do not die away, or the system does not return to its original output when an impulse is applied, the system is instead marginally stable.

System that is linear can be analysed by BIBO method, but a non linear system's stability is examined by a different method ( based on energy) called as " Lyapunov Method" ( developed by a Russian mathematician Aleksandr Lyapunov )

for more detail: visit

http://en.wikipedia.org/wiki/Lyapunov_stability

According to this " If the system loses energy over time and the energy is never restored then eventually the system must grind to a stop and reach some final resting state. This final state is called the attractor. However, finding a function that gives the precise energy of a physical system can be difficult, and for abstract mathematical systems, economic systems or biological systems, the concept of energy may not be applicable. Lyapunov's realization was that stability can be proven without requiring knowledge of the true physical energy, provided a Lyapunov function can be found to satisfy the above constraints."

Theoretically it sounds very good...

@Salvatore:

Sometime theoretical explanation doesn't meet the practical phenomenon.

If you pass a Square wave through a High pass filter, then theoretically output should be unbounded at the point of discontinuity, because transition time is of order ns at the point of discontinuity but practically if you see the output, it is saturated to +Vsat or -Vsat of op-amp.

Similarly in case of step input to the integrator circuit, theoretically output will be unbounded but practically it is saturated to the Maximum input DC voltage i.e. it hits the power supply rails and output is saturated.

In practical case, energy is always finite and to produce a unbounded output we need an infinite energy, which is not possible in any practical scenario. That's why we never get an unbounded output

I think if the output is unbounded in the steady state then the system is unstable. Accordingly integrator is unstable whereas differentiator is stable.

It is interesting to see how a real integrator can be designed so that to oscillate?

Most amplifiers use negative feedback to trade gain for other desirable properties, such as decreased distortion or improved noise reduction. Ideally, the phase characteristic of an amplifier's frequency response would be linear; however, device limitations make this goal physically unattainable. More particularly, capacitance within the amplifier's gain stages cause the output signal to lag behind the input signal by 90° for each pole they create.

It should be expected that the outputs of even the simplest operational amplifiers will have at least two poles. An unfortunate consequence of this is that at some critical frequency, the phase of the amplifier's output = -180° compared to the phase of its input signal. The amplifier will oscillate if it has a gain of one or more at this critical frequency. This is because

1. The feedback is implemented through the use of an inverting input that adds an additional -180° to the output phase making the total phase shift -360° and

2. The gain is sufficient to induce oscillation.

Feeding back any portion of this output signal to the input when the gain of the amplifier is sufficient will cause the amplifier to oscillate. This is because the feedback signal will reinforce the input signal. That is, the feedback is then positive rather than negative.

Frequency compensation is implemented to avoid this result.

I answered the question from control systems view. I felt it is what Nandan Kumar is looking for.

Nandan, it is interesting to see why we apply a negative feedback when design oscillator circuits with positive feedback. Can we directly use a positive feedback?

In oscillator, the general problem is harmonic distortion which is due to the clipping of oscillation amplitude in saturation region (amplifying elements goes into saturation). so if you want to design a oscillator of very low distortion then you must ensure that the amplifying element of oscillator must be working in linear region. To work properly in linear region, amplitude stabilization is mandatory and hence negative feedback used in that low distortion oscillators to control the amplitude of oscillation in linear region. this method of controlling the amplitude in linear region used in wein bridge oscillator.

But I mean the initial negative feedback (180 degree) that then becomes positive (e.g., by connecting a phase shifting circuit in the feedback loop)... Why we do not simply use a positive feedback?

Negative feedback is used to ensure proper DC biasing!

You can not put the amplifier in linear operating point (when the gain is large) with positive feedback, because the amplifier will be saturated to one of the supply rails!

Ring oscillators (3 or 5 or.. inverters in a ring), Schidt trigger oscillators etc., use also negative feedback for this reason. If the numbers of inverters are even, the ring oscillator becomes latch.

About stability:

stability make sense only in a feedback systems.

Considering integrators and differentiators alone is nonsense!

Op amp differentiator has DC bias, so if the input is open it is stable if the used op amp is unity gain stable. But, if you connect the input capacitor to ground the circuit becomes unstable, because the capacitor introduces additional pole in the negative op amp feedback.

Op amp integrator alone has no DC bias, and alone the output will be saturated to one of the supply rails, so it inherently needs external DC negative feedback for biasing! Depending on the phase lag in this external bias it can be stable or not.

The integrator characteristic characterizes with two points – unity gain frequency and DC gain.

Each op amp is an integrator!

Ideal integrator tr. function is: 1/(tau_u*s) , tau_u defines unity gain frequency.

One pole op amp tr. function is : Aol/(1+tau_p*s)=1/(1/Aol+tau_p*s/Aol) ~1/(tau_u*s), where tau_p defines dominant pole frequency, and again tau_u=tau_p/Aol defines unity gain frequency.

The op amp without feedback operates as comparator.

I think, you should read some basic books about op amps, circuit theory and stability!

bye

Here is the Barrie Gilbert's "Integrator inside":

http://www.edn.com/electronics-blogs/anablog/4311648/Op-amp-myths-ndash-by-Barrie-Gilbert

Please refer to this article:

https://www.researchgate.net/publication/309655805_On_the_Improved_Nonlinear_Tracking_Differentiator_based_Nonlinear_PID_Controller_Design

Article On the Improved Nonlinear Tracking Differentiator based Nonl...

RC integrator obeys asymptotic stability criteria with zero input(it is stable).When step input is provided to integrator ,then it becomes unstable.

"When step input is provided to integrator ,then it becomes unstable. "

No that is not true!

Saturation is not identical to instability!

In control theory, stability is often defined by "BIBO", bounded input bounded output stability, a form of stability for linear signals and systems that take inputs. If a system is BIBO stable, then the output will be bounded for every input to the system that is bounded.

https://en.wikipedia.org/wiki/BIBO_stability

Of course, then, there are bounded inputs that provide an unbounded output from both a integrator and differentiator.

Then again, stability theory is a wide realm

https://en.wikipedia.org/wiki/Stability_theory

Try and separate mathematics from the real world.

If you use software to differentiate a real-world rectangular signal, this will have a finite rise-time and fall time. so dv/dt will not be infinite. All algorithms avoid the division by zero at the zero-crossing points of the waveform by assuming that anything divided by zero equals zero.

Also, differentiating with a capacitor and resistor will further distance the output from the ideal, due to ESR and parasitic inductance.

A software-based integration of the same signal will effectively calculate the area under the waveform, which will be finite, since the algorithm effectively uses Simpson's Rule. Again, a resistor and capacitor will produce worse results.

Stability doesn't enter into it.