2. Introduction

Starting from initial steady state, let us consider that the set point is varied sinusoidally with, \(Y_{sp}(t) = \sin(\omega t)\), for a long period of time. Assume that during this period the measured output, \(Y_m\), is disconnected so that the feedback loop is broken before the comparator.

After the initial transient dies out, \(Y_m\) will oscillate at the excitation frequency \(\omega\) because the response of a linear system to a sinusoidal input is a sinusoidal output at the same frequency.

If the input frequency is corresponding to a phase shift of \(-180^\circ\) for the open-loop, then for a setpoint of \(Y_{sp}(t)=\sin(\omega t)\), we get \(Y_m(t)=\sin(\omega t-180^\circ) = -\sin(\omega t)\).

The frequency where the phase lag is equal to \(180^\circ\) is called the crossover frequency and is denoted by \(\omega_{co}\).


Suppose that two events occur simultaneously: (i) the set point is set to zero and, (ii) \(Y_m\) is reconnected. Under these conditions, the comparator inverts the sign of the \(Y_m\), which now plays the same role as that played by the setpoint in the ‘open-loop’. Now the error (\(\epsilon\)) remains the same.

Theoretically, the response of the system will continue to oscillate with constant amplitude, for \(\text{AR} = 1\), despite the fact that both the load and setpoint do not change.

If \(\text{AR}>1\) when \(\phi=-180^\circ\), the closed-loop system will exhibit oscillations with ever-increasing amplitude leading to an unstable system.

On the contrary, if \(\text{AR}<1\), when \(\phi=-180^\circ\), the oscillating response of the closed-loop will exhibit a continuously decreasing amplitude, leading to eventual dying out of the oscillation.