# Why “-1 + 0j” is critical point on Nyquist plot?

I am reading about Nyquist plot from this (http://www.cds.caltech.edu/~murray/books/AM08/pdf/fbs-loopanal_11Nov18.pdf) chapter. On page no. 10-2 its mentioned that "-1 + 0j" is critical point on Nyquist plot. Can someone explain why is it critical point?

It's the point where the open-loop gain is unity, and phase angle is -180 deg. So if the Nyquist trajectory passes through this point, the closed-loop will be marginally stable (continuous oscillation).

In general, the relative stability of the closed-loop can be determined from the locus of the Nyquist trajectory in the neighbourhood of the critical point.

According to the Barkausen oscillation criterion, the loop gain LG must be unity (magnitude unity and zero phase) at a certain (desired) oscillation frequency. Hence, this condition (unity loop gain, LG=+1) is the stability limit.

On the other hand, the DC operational point of the active device requires NEGATIVE feedback for DC (factor "-1"). That means: If the circuit (open feedback loop) reaches the stability limit, the frequency-determining part of the whole loop must exhibit an aditional phase shift of -180deg.

It is common practice to create a Nyquist plot for this part of the loop gain only (without the minus sign at the summing node for negative feedback).

Hence, the stability limit in the Nyquist plot for this part of the loop (without the minus sign) is at unity magnitude(1) and -180deg (on the negative real axis of the s-plane):

Critical point in the s-plane: "-1".

Comment: In analogsystemsrf$$`$$s answer the product GH is identical to product of the transfer functions of the whole loop - WITHOUT the negative sign (-1) at the summing junction. Hence, the complete loop gain of the whole loop is LG=-G*H.

If we examine the standard feedback equation (only one path --- not PID --- thru the system; only one node of comparison) often written as

G/(1 + G*H),

then for G*H = -1+J0 the denominator becomes ZERO.