Find the stability where:
Find the stability: \$ G(s)= \frac{(s+3)(s+5)}{(s-2)(s-4)}; \$ \$ H(s)= 1; \$
I run this code in matlab:

clc;close all;clear all;
G = tf([1 8 15],[1 -6 8]);                  
H = 1;
Cloop = feedback(G,H)

Output is:

enter image description here enter image description here

From the two Graph What I understand is:
\$ N= 0 \$ as \$ -1 \$ is not encircled and \$ P=2 \$ as there are two right-hand side poles.
So, \$ Z=N+P=2 \$ From this, the system is Unstable as there are two zeros in the Right-hand side.
But from the step response, it is obvious that it is stable.
So what's the problem here; Which is correct??

  • \$\begingroup\$ Try this and use the open loop nyquist plot to determine stability. \$\endgroup\$ – Andy aka Dec 9 '18 at 11:35

The Nyquist criterion requires to plot the response of the LOOP GAIN (feedback loop open!) and to apply the stability check. This check tells you if the CLOSED loop will be stable or not.

Question 1: You have shown the step response - is it the response of the closed loop or of the open loop?

Question 2: What is the sign at the summing node for closing the loop ? Please note that the "critical point" for checking the stability is "-1" only for the simple product G*H (that means: not including the minus sign for negative feedback). Otherwise you have to investigate the point "+1".

My result: If the loop is closed (negative feedback) the system is stable. Your Nyquist plot starts in the vicinity of "+4" and circles the point "+1" twice counterclockwise.

(It starts at "+4" because the critical pint in your plot is not at "-1" but at "+1". Hence, a shift to the right.)

  • \$\begingroup\$ Yes, the respnse is for closed loop.And thanks for the reply.It's vey helpful. \$\endgroup\$ – Fazla Rabbi Mashrur Dec 9 '18 at 12:07

You are drawing the nyquist diagram for the closed loop system but you need to do it for the open loop system (i.e. G(s)H(s)). The correct nyqyist is then

G=zpk([-3 -5],[2 4],1);

It shows two encirclements of -1 (i.e. in counterclockwise direction,hence N=2).

enter image description here

The number of closed-loop poles in the right-half plane is Z=P-N=2-2=0. Since there is no closed-loop poles in the right-half plane, the system is stable.


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