Kriteriyê Stabilitetê ya Nyquist: Yê ne? (Bîgire Hejmarên Matlab)
Nyquist Stability Criterion Yehiye Çiye
Nyquist stability criterion (yane Nyquist criteria) kontrol mühendisiyê de dinamik bir sistemîn istikrarina pîşandekirinê wekheviya grafikiyê taybetand. Nyquist stability criteria tevahî ya open-loop control systemsê an jî Nyquist plotê biceribîne, ji bo ku bi serfirazê pol û zeros ê closed-loop yane open-loop systemê hesab bike.
Vê yola, Nyquist criteria tevahîên non-rational functions bi (waşte systems û delays) pêkide. Bode plots derbasdar transfer functions bi singularities di herêmî sê yê dawî de nekar hatine.
Nyquist Stability Criterion wekheviya bikar îne:
Z = N + P
Yê:
Z = roots of 1+G(s)H(s) di herêmî sê yê dawî de (It is also called zeros of characteristics equation)
N = number of encirclement of critical point 1+j0 in the clockwise direction
P = poles of open loop transfer function (OLTF) [i.e. G(s)H(s)] di herêmî sê yê dawî de.
The above condition (i.e. Z=N+P) is valid for all the systems whether stable or unstable.
Now we will explain this criterion with examples of Nyquist stability criterion.
Nyquist Stability Criterion Examples
Nyquist Criterion Example 1
Consider an open-loop transfer function (OLTF) as 
Is it a stable system or an unstable. Perhaps most of you will say it is an unstable system because one pole is at +2. However, note that stability depends on the denominator of the closed-loop transfer function.
If any root of the denominator of the closed-loop transfer function (also called characteristics equation) is at RHS of the s-plane then the system is unstable. So in the case above, a pole at +2 will try to bring the system towards instability, but the system may be stable. Here Nyquist plot is useful to find stability.
According to Nyquist theory Z=N+P (for any system, whether it is stable or unstable).
For the stable system, Z=0, i.e. No roots of characteristics equation should be at RHS.
So for the stable system N = –P.
The Nyquist plot of the above system is as shown below
Nyquist Plot Matlab Code
s = tf('s')
G1 = 120 / ((s-2)*(s+6)*(s+8))
nyquist(G1, 'red')
As per the diagram, Nyquist plot encircle the point –1+j0 (also called critical point) once in a counter clock wise direction. Therefore N= –1, In OLTF, one pole (at +2) is at RHS, hence P =1. You can see N= –P, hence system is stable.
If you will find roots of characteristics equation, it will be –10.3, –0.86±j1.24. (i.e. system is stable), and Z=0. One question can be asked, if roots of characteristics equation can be found, so we can comment on the stability on that basis, then what is the need of Nyquist plot. The answer is, when software’s were not available, in those days Nyquist plot was very useful.
Nyquist Criterion Example 2
Now take another example: 
Nyquist plot is as follows:
Nyquist Plot Matlab Code
Pêşniyariyek
