# Nyquist Plot

The stability analysis of a feedback control system is based on identifying the location of the roots of the characteristic equation on s-plane. The system is stable if the roots lie on left hand side of s-plane. Relative stability of a system can be determined by using frequency response methods like **Nyquist plot** and Bode plot. Nyquist criterion is used to identify the presence of roots of a characteristic equation in a specified region of s-plane. To understand **Nyquist plot** we need to know about some of the terminologies.

Contour : Closed path in a complex plane is called contour.

## Nyquist path or Nyquist contour

The**Nyquist contour**is a closed contour in the s-plane which completely encloses the entire right hand half of s-plane. In order to enclose the complete RHS of s-plane a large semicircle path is drawn with diameter along jω axis and centre at origin. The radius of the semicircle is treated as Nyquist Encirclement.

## Nyquist Encirclement

A point is said to be encircled by a contour if it is found inside the contour.## Nyquist Mapping

The process by which a point in s-plane is transformed into a point in F(s) plane is called mapping and F(s) is called mapping function.### Steps of drawing the Nyquist path

- Step 1 - Check for the poles of G(s) H(s) of jω axis including that at origin.
- Step 2 - Select the proper Nyquist contour – a) Include the entire right half of s-plane by drawing a semicircle of radius R with R tends to infinity.
- Step 3 - Identify the various segments on the contour with reference to
**Nyquist path** - Step 4 - Perform the mapping segment by segment by substituting the equation for respective segment in the mapping function. Basically we have to sketch the polar plots of the respective segment.
- Step 5 - Mapping of the segments are usually mirror images of mapping of respective path of +ve imaginary axis.
- Step 6 - The semicircular path which covers the right half of s plane generally maps into a point in G(s) H(s) plane.
- Step 7- Interconnect all the mapping of different segments to yield the required
**Nyquist diagram**. - Step 8 - Note the number of clockwise encirclement about (-1, 0) and decide stability by N = Z – P

is the Closed loop transfer function (C.L.T.F)
N(s) = 0 is the open loop zero and D(s) is the open loop pole
From stability point of view no closed loop poles should lie in the RH side of s-plane. Characteristics equation 1 + G(s) H(s) = 0 denotes closed loop poles .
Now as 1 + G(s) H(s) = 0 hence q(s) should also be zero.
Therefore , from the stability point of view zeroes of q(s) should not lie in RHP of s-plane.

To define the stability entire RHP (Right Hand Plane) is considered. We assume a semicircle which encloses all points in the RHP by considering the radius of the semicircle R tends to infinity. [R → ∞].

The first step to understand the application of **Nyquist criterion** in relation for determination of stability of control systems is mapping from s-plane to G(s) H(s) - plane. s is considered as independent complex variable and corresponding value of G(s) H(s) being the dependent variable plotted in another complex plane called G(s) H(s) - plane. Thus for every point in s-plane there exists a corresponding point in G(s) H(s) - plane. During the process of mapping the independent variable s is varied along a specified path in s - plane and the corresponding points in G(s)H(s) plane are joined. This completes the process of mapping from s-plane to G(s)H(s) - plane.

**Nyquist stability criterion** says that N = Z - P. Where, N is the total no. of encirclement about the origin, P is the total no. of poles and Z is the total no. of zeroes.

Case 1 :- N = 0 (no encirclement), so Z = P = 0 and Z = P

If N = 0, P must be zero therefore system is stable.

Case 2 :- N > 0 (clockwise encirclement), so P = 0, Z ≠0 and Z > P

For both cases system is unstable.

Case 3 :- N < 0 (counter clockwise encirclement), so Z = 0, P ≠0 and P > Z

System is stable.

A | ALAA commented on 12/05/2018folish |

S | SRINIVAS commented on 09/04/2018It will be easy if you have given an example problem |