# Nyquist stability criterion

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The Nyquist stability criterion, named for Harry Nyquist, provides a simple test for stability of a closed-loop control system by examining the open-loop system's Nyquist plot. Stability of the closed-loop control system may be determined directly by computing the poles of the closed-loop transfer function. In contrast, the Nyquist stability criterion allows stability to be determined without computing the closed-loop poles.

## BackgroundEdit

We consider a system whose open loop transfer function (OLTF) is G(s); when placed in a closed loop with feedback H(s), the closed loop transfer function (CLTF) then becomes G/(GH+1). The case where H=1 is usually taken, when investigating stability, and then the "Characteristic Equation', used to predict stability, becomes G+1=0. Stability can be determined by examining the roots of this equation eg using the Routh array, but this method is somewhat tedious. Conclusions can also be reached by examining the OLTF, using its Bode plots or, as here, polar plot of the OLTF using the Nyquist criterion , as follows.

Any Laplace domain transfer function $\mathcal{T}(s)$ can be expressed as the ratio of two polynomials

$\mathcal{T}(s) = \frac{N(s)}{D(s)}$.

We define:

• Zero: the zeros of $\mathcal{T}(s)$ are the roots of $N(s) = 0$, and
• Pole: the poles of $\mathcal{T}(s)$ are the roots of $D(s) = 0$.

Stability of $\mathcal{T}(s)$ is determined by its poles or simply the roots of the characteristic equation: $D(s) = 0$. For stability, the real part of every pole must be negative. If $\mathcal{T}(s)$ is formed by closing a negative feedback loop around the open-loop transfer function $\mathcal{F}(s) = \frac{A(s)}{B(s)}$, then the roots of the characteristic equation are also the zeros of $1 + \mathcal{F}(s)$, or simply the roots of $A(s) + B(s)$.

## Cauchy's Argument principleEdit

From complex analysis, we know that a contour $\Gamma_s$ drawn in the complex $s$ plane, encompassing but not passing through any number of non-analytic points, can be mapped to another plane (the $F(s)$ plane) by a function $F(s)$. The resulting contour $\Gamma_{F(s)}$ will encircle the origin of the $F(S)$ plane $N$ times, where $N = Z - P$. $Z$ and $P$ are the number of zeros and poles of $F(s)$, respectively. Note that we count encirlements in the $F(s)$ plane in the same sense as the contour $\Gamma_s$ and that encirclements in the opposite direction are negative encirclements.

## The Nyquist criterionEdit

We first construct The Nyquist Contour, a contour that encompasses the right-half of the complex plane:

• a path traveling up the $j\omega$ axis, from $0 - j\infty$ to $0 + j\infty$.
• a semicircular arc, with radius $r \to \infty$, that starts at $0 + j\infty$ and travels clock-wise to $0 - j\infty$.

The Nyquist Contour mapped through the open-loop transfer function $F(s)$ yields a Nyquist plot for $F(s)$. By the Argument Principle, the number of clock-wise encirclements of the origin must be the number of zeros of $F(s)$ in the right-half complex plane minus the poles of $F(s)$ in the right-half complex plane. If we look at the contour's encirclements of -1 instead of the origin, we find the difference between the number of poles and zeros in the right-half complex plane of $1+F(s)$. Recalling that the zeros of $1+F(s)$ are the poles of the close-loop system, and noting that the poles of $1+F(s)$ are same as the poles of $F(s)$, we now state The Nyquist Criterion:

Given a Nyquist contour $\Gamma_s$, let $P$ be the number of poles of $F(s)$ encircled by $\Gamma_s$, and $Z$ be the number of zeros of $F(s)$ encircled by $\Gamma_s$ -- therefore the number of poles of $\mathcal{T}(s)$ enclosed by $\Gamma_s$. The resultant contour in the $F(s)$-plane, $\Gamma_{F(s)}$ shall, for a stable feedback system, encircle (clock-wise) the point (-1 + j0) $N$ times such that $N = Z - P$.

Summary:

• If the open-loop transfer function $F(s)$ is stable, then the closed-loop system is unstable for any encirclement of the point -1.
• If the open-loop transfer function $F(s)$ is unstable, then there must be one counter clock-wise encirclement of -1 for each pole of $F(s)$ in the right-half of the complex plane.
• The number of surplus encirclements (greater than N+P) is exactly the number of unstable poles of the closed-loop system