The Nyquist criterion is an important stability test with applications to systems, circuits, and networks [1]. Legal. We will be concerned with the stability of the system. Additional parameters G s {\displaystyle F(s)} F {\displaystyle G(s)} (There is no particular reason that \(a\) needs to be real in this example. H G ( In Cartesian coordinates, the real part of the transfer function is plotted on the X-axis while the imaginary part is plotted on the Y-axis. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. (Using RHP zeros to "cancel out" RHP poles does not remove the instability, but rather ensures that the system will remain unstable even in the presence of feedback, since the closed-loop roots travel between open-loop poles and zeros in the presence of feedback. ) . s r The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. + ( {\displaystyle G(s)} denotes the number of zeros of ) To be able to analyze systems with poles on the imaginary axis, the Nyquist Contour can be modified to avoid passing through the point ) Cauchy's argument principle states that, Where {\displaystyle N} Here, \(\gamma\) is the imaginary \(s\)-axis and \(P_{G, RHP}\) is the number o poles of the original open loop system function \(G(s)\) in the right half-plane. That is, setting (0.375) yields the gain that creates marginal stability (3/2). The poles of the closed loop system function \(G_{CL} (s)\) given in Equation 12.3.2 are the zeros of \(1 + kG(s)\). We regard this closed-loop system as being uncommon or unusual because it is stable for small and large values of gain \(\Lambda\), but unstable for a range of intermediate values. Look at the pole diagram and use the mouse to drag the yellow point up and down the imaginary axis. The most common use of Nyquist plots is for assessing the stability of a system with feedback. ) = s j Hence, the number of counter-clockwise encirclements about Its system function is given by Black's formula, \[G_{CL} (s) = \dfrac{G(s)}{1 + kG(s)},\]. ) This case can be analyzed using our techniques. Non-linear systems must use more complex stability criteria, such as Lyapunov or the circle criterion. -plane, . ( (iii) Given that \ ( k \) is set to 48 : a. are, respectively, the number of zeros of We suppose that we have a clockwise (i.e. The closed loop system function is, \[G_{CL} (s) = \dfrac{G}{1 + kG} = \dfrac{(s - 1)/(s + 1)}{1 + 2(s - 1)/(s + 1)} = \dfrac{s - 1}{3s - 1}.\]. Z {\displaystyle P} ( {\displaystyle N=Z-P} {\displaystyle F(s)} s and poles of is the multiplicity of the pole on the imaginary axis. a clockwise semicircle at L(s)= in "L(s)" (see, The clockwise semicircle at infinity in "s" corresponds to a single While Nyquist is one of the most general stability tests, it is still restricted to linear, time-invariant (LTI) systems. Lecture 2 2 Nyquist Plane Results GMPM Criteria ESAC Criteria Real Axis Nyquist Contour, Unstable Case Nyquist Contour, Stable Case Imaginary Refresh the page, to put the zero and poles back to their original state. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. , let . It is perfectly clear and rolls off the tongue a little easier! As a result, it can be applied to systems defined by non-rational functions, such as systems with delays. The Nyquist criterion allows us to assess the stability properties of a feedback system based on P ( s) C ( s) only. times such that ( We present only the essence of the Nyquist stability criterion and dene the phase and gain stability margins. The following MATLAB commands calculate and plot the two frequency responses and also, for determining phase margins as shown on Figure \(\PageIndex{2}\), an arc of the unit circle centered on the origin of the complex \(O L F R F(\omega)\)-plane. + s s s ( 20 points) b) Using the Bode plots, calculate the phase margin and gain margin for K =1. + ( This has one pole at \(s = 1/3\), so the closed loop system is unstable. %PDF-1.3 % Lecture 1: The Nyquist Criterion S.D. The significant roots of Equation \(\ref{eqn:17.19}\) are shown on Figure \(\PageIndex{3}\): the complete locus of oscillatory roots with positive imaginary parts is shown; only the beginning of the locus of real (exponentially stable) roots is shown, since those roots become progressively more negative as gain \(\Lambda\) increases from the initial small values. Draw the Nyquist plot with \(k = 1\). encircled by 0000001210 00000 n The poles of \(G(s)\) correspond to what are called modes of the system. In fact, we find that the above integral corresponds precisely to the number of times the Nyquist plot encircles the point T We first note that they all have a single zero at the origin. The poles of \(G\). Another aspect of the difference between the plots on the two figures is particularly significant: whereas the plots on Figure \(\PageIndex{1}\) cross the negative \(\operatorname{Re}[O L F R F]\) axis only once as driving frequency \(\omega\) increases, those on Figure \(\PageIndex{4}\) have two phase crossovers, i.e., the phase angle is 180 for two different values of \(\omega\). An approach to this end is through the use of Nyquist techniques. . A linear time invariant system has a system function which is a function of a complex variable. {\displaystyle T(s)} {\displaystyle 1+GH} The oscillatory roots on Figure \(\PageIndex{3}\) show that the closed-loop system is stable for \(\Lambda=0\) up to \(\Lambda \approx 1\), it is unstable for \(\Lambda \approx 1\) up to \(\Lambda \approx 15\), and it becomes stable again for \(\Lambda\) greater than \(\approx 15\). ) 0. If the answer to the first question is yes, how many closed-loop poles are outside the unit circle? \(G(s) = \dfrac{s - 1}{s + 1}\). s A ( 0000000608 00000 n {\displaystyle (-1+j0)} \[G_{CL} (s) = \dfrac{1/(s + a)}{1 + 1/(s + a)} = \dfrac{1}{s + a + 1}.\], This has a pole at \(s = -a - 1\), so it's stable if \(a > -1\). T \nonumber\]. in the right-half complex plane minus the number of poles of Alternatively, and more importantly, if If the number of poles is greater than the number of zeros, then the Nyquist criterion tells us how to use the Nyquist plot to graphically determine the stability of the closed loop system. Is the open loop system stable? {\displaystyle 0+j\omega } ) The range of gains over which the system will be stable can be determined by looking at crossings of the real axis. Note that the phase margin for \(\Lambda=0.7\), found as shown on Figure \(\PageIndex{2}\), is quite clear on Figure \(\PageIndex{4}\) and not at all ambiguous like the gain margin: \(\mathrm{PM}_{0.7} \approx+20^{\circ}\); this value also indicates a stable, but weakly so, closed-loop system. {\displaystyle {\mathcal {T}}(s)={\frac {N(s)}{D(s)}}.}. Moreover, we will add to the same graph the Nyquist plots of frequency response for a case of positive closed-loop stability with \(\Lambda=1 / 2 \Lambda_{n s}=20,000\) s-2, and for a case of closed-loop instability with \(\Lambda= 2 \Lambda_{n s}=80,000\) s-2. ) (3h) lecture: Nyquist diagram and on the effects of feedback. s If, on the other hand, we were to calculate gain margin using the other phase crossing, at about \(-0.04+j 0\), then that would lead to the exaggerated \(\mathrm{GM} \approx 25=28\) dB, which is obviously a defective metric of stability. the number of the counterclockwise encirclements of \(1\) point by the Nyquist plot in the \(GH\)-plane is equal to the number of the unstable poles of the open-loop transfer function. The Nyquist stability criterion has been used extensively in science and engineering to assess the stability of physical systems that can be represented by sets of Precisely, each complex point The Nyquist plot is named after Harry Nyquist, a former engineer at Bell Laboratories. \(\PageIndex{4}\) includes the Nyquist plots for both \(\Lambda=0.7\) and \(\Lambda =\Lambda_{n s 1}\), the latter of which by definition crosses the negative \(\operatorname{Re}[O L F R F]\) axis at the point \(-1+j 0\), not far to the left of where the \(\Lambda=0.7\) plot crosses at about \(-0.73+j 0\); therefore, it might be that the appropriate value of gain margin for \(\Lambda=0.7\) is found from \(1 / \mathrm{GM}_{0.7} \approx 0.73\), so that \(\mathrm{GM}_{0.7} \approx 1.37=2.7\) dB, a small gain margin indicating that the closed-loop system is just weakly stable. Gain margin (GM) is defined by Equation 17.1.8, from which we find, \[\frac{1}{G M(\Lambda)}=|O L F R F(\omega)|_{\mid} \mid \text {at }\left.\angle O L F R F(\omega)\right|_{\Lambda}=-180^{\circ}\label{eqn:17.17} \]. In contrast to Bode plots, it can handle transfer functions with right half-plane singularities. Open the Nyquist Plot applet at. Graphical method of determining the stability of a dynamical system, The Nyquist criterion for systems with poles on the imaginary axis, "Chapter 4.3. ( Does the system have closed-loop poles outside the unit circle? Any Laplace domain transfer function We will look a The algebra involved in canceling the \(s + a\) term in the denominators is exactly the cancellation that makes the poles of \(G\) removable singularities in \(G_{CL}\). s Note on Figure \(\PageIndex{2}\) that the phase-crossover point (phase angle \(\phi=-180^{\circ}\)) and the gain-crossover point (magnitude ratio \(MR = 1\)) of an \(FRF\) are clearly evident on a Nyquist plot, perhaps even more naturally than on a Bode diagram. We can visualize \(G(s)\) using a pole-zero diagram. H , which is to say. Rule 1. 0 k u ) in the right-half complex plane. That is, we consider clockwise encirclements to be positive and counterclockwise encirclements to be negative. A Nyquist plot is a parametric plot of a frequency response used in automatic control and signal processing. { s - 1 } { s + 1 } { s - 1 } \.... 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