F D are, respectively, the number of zeros of The condition for the stability of the system in 19.3 is assured if the zeros of 1 + L are all in the left half of the complex plane. G G ( + G are called the zeros of The same plot can be described using polar coordinates, where gain of the transfer function is the radial coordinate, and the phase of the transfer function is the corresponding angular coordinate. Sudhoff Energy Sources Analysis Consortium ESAC DC Stability Toolbox Tutorial January 4, 2002 Version 2.1. These are the same systems as in the examples just above. The theorem recognizes these. Thus, this physical system (of Figures 16.3.1, 16.3.2, and 17.1.2) is considered a common system, for which gain margin and phase margin provide clear and unambiguous metrics of stability. ( ). N Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. Nyquist stability criterion is a general stability test that checks for the stability of linear time-invariant systems. The Nyquist bandwidth is defined to be the frequency spectrum from dc to fs/2.The frequency spectrum is divided into an infinite number of Nyquist zones, each having a width equal to 0.5fs as shown. Nyquist Plot Example 1, Procedure to draw Nyquist plot in There is one branch of the root-locus for every root of b (s). F ) 0 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. P G times such that F We know from Figure \(\PageIndex{3}\) that this case of \(\Lambda=4.75\) is closed-loop unstable. There are 11 rules that, if followed correctly, will allow you to create a correct root-locus graph. s are the poles of the closed-loop system, and noting that the poles of s F right half plane. As \(k\) goes to 0, the Nyquist plot shrinks to a single point at the origin. On the other hand, the phase margin shown on Figure \(\PageIndex{6}\), \(\mathrm{PM}_{18.5} \approx+12^{\circ}\), correctly indicates weak stability. l {\displaystyle F(s)} . ( {\displaystyle G(s)} The MATLAB commands follow that calculate [from Equations 17.1.7 and 17.1.12] and plot these cases of open-loop frequency-response function, and the resulting Nyquist diagram (after additional editing): >> olfrf01=wb./(j*w.*(j*w+coj). Now refresh the browser to restore the applet to its original state. {\displaystyle 0+j\omega } The poles of the closed loop system function \(G_{CL} (s)\) given in Equation 12.3.2 are the zeros of \(1 + kG(s)\). \[G_{CL} (s) \text{ is stable } \Leftrightarrow \text{ Ind} (kG \circ \gamma, -1) = P_{G, RHP}\]. {\displaystyle 1+GH(s)} Answer: The closed loop system is stable for \(k\) (roughly) between 0.7 and 3.10. Legal. 1 Now how can I verify this formula for the open-loop transfer function: H ( s) = 1 s 3 ( s + 1) The Nyquist plot is attached in the image. = the same system without its feedback loop). {\displaystyle {\mathcal {T}}(s)} The stability of If s On the other hand, a Bode diagram displays the phase-crossover and gain-crossover frequencies, which are not explicit on a traditional Nyquist plot. 0 ( , where s Suppose \(G(s) = \dfrac{s + 1}{s - 1}\). The Nyquist criterion is widely used in electronics and control system engineering, as well as other fields, for designing and analyzing systems with feedback. + + 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. Then the closed loop system with feedback factor \(k\) is stable if and only if the winding number of the Nyquist plot around \(w = -1\) equals the number of poles of \(G(s)\) in the right half-plane. 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. {\displaystyle 0+j\omega } s In this case, we have, \[G_{CL} (s) = \dfrac{G(s)}{1 + kG(s)} = \dfrac{\dfrac{s - 1}{(s - 0.33)^2 + 1.75^2}}{1 + \dfrac{k(s - 1)}{(s - 0.33)^2 + 1.75^2}} = \dfrac{s - 1}{(s - 0.33)^2 + 1.75^2 + k(s - 1)} \nonumber\], \[(s - 0.33)^2 + 1.75^2 + k(s - 1) = s^2 + (k - 0.66)s + 0.33^2 + 1.75^2 - k \nonumber\], For a quadratic with positive coefficients the roots both have negative real part. are also said to be the roots of the characteristic equation All the coefficients of the characteristic polynomial, s 4 + 2 s 3 + s 2 + 2 s + 1 are positive. Lecture 1 2 Were not really interested in stability analysis though, we really are interested in driving design specs. {\displaystyle 1+G(s)} But in physical systems, complex poles will tend to come in conjugate pairs.). G Choose \(R\) large enough that the (finite number) of poles and zeros of \(G\) in the right half-plane are all inside \(\gamma_R\). {\displaystyle \Gamma _{s}} ( s The \(\Lambda=\Lambda_{n s 1}\) plot of Figure \(\PageIndex{4}\) is expanded radially outward on Figure \(\PageIndex{5}\) by the factor of \(4.75 / 0.96438=4.9254\), so the loop for high frequencies beneath the negative \(\operatorname{Re}[O L F R F]\) axis is more prominent than on Figure \(\PageIndex{4}\). 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. Does the system have closed-loop poles outside the unit circle? H ) + Check the \(Formula\) box. To use this criterion, the frequency response data of a system must be presented as a polar plot in {\displaystyle Z=N+P} ( plane, encompassing but not passing through any number of zeros and poles of a function = + + ( Any clockwise encirclements of the critical point by the open-loop frequency response (when judged from low frequency to high frequency) would indicate that the feedback control system would be destabilizing if the loop were closed. Is the system with system function \(G(s) = \dfrac{s}{(s + 2) (s^2 + 4s + 5)}\) stable? G ) ( Matrix Result This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. F F \[G(s) = \dfrac{1}{(s - s_0)^n} (b_n + b_{n - 1} (s - s_0) + \ a_0 (s - s_0)^n + a_1 (s - s_0)^{n + 1} + \ ),\], \[\begin{array} {rcl} {G_{CL} (s)} & = & {\dfrac{\dfrac{1}{(s - s_0)^n} (b_n + b_{n - 1} (s - s_0) + \ a_0 (s - s_0)^n + \ )}{1 + \dfrac{k}{(s - s_0)^n} (b_n + b_{n - 1} (s - s_0) + \ a_0 (s - s_0)^n + \ )}} \\ { } & = & {\dfrac{(b_n + b_{n - 1} (s - s_0) + \ a_0 (s - s_0)^n + \ )}{(s - s_0)^n + k (b_n + b_{n - 1} (s - s_0) + \ a_0 (s - s_0)^n + \ )}} \end{array}\], which is clearly analytic at \(s_0\). and However, the gain margin calculated from either of the two phase crossovers suggests instability, showing that both are deceptively defective metrics of stability. 1 ) ) {\displaystyle G(s)} B + From now on we will allow ourselves to be a little more casual and say the system \(G(s)\)'. This assumption holds in many interesting cases. ( {\displaystyle G(s)} Is the system with system function \(G(s) = \dfrac{s}{(s + 2) (s^2 + 4)}\) stable? So the winding number is -1, which does not equal the number of poles of \(G\) in the right half-plane. ( This is a case where feedback destabilized a stable system. ), Start with a system whose characteristic equation is given by {\displaystyle \Gamma _{s}} {\displaystyle v(u)={\frac {u-1}{k}}} When \(k\) is small the Nyquist plot has winding number 0 around -1. G around D T Phase margins are indicated graphically on Figure \(\PageIndex{2}\). Mark the roots of b + G 0 For gain \(\Lambda = 18.5\), there are two phase crossovers: one evident on Figure \(\PageIndex{6}\) at \(-18.5 / 15.0356+j 0=-1.230+j 0\), and the other way beyond the range of Figure \(\PageIndex{6}\) at \(-18.5 / 0.96438+j 0=-19.18+j 0\). That is, the Nyquist plot is the image of the imaginary axis under the map \(w = kG(s)\). {\displaystyle G(s)} When plotted computationally, one needs to be careful to cover all frequencies of interest. If the answer to the first question is yes, how many closed-loop poles are outside the unit circle? u , then the roots of the characteristic equation are also the zeros of {\displaystyle Z} T {\displaystyle A(s)+B(s)=0} s . Routh Hurwitz Stability Criterion Calculator I learned about this in ELEC 341, the systems and controls class. ) s (Actually, for \(a = 0\) the open loop is marginally stable, but it is fully stabilized by the closed loop.). D Hb```f``$02 +0p$ 5;p.BeqkR ) have positive real part. The Nyquist plot is the trajectory of \(K(i\omega) G(i\omega) = ke^{-ia\omega}G(i\omega)\) , where \(i\omega\) traverses the imaginary axis. + The poles are \(-2, -2\pm i\). Pole-zero diagrams for the three systems. yields a plot of ) Since on Figure \(\PageIndex{4}\) there are two different frequencies at which \(\left.\angle O L F R F(\omega)\right|_{\Lambda}=-180^{\circ}\), the definition of gain margin in Equations 17.1.8 and \(\ref{eqn:17.17}\) is ambiguous: at which, if either, of the phase crossovers is it appropriate to read the quantity \(1 / \mathrm{GM}\), as shown on \(\PageIndex{2}\)? 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 roots of s The Nyquist criterion is an important stability test with applications to systems, circuits, and networks [1]. , the closed loop transfer function (CLTF) then becomes {\displaystyle {\mathcal {T}}(s)} ( *(26- w.^2+2*j*w)); >> plot(real(olfrf007),imag(olfrf007)),grid, >> hold,plot(cos(cirangrad),sin(cirangrad)). ) ) Let us continue this study by computing \(OLFRF(\omega)\) and displaying it as a Nyquist plot for an intermediate value of gain, \(\Lambda=4.75\), for which Figure \(\PageIndex{3}\) shows the closed-loop system is unstable. While Nyquist is one of the most general stability tests, it is still restricted to linear, time-invariant (LTI) systems. It can happen! ( ( , as evaluated above, is equal to0. That is, setting G Rearranging, we have The Nyquist criterion is a frequency domain tool which is used in the study of stability. The portion of the Nyquist plot for gain \(\Lambda=4.75\) that is closest to the negative \(\operatorname{Re}[O L F R F]\) axis is shown on Figure \(\PageIndex{5}\). The formula is an easy way to read off the values of the poles and zeros of \(G(s)\). F Techniques like Bode plots, while less general, are sometimes a more useful design tool. If the counterclockwise detour was around a double pole on the axis (for example two Another unusual case that would require the general Nyquist stability criterion is an open-loop system with more than one gain crossover, i.e., a system whose frequency Figure 19.3 : Unity Feedback Confuguration. Nyquist plot of the transfer function s/(s-1)^3. We will just accept this formula. ( Z It is more challenging for higher order systems, but there are methods that dont require computing the poles. {\displaystyle G(s)} We can measure phase margin directly by drawing on the Nyquist diagram a circle with radius of 1 unit and centered on the origin of the complex \(OLFRF\)-plane, so that it passes through the important point \(-1+j 0\). in the complex plane. 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 linear equations. s 0000002847 00000 n We will make a standard assumption that \(G(s)\) is meromorphic with a finite number of (finite) poles. s For this topic we will content ourselves with a statement of the problem with only the tiniest bit of physical context. 1 G(s)= s(s+5)(s+10)500K slopes, frequencies, magnitudes, on the next pages!) ( Recalling that the zeros of For what values of \(a\) is the corresponding closed loop system \(G_{CL} (s)\) stable? {\displaystyle N=P-Z} Gain \(\Lambda\) has physical units of s-1, but we will not bother to show units in the following discussion. In this case the winding number around -1 is 0 and the Nyquist criterion says the closed loop system is stable if and only if the open loop system is stable. While Nyquist is one of the most general stability tests, it is still restricted to linear time-invariant (LTI) systems. 1 ) The closed loop system function is, \[G_{CL} (s) = \dfrac{G}{1 + kG} = \dfrac{(s + 1)/(s - 1)}{1 + 2(s + 1)/(s - 1)} = \dfrac{s + 1}{3s + 1}.\]. j Set the feedback factor \(k = 1\). ) s = using the Routh array, but this method is somewhat tedious. Z D = be the number of zeros of The negative phase margin indicates, to the contrary, instability. The most common use of Nyquist plots is for assessing the stability of a system with feedback. = Based on analysis of the Nyquist Diagram: (i) Comment on the stability of the closed loop system. Since the number of poles of \(G\) in the right half-plane is the same as this winding number, the closed loop system is stable. ) represents how slow or how fast is a reaction is. Routh-Hurwitz and Root-Locus can tell us where the poles of the system are for particular values of gain. / {\displaystyle F(s)} Here N = 1. {\displaystyle F(s)} The poles of \(G\). + + in the contour 0000001503 00000 n Since there are poles on the imaginary axis, the system is marginally stable. The mathematics uses the Laplace transform, which transforms integrals and derivatives in the time domain to simple multiplication and division in the s domain. + \[G_{CL} (s) = \dfrac{1/(s + a)}{1 + 1/(s + a)} = \dfrac{1}{s + a + 1}.\], This has a pole at \(s = -a - 1\), so it's stable if \(a > -1\). 0 The frequency-response curve leading into that loop crosses the \(\operatorname{Re}[O L F R F]\) axis at about \(-0.315+j 0\); if we were to use this phase crossover to calculate gain margin, then we would find \(\mathrm{GM} \approx 1 / 0.315=3.175=10.0\) dB. Is the closed loop system stable when \(k = 2\). Let us consider next an uncommon system, for which the determination of stability or instability requires a more detailed examination of the stability margins. {\displaystyle P} This is a diagram in the \(s\)-plane where we put a small cross at each pole and a small circle at each zero. ( Z (0.375) yields the gain that creates marginal stability (3/2). (At \(s_0\) it equals \(b_n/(kb_n) = 1/k\).). Refresh the page, to put the zero and poles back to their original state. In \(\gamma (\omega)\) the variable is a greek omega and in \(w = G \circ \gamma\) we have a double-u. be the number of poles of If we have time we will do the analysis. ) entire right half plane. ) {\displaystyle -1+j0} 1This transfer function was concocted for the purpose of demonstration. If we set \(k = 3\), the closed loop system is stable. To simulate that testing, we have from Equation \(\ref{eqn:17.18}\), the following equation for the frequency-response function: \[O L F R F(\omega) \equiv O L T F(j \omega)=\Lambda \frac{104-\omega^{2}+4 \times j \omega}{(1+j \omega)\left(26-\omega^{2}+2 \times j \omega\right)}\label{eqn:17.20} \]. {\displaystyle T(s)} , which is to say our Nyquist plot. Give zero-pole diagrams for each of the systems, \[G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s^2 - 4) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4)}\]. H (2 h) lecture: Introduction to the controller's design specifications. If instead, the contour is mapped through the open-loop transfer function {\displaystyle D(s)} s We know from Figure \(\PageIndex{3}\) that the closed-loop system with \(\Lambda = 18.5\) is stable, albeit weakly. G If we were to test experimentally the open-loop part of this system in order to determine the stability of the closed-loop system, what would the open-loop frequency responses be for different values of gain \(\Lambda\)? negatively oriented) contour {\displaystyle 0+j\omega } , we now state the Nyquist Criterion: Given a Nyquist contour -plane, + We can show this formally using Laurent series. The Bode plot for We will be concerned with the stability of the system. Given our definition of stability above, we could, in principle, discuss stability without the slightest idea what it means for physical systems. {\displaystyle 1+kF(s)} Yes! %PDF-1.3 % 1 ) . s {\displaystyle G(s)} in the right half plane, the resultant contour in the {\displaystyle F(s)} s Lets look at an example: Note that I usually dont include negative frequencies in my Nyquist plots. P s -P_PcXJ']b9-@f8+5YjmK p"yHL0:8UK=MY9X"R&t5]M/o 3\\6%W+7J$)^p;% XpXC#::` :@2p1A%TQHD1Mdq!1 If the system is originally open-loop unstable, feedback is necessary to stabilize the system. ( s s G ( s ) and travels anticlockwise to the clockwise direction. 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