Any Laplace domain transfer function N G(s)= s(s+5)(s+10)500K slopes, frequencies, magnitudes, on the next pages!) It applies the principle of argument to an open-loop transfer function to derive information about the stability of the closed-loop systems transfer function. = L is called the open-loop transfer function. (At \(s_0\) it equals \(b_n/(kb_n) = 1/k\).). (ii) Determine the range of \ ( k \) to ensure a stable closed loop response. However, the actual hardware of such an open-loop system could not be subjected to frequency-response experimental testing due to its unstable character, so a control-system engineer would find it necessary to analyze a mathematical model of the system. {\displaystyle s} The most common use of Nyquist plots is for assessing the stability of a system with feedback. P ( ) does not have any pole on the imaginary axis (i.e. = D Let us consider next an uncommon system, for which the determination of stability or instability requires a more detailed examination of the stability margins. 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. L is called the open-loop transfer function. {\displaystyle v(u)={\frac {u-1}{k}}} In control system theory, the RouthHurwitz stability criterion is a mathematical test that is a necessary and sufficient condition for the stability of a linear time-invariant (LTI) dynamical system or control system.A stable system is one whose output signal is bounded; the position, velocity or energy do not increase to infinity as time goes on. There are no poles in the right half-plane. s ( ( So the winding number is -1, which does not equal the number of poles of \(G\) in the right half-plane. ) has zeros outside the open left-half-plane (commonly initialized as OLHP). 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. 0 0000039933 00000 n
If we have time we will do the analysis. 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. ) To get a feel for the Nyquist plot. Given our definition of stability above, we could, in principle, discuss stability without the slightest idea what it means for physical systems. There are two poles in the right half-plane, so the open loop system \(G(s)\) is unstable. ) H Nyquist stability criterion is a general stability test that checks for the stability of linear time-invariant systems. 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G We can show this formally using Laurent series. Suppose \(G(s) = \dfrac{s + 1}{s - 1}\). 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 \(\Lambda\)? {\displaystyle H(s)} right half plane. Nyquist Plot Example 1, Procedure to draw Nyquist plot in In fact, the RHP zero can make the unstable pole unobservable and therefore not stabilizable through feedback.). ( = Determining Stability using the Nyquist Plot - Erik Cheever 1 {\displaystyle Z} 0000002305 00000 n
The approach explained here is similar to the approach used by Leroy MacColl (Fundamental theory of servomechanisms 1945) or by Hendrik Bode (Network analysis and feedback amplifier design 1945), both of whom also worked for Bell Laboratories. 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. P ( For example, the unusual case of an open-loop system that has unstable poles requires the general Nyquist stability criterion. That is, if all the poles of \(G\) have negative real part. This is possible for small systems. For a SISO feedback system the closed-looptransfer function is given by where represents the system and is the feedback element. F The poles are \(\pm 2, -2 \pm i\). 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 \(G_{CL}\). ) The Nyquist criterion is a graphical technique for telling whether an unstable linear time invariant system can be stabilized using a negative feedback loop. Typically, the complex variable is denoted by \(s\) and a capital letter is used for the system function. ( Suppose that the open-loop transfer function of a system is1, \[G(s) \times H(s) \equiv O L T F(s)=\Lambda \frac{s^{2}+4 s+104}{(s+1)\left(s^{2}+2 s+26\right)}=\Lambda \frac{s^{2}+4 s+104}{s^{3}+3 s^{2}+28 s+26}\label{eqn:17.18} \]. T Looking at Equation 12.3.2, there are two possible sources of poles for \(G_{CL}\). {\displaystyle {\mathcal {T}}(s)} If {\displaystyle D(s)=1+kG(s)} Choose \(R\) large enough that the (finite number) of poles and zeros of \(G\) in the right half-plane are all inside \(\gamma_R\). *(26- w.^2+2*j*w)); >> plot(real(olfrf007),imag(olfrf007)),grid, >> hold,plot(cos(cirangrad),sin(cirangrad)). Since they are all in the left half-plane, the system is stable. Nyquist plot of \(G(s) = 1/(s + 1)\), with \(k = 1\). s Mark the roots of b . where \(k\) is called the feedback factor. = j To begin this study, we will repeat the Nyquist plot of Figure 17.2.2, the closed-loop neutral-stability case, for which \(\Lambda=\Lambda_{n s}=40,000\) s-2 and \(\omega_{n s}=100 \sqrt{3}\) rad/s, but over a narrower band of excitation frequencies, \(100 \leq \omega \leq 1,000\) rad/s, or \(1 / \sqrt{3} \leq \omega / \omega_{n s} \leq 10 / \sqrt{3}\); the intent here is to restrict our attention primarily to frequency response for which the phase lag exceeds about 150, i.e., for which the frequency-response curve in the \(OLFRF\)-plane is somewhat close to the negative real axis. To use this criterion, the frequency response data of a system must be presented as a polar plot in 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. Since \(G_{CL}\) is a system function, we can ask if the system is stable. The frequency is swept as a parameter, resulting in a plot per frequency. {\displaystyle 1+G(s)} ( This happens when, \[0.66 < k < 0.33^2 + 1.75^2 \approx 3.17. s This results from the requirement of the argument principle that the contour cannot pass through any pole of the mapping function. travels along an arc of infinite radius by P . If the system is originally open-loop unstable, feedback is necessary to stabilize the system. The Nyquist criterion gives a graphical method for checking the stability of the closed loop system. For our purposes it would require and an indented contour along the imaginary axis. = Assessment of the stability of a closed-loop negative feedback system is done by applying the Nyquist stability criterion to the Nyquist plot of the open-loop system (i.e. This approach appears in most modern textbooks on control theory. G ) We dont analyze stability by plotting the open-loop gain or ) {\displaystyle F(s)} s F . s 0.375=3/2 (the current gain (4) multiplied by the gain margin Give zero-pole diagrams for each of the systems, \[G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s^2 - 4) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4)}\]. {\displaystyle G(s)} G One way to do it is to construct a semicircular arc with radius When \(k\) is small the Nyquist plot has winding number 0 around -1. {\displaystyle 0+j(\omega -r)} Cauchy's argument principle states that, Where s Routh Hurwitz Stability Criterion Calculator I learned about this in ELEC 341, the systems and controls class. In the case \(G(s)\) is a fractional linear transformation, so we know it maps the imaginary axis to a circle. as defined above corresponds to a stable unity-feedback system when We first construct the Nyquist contour, a contour that encompasses the right-half of the complex plane: The Nyquist contour mapped through the function Static and dynamic specifications. the clockwise direction. as the first and second order system. v The assumption that \(G(s)\) decays 0 to as \(s\) goes to \(\infty\) implies that in the limit, the entire curve \(kG \circ C_R\) becomes a single point at the origin. ( To use this criterion, the frequency response data of a system must be presented as a polar plot in which the magnitude and the phase angle are expressed as The left hand graph is the pole-zero diagram. Lecture 2 2 Nyquist Plane Results GMPM Criteria ESAC Criteria Real Axis Nyquist Contour, Unstable Case Nyquist Contour, Stable Case Imaginary G Transfer Function System Order -thorder system Characteristic Equation ) If the counterclockwise detour was around a double pole on the axis (for example two ( ). 1 We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. 0000000701 00000 n
Is the closed loop system stable when \(k = 2\). The factor \(k = 2\) will scale the circle in the previous example by 2. It does not represent any specific real physical system, but it has characteristics that are representative of some real systems. Z and travels anticlockwise to >> olfrf01=(104-w.^2+4*j*w)./((1+j*w). k If the system with system function \(G(s)\) is unstable it can sometimes be stabilized by what is called a negative feedback loop. 1 The following MATLAB commands calculate [from Equations 17.1.12 and \(\ref{eqn:17.20}\)] and plot the frequency response and an arc of the unit circle centered at the origin of the complex \(OLFRF(\omega)\)-plane. I'm confused due to the fact that the Nyquist stability criterion and looking at the transfer function doesn't give the same results whether a feedback system is stable or not. ) {\displaystyle G(s)} Hence, the number of counter-clockwise encirclements about While Nyquist is one of the most general stability tests, it is still restricted to linear, time-invariant (LTI) systems. 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 Z Let \(G(s)\) be such a system function. It is certainly reasonable to call a system that does this in response to a zero signal (often called no input) unstable. {\displaystyle D(s)=0} + {\displaystyle \Gamma _{s}} {\displaystyle D(s)} ( In contrast to Bode plots, it can handle transfer functions with right half-plane singularities. ( plane) by the function s In the previous problem could you determine analytically the range of \(k\) where \(G_{CL} (s)\) is stable? The poles are \(-2, \pm 2i\). We will be concerned with the stability of the system. D The reason we use the Nyquist Stability Criterion is that it gives use information about the relative stability of a system and gives us clues as to how to make a system more stable. s have positive real part. Does the system have closed-loop poles outside the unit circle? Natural Language; Math Input; Extended Keyboard Examples Upload Random. (3h) lecture: Nyquist diagram and on the effects of feedback. The Nyquist criterion allows us to answer two questions: 1. {\displaystyle 1+G(s)} {\displaystyle N=Z-P} 0 {\displaystyle 1+G(s)} , and the roots of By the argument principle, the number of clockwise encirclements of the origin must be the number of zeros of s Nyquist Stability Criterion A feedback system is stable if and only if \(N=-P\), i.e. {\displaystyle GH(s)} s 0 We begin by considering the closed-loop characteristic polynomial (4.23) where L ( z) denotes the loop gain. From now on we will allow ourselves to be a little more casual and say the system \(G(s)\)'. Precisely, each complex point For example, Brogan, 1974, page 25, wrote Experience has shown that acceptable transient response will usually require stability margins on the order of \(\mathrm{PM}>30^{\circ}\), \(\mathrm{GM}>6\) dB. Franklin, et al., 1991, page 285, wrote Many engineers think directly in terms of \(\text { PM }\) in judging whether a control system is adequately stabilized. ( 20 points) b) Using the Bode plots, calculate the phase margin and gain margin for K =1. , let {\displaystyle P} {\displaystyle D(s)} ( {\displaystyle (-1+j0)} Note that a closed-loop-stable case has \(0<1 / \mathrm{GM}_{\mathrm{S}}<1\) so that \(\mathrm{GM}_{\mathrm{S}}>1\), and a closed-loop-unstable case has \(1 / \mathrm{GM}_{\mathrm{U}}>1\) so that \(0<\mathrm{GM}_{\mathrm{U}}<1\). The mathlet shows the Nyquist plot winds once around \(w = -1\) in the \(clockwise\) direction. 1 {\displaystyle v(u(\Gamma _{s}))={{D(\Gamma _{s})-1} \over {k}}=G(\Gamma _{s})} . {\displaystyle P} 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. Thus, for all large \(R\), \[\text{the system is stable } \Leftrightarrow \ Z_{1 + kG, \gamma_R} = 0 \ \Leftrightarow \ \text{ Ind} (kG \circ \gamma_R, -1) = P_{G, \gamma_R}\], Finally, we can let \(R\) go to infinity. 0000002847 00000 n
{\displaystyle P} ) *(j*w+wb)); >> olfrf20k=20e3*olfrf01;olfrf40k=40e3*olfrf01;olfrf80k=80e3*olfrf01; >> plot(real(olfrf80k),imag(olfrf80k),real(olfrf40k),imag(olfrf40k),, Gain margin and phase margin are present and measurable on Nyquist plots such as those of Figure \(\PageIndex{1}\). N + 0000001210 00000 n
s 1 Nyquist plot of the transfer function s/(s-1)^3. ) Does the system have closed-loop poles outside the unit circle? F Thus, we may finally state that. ( G {\displaystyle N=P-Z} {\displaystyle G(s)} 1 encirclements of the -1+j0 point in "L(s).". {\displaystyle 1+kF(s)} D + {\displaystyle u(s)=D(s)} The Nyquist criterion is an important stability test with applications to systems, circuits, and networks [1]. A pole with positive real part would correspond to a mode that goes to infinity as \(t\) grows. The Routh test is an efficient The roots of b (s) are the poles of the open-loop transfer function. We will look a little more closely at such systems when we study the Laplace transform in the next topic. domain where the path of "s" encloses the Non-linear systems must use more complex stability criteria, such as Lyapunov or the circle criterion. is the number of poles of the open-loop transfer function are also said to be the roots of the characteristic equation Is the open loop system stable? The system with system function \(G(s)\) is called stable if all the poles of \(G\) are in the left half-plane. Look at the pole diagram and use the mouse to drag the yellow point up and down the imaginary axis. F ( ( must be equal to the number of open-loop poles in the RHP. The system is called unstable if any poles are in the right half-plane, i.e. = ) ( 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. 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