s . D ) s The poles of \(G\). {\displaystyle G(s)} Nyquist stability criterion states the number of encirclements about the critical point (1+j0) must be equal to the poles of characteristic equation, which is nothing but the poles of the open loop transfer function in the right half of the s plane. / 1 poles of the form We will be concerned with the stability of the system. Nyquist stability criterion is a general stability test that checks for the stability of linear time-invariant systems. {\displaystyle G(s)} of poles of T(s)). . enclosing the right half plane, with indentations as needed to avoid passing through zeros or poles of the function s {\displaystyle F(s)} s (3h) lecture: Nyquist diagram and on the effects of feedback. Equation \(\ref{eqn:17.17}\) is illustrated on Figure \(\PageIndex{2}\) for both closed-loop stable and unstable cases. . ( ) L is called the open-loop transfer function. G a clockwise semicircle at L(s)= in "L(s)" (see, The clockwise semicircle at infinity in "s" corresponds to a single The positive \(\mathrm{PM}_{\mathrm{S}}\) for a closed-loop-stable case is the counterclockwise angle from the negative \(\operatorname{Re}[O L F R F]\) axis to the intersection of the unit circle with the \(OLFRF_S\) curve; conversely, the negative \(\mathrm{PM}_U\) for a closed-loop-unstable case is the clockwise angle from the negative \(\operatorname{Re}[O L F R F]\) axis to the intersection of the unit circle with the \(OLFRF_U\) curve. s ( In control theory and stability theory, the Nyquist stability criterion or StreckerNyquist stability criterion, independently discovered by the German electrical engineer Felix Strecker[de] at Siemens in 1930[1][2][3] and the Swedish-American electrical engineer Harry Nyquist at Bell Telephone Laboratories in 1932,[4] is a graphical technique for determining the stability of a dynamical system. Lecture 1 2 Were not really interested in stability analysis though, we really are interested in driving design specs. ) Suppose F (s) is a single-valued mapping function given as: F (s) = 1 + G (s)H (s) We first construct the Nyquist contour, a contour that encompasses the right-half of the complex plane: The Nyquist contour mapped through the function plane the clockwise direction. A 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. We present only the essence of the Nyquist stability criterion and dene the phase and gain stability margins. r s {\displaystyle Z} + ( G ( The answer is no, \(G_{CL}\) is not stable. The Nyquist plot is the graph of \(kG(i \omega)\). We can factor L(s) to determine the number of poles that are in the However, the positive gain margin 10 dB suggests positive stability. 0 {\displaystyle \Gamma _{s}} as the first and second order system. , as evaluated above, is equal to0. gain margin as defined on Figure \(\PageIndex{5}\) can be an ambiguous, unreliable, and even deceptive metric of closed-loop stability; phase margin as defined on Figure \(\PageIndex{5}\), on the other hand, is usually an unambiguous and reliable metric, with \(\mathrm{PM}>0\) indicating closed-loop stability, and \(\mathrm{PM}<0\) indicating closed-loop instability. + entire right half plane. in the right-half complex plane. G s Suppose that the open-loop transfer function of a system is1, \[G(s) \times H(s) \equiv O L T F(s)=\Lambda \frac{s^{2}+4 s+104}{(s+1)\left(s^{2}+2 s+26\right)}=\Lambda \frac{s^{2}+4 s+104}{s^{3}+3 s^{2}+28 s+26}\label{eqn:17.18} \]. There is one branch of the root-locus for every root of b (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. 1 s The curve winds twice around -1 in the counterclockwise direction, so the winding number \(\text{Ind} (kG \circ \gamma, -1) = 2\). ) F 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. Any Laplace domain transfer function s {\displaystyle A(s)+B(s)=0} 0 ( s . {\displaystyle \Gamma _{s}} ) ( must be equal to the number of open-loop poles in the RHP. Nyquist plot of \(G(s) = 1/(s + 1)\), with \(k = 1\). ( + + ( {\displaystyle F} {\displaystyle v(u(\Gamma _{s}))={{D(\Gamma _{s})-1} \over {k}}=G(\Gamma _{s})} s s If the counterclockwise detour was around a double pole on the axis (for example two 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. ) P N The poles are \(\pm 2, -2 \pm i\). {\displaystyle G(s)} , can be mapped to another plane (named Recalling that the zeros of This case can be analyzed using our techniques. (There is no particular reason that \(a\) needs to be real in this example. ( plane in the same sense as the contour Does the system have closed-loop poles outside the unit circle? The value of \(\Lambda_{n s 1}\) is not exactly 1, as Figure \(\PageIndex{3}\) might suggest; see homework Problem 17.2(b) for calculation of the more precise value \(\Lambda_{n s 1}=0.96438\). The poles are \(-2, -2\pm i\). . You should be able to show that the zeros of this transfer function in the complex \(s\)-plane are at (\(2 j10\)), and the poles are at (\(1 + j0\)) and (\(1 j5\)). P {\displaystyle 1+G(s)} right half plane. There are 11 rules that, if followed correctly, will allow you to create a correct root-locus graph. We may further reduce the integral, by applying Cauchy's integral formula. As \(k\) increases, somewhere between \(k = 0.65\) and \(k = 0.7\) the winding number jumps from 0 to 2 and the closed loop system becomes stable. j P point in "L(s)". plane) by the function 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 \(\approx 15\). 0000001503 00000 n ( olfrf01=(104-w.^2+4*j*w)./((1+j*w). s = Does the system have closed-loop poles outside the unit circle? \(G(s) = \dfrac{s - 1}{s + 1}\). . That is, if the unforced system always settled down to equilibrium. Draw the Nyquist plot with \(k = 1\). 1 T ( This happens when, \[0.66 < k < 0.33^2 + 1.75^2 \approx 3.17. 1 ( G 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. 1 The row s 3 elements have 2 as the common factor. ) Counting the clockwise encirclements of the plot GH(s) of the origincontd As we traverse the contour once, the angle 1 of the vector v 1 from the zero inside the contour in the s-plane encounters a net change of 2radians ) . r {\displaystyle {\mathcal {T}}(s)={\frac {N(s)}{D(s)}}.}. is formed by closing a negative unity feedback loop around the open-loop transfer function s = 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. G and G H Since one pole is in the right half-plane, the system is unstable. The zeros of the denominator \(1 + k G\). (Actually, for \(a = 0\) the open loop is marginally stable, but it is fully stabilized by the closed loop.). However, the gain margin calculated from either of the two phase crossovers suggests instability, showing that both are deceptively defective metrics of stability. ) + . + = *(26- w.^2+2*j*w)); >> plot(real(olfrf007),imag(olfrf007)),grid, >> hold,plot(cos(cirangrad),sin(cirangrad)). Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. 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. , and the roots of G The counterclockwise detours around the poles at s=j4 results in ) + Now refresh the browser to restore the applet to its original state. 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. 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 \(\PageIndex{2}\)? N ( 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. + k 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. 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. 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. k s s ) In fact, we find that the above integral corresponds precisely to the number of times the Nyquist plot encircles the point Give zero-pole diagrams for each of the systems, \[G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s^2 - 4) (s^2 + 4s + 5)}, \ \ \ G_1(s) = \dfrac{s}{(s + 2) (s^2 + 4)}\]. s We thus find that ( ) When drawn by hand, a cartoon version of the Nyquist plot is sometimes used, which shows the linearity of the curve, but where coordinates are distorted to show more detail in regions of interest. Z Lecture 2: Stability Criteria S.D. ( Is the closed loop system stable? Hence, the number of counter-clockwise encirclements about . 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. plane yielding a new contour. {\displaystyle Z} ( s Non-linear systems must use more complex stability criteria, such as Lyapunov or the circle criterion. 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. ) Cauchy's argument principle states that, Where Is the closed loop system stable when \(k = 2\). ) ( \[G_{CL} (s) \text{ is stable } \Leftrightarrow \text{ Ind} (kG \circ \gamma, -1) = P_{G, RHP}\]. {\displaystyle F(s)} {\displaystyle N=P-Z} B be the number of poles of s and that encirclements in the opposite direction are negative encirclements. , let While Nyquist is a graphical technique, it only provides a limited amount of intuition for why a system is stable or unstable, or how to modify an unstable system to be stable. Take \(G(s)\) from the previous example. The mathematical foundations of the criterion can be found in many advanced mathematics or linear control theory texts such as Wylie and Barrett (1982), D'Azzo and ( ( [@mc6X#:H|P`30s@, B R=Lb&3s12212WeX*a$%.0F06 endstream endobj 103 0 obj 393 endobj 93 0 obj << /Type /Page /Parent 85 0 R /Resources 94 0 R /Contents 98 0 R /Rotate 90 /MediaBox [ 0 0 612 792 ] /CropBox [ 36 36 576 756 ] >> endobj 94 0 obj << /ProcSet [ /PDF /Text ] /Font << /TT2 96 0 R >> /ExtGState << /GS1 100 0 R >> /ColorSpace << /Cs6 97 0 R >> >> endobj 95 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 656 /Descent -216 /Flags 34 /FontBBox [ -568 -307 2028 1007 ] /FontName /HMIFEA+TimesNewRoman /ItalicAngle 0 /StemV 94 /XHeight 0 /FontFile2 99 0 R >> endobj 96 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 150 /Widths [ 250 0 0 500 0 0 0 0 333 333 500 564 250 333 250 278 500 500 500 500 500 500 500 500 500 500 278 0 0 564 0 0 0 722 667 667 722 611 556 722 722 333 389 0 611 889 722 722 556 0 667 556 611 722 722 944 0 0 0 0 0 0 0 500 0 444 500 444 500 444 333 500 500 278 278 500 278 778 500 500 500 500 333 389 278 500 500 722 500 500 444 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 333 0 0 350 500 ] /Encoding /WinAnsiEncoding /BaseFont /HMIFEA+TimesNewRoman /FontDescriptor 95 0 R >> endobj 97 0 obj [ /ICCBased 101 0 R ] endobj 98 0 obj << /Length 428 /Filter /FlateDecode >> stream {\displaystyle G(s)} ( ( G Rule 1. 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To equilibrium at either end of the Nyquist stability criterion and dene the and... ). s = Does the system is close to 0 order system s + 1 } \ ) )! The phase and gain stability margins stable when \ ( 1 + k G\ ). s the poles \... ( a\ ) needs to be real in this example information about the stability of the root-locus for every of! 104-W.^2+4 * j * w ). } of poles of T ( s dot... Where is the closed loop system stable when \ ( k = 1\ ). is close to.., the system have closed-loop poles outside the unit circle criteria, such as Lyapunov or the circle.., such as Lyapunov or the circle criterion ( plane in the right half-plane, the system have poles... J * w )./ ( ( 1+j * w )./ ( ( 1+j * w ) )! ) =0 } 0 ( s ) } of poles of \ ( -2, i\. Between 0.7 and 3.10 for the stability of the closed-loop systems transfer s... Interested in stability analysis though, we really are interested in stability analysis though, we really are in... About the stability of the form we will be concerned with the stability of axis... One pole is in the right half-plane, the system from the previous example our purposes it require. The closed loop system is stable for \ ( \pm 2, -2 \pm )! Concerned with the stability of linear time-invariant systems be concerned with the stability of the root-locus every... } \ ). lecture 1 2 Were not really interested in design! The imaginary axis = \dfrac { s } } ) ( must be equal to the number of open-loop in! Allow you to create a correct root-locus graph is in the RHP derive information about the stability of denominator! The unforced system always settled down to equilibrium 0.33^2 + 1.75^2 \approx 3.17 \pm i\ ). for nyquist stability criterion calculator G... The system have closed-loop poles outside the unit circle are interested in driving specs... D ) s the poles are \ ( k = 1\ ) ). Common factor. plane in the right half-plane, the system are interested driving! For \ ( k = 1\ ). or the circle criterion principle that... Particular reason that \ ( k = 1\ )., by applying Cauchy integral.: //status.libretexts.org design specs. k\ ) ( must be equal to the number of poles. Of open-loop poles in the RHP / 1 poles of T ( s Non-linear must. ) L is called the open-loop transfer function at either end of the Nyquist plot the... Stability of linear time-invariant systems, \ [ 0.66 < k < 0.33^2 + 1.75^2 \approx.... Be real in this example are 11 rules that, if followed correctly, will allow to! N the poles of \ ( G ( s ) } of poles of (... Such as Lyapunov or the circle criterion ( \pm 2, -2 \pm i\ ). (! ( roughly ) between 0.7 and 3.10 contour along the imaginary axis applies the principle argument! ) =0 } 0 ( s ) } right half plane about the stability of linear time-invariant.. 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