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Nyquist Plot (Problems) – Frequency Response Analysis – Control System

Nyquist Plot (Problems) – Frequency Response Analysis – Control System
Nyquist Plot (Problems) – Frequency Response Analysis – Control System

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Nyquist Plot (Problems) - Frequency Response Analysis - Control System
Nyquist Plot (Problems) – Frequency Response Analysis – Control System

In my previous post we saw how to analyze a system using polar plot. Now we need to learn about Nyquist plot (which is basically an extension of the polar plot). The ‘Nyquist plot’  method handles systems with time delay without the necessity of approximations and give an exact result about both absolute and relative stability of the system. Now , we should learn one important concept i.e ‘principle of argument’. If let say ‘p’ poles and ‘z’ zeros of q(s) are confined by the s-plane contour, then the q(s) plane contour must enclose the origin ‘z’ times in the clockwise direction and ‘p’ times in the anti-clockwise direction (since generally p > z). This relationship between the enclosure of poles and zeros by the s-plane contour (an enclosed path in a complex plane is known as a contour) and the encirclement of the origin by the q(S) plane contour is known as ‘principles of argument’ . We should know the following terms before we look into the steps to solve problems. Encirclement: Any specific point is declared to be encircled by a closed path if it lies inse the closed path. Mapping theorem: For any closed-loop system to be stable , the Nyquist plot of G(s)H(s) should encircle the point -1+ j0 as many times as the number of poles G(s)H(s) that are in the right half of s plane. Number of encirclements is given as : N = P – Z Where N = number of encirclement of point -1 + j0 P = Number of poles  G(s)H(s) that are on right half of s plane Z = Number of zeros G(s) H(s) For stability of closed loop, Z should be zero , i.e N = P Nyquist plot analysis: Pole – zero configuration Conser closed loop transfer function    as Poles of   1+ G(s)H(s) = open loop poles of a system Zeros of 1 + G(s)H(s) =  Closed loop poles of a system From Nyquist point of view, the system is absolutely stable if all zeros of 1 + G(s)H(s) , i.e closed loop poles of the system are located in the left half of s plane. Steps to solving problems: Count how many number of poles of G(s)H(s) are on the right half of  s – plane. Dece the stability criterion as N = -P i.e how many times Nyquist plot should encircle -1 + j0 point for absolute stability . Select Nyquist path as per the function G(s)H(s) . Analyze the sections as a starting point and terminating point . Mathematically find out and the intersection of  Nyquist plot with a negative real axis by rationalizing  G(j)H(j ). With the knowledge of steps 4 and 5 , sketch the Nyquist plot Count the number of encirclements N of -1 + j0 by Nyquist plot . If this matches the criterion deced in step 2, system is stable otherwise unstable. After plotting the Nyquist plot, we can then obtain the absolute stability of this loop that is a closed control process implementing the Nyquist stability criterion. Then the given closed-loop controller process is completely stable if the critical point that is the point -1+j0 falls outse the enclosure. Now we should look into one simple example to understand this topic in a luc manner. Given    G(s)H(S) =   , comment on the stability using Nyquist plot. Step 1 : Position of poles , s = -1 Hence, there are no poles in the right half plane(RHP) Thus, P = 0 Hence for stability , number of encirclements in the anticlockwise direction about the  (-1 + 0j) point should be zero, i.e N = 0 . We need to check if we get  N = 0. Step 2 : The Nyquist contour is shown below, this encloses entire left half plane. The Nyquist paths are Path a – b    s = jω  ,. Path b – c – d      s =         -90    θ   90 ,. Path d – a    s = – jω ,. We shall draw each path. Step 3 : Path a – b Here s = jω  , this path is the same as the polar plot, G(s)H(S) =   , Now we have to convert  ‘s’ by ‘jω’ , G(jω)H(jω) = We now find the magnitude and phase |G(jω)H(jω)| =  ∠ G(jω)H(jω)  =     = , ∠ G(jω)H(jω)  =  –  ,   Vary ‘ω’ from  0  to ∞ Now instead of taking various values for ‘ω’ ,we can easily conser the two extreme values of ω i.e ω = 0 and ω = ∞ At  ω = 0,        |G(jω)H(jω)| =  1 ∠ G(jω)H(jω)  =  –   =    0  , At  ω = ∞ ,   |G(jω)H(jω)|  =  0  , ∠ G(jω)H(jω)  =  –   =    -90  ,    The polar plot is shown below, Step 4 : Path b – c – d Here s =   , Hence, G(s)H(s) =    , Since R is ∞ , |G(jω)H(jω)|  =  0 for all values |G(jω)H(jω)|  =    = 0  , Since magnitude is 0, no use of finding the angle values. Therefore, path b – c- d maps on to the center. ( Note : as long as system has more poles , section b – c – d will always be 0) Step 5 : Path d – a Here s = -jω  ,    This is the mirror image of path a – b , hence the final Nyquist plot is shown below. Since there are no encirclements about the  (-1 + 0j) point. N = 0 We know, N = P – Z 0 = 0 – Z ( since P = 0) Hence , Z = 0 Z = 0 implies that there are no zeros of  q(s) (no poles of the characteristic equation) lying on the Right half plane . Hence the system is stable. Nyquist stability criterion : It states  that the value of N = Z – P. Where, ‘N’ is simply the total number of the encirclements about the center point i.e the origin, ‘Z’ is the total number of zeroes and ‘P’ is total number of the poles. Case 1 : If the value of ‘N’ is equal to ‘0’ (no encirclement), so Z = P = 0 and Z = P If  N = 0, P must be zero hence the system is sa to be stable. Case 2 :  If value of ‘N’ is less than ‘0’ (counter clockwise encirclement), so Z = 0, P ≠0 and P > Z System is stable. Case 3 : If the value of ‘N’ is greater than ‘0’ (clockwise encirclement), so P = 0, Z ≠0 and Z > P For both cases system is sa to be unstable. Let us now see some of the advantages of using Nyquist plots: The stability of closed-loop system with a pure time delay can be studied using Nyquist plot. It proves information about the absolute stability and also the relative stability. We can determine stability of closed-loop system from open-loop transfer function without knowing the roots of the characteristic equation. Frequency domain characteristics can be easily obtained(easy to design hence much faster). Hope you really enjoyed this post, see you soon in my next post. report this ad

What is Nyquist plot in control system?

A Nyquist plot is a parametric plot of a frequency response used in automatic control and signal processing. The most common use of Nyquist plots is for assessing the stability of a system with feedback. In Cartesian coordinates, the real part of the transfer function is plotted on the X-axis.

How do you analyze a Nyquist plot?

With a Nyquist plot, you can simply observe the distance between (–1, 0) and the point at which the curve crosses the negative real axis. More distance between these two points corresponds to a larger gain margin and, consequently, to a circuit that is more reliably stable.

How do you find the number of encirclement in a Nyquist plot?

1.15. 2 Determine N and P
  1. For the open-loop system, the poles are at −1.47,0.23±0.79j. Therefore, P=2.
  2. If K<2, then −0.5K>−1, the Nyquist plot does not encircle −1. Therefore N=0,Z=2. The system has 2 unstable poles.
  3. If K>2, then −0.5K<−1, the Nyquist plot encircle −1 counterclockwise twice. Therefore N=−2 and Z=0.

How do you plot a polar plot in a control system?

Follow these rules for plotting the polar plots.
  1. Substitute, s=jω in the open loop transfer function.
  2. Write the expressions for magnitude and the phase of G(jω)H(jω).
  3. Find the starting magnitude and the phase of G(jω)H(jω) by substituting ω=0. …
  4. Find the ending magnitude and the phase of G(jω)H(jω) by substituting ω=∞.

What are Nyquist and Bode plot?

There are two Bode plots one for gain (or magnitude) and one for phase. The amplitude response curves given above are examples of the Bode gain plot. The Nyquist plot combines gain and phase into one plot in the complex plane. It is drawn by plotting the complex gain g(iw) for all frequencies w.

What are the advantages of Nyquist plot?

Advantages of the Nyquist plot

It can determine the stability of the control system. It is better than the root locus in terms of time delay. It means that the Nyquist plot can easily handle the time delay in the system. It provides a way to use the bode plots.

How is Nyquist plot used to determine stability?

Nyquist Stability Criterion
  1. The Nyquist stability criterion works on the principle of argument. It states that if there are P poles and Z zeros are enclosed by the ‘s’ plane closed path, then the corresponding G(s)H(s) plane must encircle the origin P−Z times. …
  2. N=P−Z.
  3. i.e.,P=0⇒N=−Z.
  4. i.e.,Z=0⇒N=P.
  5. PM=1800+ϕgc.

How do you read a Nyquist plot impedance?

At low frequencies, the reactants have to diffuse farther, increasing the Warburg-impedance. On a Nyquist Plot the Warburg impedance appears as a diagonal line with an slope of 45°. On a Bode Plot, the Warburg impedance exhibits a phase shift of 45°.

ω = radial frequency
n = number of electrons involved

What is Nyquist criteria for sampling?

Simply stated, the Nyquist criterion requires that the sampling frequency be at least twice the highest frequency contained in the signal, or information about the signal will be lost. If the sampling frequency is less than twice the maximum analog signal frequency, a phenomenon known as aliasing will occur.

What is encirclement in Nyquist plot?

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. If you will find roots of characteristics equation, it will be –10.3, –0.86±j1.

What is GM and PM in Bode plot?

Gm and Pm of a system indicate the relative stability of the closed-loop system formed by applying unit negative feedback to sys , as shown in the following figure. Gm is the amount of gain variance required to make the loop gain unity at the frequency Wcg where the phase angle is –180° (modulo 360°).

What is the difference between polar plot and Bode plot?

In the Bode plot, the frequency response is sketched using a logarithmic scale. So, in a polar plot, a sketch between the magnitude and phase angle of the transfer function G(jω) is formed for different values of ω. So, with the variation in ω from 0 to ∞, the values of M and φ can be determined.

Why do we use polar plots?

The polar plot of a sinusoidal transfer function is the plot of the magnitude G(jω) versus the phase angle of G(jω) on the polar coordinates. The frequency in the polar plot is varied from zero to infinity.

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