Module-9:Frequency Domain Analysis of Linear Systems
Nyquist Plot
Akshya Swain
Department of Electrical, Computer & Software Engineering, The University of Auckland,
Auckland, New Zealand.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 1 / 60
Learning Outcome
After completion of this module the students would have learned following:
1. How to apply principle of argument to study stability of a system ? 2. How to map the Nyquist path to Nyquist contour
3. How to investigate the relative stability from the Nyquist plot ?
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 2 / 60
Nyquist Stability Criterion
I The Nyquist stability criterion relates the stability of the closed-loop system to the frequency response of the open loop system.
I The Nyquist criterion possesses the following features that make it desirable for the analysis as well as the design of control systems
1. It provides the same amount of information on the absolute stability of a control system as the Routh-Hurwitz criterion.
2. In addition to absolute system stability, the Nyquist criterion indicates the degree of stability of a stable system and gives an indication of how the system stability may be improved, if needed.
3. It gives information on the frequency-domain response of the system.
4. It can be used for a stability study of systems with time delay.
5. It can be modified for nonlinear systems.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 3 / 60
Mapping of Analytic Function of a Complex Variable
I Let F (s) be a function of the complex variable s = σ + jω.
I The function F(s), in general, will be complex. Let us denote F(s) as:
F (s) = u(σ, ω) + jv(σ, ω)
where u is the real part of F (s) and v is the imaginary part of F (s).
I Just as the complex variable s is shown in a plane ( called as s-plane) with the
real axis σ and the imaginary axis ω, it is possible to illustrate F in a plane ( say F-plane) with real axis u and the imaginary axis v.
Analytic Conditions of a Function
– An F(s), defined in a domain D in the s− plane, is said to be analytic in D if the derivative dF/ds is continuous in D.
– It can be shown that F(s) is analytic in D if and only if the Cauchy-Riemann equations
∂u=∂v, and ∂u=−∂v ∂σ∂ω ∂ω∂σ
are satisfied in the domain D.
It can be proved that all rational functions of s are analytic everywhere in the s-plane, except at singular points. Thus, all transfer functions are analytic in the s-plane, except at their poles.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 4 / 60
Example-1: Mapping of Analytic Function of a Complex Variable
Mapping of a Point in s-plane
I First let us map any point in the s-plane into the F-plane.
I This is carried out by locating the values of u and v in the F-plane, for the given
values of σ and ω, in the s-plane. I Consider the function
F (s) = 2s + 3 s+5
I Fors1=1+j2,weget
F(1+j2) = 2(1+j2)+3 = 5+j4 = (5+j4)(6−j2) = 38+j14 = 0.95+j0.33 1+j2+5 6+j2 (6+j2)(6−j2) 40
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 5 / 60
Example-1: Mapping of Contours in the s-Plane
I Consider a function
F(s) = 2s+1 = 2(s+ 1) 2
and a contour in the s-plane as shown in the Figure-a.
The mapping of the unit square in the s-plane to the F (s) plane is accomplished through the relation F(s). This gives
u+jv=F(s)=2s+2=2(σ+jω)+1, =⇒ u=2σ+1, v=2ω
I Note that this type of mapping, which retains the angles of the s-plane contour on the F(s)-plane, is called conformal mapping.
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 6 / 60
Example-2: Mapping of Contours in the s-Plane
Consider mapping a unit square contour shown in Fig-a to F(s) plane where
F(s)= s s+2
I Several values of F(s) as s traverses the square are given in Table.
s=σ+ω
F (s) = u + jv
1+j1 1 1−j1 −j1 −1−j1 −1 −1+j1 j1
4+2j 2
1 3
4−2j 10
1−2j 5
−j
−1
+j
1+2j 5
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 7 / 60
Example-2 Mapping of Contours in the s-Plane
I The mapping is shown in the Figure.
I Note that the s-plane contour encircles the zero of F(s) which is located at the origin; but does not encicle the pole of F(s) located at s = −2
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 8 / 60
Example-3 : Mapping of Contours in the s-Plane
I The mapping of the contour in the s-plane ( see Fig-a) to the F (s) plane where F(s)= s
s+1 2
Note: The s-plane contour in example-1 and 2 encloses one zero of the function F(s). The corresponding contour in the F(s) plane encloses the origin once.
I However, the s-plane contour in example-3 encloses both the pole and zero ( one each) of the function F(s). The corresponding contour in the F(s) plane does not enclose the origin. This is due to Principle of Argument ( Cauchy’s Theorem)(discussed next).
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 9 / 60
Principle of Argument; Foundation of Nyquist Stability Criterion
Principle of Argument:
If a contour Γs in the s-plane encloses Z zeros and P poles of F(s) and does not pass through any poles or zeros of F(s), then with the traversal along the contour Γs in the clock-wise direction, the corresponding contour ΓF in the F(s)-plane encircles the origin of the F(s) plane N-times in the clockwise direction where
N=Z−P
Case-1: N > 0 (Z > P ). If the s-plane locus encircles more zeros than poles of F(s) in a certain prescribed direction (clockwise or counterclockwise), N is a positive integer. In this case the F(s)-plane locus will encircle the origin of the F(s)-plane N times in the same direction as that of Γs.
Case-2: N = 0 (Z = P ). If the s-plane locus encircles as many poles as zeros, or no poles and zeros, of F(s), the F(s)-plane locus ΓF will not encircle the origin of the F(s)-plane.
Case-3: N < 0 (Z < P ). If the s-plane locus encircles more poles than zeros of F (s) in a certain direction, N is a negative integer. In this case the F (s)-plane locus, ΓF , will encircle the origin N times in the opposite direction from that of Γs .
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 10 / 60
Example-Case-1: (N > 0) Principle of Argument
I The pole-zero pattern of a function F(s) is shown in Fig-a.
I The s-plane contour Γs encloses three zeros and one pole of F (s) i.e. Z = 3 and
P = 1 ( in the clock-wise direction ).Thus
N = 3 − 1 = +2
I Therefore, the F-plane contour ΓF encloses the origin 2-times in the clockwise direction. (see Fig b)
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 11 / 60
Example-Case-2: (N = 0) Principle of Argument
I The pole-zero pattern of a function F(s) is shown in Fig-a.
I The s-plane contour Γs encloses one pole and one zero of F (s) i.e. Z = 1 and P = 1.Thus
N=1−1=0
I Therefore, the F-plane contour ΓF does not enclose the origin(see Fig b).
Akshya Swain
Module-9:Frequency Domain Analysis of
Linear Systems 12 / 60
Example-Case-3: (N < 0) Principle of Argument I The pole-zero pattern of a function F(s) is shown in Fig-a. I The s-plane contour Γs encloses one pole and no zeros of F (s) i.e. Z = 0 and P = 1 ( in the clock-wise direction ).Thus N = 0 − 1 = −1 I Therefore, the F-plane contour ΓF encloses the origin 1-times in the counterclockwise direction ( because N is negative). (see Fig b). Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 13 / 60 Nyquist Stability:Implications of Cauchy’s Principle for Stability I After illustrating the concept of mapping of contours through a function F (s), let us answer the following: I Harry Nyquist found that the the principle of the argument could be used to solve the stability problems Question ? Can the Principle of Argument be applied to investigate the stability of a system ? Answer: YES. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 14 / 60 Philosophy of Nyquist Stability Criterion I Let the closed loop transfer function of a system be T (s) = G(s) 1 + G(s)H(s) I Let the closed loop characteristic equation of the system is expressed as: F(s) = 1 + G(s)H(s) = 0 Note: a. The zeroes of F(s) are the poles of T(s). b. The poles of F(s) are the poles of open-loop transfer function G(s)H(s); however, the zeros of F(s) are different from the zeros of G(s)H(s) and are not known. a. We know that for a system to be stable, all the poles of T(s) must lie in the left half of s-plane. b. This problem, therefore, reduces to determine whether exist any zeros of F (s) in the right half of s-plane. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 15 / 60 Philosophy of Nyquist Stability Criterion Note: We are interested to know if there exist any zeros of F (s) in the right half of s-plane, I Nyquist discovered that the principle of the argument could be used to solve the stability problems. Therefore, he did the following: Idea : 1. Define a contour Γs which encloses the entire right half of s-plane including the jω axis. This is called Nyquist Contour or Nyquist Path. 2. If the function F(s) has a pole or zero at the the origin or on the jω axis, make a detour along an infinitesimal semicircle. 3. Map the contour Γs in the s-plane into the contour ΓF in the F (s) plane using the function F(s). 4. Determine the number of encirclements of the origin of the F-plane. Unless, otherwise stated, in this study it is assumed that the the Nyquist path Γs traverses in the clockwise direction. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 16 / 60 Nyquist Stability Criterion Note: Since F(s) = 1+G(s)H(s) = 1+L(s) the origin of the F plane is the point −1 + j0 in the GH plane. I Let F(s) have P poles and Z zeros within the Nyquist contour (path). I From the principle of argument, a map of Nyquist contour in the F-plane will encircle the origin of the F-plane ( or −1 + j0 point in GH-plane) N times in the clockwise direction ( assuming that the direction of Γs contour is clockwise), where N=Z−P I For the system to be stable, the characteristic equation must not have any root within the Nyquist contour i.e we must have Z = 0. This imples N = −P Statement: Nyquist Stability Criterion A feedback control system is stable, if and only if, for the contour ΓF , the number of counterclockwise encirclements of the point −1 + j0 is equal to the number of poles of G(s)H(s) within the Nyquist contour Γs in the s-plane. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 17 / 60 Nyquist Path/Contour: No Poles on jω-Axis Case-1: When the open loop transfer function G(s)H(s) has neither poles nor zeros on the jω-axis including the origin, then the Nyquist path consists of the following 3-sections: I Example: The Nyquist path for the system with open loop transfer function G(s)H(s) = 1 s+3 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 18 / 60 Nyquist Path/Contour: Poles on jω-axis Case-2: When the open loop transfer function G(s)H(s) has either poles or zeros on the jω-axis including the origin, then the Nyquist path consists of the following 4-sections: a. A Nyquist path encloses the entire right half of the s-plane. b. If any pole(s) or zero(s) lie on the jω-axis then a detour is taken so as to exclude them from the path as shown in the figure below . c. Path components 2, 3 and 4 are common to any Nyquist Path but the number of appearances of path component 1 depends on the number of poles or zeros on the jω-axis. d. Nyquist Path is mapped into plane to obtain Nyquist Plot. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 19 / 60 Nyquist Path/Contour: No poles on jω-axis Example: The Nyquist paty for system with open-loop transfer function G(s)H(s) = s − 2 s(s+1)(s2 +4)(s−4) I Example: The Nyquist path for the system with open loop transfer function G(s)H(s) = 1 s+3 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 20 / 60 Various Steps for Plotting Nyquist Plot I Step 1: Check for the poles of G(s)H(s) of jω- axis including that at origin. I Step 2: Select the proper Nyquist contour – a) Include the entire right half of s-plane by drawing a semicircle of radius r with r tends to infinity. I Step 3: Identify the various segments on the contour with reference to Nyquist path I Step 4: Perform the mapping segment by segment by substituting the equation for the respective segment in the mapping function. Basically, we have to sketch the polar plots of the respective segment. I Step 5: Mapping of the segments are usually mirror images of mapping of the respective path of +ve imaginary axis. I Step 6: The semicircular path which covers the right half of s plane generally maps into a point in G(s)H(s) plane. I Step 7: Interconnect all the mapping of different segments to yield the required Nyquist diagram. I Step 8: Note the number of clockwise encirclement about (−1, 0) and decide stability by N = Z − P . Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 21 / 60 Example-1: First Order System with Real Pole Consider a unity feedback system with open-loop transfer function i G(s)= K , =⇒ G(jω)= K 1+τs 1+jωτ |G(jω)| = √ K 1+ω2τ2 I The Nyquist path is shown below. , φ = −tan−1ωτ Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 22 / 60 Example-1:First Order System with Real Pole(Contd) Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. lim G(s) r→∞ K = lim K − j φ − j φ = lim e =0e s=rejφ r→∞ sτ s=rejφ r→∞ τr I Thus, as the phasor in Γs is traversed from +900 to −900 ( i.e from ω = +∞ to ω = −∞) through 1800 clockwise,the corresponding contour ΓF i.e Nyquist plot of G(s), is traced by a phasor of practically zero length from −900 to +900 in the counter clockwise sense and satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −900 At ω = −∞ |G(jω)| = 0 \G(jω) = +900 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 23 / 60 Example-1: First Order System with Real Pole(Contd) Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The starting point of this part of the contour ΓF , is computed as: lim G(jω)= lim K\−900=0\−900 ω→∞ ω→∞τω iii. For other values of ω, this is computed from ω= 1 , φ=−5.70,ω= 2, φ=−63.40,ω= 10, φ=−84.30 10τ τ τ lim G(jω) = lim K ω→0 ω→0 (1 + jωτ) = K\00 iii. The termination (end) point of this part of the contour ΓF , is computed as: I The beginning and end of ΓF contour in this section of Γs satisfies the following At ω = 0+ |G(jω)| = K \G(jω) = 00 At ω = +∞ |G(jω)| = 0 \G(jω) = −900 |G(jω)| = √ K 1+ω2τ2 , φ = \G(jω) = −tan−1ωτ Forinstanceatω= 1, φ=−450,ω= 1 , φ=−26.60 τ 2τ Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 24 / 60 Example-1: First Order System with Real Pole Mapping Section III: ω varies from −∞ to 0− I This portion of the contour is the mirror image of the polar plot. I Figure below shows the Nyquist plot for the unity feedback system with feed-forward transfer function G(s)= 5 = 2.5 = K , K=2.5, τ=0.5 s+2 1+0.5s 1+τs Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 25 / 60 Example-2: Second Order System with Two Real Poles Consider the unity feedback system with open-loop transfer function i G(s)= K , =⇒ G(jω)= K (1 + τ1s)(1 + τ2s) (1 + jωτ1)(1 + jωτ2) K −1 −1 |G(jω)| = (1 + ω2τ12)(1 + ω2τ2 , φ = −tan ωτ1 − −tan ωτ2 I The Nyquist path is shown below. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 26 / 60 Example-2:Second Order System with Two Real Poles(Contd) Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. K −2jφ −2jφ =lim 2e =0e I Thus, as the phasor in Γs is traversed from +900 to −900 ( i.e from ω = +∞ to ω = −∞) through 1800 clockwise,the corresponding contour ΓF i.e Nyquist plot of G(s), is traced by a phasor of practically zero length from −1800 to +1800 in the counter clockwise sense and satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 At ω = −∞ |G(jω)| = 0 \G(jω) = +1800 limG(s) r→∞ s=rejφ K =lim 2 r→∞ s τ1τ2 s=rejφ r→∞ τ1τ2r Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 27 / 60 Example-2:Second Order System with Two Real Poles (Contd) Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The starting point of this part of the contour ΓF , is computed as: G(s)= K , =⇒ G(jω)= K (1 + τ1s)(1 + τ2s) (1 + jωτ1)(1 + jωτ2) K −1−1 |G(jω)|= 1+ω2τ121+ω2τ2, φ=\G(jω)=−tan ωτ1 −tan ωτ2 iv. The magnitude and phase of ΓF contour at ω = +∞ is given by lim G(jω) = lim −K = 0\−1800 ω→∞ ω→∞ τ1τ2ω2 G(s). lim G(jω) = lim K ω→0 ω→0 (1 + jωτ1)(1 = jωτ2) = K\00 iii. For other values of ω, this is computed from the magnitude and phase angle of I The ΓF contour satisfies the following At ω = 0+ |G(jω)| = K \G(jω) = 00 At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 28 / 60 Example-2: Second Order System with Two Real Poles I The Nyquist plot is shown below Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 29 / 60 Example-3:Second Order System with Single Pole at the Origin Consider the unity feedback system with feedforward path transfer function G(s)= K , =⇒ G(jω)= K s(1 + τs) |G(jω)| = √ K , ω 1+ω2τ2 jω(1 + jωτ) θ = \G(jω) = −900 −tan−1ωτ I The Nyquist path for the transfer function is shown below Mapping Section-IV: s = εejφ limG(s)=lim K =limKe−jφ ε→0 ε→0 εejφ ε→0 ε Mapping Section-II: s = rejφ lim G(s) = lim K = lim K e−jφ r→∞ r→∞ ejφ ε→0 ε Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 30 / 60 Example-3:Second Order System with Single Pole at the Origin (Contd) Mapping Section IV: s = εejφ i. The Nyquist path is a small semicircular detour around the pole at the origin. ii. The equation of the semicircle ( detour) is given by s=εejφ, −900 ≤φ≤+900 where ε is infinitesimally small and is traversed from points j0− at φ = −900 to j0+ at φ = +900. limG(s) =lim K =limKe−jφ ε→0 s=εejφ ε→0 ejφ ε→0 ε I Forthes-planelocusΓs,atω=0−,φ=−900 andatω=0+,φ=+900 . Hence, the map of this Γs into the G(s)H(s) plane ΓF satisfies the following: At ω = 0− |G(jω)| = ∞ \G(jω) = +900 At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 31 / 60 Example-3:Second Order System with Single Pole at the Origin (Contd) Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. lim G(s) r→∞ K K − 2 j φ − 2 j φ = lim 2e =0e s=rejφ = lim 2 r→∞ s τ s=rejφ r→∞ τr I This portion of s-plane locus Γs,at ω = +∞, φ = +900 and at ω = −∞, φ = −900 is mapped into the G(s)H(s) plane ΓF which satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 At ω = −∞ |G(jω)| = 0 \G(jω) = +1800 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 32 / 60 Example-3:Second Order System with Single Pole at the Origin (Contd) Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The beginning and end point of the contour ΓF , for this portion of the Γs, has been obtained before. iii. For other values of ω, this computed from the magnitude and phase angle of G(s)H(s). G(s)= K , =⇒ G(jω)= K s(1 + τs) jω(1 + jωτ) |G(jω)| = √ K , φ = \G(jω) = −900 −tan−1ωτ ω 1+ω2τ2 I This portion of s-plane locus Γs begins from at ω = 0+, φ = +900 and ends atω=+∞,φ=+900 . I This is essentially the polar plot of the G(s)H(s). At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 33 / 60 Example-3:Second Order System with Single Pole at the Origin(Contd) Mapping Section III: ω varies from −∞ to 0− I This portion of the contour is the mirror image of the polar plot. I The magnitude and phase angle of Nyquist contour ΓF for different sections of Nyquist path Γs are summarised below. Section-I At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 Section-III At ω = −∞− |G(jω)| = 0 \G(jω) = +1800 At ω = 0− |G(jω)| = 0 \G(jω) = −900 Section-II At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 At ω = −∞ |G(jω)| = 0 \G(jω) = +1800 Section-IV At ω = 0− |G(jω)| = ∞ \G(jω) = +900 At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 34 / 60 Example-3:Second Order System with Single Pole at the Origin The Nyquist path and the Nyquist contour is shown in the Figure. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 35 / 60 Example-4: Third Order System with Single Pole at the Origin Consider the unity feedback system with feedforward path transfer function G(s)= K , =⇒ G(jω)= K s(1 + τs)(1 + τ2s) jω(1 + jωτ1)(1 + jωτ2) K 0−1−1 |G(jω)|= 2 2 2 2, θ=\G(jω)=−90 −tan ωτ1 −tan ωτ2 ω 1+ω τ1 1+ω τ2 I The Nyquist path for the transfer function is shown below Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 36 / 60 Example-4:Third Order System with Single Pole at the Origin (Contd) Mapping Section IV: s = εejφ i. The Nyquist path is a small semicircular detour around the pole at the origin. ii. The equation of the semicircle ( detour) is given by s=εejφ, −900 ≤φ≤+900 where ε is infinitesimally small and is traversed from points j0− at φ = −900 to j0+ at φ = +900. limG(s) =lim K =limKe−jφ ε→0 s=εejφ ε→0 ejφ ε→0 ε I Forthes-planelocusΓs,atω=0−,φ=−900 andatω=0+,φ=+900 . Hence, the map of this Γs into the G(s)H(s) plane ΓF satisfies the following: At ω = 0− |G(jω)| = ∞ \G(jω) = +900 At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 37 / 60 Example-4: Third Order System with Single Pole at the Origin (Contd) Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. K =lim 3 K −3jφ −3jφ =lim 3e =0e I This portion of s-plane locus Γs,at ω = +∞, φ = +900 and at ω = −∞, φ = −900 is mapped into the G(s)H(s) plane ΓF which satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −2700 At ω = −∞ |G(jω)| = 0 \G(jω) = +2700 limG(s) r→∞ s=rejφ r→∞ s τ1τ2 s=rejφ r→∞ τ1τ2r Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 38 / 60 Example-4:Third Order System with Single Pole at the Origin (Contd) Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The beginning and end point of the contour ΓF , for this portion of the Γs, has been obtained before. iii. For other values of ω, this computed from the magnitude and phase angle of G(s)H(s). G(s)= K , =⇒ G(jω)= K s(1 + τ1s)(1 + τ2s) jω(1 + jωτ1)(1 + jωτ2) K 0−1−1 |G(jω)|= 2 2 2 2, φ=\G(jω)=−90 −tan ωτ1 −tan ωτ2 ω 1+ωτ1 1+ωτ2 I This portion of s-plane locus Γs begins from at ω = 0+, φ = +900 and ends atω=+∞,φ=+900 . I This is essentially the polar plot of the G(s)H(s). At ω = 0+ |G(jω)| = ∞ \G(jω) = −900 At ω = +∞ |G(jω)| = 0 \G(jω) = −2700 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 39 / 60 Example-4:Third Order System with Single Pole at the Origin (Contd) I From the mapping of section-I, it is observed that ΓF contour approaches a magnitude of zero at an angle of −2700. I Note that to approach at an angle of −2700, the locus must cross the real axis (u-axis) in the G(s)H(s) plane. Determination of Real-Axis Cross Over Point of G(s)H(s) Locus: Now G(jω) = K jω(1 + jωτ1)(1 + jωτ2) I This can be expressed by the form u + jv as G(jω) = K(−jω)(1 − jωτ1)(1 − jωτ2) (jω)(−jω)(1 + jωτ1)(1 + jωτ2)(1 − jωτ1)(1 − jωτ2) = −Kω2(τ1 + τ2) − jKω(1 − ω2τ1τ2) ω2(1 + ω2τ12)(1 + ω2τ2) = −K(τ1 + τ2) − j (K/ω)(1 − ω2τ1τ2) 1 + ω2(τ12 + τ2) + ω4τ12τ2 1 + ω2(τ12 + τ2) + ω4τ12τ2 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 40 / 60 Example-4: Third Order System with Single Pole at the Origin (Contd) I The polar plot intersects the real axis where the imaginary part of G(jω) = 0. This is computed by solving −(K/ω)(1 − ω2τ1τ2) 1 + ω2(τ12 + τ2) + ω4τ12τ2 ω=√1 τ1 τ2 = 0 This gives I The magnitude of the real part of G(jω)H(jω) at this frequency is obtained by computing the real part at this frequency. This gives −K(τ1+τ2) =− Kτ1τ2 1 + ω2(τ12 + τ2) + ω4τ12τ2 (τ1 + τ2) I Thus for the system to be stable Kτ1τ2 <1, =⇒ K<τ1+τ2 τ1 +τ2 τ1τ2 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 41 / 60 Example-4: Third Order System with Single Pole at the Origin (Contd) Mapping Section III: ω varies from −∞ to 0− I This portion of the contour is the mirror image of the polar plot. I The Nyquist contour is shown below Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 42 / 60 Example-5:Third Order System with Two Poles at the Origin Consider the unity feedback system with feed forward path transfer function G(s)= K , =⇒ G(jω)= K s2(1 + τs) −ω2(1 + jωτ) |G(jω)| = √ K , θ = \G(jω) = −1800 − tan−1ωτ ω2 1+ω2τ2 I The Nyquist path for the transfer function is shown below Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 43 / 60 Example-5:Third Order System with Two Poles at the Origin(Contd) Mapping Section IV: s = εejφ i. The Nyquist path is a small semicircular detour around the pole at the origin. ii. The equation of the semicircle ( detour) is given by s=εejφ, −900 ≤φ≤+900 where ε is infinitesimally small and is traversed from points j0− at φ = −900 to j0+ at φ = +900. lim G(s) = lim K = lim K e−2jφ = ∞e−2jφ ε→0 s=εejφ ε→0 ε2ej2φ ε→0 ε2 I Forthes-planelocusΓs,atω=0−,φ=−900 andatω=0+,φ=+900 . Hence, the map of this Γs into the G(s)H(s) plane ΓF satisfies the following: At ω = 0− |G(jω)| = ∞ \G(jω) = +1800 At ω = 0+ |G(jω)| = ∞ \G(jω) = −1800 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 44 / 60 Example-5: Third Order System with Two Poles at the Origin (Contd) Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. lim G(s) r→∞ K K − 3 j φ − 3 j φ = lim 3e =0e s=rejφ = lim 3 r→∞ s τ s=rejφ r→∞ τr I This portion of s-plane locus Γs,at ω = +∞, φ = +900 and at ω = −∞, φ = −900 is mapped into the G(s)H(s) plane ΓF which satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −2700 At ω = −∞ |G(jω)| = 0 \G(jω) = +2700 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 45 / 60 Example-5:Third Order System with Two Poles at the Origin (Contd) Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The beginning and end point of the contour ΓF , for this portion of the Γs, has been obtained before. iii. For other values of ω, this computed from the magnitude and phase angle of G(s)H(s). G(s)= K , =⇒ G(jω)= K s2(1 + τs) −ω2(1 + jωτ) |G(jω)| = √ K , θ = \G(jω) = −1800 − tan−1ωτ ω2 1+ω2τ2 I This portion of s-plane locus Γs begins from at ω = 0+, φ = +900 and ends atω=+∞,φ=+900 . I This is essentially the polar plot of the G(s)H(s). At ω = 0+ |G(jω)| = ∞ \G(jω) = −1800 At ω = +∞ |G(jω)| = 0 \G(jω) = −2700 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 46 / 60 Example-5:Third Order System with Two Poles at the Origin (Contd) Mapping Section III: ω varies from −∞ to 0− I This portion of the contour is the mirror image of the polar plot. I The Nyquist contour is shown below Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 47 / 60 Example-6: Second Order System with Real Pole in the Right Half Consider a unity feedback system with open-loop transfer function G(s) = K1 = (−K1/ab) (s+a)(s−b) (1+s/a)(1−s/b) = −K ,K=K1/ab, =⇒ G(jω)= −K (1 + τ1s)(1 − τ2s) (1 + jωτ1)(1 − jωτ2) K 0 −1 −1 |G(jω)|= 1+ω2τ121+ω2τ2, φ=−180 −tan ωτ1 +tan ωτ2 I The Nyquist path is shown below. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 48 / 60 Example-6:Second Order System with One Real Pole in Right Half Mapping Section II: s = rejφ, r → ∞ i. The Nyquist path is a semicircle of infinite radius r tending to infinity, which includes the entire right half of s-plane. ii. The equation of the semicircle is given by s = rejφ, r → ∞ where r tends to infinity and is traversed from points +j∞ at φ = +900 to −j∞ at φ = −900. K −j2φ −j2φ =lim 2e =0e I Thus, as the phasor in Γs is traversed from +900 to −900 ( i.e from ω = +∞ to ω = −∞) through 1800 clockwise,the corresponding contour ΓF i.e Nyquist plot of G(s), is traced by a phasor of practically zero length from −900 to +900 in the counter clockwise sense and satisfies the following: At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 At ω = −∞ |G(jω)| = 0 \G(jω) = +1800 limG(s) r→∞ s=rejφ K =lim 2 r→∞ s τ1τ2 s=rejφ r→∞ τ1τ2r Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 49 / 60 Example-6: Second Order System with One Real Pole in Right Half Mapping Section I: ω varies from 0+ to +∞ i. The Nyquist path is the positive jω-axis where ω varies from 0+ to +∞. ii. The starting point of this part of the contour ΓF , is computed as: lim G(jω) = lim −K ω→0 ω→0 (1 + jωτ1)(1 − jωτ2) = K\−1800 iii. The termination (end) point of this part of the contour ΓF , is computed before and is 0\−1800: I The beginning and end of ΓF contour in this section of Γs satisfies the following At ω = 0+ |G(jω)| = K \G(jω) = −1800 At ω = +∞ |G(jω)| = 0 \G(jω) = −1800 iii. For other values of ω, this is computed from K 0 −1 −1 |G(jω)|= 1+ω2τ121+ω2τ2, φ=−180 −tan ωτ1 +tan ωτ2 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 50 / 60 Example-6: Second Order System with one Real Pole in Right Half Mapping Section III: ω varies from −∞ to 0− I This portion of the contour is the mirror image of the polar plot. I Figure below shows the Nyquist plot for the unity feedback system with feed-forward transfer function G(s) = 5 = 2.5 (s+2)(s−1) (1+0.5s)(1−s) = K1 , K=5,K1 =K/2, τ1 =0.5,τ2 =1 (1 + τ1s)(1 + τ2s) Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 51 / 60 Stability Investigation from Nyquist Plot I Consider the unity feedback system shown in the Figure. I If all the open-loop poles of a system lie in the left half of s-plane, then the closed loop system will be asymptotically stable if and only if the Nyquist plot makes no encirclements around the point (−1, 0) I As K increases over a critical value, the closed loop system goes from asymptotically stable to unstable. I In this context, the proximity of F(jω) contour to the critical ;point (-1,0) is an indicator of the proximity of the closed loop system to instability. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 52 / 60 Stability Investigation from Nyquist Plot-2: Gain Margin I The gain margin kGM is the smallest amount that the closed loop system can tolerate (strictly) before it becomes unstable. I Intheplotωπ =ωpc isthe phase-cross over frequency. At this frequency, the phase angle of F(jω) equals to −π i.e ωπ :\F(jωπ =−π kGM|dB =−|F(jωπ)|dB I Note that if we multiply F(jω) by the the quantity kGM , the Nyquist diagram will pass through the critical point. kGM= 1 |F (jωπ )| Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 53 / 60 Stability Investigation from Nyquist Plot-3: Phase Margin Phase Margin PM(φ): I The phase margin margin PM is the amount of lag the closed loop system can tolerate (strictly) before it becomes unstable. I Let the frequency at which the gain is unity be denoted as ωc . This frequency is called gain crossover frequency of the system. ωc : |F(jωc)| = 1 ωc :|F(jωc)|dB =0dB PM =π+\F(jωc) Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 54 / 60 Stability Investigation from Nyquist Plot-4 I The Nyquist plot of three systems A,B, and C are shown in the figure. 1. The gain margin of both the systems B and A are same; however, their phase margins are different. 2. The phase margin of both the systems C and A are same; however,their gain margins are different. Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 55 / 60 Stability Investigation from Nyquist Plot-5: Systems with Infinite Gain Margin, Phase Margin System : Infinite Gain Margin System: Infinite Phase Margin Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 56 / 60 Example-1: First Order System with Real Pole I The Nyquist plot for the unity feedback system with feed-forward transfer function F(s)= 1 s+1 function [] = circle(r,c) % This function plots a circle of radius r n=100;t=linspace(0,2*pi,n)’; x=c(1)+r*sin(t);y=c(2)+r*cos(t); plot(x,y,’.’,’LineWidth’,1.5) end Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 57 / 60 Example-1: First Order System with Real Pole MATLAB Source Code clear K=1;num=[K];den=[1 1]; G=tf(num,den); h=nyquistplot(G); xmin=-1.5; xmax=1.5; ymin=-1.2; ymax=1.2; axis ([xmin xmax ymin ymax]) xreal=linspace(xmin,xmax,200)’;yreal=zeros(200,1); yimag=linspace(ymin,ymax,200)’;ximag=zeros(200,1); hold on plot(xreal,yreal,’LineWidth’,1.5) hold on plot(ximag,yimag,’LineWidth’,1.5) hold on circle(1,[0 0]); title(’First Order System’) % gtext(’ω = 0+’,’FontSize’,18) % gtext(’ω = 0−’,’FontSize’,18) % gtext(’ω = +∞’,’FontSize’,18) % gtext(’ω = −∞’,’FontSize’,18) % gtext(’(K)’,’FontSize’,18) Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 58 / 60 Example-2:Nyquist Plot I The Nyquist plot for the unity feedback system with feed-forward transfer function F (s) = 100 (s + 10)2 num=[100];den=[1 20 100]; G=tf(num,den); h=nyquistplot(G); Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 59 / 60 Akshya Swain Module-9:Frequency Domain Analysis of Linear Systems 60 / 60