Unit - 4
Stability Criterion Continued
Q1) Plot polar plot for T(S) = 1/S + 1
A1) (1) For polar plot substitute S=jw.
TF = 1/1 + jw
(2). Magnitude M = 1 + 0j / 1 + jw = 1/√1 + w2
(3). Phase φ = tan-1(0)/ tan-1w = - tan-1w
W M φ
0 1 00
1 0.707 -450
∞ 0 -900
The plot is shown in fig. 1
Fig 1 Polar Plot T(S) = 1/S + 1
Q2) Plot polar plot for T(S) = 1/(S+1)(S+2)
A2)
(1). S = jw
TF = 1/(1+jw)(2+jw)
(2). M = 1/(1+jw)(2+jw) = 1/-w2 + 3jw + 2
M = 1/√1 + w2√4 + w2
(3). Φ = - tan-1 w - tan-1(w/2)
W M Φ
0 0.5 00
1 0.316 -71.560
2 0.158 -108.430
∞ 0 -1800
The plot is shown in fig 2
Fig 2 Polar Plot for T(S) = 1/(S+1)(S+2)
Intersection of polar plot with imaginary axis will be when realpart of Transfer function = 0
M = 1/(jw + 1)(jw + 2)
= 1/-w2 + j3w + 2
Real part
Re(M) = 1/(2-w2)+j3w x (2-w2)-3jw/(2-w2)-3jw
Re(M) = (2-w2)/(2-w2+9w2) - 3jw/(2-w2) +9w2
Equating Real part = 0
(2-w2)/(2-w2)+ 9w2 = 0
W = +-√2
For w=√2 real part on the polar plot becomes zero.
So, polar plot intersects imaginary axis at w=√2 at φ = -900
Q3) Plot polar plot for T(S) = 1/(S+1)(S+2)(S+3)
A3)
(1). Substitute S =jw
(2). M = 1/√1+w2 √4+w2 √9+w2
(3). Φ = -tan-1w – tan-1 w/2 – tan-1w/3
W M φ
0 0.16 0
1 0.1 -900
2 0.04 -142.10
∞ 0 -270
(4). Intersection of Polar plot with Real axis
M = 1/(S+1)(S+2)(S+3)
= 1/S3+6S2+11S+6
=1/(jw)2+6(jw)2+11jw+6
= 1/(6-6w2)+j(11w-w3)
Re(M) = 0 [Intersection with imaginary axis ]
M = 6-6w2/(6-6w2)2+(11w-w2)2 - j(11w-w3)/(6-6w2)2+(11w-w2)2
Re(M) = 0
6-6w2 = 0
w=1
Im(M) = 0[Intersection with Real axis]
11w = w3
The plot is shown in fig 3
Fig 3 Polar Plot for T(S) = 1/(S+1)(S+2)(S+3)
Q4) For T(S) = 1/S(S+1) plot polar plot?
A4)
(1). M = 1/W√1+w2
(2). Φ = -900 - tan -1(W/T)
W M φ
0 ∞ -900
1 0.707 -1350
2 0.45 -153.40
∞ 0 -1800
The plot is shown in fig.4
Fig 4 Polar Plot for T(S) = 1/S(S+1)
Q5) For T(s) = 1/S2(S+1) plot polar plot?
A5)
(1). M = 1/w2√1+jw
(2). Φ = -1800 – tan-1W/T
The plot is shown in fig. 5
Fig 5 Polar Plot for T(s) = 1/S2(S+1)
Q6) For T(S) = 1/S plot polar plot?
A6)
(1). S = jw
(2). M = 1/W
(3). Φ = -tan-1(W/O) = -900
W M φ
0 ∞ -900
1 1 -900
2 0.5 -900
∞ 0 -900
The plot is shown in fig. 6
Fig 6 Polar plot for T(S) = 1/S
Q7) For the transfer function below plot the Nyquist plot and also comment on stability? G(S) = 1/S+1
A7)
N = Z – P (No pole of right half of S plane P = 0)
P = 0, N = Z
NYQUIST PATH:
P1 = W – (0 to - ∞)
P2 = ϴ( - π/2 to 0 to π/2 )
P3 = W(+∞ to 0)
Substituting S = jw
G(jw) = 1/jw + 1
M = 1/√1+W2
Φ = -tan-1(W/I)
For P1: W(0 to -∞)
W M φ
0 1 0
-1 1/√2 +450
-∞ 0 +900
Path P2:
W = Rejϴ R ∞ϴ -π/2 to 0 to π/2
G(jw) = 1/1+jw
= 1/1+j(Rejϴ) (neglecting 1 as R ∞)
M = 1/Rejϴ = 1/R e-jϴ
M = 0 e-jϴ = 0
Path P3:
W = -∞ to 0
M = 1/√1+W2, φ = -tan-1(W/I)
W M φ
∞ 0 -900
1 1/√2 -450
0 1 00
The Nyquist Plot is shown in fig 7
Fig 7 Nyquist plot for G(S) = 1/S+1
From plot we can see that -1 is not encircled so, N = 0
But N = Z, Z = 0
So, system is stable.
Q8) For the transfer function below plot the Nyquist Plot and comment on stability G(S) = 1/(S + 4)(S + 5)
A8)
N = Z – P , P = 0, No pole on right half of S-plane
N = Z
NYQUIST PATH
P1 = W(0 to -∞)
P2 = ϴ(-π/2 to 0 to +π/2)
P3 = W(∞ to 0)
Path P1 W(0 to -∞)
M = 1/√42 + w2 √52 + w2
Φ = -tan-1(W/4) – tan-1(W/5)
W M Φ
0 1/20 00
-1 0.047 25.350
-∞ 0 +1800
Path P3 will be the mirror image across the real axis.
Path P2: ϴ(-π/2 to 0 to +π/2)
S = Rejϴ
G(S) = 1/(Rejϴ + 4)( Rejϴ + 5)
R∞
= 1/ R2e2jϴ = 0.e-j2ϴ = 0
The plot is shown in fig 8. From plot N=0, Z=0, system stable.
Fig 8. Nyquist plot for G(S) = 1/(S + 4)(S + 5)
Q9) For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?
A9)
As the poles exists at origin. So, first time we do not include poles in Nyquist plot. Then check the stability for second case we include the poles at origin in Nyquist path. Then again check the stability.
PART – 1: Not including poles at origin in the Nyquist Path.
P1 W(∞ Ɛ) where Ɛ 0
P2 S = Ɛejϴ ϴ(+π/2 to 0 to -π/2)
P3 W = -Ɛ to -∞
P4 S = Rejϴ, R ∞, ϴ = -π/2 to 0 to +π/2
For P1
M = 1/w.w√102 + w2 = 1/w2√102 + w2
Φ = -1800 – tan-1(w/10)
W M Φ
∞ 0 -3 π/2
Ɛ ∞ -1800
Path P3 will be mirror image of P1 about Real axis.
G(Ɛ ejϴ) = 1/( Ɛ ejϴ)2(Ɛ ejϴ + 10)
Ɛ 0, ϴ = π/2 to 0 to -π/2
= 1/ Ɛ2 e2jϴ(Ɛ ejϴ + 10)
= ∞. e-j2ϴ [ -2ϴ = -π to 0 to +π ]
Path P2 will be formed by rotating through -π to 0 to +π
Path P4 S = Rejϴ R ∞ ϴ = -π/2 to 0 to +π/2
G(Rejϴ) = 1/ (Rejϴ)2(10 + Rejϴ)
= 0
N = Z – P
No poles on right half of S plane so, P = 0
N = Z – 0
Fig 9 Nyquist plot for G(S) = k/S2(S + 10)
But from plot shown in fig 9. It is clear that number of encirclements in Anticlockwise direction. So,
N = 2
N = Z – P
2 = Z – 0
Z = 2
Hence, system unstable.
PART 2 Including poles at origin in the Nyquist Path.
P1 W(∞ to Ɛ) Ɛ 0
P2 S = Ɛejϴ Ɛ 0 ϴ(+π/2 to +π to +3π/2)
P3 W(-Ɛ to -∞) Ɛ 0
P4 S = Rejϴ, R ∞, ϴ(3π/2 to 2π to +5π/2)
M = 1/W2√102 + W2 , φ = - π – tan-1(W/10)
P1 W(∞ to Ɛ)
W M φ
∞ 0 -3 π/2
Ɛ ∞ -1800
P3( mirror image of P1)
P2 S = Ɛejϴ
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10 + Ɛejϴ)
Ɛ 0
G(Ɛejϴ) = 1/ Ɛ2e2jϴ(10)
= ∞. e-j2ϴϴ(π/2 to π to 3π/2)
-2ϴ = (-π to -2π to -3π)
P4 = 0
Fig 10 Nyquist Plot for G(S) = k/S2(S + 10)
The plot is shown in fig. From the plot it is clear that there is no encirclement of -1 in Nyquist path. (N = 0). But the two poles at origin lies to the right half of S-plane in Nyquist path.(P = 2)[see path P2]
N = Z – P
0 = Z – 2
Z = 2
Hence, system is unstable.
Path P2 will be formed by rotating through -π to -2π to -3π
Q10) Explain relation between time and frequency response?
A10) The calculation of system parameters for higher order systems is difficult in time domain analysis. This can be overcome by frequency domain analysis. As the frequency response of a linear time invariant system is independent of amplitude and phase of the input test signal. The effect of noise and parameter variation in frequency domain can be easily computed.
In frequency response can be determine by calculating the phase and amplitude oof the given system transfer function. If a sinusoidal signal XG(j
)sin
t then its amplitude will be
C(t) = XG(j
)sin
t
= X|G(j
)|
sin
t
Magnitude = |G(j
)|
Phase = 
In polar form = |G(j
)|
The transfer function of second order system is shown as
C(S)/R(S) = W2n / S2 + 2ξWnS + W2n - - (1)
ξ = Ramping factor
Wn = Undamped natural frequency for frequency response let S = jw
C(jw) / R(jw) = W2n / (jw)2 + 2 ξWn(jw) + W2n
Let U = W/Wn above equation becomes
T(jw) = W2n / 1 – U2 + j2 ξU
So,
| T(jw) | = M = 1/√(1 – u2)2 + (2ξU)2 - - (2)
T(jw) = φ = -tan-1[ 2ξu/(1-u2)] - - (3)
For sinusoidal input the output response for the system is given by
C(t) = 1/√(1-u2)2 + (2ξu)2Sin[wt - tan-1 2ξu/1-u2] - - (4)
The frequency where M has the peak value is known as Resonant frequency Wn. This frequency is given as (from eqn (2)).
DM/du|u=ur = Wr = Wn√(1-2ξ2) - - (5)
From equation(2) the maximum value of magnitude is known as Resonant peak.
Mr = 1/2ξ√1-ξ2 - - (6)
The phase angle at resonant frequency is given as
Φr = - tan-1 [√1-2ξ2/ ξ] - - (7)
As we already know for step response of second order system the value of damped frequency and peak overshoot are given as
Wd = Wn√1-ξ2 - - (8)
Mp = e- πξ2|√1-ξ2 - - (9)
Fig 11 Frequency Domain Specification
The comparison of Mr and Mp is shown in figure. The two performance indices are correlated as both are functions of the damping factor ξ only. When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.
Similarly the correlation of Wr and Wd is shown in figure for the given input step response [ from eqn(5) & eqn(8) ]
Wr/Wd = √(1- 2ξ2)/(1-ξ2)
Mp = Peak overshoot of step response
Mr = Resonant Peak of frequency response
Wr = Resonant frequency of Frequency response
Wd = Damping frequency of oscillation of step response.
From fig(17) it is clear that for ξ> 1/2, value of Mr does not exists.
Q11) For the following system shown below find the phase margin and gain margin if the value of k=500by using Nyquist plot.
A11) G(s)H(s) = 500/s(s+6)(s+9)
The polar plot for above OLTF is shown in figure below.
Fig 12 Polar Plot for given system
From figure we see A=0.6,
The GM in dB = -20logA.
= 4.43dB
PM = 1800-1650
= 150
Q12) For the system shown in figure, find the GM and PM using Nyquist plot for k=10. Find also range of k for stability?
A12) The transfer function is G(s) H(s) = 10(s+2(s+4)/s(s2 – 3s+10)
As the OLTF is unstable system so PM and GM concepts cannot be applied.
G(jω) H(jω) = 10(jω+2) (jω+4)/ jω(-ω2 -3 jω+10)
| G(jω) H(jω)|= 10(√4+ω2)( √16+ω2)/ ω√[(10-ω)2+9 ω2]
G(jω)H(jω) = -900 +tan-1(ω/2) + tan-1(ω/4) - tan-1 (-3ω/10- ω2)
The polar plot for above system is shown below.
Fig 13 Polar plot for the given system
From figure we have A=2.6 and ɸ =-1350
The GM in dB = -20logA.
= -8.3dB
As GM is negative it indicates that system is open loop unstable.
PM = 1800-1650 = 450
Q13) Write short note on Nicholas chart?
A13) The Nichols plot is named after the American engineer N.B Nichols who formulated this plot. Compensators can be designed using Nichols plot. Nichols plot technique is however also used in designing of dc motor. This is used in signal processing and control design. Nyquist plot in complex plane shows how phase of transfer function and frequency variation of magnitude are related. • Angle of positive real axis determines the phase and distance from origin of complex plane determines the gain.
Advantages
- Gain and phase margin can be determined easily and also graphically.
- Closed loop frequency response is obtained from open loop frequency response.
- Gain of the system can be adjusted to suitable values.
- Nichols chart provides frequency domain specifications.
Disadvantage
Using Nichols plot small changes in gain cannot be encountered easily.
Q14) What are M circles? Derive its equation for forming circle?
A14) Constant magnitude loci that are M-circles and constant phase angle loci that are N-circles are the fundamental components. The constant M and constant N circles in G (jω) plane can be used for the analysis and design of control systems. However, the constant M and constant N circles in gain phase plane are prepared for system design and analysis as these plots supply information with fewer manipulations. Gain phase plane is the graph having gain in decibel along the ordinate (vertical axis) and phase angle along the abscissa (horizontal axis).
For Constant M Circle
For a system with unity feedback having transfer function
T(s)= G(s)/1+G(s)
Replace s=jω
T(jω) = G(jω)/1+G(jω)
But we can write G(jω)= X(jω)+jY(jω)
The closed loop transfer function T(s) is
T(jω) = X+jY/1+X+jY= Mejα
M= |X+jY|/|1+X+jY|
M2=X2+Y2/(1+X2) + Y2
Solving above equation we get
[X-(M2/1-M2)]2 + Y2 = ([M/1-M2)]2
This is equation of circle with centre (M2/1-M2),0 and radius M/1-M2
When M<1, M decreases the radius of the circle decreases and centres located on positive real axis shifts towards origin. When M=0 the radius becomes 0 and the centre (0,0). This is shown in figure above.
Fig 14 Constant M circle
Q15) Explain the Nyquist stability criterion of closed loop system?
A15) Nyquist Stability Criteria:
The Nyquist criteria is a semi graphical method that determines stability of CL system investigating the properties of the frequency domain plot (Polar plot), the Nyquist plot of the OLTF G(S) H(S) is represented as L(S)
L(S) = G(S)H(S)
Specially the Nyquist plot of L(S) is a plot drawn by substituting S=jw and varying the value of w as per in polar plot. In polar plot we take one sided frequency response ( 0 - ∞) in Nyquist plot we will vary the frequency in entire range possible from ( -∞ to 0 ) and (0 to ∞ )
Nyquist Criteria also gives:
(1). In addition to providing the absolute stability like other plots, the Nyquist criteria also gives information on the relative stability of a stable system and the degree of instability of an unstable system.
(2). It also gives indications on how the system stability can be improved.
(3). The Nyquist plot of G(S) H(S) is the polar plot of G(S) H(S) drawn with wider range of frequency ( -∞ to ∞ ) and along the Nyquist path.
(4). The Nyquist plot of G(S) H(S) gives information on frequency domain characteristics such as B.W, gain margin and phase margin.
Construction of Nyquist Plot
Encircled: A point or region in a complex function phase i.e. S-plane is said to be encircled by a closed path if it is found inside the path.
Assumption:
Fig 15 Encirclement
In this example point A is encircled by the closed path Y. Since, A is inside the closed path point B is not encircled by y. It is outside the path. Furthermore when the closed path Y, has a direction assign to it, encirclement, if made can be in the clockwise direction or in the anti-clockwise direction.
Point A is encircled by Y by anticlockwise direction. We can say that the region inside the path is encircled in the prescribed direction and the region outside the path is not encircled.
Enclosed:
A point or region is said to be enclosed by a closed path if it is encircled in the counter clockwise direction, or the point or region lies to the left of the path (always), when the path is traveling in the prescribed direction.
The concept of enclosure is particularly useful, if only a portion of a closed path is shown.
In this example the shaded region are
Fig 16. Enclosure
Considered to be enclosed by the closed path Y. In other words, point A is enclosed by Y in fig a. But is not enclosed by Y in fig b. And for point B it is viceversa.
No of encirclements and enclosures:
For A line is cut once
For B line is cut twice
Fig 17. Encirclement and Enclosure with example
As it’s overlapping but 2 times in Same direction
When a point is encircled by a closed path Y, a no. N can be assigned to the no. Of times it is encircled. The magnitude of N can be determined by drawing an arrow around the closed path Y.
Taking an arbitrary point S, and moving around in clockwise direction and anti-clockwise direction respectively. We are getting a direction.
The path followed by S1 gives us the direction and this path which covers the total number of revolution travelled by this point S1 is N or the net angle is ‘ 2 π N ’.
For B = 2 = N for A = 1 = N
In this eg. Point A is encircled ones (or 2 π radians) by function Y and point B is encircled twice (or 4 π radians)all in clockwise direction.
In diagram b again A and B are encircled but in counter clockwise direction thus for this diagram A is enclosed one’s and B is enclosed twice.
By definition M is +ve for anticlockwise(direction) encirclement and –ve for clockwise encirclement.
Q16) What is gain and phase margin in polar plots? How it is determined?
A16) Gain margin (GM): The gain margin is the change in open-loop gain, expressed in decibels (dB), required at 180◦ of phase shift to make the closed-loop system unstable.
Phase margin (PM): The phase margin is the change in open-loop phase shift, required at unity gain to make the closed-loop system unstable.
GM = 1/|G(jw)H(jw)| for w=wpc
=20log10(1/|G(jw)H(jw)|)
=-20 log10 (|G(jw)H(jw)|)
For PM refer figure
PM = 1800+ɸ
Gain crossover frequency (ωgc):
The frequency at which magnitude of G(wj)H(wj) is unity (i.e. |G(wj)H(wj)| =1) called gain frequency crossover.
ωpc > ωgc
GM & PM are stable (+ve) Stable system
Phase crossover frequency (ωpc):
The frequency at which phase angle of G(wj)H(wj) is -180o
ωgc > ωpc GM & PM are stable (-ve) Unstable system.
Q17) What are the advantages and disadvantage of frequency response?
A17) The advantages of frequency response
- The absolute and relative stability of the closed loop system can be estimated from the knowledge of open loop frequency response.
- The practical testing of the system can be easier carried with available sinusoidal signal generators and precise measurement equipment.
- The transfer function of complicated functions can be determined experimentally by frequency response test.
- The design and parameter adjustment can be carried more easily.
- The corrective measure for noise disturbance and parameter variation can be easily carried.
- It can be extended to certain non-linear system.
The disadvantages of frequency response analysis are:
- Frequency response analysis is not recommended for the system with very large time constants.
- It is not useful for non-interruptible systems.
- It can generally be applied only to linear systems. When this approach is applied to a non-linear system, the result obtained is not exact.
- It is considered outdated when compared with the methods developed for digital computers and modelling.
