Unit 4

Frequency Domain Analysis

4.1 Frequency response and frequency domain specifications

1>. Resonant Peak (Mr): The maximum value of magnitude is known as Resonant peak. The relative stability of the system can be determined by Mr. The larger the value of Mr the undesirable is the transient response.

2>. Resonant Frequency (Wr): The frequency at which magnitude has maximum value.

3>. Bandwidth: The band of frequencies lying between -3db points.

4>. Cut-off frequency –The frequency at which the magnitude is 3db below its zero frequency.

5>. Cut-off Rate – It is the slope of the log magnitude curve near the cut off frequency.

Fig 1 Frequency Domain Specification

4.2 Correlation between time domain and frequency domain specifications

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)

The comparison of Mr and Mp is shown in figure(1). 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 fig(2) 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(1) it is clear that for ξ> 1/2, value of Mr does not exists.

Key takeaway

i)                   Mr and Mp are correlated as both are functions of the damping factor ξ only

Ii)                 When subjected to step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

4.3 Polar plot

It gives the frequency response of the system. If the transfer function is given, then from the plot number of poles and zeros can be calculated.

Polar plot of some standard functions:

# TYPE ‘O’

Ex: 1 T(S) = 1/S + 1

(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. 3(a)

Fig 3(a) Polar Plot T(S) = 1/S + 1

Ex.2>.  T(S) = 1/(S+1)(S+2)

(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 fig3(b)

Fig3(b) Polar Plot T(S) = 1/(S+1)(S+2)

Intersection of polar plot with imaginary axis will be when real part of Transfer function = 0

M = 1/(jw + 1)(jw + 2)

= 1/-w2 + j3w + 2

TYPE ‘1’

Ex.1 T(S) = 1/S

(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.4(a)

Fig 4(a) Polar Plot T(S) = 1/S

Ex.2 T(S) = 1/S2

(1). S = jw

(2). M = 1/w2

(3). Φ = -tan-1(W/O)-tan-1(W/O) = -1800

W  M  Φ

0  ∞  -1800

1  1  -1800

2  0.25  -1800

∞  0  -1800

The plot is shown in fig.4(b)

Fig 4(b) Polar Plot T(S) = 1/S2

Key takeaway

• For Polar Plot starting point depends upon type of system
• The terminating phase depends on order of the system.

TYPE 1 ORDER 2

Ex.1  T(S) = 1/S(S+1)

(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.5(a)

Fig 5(a) Polar Plot T(S) = 1/S(S+1)

Ex.2TYPE 2 ORDER 3

T(s) = 1/S2(S+1)

(1). M = 1/w2√1+jw

(2). Φ = -1800 – tan-1W/T

The plot is shown in fig.5(b)

Fig 5(b) Polar Plot T(s) = 1/S2(S+1)

4.4 Nyquist stability criterion and construction of Nyquist plot

It gives the frequency response as well as comments on the stability and the relative stability of the system.

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 6 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 is

Fig 7 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 vice versa.

No of encirclements and enclosures:

 For A line is cut once

For B line is cut twice

Fig 8 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.

OLTF  G(S) = (S + Z1)(S + Z2)/(S + P1)(S + P2)  H(S) = 1  - - (1)

Fig 9 Unity feedback control system

CLTF  G(S)/1 + G(S)

CE = 1 + G(S)

= 1 + (S + Z1)(S + Z2)/(S + P1)(S + P2)

CE = (S + P1)(S + P2) + (S + Z1)(S + Z2)/ (S + P1)(S + P2)  - - (2)

Key takeaway:

# OLTF poles is equal to CE poles.

CE = (S + Z’1)( S + Z’2)/( S + P1)( S + P2)  - - (3)

CLTF = G(S) –(1) / 1 + G(S) –(3)

= (S + Z1)( S + Z2) / (S + Z’1)( S + Z’2)  - - (4)

Key takeaway:

# Zeros of characteristic equation is poles of CLTF (3 and 4).

For the closed loop system to be stable zeros of CE(i.e. poles of CLTF) should not be located at right half of the S-plane.

Consider a contour, which covers the entire right half of S-plane.

Fig 10 Contour for Right half of S-plane

P1 W(0  - ∞)

P2 RejR  ∞

ϴ - π/2 to 0 to + π/2

P3 W(∞ to 0 )

If each and every point along the boundary of contour is mapped in q(S) where q(S) is 1+G(S)H(S)[CE]. The CE is drawn in S-domain. Now, as the CE: q(s) = 1 + G(S)H(S) contour is drawn into S-plane.

This q(S) contour may encircle the origin. Thus, the number of encirclements of q(S) contour with respect to origin is given by

N = Z – P

Where: Z1P  zeros and poles of q(S)[CE]

N  Total no of encirclement of origin

• Z1P  Zeros and poles of CE in the right half of S-plane

** for the CL system to be stable Z=0 always.

Important:

# Open loop System(stable): When OL system is stable P=0 i.e. no of poles on right half

N = Z – P

If P = 0

N = Z

• Now for CL system to be stable Z = 0
• N = 0
• i.e. q(s) contour should not encircle the origin.

# Open loop system(unstable): Let P = 1 i.e. one OL pole is located in right half of S-plane i.e. OLTF is unstable.

As N = Z – P

N = Z – 1

For CL system to be stable the only criteria is (Z=0) i.e.

N = -1

Which means q(S) contour should encircle the origin one’s in CW direction.

Key takeaway:

(1). When OL system is unstable then corresponding CL will be stable only when q(S) contour will encircle origin in CW direction.

(2). The no of encirclements should be equal to no of open loop poles located in right half of S-plane.

** the no of encirclements(N) can also be calculated by using G(S) contour (instead of q(S) contour) but the reference is -1+j0 instead of 0+j0 i.e. the no of encirclements should be considered w.r.t -1+j0 and not with the origin.

Explanation Mapping

q(S) = 1 + G(S)

G(S) is always given to us, so we can relate G(S) with q(S).

G(S) = q(S) – 1

But q(S) can be drawn by adding 1 real part to the q(S).

Fig 11 Mapping

G(S) given then q(S) shift to right side.

Q1. For the transfer function below plot the Nyquist plot and also comment on stability?

G(S) = 1/S+1

Sol:- 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)

Fig 12 Nyquist path

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 13

Fig 13 Nyquist Plot 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.

Q.2. For the transfer function below plot the Nyquist Plot and comment on stability G(S) = 1/(S + 4)(S + 5)

Soln :-  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)

Fig 14 Nyquist Path

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 15. From plot N=0, Z=0, system stable.

Fig 15 Nyquist Plot G(S) = 1/(S + 4)(S + 5)

Q.3. For the given transfer function, plot the Nyquist plot and comment on stability G(S) = k/S2(S + 10)?

Soln: 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.

Fig 16 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

= Z – P

No poles on right half of S plane so, P = 0

N = Z – 0

Fig 17 Nyquist Plot for G(S) = k/S2(S + 10)

But from plot shown in fig 16. 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.

Fig 18 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 19 Nyquist Plot G(S) = k/S2(S + 10)

The plot is shown in fig 19. 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

NOTE: The sign convention for angles is shown

Fig 20 Sign for Angle Directions

Angles are considered –ve for anticlockwise directions and +ve for clockwise directions.

4.5 Bode plot

In polar plot any point gives the magnitude phase of the transfer function in bode we split magnitude and plot.

Advantages

1. By looking at bode plot we can write the transfer function of system

Q. G(S) =

1. Substitute S = j

G(j) = 

M =

= tan-1 = -900

Magnitude varies with ‘w’ but phase is constant.

MdB = +20 log10

Decade frequency:

W present = 10 past

Then present is called decade frequency of past

2 = 10 1

2 is decade frequency of 1

 MdB

0.01 40

0.1 20

1 0 (shows pole at origin)

0 -20

10 -40

100 -60

Slope  = (20db/decade)

Fig 21 MAGNITUDE PLOT

Fig 22 PHASE PLOT

Q.G(S) =

G(j) =

M =   ;  = -1800 (-20tan-1)

MdB = +20 log -2

MdB = -40 log10

 MdB

0.01 80

0.1 40

1 0 (pole at origin)

10 -40

100 -80

Slope = 40dbdecade

Q. G(S) = S

M= W

= 900

MdB = 20 log10

 MdB

0.01 -40

0.1 -20

1 0 

10 20

100 90

1000 60

Fig 23 Bode Plot G(S) = S

Q. G(S) = S2

M= 2   MdB = 20 log102

= 1800         = 40 log10

W MdB

0.01 -80

0.1 -40

1 0 

10 40

100 80

Fig 24 Magnitude Plot G(S) = S2

Q. G(S) =

G(j) =

M =

MdB = 20 log10 K-20 log10

= tan-1() –tan-1()

= 0-900 = -900

K=1   K=10

  MDb   MdB

=-20 log10  =20 -20 log10

0.01  40   60

0.1  20   40

1  0   20  

10  -20   0

100  -40   -20

Fig 25 Bode Plot G(S) =

Fig 25 All bode plots in one plot

Fig 26 Variation in K shifts magnitude plot by +20db

As we vary K then plot shift by 20 log10K

i.e adding a d.c. To a.c. Quantity

Approximation of Bode Plot:

IF poles and zeros  are not located at origin

G(S) =

TF =

M =

MdB = -20 log10 (

= -tan-1

Approximation: T >> 1. So, we can neglect 1.

MdB = -20 log10

MdB = -20 log10T . ; = -tan-1(T)

Approximation: T << 1. So, we can neglecting T.

MdB= 0dB, = 00

At a point both meet so equal i.e a time will come hence both approx become equal

-20 log10T= 0

T= 1

  corner frequency

At this frequency both the cases are equal

MdB = -20 log10

Now for

MdB = -20 log10

= -20 log10

= -10 log102

MdB = 10

Fig 27 Approximation in bode plot

When we increase the value of in app 2 and decrease the of app 1 so a RT comes when both cases are equal and hence for that value of where both app are equal gives max. Error we found above and is equal to 3dB

At corner frequency we have max error of -3dB

Q. G(S) =

TF =

M =

MdB = -20 log10 (     at T=2

  MdB

1  -20 log10

10  -20 log10

100 -20 log10

  MdB  = =

0.1  -20 log10 = 1.73 10-3

0.1  -20 log10 = -0.1703

0.5  -20 log10 = -3dB

1  -20 log10 = -6.98

10  -20 log10 = -26.03

100 -20 log10 = -46.02

Fig 28 Magnitude Plot with approximation

Without approximation

For second order system

TF =

TF =

=

=

=

M=

MdB=

Case 1 <<

<< 1

MdB= 20 log10 = 0 dB

Case 2  >>

>> 1

MdB = -20 log10

= -20 log10

= -20 log10

< 1 is very large so neglecting other two terms

MdB = -20 log10

= -40 log10

Case 3 . When case 1 is equal to case 2

-40 log10 = 0

= 1

The natural frequency is our corner frequency

Fig 29 Magnitude Plot

Max error at i.e at corner frequency

MdB = -20 log10

For

MdB = -20 log10

error for

Completely the error depends upon the value of (error at corner frequency)

The maximum error will be

MdB = -20 log10

M = -20 log10

= 0

is resonant frequency and at this frequency we are getting the maximum error so the magnitude will be

M = -+

=

Mr =

MdB = -20 log10

MdB = -20 log10 

= tan-1

Mr =

4.6 Determination of frequency domain specifications and stability analysis using Nyquist plot and Bode plot

4.6.1 Determination of frequency domain specifications

Stability Condition in Bode Plot:

     Gain Cross Over Frequency:

The frequency at which the bode plot culls the 0db axis is called as Gain Cross Over Frequency.

Fig 30 Gain cross over frequency

     Phase Cross Over Frequency

The Frequency at which the phase plot culls the -1800 axis.

Fig 31 Phase Margin and Gain Margin

GM=MdB= -20 log [ G (jw)]

                   .:    

               .:

1) When gain cross over frequency is smaller than phase curves over frequency the system is stable and vice versa.

Key takeaway:

i)                   More the difference b/WPC and WGC core are the stability of system

Ii)                 If GM is below 0dB axis than take ilb +ve and stable. If GM above 0dB axis, that is take -ve

GM= ODB - 20 log M 

Iii)              The IM should also lie above -1800 for making the system (i.e. pm=+ve

Iv)               For a stable system GM and PM should be -ve

v)                 GM and PM both should be +ve more the value of GM and PM more the system is stable.

Vi)               If Wpc and Wgc are in same line Wpc= Wgc  than system is marginally stable . As we get GM=0dB.

4.6.2 Stability analysis using 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 32 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 is

Fig 33 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 vice versa.

No of encirclements and enclosures:

 For A line is cut once

For B line is cut twice

Fig 34 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.

OLTF  G(S) = (S + Z1)(S + Z2)/(S + P1)(S + P2)  H(S) = 1  - - (1)

Fig 35 Unity feedback control system

CLTF  G(S)/1 + G(S)

CE = 1 + G(S)

= 1 + (S + Z1)(S + Z2)/(S + P1)(S + P2)

CE = (S + P1)(S + P2) + (S + Z1)(S + Z2)/ (S + P1)(S + P2)  - - (2)

Key takeaway:

# OLTF poles is equal to CE poles.

CE = (S + Z’1)( S + Z’2)/( S + P1)( S + P2)  - - (3)

CLTF = G(S) –(1) / 1 + G(S) –(3)

= (S + Z1)( S + Z2) / (S + Z’1)( S + Z’2)  - - (4)

Key takeaway:

# Zeros of characteristic equation is poles of CLTF (3 and 4).

For the closed loop system to be stable zeros of CE(i.e. poles of CLTF) should not be located at right half of the S-plane.

Consider a contour, which covers the entire right half of S-plane.

Fig 36 Nyquist Path

P1 W(0  - ∞)

P2RejR  ∞

ϴ - π/2 to 0 to + π/2

P3 W(∞ to 0 )

If each and every point along the boundary of contour is mapped in q(S) where q(S) is 1+G(S)H(S)[CE]. The CE is drawn in S-domain. Now, as the CE : q(s) = 1 + G(S)H(S) contour is drawn into S-plane.

This q(S) contour may encircle the origin. Thus, the number of encirclement of q(S) contour with respect to origin is given by

N = Z – P

Where : Z1P  zeros and poles of q(S)[CE]

N  Total no of encirclement of origin

• Z1P  Zeros and poles of CE in the right half of S-plane

** for the CL system to be stable Z=0 always

4.6.3 Stability analysis using Bode plot

1)     The more the value of gain margin more is the stability.

2)     The more the value of phase margin more is the stability.

3)     The gain cross over frequency, phase cross over frequency are used in determining the stability of system.

4)     The frequency at which the two asymptotes cuts or meet each other is known as break frequency or corner frequency.

References:

1. Benjamin C. Kuo, “Automatic control systems”, Prentice Hall of India, 7th Edition.

2. M. Gopal, “Control System – Principles and Design”, Tata McGraw Hill, 4th Edition.

3. N. J. Nagrath and M. Gopal, “Control System Engineering”, New Age International Publishers, 5th Edition.

4. K. Ogata, “Modern Control Engineering”, Prentice Hall India Learning Private Limited; 5th Edition.