Unit 3 | unit 3 relationship between time and frequency response

Unit - 3

Relationship between time and frequency response

3.1 Relationship between time and frequency response

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)sint then its amplitude will be

C(t) = XG(j)sint= X|G(j)|sint

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 1 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(1) it is clear that for ξ> 1/2, value of Mr does not exists.

Key takeaways:

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.

3.2 Polar plots

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: 1T(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. 2(a)

Fig 2(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 fig2(b)

Fig2(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.3(a)

Fig 3(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.3(b)

Fig 3(b) Polar Plot T(S) = 1/S2

Key takeaways:

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.4(a)

Fig 4(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.4(b)

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

3.3 Bode plots

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

Advantages

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

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 5 MAGNITUDE PLOT

Fig 6 PHASE PLOT

Que.G(S) =

Sol: 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

Que. G(S) = S

Sol: M= W

= 900

MdB = 20 log10

MdB

0.01 -40

0.1 -20

1 0

10 20

100 90

1000 60

Fig 7 Bode Plot G(S) = S

Que. G(S) = S2

Sol: M= 2

MdB = 20 log102 == 40 log10

= 1800

W MdB

0.01 -80

0.1 -40

1 0

10 40

100 80

Fig 8 Magnitude Plot G(S) = S2

Que. G(S) =

Sol: 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 9 Bode Plot G(S) =

Fig 10 All bode plots in one plot

Fig 11 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 neglect 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 12 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

Que. G(S) =

Sol: 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 13 Magnitude Plot with approximation

Without approximation

For second order system

TF =

MdB=

Case 1 <<

<< 1

MdB= 20 log10 = 0 dB

Case 2 >>

>> 1

MdB = -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 14 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

= tan-1

Mr =

Type of system	Initial slope	Intersection
0	0 dB/decade	Parallel to 0 axis
1	-20 dB/decade	=K1
2	-40 dB/decade	=K1/2
3	-60 dB/decade	=K1/3
.	.	1
.	.	1
.	.	1
N	-20N dB/decade	=K1/N

Gain Cross Over Frequency

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

Fig 15 Gain cross over frequency

Phase Cross Over Frequency

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

Fig 16 Phase Margin and Gain Margin

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

Key takeaways:

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

Ii) If GM is below 0dB axis than take +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.

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

Effect of addition of poles & zeros on bode plot

The effect can be understood by plotting few bode plots as shown below

Q.1 sketch the bode plot for transfer function

G(S) =

Replace S = j

G(j=

This is type 0 system . So initial slope is 0 dB decade. The starting point is given as

20 log10 K = 20 log10 1000

= 60 dB

Corner frequency 1 = = 10 rad/sec

2 = = 1000 rad/sec

Slope after 1 will be -20 dB/decadetill second corner frequency i.e 2 after 2 the slope will be -40 dB/decade (-20+(-20)) as there are poles

2. For phase plot

= tan-1 0.1 - tan-1 0.001

For phase plot

100 -900

200 -9.450

300 -104.80

400 -110.360

500 -115.420

600 -120.00

700 -124.170

800 -127.940

900 -131.350

1000 -134.420

The plot is shown in figure 17

Fig 17 Magnitude Plot for G(S) =

Q.2 For the given transfer function determine

G(S) =

Gain cross over frequency phase cross over frequency phase mergence and gain margin

Initial slope = 1

N = 1, (K)1/N = 2

K = 2

Corner frequency

1 = = 2 (slope -20 dB/decade

2 = = 20 (slope -40 dB/decade

2. phase

= tan-1 - tan-1 0.5 - tan-1 0.05

= 900- tan-1 0.5 - tan-1 0.05

1 -119.430

5 -172.230

10 -195.250

15 -209.270

20 -219.30

25 -226.760

30 -232.490

35 -236.980

40 -240.570

45 -243.490

50 -245.910

Finding gc (gain cross over frequency

M =

4 = 2 ( (

6 (6.25104) + 0.2524 +2 = 4

Let 2 = x

X3 (6.25104) + 0.2522 + x = 4

X1 = 2.46

X2 = -399.9

X3 = -6.50

For x1 = 2.46

gc = 3.99 rad/sec(from plot )

For phase margin

PM = 1800 -

= 900 – tan-1 (0.5×gc) – tan-1 (0.05 × gc)

= -164.50

PM = 1800 -164.50

= 15.50

For phase cross over frequency (pc)

= 900 – tan-1 (0.5 ) – tan-1 (0.05 )

-1800 = -900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

-900 – tan-1 (0.5 pc) – tan-1 (0.05 pc)

Taking than on both sides

Tan 900 = tan-1

Let tan-1 0.5 pc = A, tan-1 0.05 pc = B

= 00

= 0

1 =0.5 pc 0.05pc

pc = 6.32 rad/sec

The plot is shown in figure 18.

Fig 18 Magnitude Plot G(S) =

Q3. For the given transfer function

G(S) =

Plot the rode plot find PM and GM

T1 = 0.5 1 = = 2 rad/sec

Zero so, slope (20 dB/decade)

T2 = 0.2 2 = = 5 rad/sec

Pole, so slope (-20 dB/decade)

T3 = 0.1 = T4 = 0.1

3 = 4 = 10 (2 pole) (-40 db/decade)

Initial slope 0 dB/decade till 1 = 2 rad/sec
From 1 to2 (i.e. 2 rad /sec to 5 rad/sec) slope will be 20 dB/decade
From 2 to 3 the slope will be 0 dB/decade (20 + (-20))
From 3,4 the slope will be -40 dB/decade (0-20-20)

Phase plot

= tan-1 0.5 - tan-1 0.2 - tan-1 0.1 - tan-1 0.1

500 -177.30

1000 -178.60

1500 -179.10

2000 -179.40

2500 -179.50

3000 -179.530

3500 -179.60

GM = 00

PM = 61.460

The plot is shown in figure 19

Fig 19 Magnitude and phase Plot for G(S) =

Q 4. For the given transfer function plot the bode plot (magnitude plot) G(S) =

Sol: Given transfer function

G(S) =

Converting above transfer function to standard from

G(S) =

As type 1 system, so initial slope will be -20 dB/decade
Final slope will be -60 dB/decade as order of system decides the final slope
Corner frequency

T1 = , 11= 5 (zero)

T2 = 1, 2 = 1 (pole)

4. Initial slope will cut zero dB axis at

(K)1/N = 10

i.e. = 10

5. finding n and

T(S) =

T(S)=

Comparing with standard second order system equation

S2+2ns +n2

n = 11 rad/sec

n = 5

11 = 5

= = 0.27

5. Maximum error

M = -20 log 2

= +6.5 dB

6. As K = 10, so whole plot will shift by 20 log 10 10 = 20 dB

The plot is shown in figure 20

Fig 20 Magnitude Plot for G(S) =

Q5. For the given plot determine the transfer function

Fig 21 Magnitude Plot

Sol: From figure 21 we can conclude that

Initial slope = -20 dB/decade so type -1
Initial slope all 0 dB axis at = 10 so

K1/N N = 1

(K)1/N = 10.

3. corner frequency

1 = = 0.2 rad/sec

2 = = 0.125 rad/sec

3. At = 5 the slope becomes -40 dB/decade, so there is a pole at = 5 as

Slope changes from -20 dB/decade to -40 dB/decade

4. At = 8 the slope changes from -40 dB/decade to -20 dB/decade hence is a zero at = 8 (-40+(+20) = 20)

5. Hence transfer function is T(S) =

3.4 Nyquist stability criterion

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 22 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 23 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 24 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 25 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 takeaways:

# 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 26 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 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.

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 takeaways:

(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 27 Mapping

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

Examples

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 28 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 29

Fig 29 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 30 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 31. From plot N=0, Z=0, system stable.

Fig 31 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 32 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 33 Nyquist Plot for G(S) = k/S2(S + 10)

But from plot shown in fig 32. 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 34 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 35 Nyquist Plot G(S) = k/S2(S + 10)

The plot is shown in fig 35. 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 36 Sign for Angle Directions

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

3.5 Relative stability using Nyquist stability criterion – gain and phase margins.

For open loop system number of poles o RHP are zero. So, the number of closed loop poles i.e zeros of Q(s) becomes

z=N+P

z=N

For open loop transfer function

G(s)H(s)= k/s(1+T1s)( 1+T2s)

The sinusoidal transfer function will be

G(jω)H(jω)=k/ jω(1+ jωT1) (1+ jωT2)

The polar plot will be

Fig 37 Polar Plot for G(s)H(s)= k/s(1+T1s)( 1+T2s)

The intersection of polar plot with real axis can be obtained by equating imaginary part of G(jω)H(jω) to zero.

G(jω)H(jω) = -k[ω(T1+T2)+j(1- ω2T1T2)]/ ω(1+ ω2T12)( 1+ ω2T22)

The imaginary part is zero when

1-ω2T1T2 = 0

ωpc = 1/√ T1T2

At ωpc the magnitude of G(jω)H(jω) is

| G(jω)H(jω)| = -k T1T2/ T1+T2

The given system is stable when number of encirclements of -1+j0 is zero.

Fig 38 Finding GM of system from polar plot

Therefore, the condition for the system to be stable is k T1T2/ T1+T2<1.

When k < T1+T2/ T1T2 the point -1+j0 is not encircled by G(jω)H(jω) and system is stable. The gain is called critical gain kc= T1+T2/ T1T2.

For k<kc system is stable and when k=kc the plot passes through -1+j0 point which indicates the system has roots on jω-axis. For k>kc system is unstable as G(jω)H(jω) encircles the point -1+j0 once. There are two relative stability measurements which are

Gain Margin

From above figure we can say that at ω= ωpc the phase angle of G(jω)H(jω) is 1800. The frequency at which the phase angle is 1800 is called as phase cross over frequency ωpc. Let A be magnitude at this frequency than if gain of the system is increased by 1/A the magnitude at ω= ωpc becomes 1 and hence G(jω)H(jω) locus passes through -1+j0 point and driving the system to the verse of instability. The gain margin can be defined as the reciprocal of the magnitude of open loop transfer function evaluated at the phase cross over frequency.

GM= 1/| G(j ωpc)H(j ωpc)|=1/A

G(jω)H(jω) = - 1800

The GM in dB = -20logA.

Phase Margin

The frequency at which | G(jω)H(jω)|=1 is known as gain cross over frequency ωgc.As shown in figure bellow point A is the intersection of G(jω)H(jω)plot and the unit circle centred at the origin.

Fig 39 Finding PM of system from polar plot

Now draw a line connecting point A and origin and measure the phase angle between OA and positive real axis. The angle is known as phase angle of system at the gain cross over frequency.

G(j ωpc)H(j ωpc)= ɸ

An additional phase lag ɸPM is introduced at this frequency then the phase angle G(j ωpc)H(j ωpc) will become 1800 and the point A coincides with (-1+j0) driving the system to the verge of instability. This additional phase lag is known as phase margin. The value of phase margin can be obtained by

ɸPM= 1800 + G(jω)H(jω) at ω= ωgc

ɸPM= 1800 + ɸ

Key takeaways:

A zero GM in dB indicates critical stability. A positive GM indicates stability and a negative GM indicates instability.

For stable system PM is positive. A PM close to zero corresponds to highly oscillatory system.

Example

Q1) For the following system shown below find the phase margin and gain margin if the value of k=500by using Nyquist plot.

A1) G(s)H(s) = 500/s(s+6)(s+9)

The polar plot for above OLTF is shown in figure below.

Fig 40 Polar Plot for given system

From figure we see A=0.6,

The GM in dB = -20logA.

= 4.43dB

PM = 1800-1650

= 150

Q2) For the system shown in figure, find the GM and PM using Nyquist plot for k=10. Find also range of k for stability?

A2) 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 41 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

3.6 Closed-loop frequency response

The closed-loop frequency response reveals important information about the relative stability and the speed of response in the time-domain. For unity-gain feedback configuration (H(s)=1), the closed-loop frequency response is computed as

To proceed further, use rectangular form for the open loop frequency response:

The above relation represents the equation of a circle with centre at

And radius; these are described as constant M circles on the

Frequency response plot

Further, as M→1, X=−1/2, which is a vertical line that separates the circles for M<1 from those with M>1.

The phase relationship similarly reveals an equation for constant phase circles:

Where N = tan . The constant phase circle iscentred at X =-, Y= , and have radius

Resonance Peak

The resonance peak in the frequency response occurs at ω=ωr for ζ<0.7. For a prototype second-order system, we have

Resonant frequency=

Reasonant peak =

When the polar plot of the loop gain, KGH(jω), is superimposed onto the constant M and constant N contours, it reveals the magnitude peak Mr in T(jω).

The magnitude peak represents a measure of relative stability, while the resonant frequency associated with it is a measure of speed of response in the time-domain. A value of Mr=1.3 (or 2.5dB) is considered a good compromise between speed and stability.

Constant M Circle, Constant N Circle

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).

The M and N circles of G (jω) in the gain phase plane are transformed into M and N contours in rectangular co-ordinates.

A point on the constant M loci in G (jω) plane is transferred to gain phase plane by drawing the vector directed from the origin of G (jω) plane to a particular point on M circle and then measuring the length in db and angle in degree. The critical point in G (jω), plane corresponds to the point of zero decibel and -180o in the gain phase plane. Plot of M and N circles in gain phase plane is known as Nichols chart /plot.

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 42 Constant M circle

Constant N Circle

The locus of constant phase of the closed loop system can be plotted in the G(jω) plane in same way as in constant M circle.

T(jω) = ɸ = (X+jY)/(1+X+jY) = tan-1 (Y/X) – tan-1 (Y/1+X)

Solving above equation we get,

[X+1/2]2 +[Y-1/2N]2 =1/4 +[1/2N]2

This is equation of circle centred at (-1/2,1/2N) and radius √ (1/4) +(1/2N)2

Fig 43 Constant N circle

Nichols Chart

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.

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