Unit 4

Frequency Response Analysis

The transfer function of second order system is given as

C(s)/ R(s) = wn2/s2+2wns+wn2 (1)

=damping factor

wn=undamped natural frequency

For frequency response let s=jw

Let u= the above equation becomes

T(j)=

|T(j)|=M=1/ (2)

=φ=-tan-1[/] (3)

For sinusoidal input the output response for the system is given by

c(t)=sin[t-tan-1[/] (4)

The frequency where M has the peak value is known as Resonant Frequency . This frequency is given as from equation (2)

(5)

From equation (2) the maximum value of magnitude is known as Resonant Peak

Mr= (6)

The phase angle at resonant frequency is given as

φr=-tan-1[ (7)

As we already know for step response of second order system the value of damped frequency and peak overshoot are given as

(8)

Mp= (9)

The comparison of Mr and Mp is shown if figure below. The two performance indices are correlated as both are function of damping factor. When subjected t step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

Similarly, the correlation of and is also shown below. For the given input step response [equation (5) and (6)]

Mp= Peak overshoot of step response

Mr= Resonant peak of frequency response

=Resonant frequency of frequency response

=Damping frequency of oscillations of step response

DIAGRAM

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

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)

MAGNITUDE PLOT

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

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

Q. (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

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

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

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

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 =

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 |

Sketching of Bode Plot:

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 1

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 2

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 3

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

G(S) =

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 4

Q5. For the given plot determine the transfer function

From figure 5 we can conclude that

- Initial slope = -20 dB/decade so type -1
- Initial slope alls 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

5. is a zero at = 8 (-40+(+20)=20)

6. Hence transfer function is

T(S) =

The system which has all the poles and zeros on the left half of s-plane are called as minimum phase transfer functions.

Let the transfer function be

G(j)=

The pole zero configuration is shown below

Fig: Minimum phase function

Fig: Phase angle characteristics of minimum phase function

The frequency response representation is either by plotting Bode plot or polar plots. Another way of frequency response representation is by plotting magnitude in db v/s phase angle with frequency ω as the running parameter. Normally, we first obtain the bode plot and then by reading the values of log-magnitude and phase angle at different frequencies from these plots the log-magnitude v/s phase plot angle can be constructed.

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.

Phase Cross Over Frequency-:

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

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

.:

.:

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

Inference 1:-

More the difference b/WPC and WGC core is the stability of system

Inference 2:-

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 * * L

Inference 3:-

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

For a stable system GM and PM should be -ve

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

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

It gives the frequency response of the system. If the transfer function is given, than 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. 3(a)

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)

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

Ex.3 T(S) = 1/(S+1)(S+2)(S+3)

(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(c)

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)

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)

- 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)

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)

Reference Books :

- Modern Control Engineering by K.Ogata, Pearson Education.

2. Control Engineering by Kuo.

Unit 4

Frequency Response Analysis

The transfer function of second order system is given as

C(s)/ R(s) = wn2/s2+2wns+wn2 (1)

=damping factor

wn=undamped natural frequency

For frequency response let s=jw

Let u= the above equation becomes

T(j)=

|T(j)|=M=1/ (2)

=φ=-tan-1[/] (3)

For sinusoidal input the output response for the system is given by

c(t)=sin[t-tan-1[/] (4)

The frequency where M has the peak value is known as Resonant Frequency . This frequency is given as from equation (2)

(5)

From equation (2) the maximum value of magnitude is known as Resonant Peak

Mr= (6)

The phase angle at resonant frequency is given as

φr=-tan-1[ (7)

As we already know for step response of second order system the value of damped frequency and peak overshoot are given as

(8)

Mp= (9)

The comparison of Mr and Mp is shown if figure below. The two performance indices are correlated as both are function of damping factor. When subjected t step input the system with given value of Mr of its frequency response will exhibit a corresponding value of Mp.

Similarly, the correlation of and is also shown below. For the given input step response [equation (5) and (6)]

Mp= Peak overshoot of step response

Mr= Resonant peak of frequency response

=Resonant frequency of frequency response

=Damping frequency of oscillations of step response

DIAGRAM

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

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)

MAGNITUDE PLOT

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

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

Q. (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

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

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

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

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 =

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 |

Sketching of Bode Plot:

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 1

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 2

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 3

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

G(S) =

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 4

Q5. For the given plot determine the transfer function

From figure 5 we can conclude that

- Initial slope = -20 dB/decade so type -1
- Initial slope alls 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

5. is a zero at = 8 (-40+(+20)=20)

6. Hence transfer function is

T(S) =

The system which has all the poles and zeros on the left half of s-plane are called as minimum phase transfer functions.

Let the transfer function be

G(j)=

The pole zero configuration is shown below

Fig: Minimum phase function

Fig: Phase angle characteristics of minimum phase function

The frequency response representation is either by plotting Bode plot or polar plots. Another way of frequency response representation is by plotting magnitude in db v/s phase angle with frequency ω as the running parameter. Normally, we first obtain the bode plot and then by reading the values of log-magnitude and phase angle at different frequencies from these plots the log-magnitude v/s phase plot angle can be constructed.

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.

Phase Cross Over Frequency-:

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

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

.:

.:

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

Inference 1:-

More the difference b/WPC and WGC core is the stability of system

Inference 2:-

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 * * L

Inference 3:-

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

For a stable system GM and PM should be -ve

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

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

It gives the frequency response of the system. If the transfer function is given, than 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. 3(a)

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)

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

Ex.3 T(S) = 1/(S+1)(S+2)(S+3)

(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(c)

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)

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)

- 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)

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)

Reference Books :

- Modern Control Engineering by K.Ogata, Pearson Education.

2. Control Engineering by Kuo.