1. Frequency Response OBJECTIVE
- Bode and Nyquist plots for control analysis
- Determination of transfer function
- Gain and Phase margins
- Stability in frequency response ~.

2. Magnitude and Phase Angle
Transfer function is
R(s)
Y(s)
By replacing we can find its magnitude
and its phase angle
Higher order transfer function
Can be presented in magnitude-phase form as

3. Example
A system with transfer function of G(s) is subjected to a sinusoidal input.
Determine the time response of the system.
Solution
The input in phasor form (manitude-phase form) can be presented as
As the transfer function is
Hence the output is by replacing
In time domain which give the time response of the system, the output is

4. First order Transfer function by replacing Frequency response
Its magnitude
and phase angle

5. Second order Frequency response by replacing Its magnitude
Transfer function
Frequency response by replacing
Its magnitude
and phase angle

6. Find the frequency response of the following transfer function
Higher Order
Cascade form
Frequency response by replacing
Or in phasor form
Example:
Find the frequency response of the following transfer function

7. Where
and
Respective magnitude and phase angle
and
and

8. Bode Plot Consider higher order system Logarithmic form In dB.
Phase angle

10. Bode Plot for constant gain
Log magnitude in dB
Its phase angle
Let K=200
>> bode([200],[1]);grid

12. Bode Plot of pole on the origin
Log magnitude in dB 1 10
-20
102
-40
103
-60

13. 1
10-1 20
10-2 40
10-3 60

14. 1 2 -6 22 -12 23 -18 Slope of –20dB/decade or –6dB/octave.2 -6 22
-12 23
-18
Slope of –20dB/decade or –6dB/octave.
Its phase angle
>>bode([1],[1 0]);grid

16. Bode Plot for Real Pole Frequency response Logarithmic magnitude in dB
Phase angle

17. Can also be defined as a tenth of the corner frequency i.e.
The corner frequency is
For low frequency,
and
Can also be defined as a tenth of the corner frequency i.e.
For high frequency,
Can also be defined as a tenth of the corner frequency i.e.
and

18. A straight line approximation for a first order system of transfer function

19. And the MATLAB command used is given by
The actual Bode plot can be obtained by using the exact equation for the log magnitude and phase angle. There are small differences between the actual and approximate as shown by the following table
Using MATLAB we can display the Bode plot of an open-loop transfer function of
And the MATLAB command used is given by
» nr=[1];
» dr=[1 1];
» sys=tf(nr,dr)
» bode(sys)

20. Actual and approximate value of log magnitude for an open-loop transfer function of for a frequency range between rad/s

21. Actual and approximate value of phase angle for an open-loop transfer function of for a frequency range between rad/s

23. Bode Plot for Complex poles

24. We define the low frequency as a tenth of the corner frequency i.e.
Is the corner frequency
For low frequency ,
and
We define the low frequency as a tenth of the corner frequency i.e.
For high frequency
and
While the high frequency as ten time of the corner frequency i.e.

25. Based on straight line approximation the Bode plots for LM and phase angle are shown

26. We can observe the Bode plots in MATLAB by considering
for
and
>>zeta=0.25;wn=1;num=[1];den1=[1 2*zeta*wn wn*wn];
sys1=tf(num,den1);
>>zeta=0.5;wn=1;num=[1];den2=[1 2*zeta*wn wn*wn];
sys2=tf(num,den2);
>>zeta=0.75;wn=1;num=[1];den3=[1 2*zeta*wn
wn*wn];sys3=tf(num,den3);
>> bode(sys1,sys2,sys3);grid

28. Example
Obtain the bode plot of the following block diagram if + _

29. 1/s
500
s+4
By replacing with s=j and making the open-loop transform function
into corner freqeuncy form

30. The intial LM and phse at strat frequency is given by

31. For the magnitude polot of the Bode plot, we build a table to tshow the contribution of gradient/slope by the zero and poles at respective corner frequencies. The pole at origin will provide a slope of -20db/decade for all range of frequency

32. - For the phase plot, we need to know the change of gradient of the phase at low and high frequencies of respective corner frequency - The pole at the origin does not contribute to gradient of the phase angle as its phase angle is a constant 90o

33. The LM vs freq. plot can be determined the LM value at the corner
frequency, this can be obtained by simple trigonometry
(dB/dec)
Using the trignometry’s formula,
(rad.s-1)
0.1 2 40
400
1000
Slope
(dB/dec)
-20
-40
-60
LM (dB) 22 -4
-56
-116
-140

34. phase angle slope at the low and high frequency for each corner
As for the LM slope, we apply a trignometry’s formula to obtain the
phase angle slope at the low and high frequency for each corner
frequency as in table below
(deg/decade)
 (rad.s-1)
0.1
0.2 4 20
400
1000
Slope
(deg/dec)
-45
-90
(degree)
-149
-212
-271

35. Bode plot for the magnitude and phase using straight line approximations

36. We can obtain the actual Bode plot using MATLAB as >>bode([ ],[ ],{10^-1,10^4});

37. Determination of transfer function
Constant gain
Example: If
dB, determine the ransfer function. dB -
 (rad.s-1) dB

38. Pole/zero at origin -  (rad.s-1) Example: If rad.s-1, dB and slope of
dB/decade. dB
dB/dekad -
 (rad.s-1)
We know
rad.s-1, dB

39. Real pole/zero 45 Example: dB dB/dekad  (rad.s-1) 0.01 40 250

41. Pair of complex poles Example: Assume damping ratio of 0.5. dB 60
dB/dekad
 (rad.s-1)
0.1 50
400
dB/dekad

43. Nyquist Plot  Nyquist plot is a plot of magnitude, and phase angle
for frequency
on s-plane.
 (rad.s-1) 
However we can obtain the sketch of the plot by obtaining the following vectors: at
(ii) at
(iii) at
, crossing on the real axis
(iv) at
, crossing on the imaginary axis

44. First order
Frequency response
Magnitude
dan
Phase angle

45. , , (i) At , and (ii) At dan . i.e. or
(iii) No crossing in the real axis as ,
(iv) No crossing in the imaginary axis as ,
is a circle.

46. Example: At Nyquist plot of Frequency response and and At and,
and , . At
and At
and
>> nyquist([5],[.25 1])

48. Second Order
Frequency response
Rearrange

49. and Magnitude Phase angle (i) at (ii) At and,
and
(ii) At ,
and .
(iii) No crossing on the real axis.
(iv) Crossing of the imaginary axis when , .
and

50. Example:
Frequency response
Magnitude
Phase angle

51. , , , , (i) At , (ii) At and (iii) Real axis crossing at dan
rad.s-1 or
Magnitude
(iv) Imaginary crossing at ,
=0.6 rad.s-1
Magnitude
»dr1=[2 1];
» dr2=[1 1 1];
» dr=conv(dr1,dr2);
» nr=1;
» nyquist(nr,dr)

53. Nyquist Stability Criteria
For Nyquist path j
s-plane r 
Contour of
which will give the closed loop poles for
by replacing
till
to become j
satah-F(j) 

54. j  However Hence on F(j)-plane, we represent GH(j)-plane to plot
satah-GH(j)
-1+j0  .
where
Z = Number of zeros
on the right-half plane of j-axis.
N = Numberof clockwise encirclement of
P = Number of zeros
on the right-half plane of j-axis.

55. Gain margin and Phase margin in Nyquist Plot
satah-GH(j) j
1/GM 
-1+j0

56. Is the measure of instability from on the -plane
Phase crossover frequency,
Frequency where the phase angle of
is –180 o
Gain crossover frequency,
Frequency where the magnitude of
Gain margin,

57.  j GH(j)-plane -1+j0 Unit circle will give phase angle of,
Unit circle
will give phase angle of
hence the phase margin
Phase margin,

58. Negative gain and phase marginsj
satah-GH(j)
1/GM 
-1+j0

59. Example:
For
, shows the Nyquist plot and its respective gain and phase margin + -
Open loop transfer function
Frequency response

60. Magnitude
Phase angle
(i) At,
and
(ii) At,
and

61. Imaginary crossing, when
(iii)
Real crossing, when .
then
(iv)
Imaginary crossing, when
then
» nr=40;
» dr1=[ 1 3];
» dr2=[1 4 7];
» dr=conv(dr1,dr2);
» nyquist(nr,dr)

64. Positive gain and Phase margins in Bode plot
LM (dB)
 (rad.s-1) GM  o PM
-180
GM – gain margin
PM – phase margin
- Gain crossover frequency
- Phase crossover frequency

65. Negative gain and Phase margins in Bode plot
LM (dB) GM
 (rad.s-1) PM 
-180 o
Note that, negative gain or phase margin means that the system is not stable

66. Example: If + - determine gain and phase crossover frequencies.
dan K=96
Consequently what is the system gain and phase margin. + -

67. Frequency response
and

68. Table of LM slope for the 3 poles
(rad.s-1)
0.1 2 40
(dB/dekad)
-20
Total slope dB/decade
-40
-60
kutub orijin
kutub hakiki
kutub hakiki

69.  (rad.s-1) 0.1 0.2 4 20 400 -45 -90 kecerunan (darjah/dekad)
-45
jumlah kecerunan
(darjah/dekad)
-90
kutub hakiki
kutub hakiki
Table for phase angle slope from the two poles, notes that the pole at
origin does not contribute to the slope as the angle is constant -90o

70. frequency, this can be obtained by simple trignometry
To obtain the LM vs freq. plot, we determine the LM value at the corner
frequency, this can be obtained by simple trignometry
(dB/dekad)
Using the trignometry’s formula,
(rad.s-1)
0.1 2 40
400
1000
jumlah kecerunan
(dB/dekad)
-20
-40
-60
LM (dB) 22 -4
-56
-116
-140

71. phase angle slope at the low and high frequency for each corner
As for the LM slope, we apply a trignometry’s formula to obtain the
phase angle slope at the low and high frequency for each corner
frequency as in table below
(deg/decade)
 (rad.s-1)
0.1
0.2 4 20
400
1000
jumlah kecerunan
(darjah/dekad)
-45
-90
(darjah)
-149
-212
-271

73. >> bode([96],[ ])

74. >>[GM,PM,Wg,Wp] = margin(sys)

75. >> num=[96];den=[1 42 80 0];
>> sys=tf(num,den);
>> [GM,PM,Wg,Wp] = margin(sys)
Gm =
Pm =
Wg =
Wp =
>> Gm_dB = 20*log10(Gm)
Gm_dB =