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

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

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

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

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

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

Where and Respective magnitude and phase angle and and

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

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

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

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

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

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

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

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

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)

Actual and approximate value of log magnitude for an open-loop transfer function of for a frequency range between 0.1-20 rad/s

Actual and approximate value of phase angle for an open-loop transfer function of for a frequency range between 0.1-20 rad/s

Bode Plot for Complex poles

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.

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

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

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

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

The intial LM and phse at strat frequency is given by

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

- 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

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

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

Bode plot for the magnitude and phase using straight line approximations

We can obtain the actual Bode plot using MATLAB as >>bode([5000 20000],[1 210 2000 0],{10^-1,10^4});

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

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

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

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

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

First order Frequency response Magnitude dan Phase angle

, , (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.

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

Second Order Frequency response Rearrange

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

Example: Frequency response Magnitude Phase angle

, , , , (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)

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) 

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.

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

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,

 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,

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

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

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

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)

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

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

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

Frequency response and

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

 (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

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

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

>> bode([96],[1 42 80 0])

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

>> num=[96];den=[1 42 80 0]; >> sys=tf(num,den); >> [GM,PM,Wg,Wp] = margin(sys) Gm = 35.0000 Pm = 60.5601 Wg = 8.9443 Wp = 1.0599 >> Gm_dB = 20*log10(Gm) Gm_dB = 30.8814