Frequency Response Analysis

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Kory Parsons
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1 Frequency Response Analysis Consider let the input be in the form Assume that the system is stable and the steady state response of the system to a sinusoidal inputdoes not depend on the initial conditions We have PFE yields Complex conjugate of

2 Inverse Laplace transform yields At steady state (when ) we have where

3 Since is a complex quantity, we have where and similarly Hence That is a sinusoidal input generates a sinusoidal output

4 Note that Amplitude ratios of the output sinusoid to the input Phase shift of the output with respect to the input Negative phase is called phase lag Positive phase is called phase lead There are many ways representing these, We will concentrate on 3 common graphical representations

5 Graphical Representations 1) Bode Plot A plot of magnitude in decibels (db) vs in semilogaritmic coordinates. The Phase angle is also plotted against in semilogarithmic coordinates 2) Polar Plot A plot of magnitude vs phase in polar coordinates as from zero to infinity 3) Magnitude vs Phase plots A plot of magnitude in db vs the phase on the rectangular coordinates with as a varying parameter on the curve

6 Bode Plot Construction Consider the transfer function of the system given in the form where are positive real values. In order to put the given tranfer function into Bode form let and normalize

7 Magnitude Plot Curve Breaks {up/down} by 20 db/dec at {a,b / c,d,e} Initial Value, Phase Plot Curve breaks {up/down} by 45 degrees one decade before {a,b / c,d,e} and breaks {down/up} by 45 degrees one decade after {a,b / c,d,e} Initial and final values For n=0 Calculate db using and convert to Plot a zero / pole plot for the original transfer function Calculate the angle from the figure at = 0 For n>0 Calculate -intercept = Calculate the initial slope -20n db /dec Final magnitude slope -20 RD db /dec

8 An illustrative example The Initial angle Final angle Imag Axis Real Axis

9 Bode Plot Example Construct the Bode Plot for Start with putting the transfer function into Bode form Magnitude Plot Initial Magnitude (note that n=0)

10 Use up arrow to denote a slope of +20 db/dec and down arrow to denote a slope of -20 db/dec Initial Magnitude Decade Decade RD = 0 Final slope is zero Semi log plot

11 Start with the initial magnitude and use arrows to draw the asymptotic plot db db/dec

12 Phase plot : Start with the pole / zero Plot Initial angle : Imag Axis Final angle : Real Axis

13 Use the negative of the poles and the zeros on the jw axis (zero at -1 and pole at -10) Use an up arrow to denote a slope of +45 deg/dec and a down arrow to denote a slope of -45 deg/dec deg 90 Decade 45 Initial Phase 45 deg/dec Final Phase - 45 deg/dec Frequency (rad/sec)

14 The actual Bode plot is obtain using the following code % (s+1)/(s+10) figure; num = [1 1]; denum = [1 10]; SYS=tf(num,den); bode(sys);

15 Bode plot construction Another example : This time sketch the Bode plot for In Bode form Magnitude Plot Initial Magnitude (note that n=1) Calculate the intercept Initial slope = -20.n db/dec = -20 db/dec

16 Plot the negative poles and zeros on the jw axis and use up arrows to denote a slope of -20 db/dec and down arrows to denote a slope of +20 db/dec Finally use the initial value and the arrows to draw the asymptotic plot Magnitude (db) db/dec db/dec -20 db/dec Frequency (rad/sec) -40 db/dec

17 Phase plot : Start with the pole / zero Plot Initial angle : Final angle :

18 Use the negative of the poles and the zeros on the jw axis (zero at -10 and poles at 0,-1,-100) Use an up arrow to denote a slope of +45 deg/dec and a down arrow to denote a slope of -45 deg/dec Phase (deg) -45 Initial Phase deg/dec Final Phase - 45 deg/dec Frequency (rad/sec)

19 The actual Bode plot figure; num = [10 10]; denum = [ ]; SYS=tf(num,den); bode(sys);

20 Bode Plot for repeated roots Sketch the Bode plot of In Bode form Magnitude Plot Initial Magnitude (note that n=0) db conversion

21 Plot the negative poles on jw axis (2 poles at ) use up arrows to denote a slope of -20 db/dec and down arrows to denote a slope of +20 db/dec Finally use the initial value and the arrows to draw the asymptotic plot n /10 Bode Diagrams n 10 n - 40 db/dec Decade 100 n -40 Initial Magnitude ( 20log 2 ) 1 n

22 Phase plot : Start with the pole / zero Plot Initial angle : Imag Axis Real Axis Final angle :

23 Use the negative of the poles the jw axis Use an up arrow to denote a slope of +45 deg/dec and a down arrow to denote a slope of -45 deg/dec 90 n 10 n 10 n 100 n 0-90 Initial Phase Final Phase

24 The actual Bode plot when figure; num = [1]; denum = [1 2 1]; SYS=tf(num,den); bode(sys);

25 Example Use the Bode plot to determine the stability of a closed loop system For stable. At Transfer Function Root locus form of the denominator the closed loop system is marginally the closed loop poles are at (from Root Locus) Imag Axis c Real Axis j d j d

26 Angle Condition Magnitude Condition Might not be easy to solve for and Bode Solution for finding Note that and

27 A Bode plot is the graph of magnitude and phase of the transfer function with respect to Hence we can directly obtain (read) the value of the desired frequency from the phase plot Using the desired frequency we can read the corresponding value of the magnitude of the transfer function and obtain the value of as or in normal units in db

28 The Bode plot gives us a graphical way of finding the values of the and Specific to our problem, the closed loop system is defined by

29 Construct the Bode Plot Then using Phase (deg) Magnitude (db) db -180 deg d 20 Frequency (rad/sec) That is for the stability of the system we need

30 Bode Compensator Design Phase (deg) Magnitude (db) Similar to the root locus design we can use the characteristic equation to examine the stability and/or the performance of the closed loop system using Bode plot KH ( jω φ ) KH ( jω p ) ω p ω φ Frequency (rad/sec) : Gain cross over freq. The value of frequency when : phase cross over freq. The value of frequency when

31 Gain margin and Phase margin These points on the Bode plot leads us to 2 new definitions Phase Margin Gain Margin The system is unstable if this value is less than zero

32 Bode plot Root locus The Bode plot parameters are related to the root locus parameters according to Phase margin is related to the damping ratio as follows for Root locus design parameters Damping ratio Natural frequency Note that damping ratio is a measure of relative stability Hence phase cross over frequency is a measure of relative stability as well

33 Bode Design Example Given with Find K so that the phase margin is 45 degrees for the closed loop system Solution : Design Steps Find the characteristic equation in the form Plot and asuming

34 From calculate the desired i.e. From the desired, and the plot, find the desired Raise or lower the magnitude plot so that it crosses the 0 db level at the desired value of Find the shift between the desired plot and the original amgnitude plot at the desired value of Desired + original = shift in db Calculate the value of using

35 Step 1 : Step 2 : Step 3 : Desired

36 Given the desired value find from the plot

37 Step 5 :

38 Step 6 : The difference between the desired magnitude plot and the actual magnitude plot is -9 db, thefore Step 7 :

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION Objectives Students should be able to: Draw the bode plots for first order and second order system. Determine the stability through the bode plots.