1. Frequency Response Analysis Section 6. E&CE 380 Copyright © 1998 by William J. Wilson. All rights reserved G(s)G(s)G(s)G(s) r(t) = A sin(  t) c(t) = |G(  t)| A sin(  t +  )

2. E&CE 380 Frequency Response Analysis Frequency Response Methods Stability Analysis The Nyquist plot provides absolute and relative stability information about the closed-loop system, based on a plot of the loop characteristics of the system.The Nyquist plot provides absolute and relative stability information about the closed-loop system, based on a plot of the loop characteristics of the system. The most important section of the Nyquist plot corresponds to section II of the Nyquist path, s = j , with  = 0 +  The most important section of the Nyquist plot corresponds to section II of the Nyquist path, s = j , with  = 0 +   This corresponds to the frequency response characteristics of the loop transfer function.This corresponds to the frequency response characteristics of the loop transfer function.

3. E&CE 380 Frequency Response Analysis Frequency Response Methods Stability Analysis In the Nyquist plot, this information is plotted in polar form, magnitude and phase angle as frequency is varied.In the Nyquist plot, this information is plotted in polar form, magnitude and phase angle as frequency is varied. This same information can be plotted in two rectangular plots of magnitude (in dB) vs. frequency and phase angle vs. frequency (Bode plots).This same information can be plotted in two rectangular plots of magnitude (in dB) vs. frequency and phase angle vs. frequency (Bode plots). The same Nyquist relative stability criteria can be interpreted in terms of these rectangular plotsThe same Nyquist relative stability criteria can be interpreted in terms of these rectangular plots

4. E&CE 380 Frequency Response Analysis Frequency Response of a System Consider the response of a system, G(s) to a sinusoidal input, u(t) = A sin(  t). Then the output isConsider the response of a system, G(s) to a sinusoidal input, u(t) = A sin(  t). Then the output is Solve for c(t) through partial fraction expansion.Solve for c(t) through partial fraction expansion. 22 )()()()(   s A sGsRsGsC j ejGA j jGA s sGA jsb j js 2 )( 2 )()( )( 22 1         pskpskjsbjsbsC)( 2 2 1 111           

5. E&CE 380 Frequency Response Analysis Frequency Response of a System )(/jG 2 )( 2 )()( )( 22 1      j ejGA j jGA s sGA jsb j js        where Now if G(s) is stable, then the remaining terms will be modes with exponentially decaying responses, and the “steady-state” response isNow if G(s) is stable, then the remaining terms will be modes with exponentially decaying responses, and the “steady-state” response is )sin()( 22 )(         tjGA j ee j ee jGAc tjjtjj ss

6. E&CE 380 Frequency Response Analysis Frequency Response of a System Thus the output is equal to the input sinusoid multiplied by the gain, |G(j  )| and shifted by the angle, /G(j  ).Thus the output is equal to the input sinusoid multiplied by the gain, |G(j  )| and shifted by the angle, /G(j  ). These characteristics can be obtained by substituting, s = j  into G(s), or from measurements on the physical system.These characteristics can be obtained by substituting, s = j  into G(s), or from measurements on the physical system. G(s)G(s)G(s)G(s) G(s)G(s)G(s)G(s) r(t) = A sin(  t) c(t) = |G(  t)| A sin(  t +  )

7. E&CE 380 Frequency Response Analysis Frequency Response Analysis Bode Plots Bode plots are a pair of rectangular plots of the magnitude and phase of a transfer function vs. the log of frequency.Bode plots are a pair of rectangular plots of the magnitude and phase of a transfer function vs. the log of frequency. Plot the magnitude or gain of the system in dB, vs. frequency on a semi-log scale.Plot the magnitude or gain of the system in dB, vs. frequency on a semi-log scale. Plot the angle, vs. frequency on a semi- log scale.Plot the angle, vs. frequency on a semi- log scale. This represents the information from Section II of the Nyquist plot.This represents the information from Section II of the Nyquist plot.)( log20 10 j G )(/j G

8. E&CE 380 Frequency Response Analysis Bode Plots: Example 1002 100 1 5 )( 2     sss sG Constructing Bode Plots Calculate the complex number G(j  ) for several frequencies and plot the points. Calculate the complex number G(j  ) for several frequencies and plot the points. Use MatLab to automatically plot the Bode plots. Use MatLab to automatically plot the Bode plots. Use asymptotic approximations. Use asymptotic approximations.

9. E&CE 380 Frequency Response Analysis Constructing Bode Plots: Asymptotic Approximations Consider the general form of transfer functions.Consider the general form of transfer functions. This may be written in Bode form by dividing through by all the constants.This may be written in Bode form by dividing through by all the constants.   )2))((( ))()(( )( 2 2 1 21 nn k jjpjj zjzjK jG   jj    ))(21)(1( ))(1)(1( )( 2 1 21        nn k B j p j j zz K jG        

10. E&CE 380 Frequency Response Analysis Constructing Bode Plots: Asymptotic Approximations Now consider the gain of G(j  ) in dB.Now consider the gain of G(j  ) in dB. where the Bode Gain is zzK 2 1 21 n B p K             dB nn dB dB k dBdB dB B dB j p j j z j z j K jGjG 2 1 21 10 2 11)( 11 )(log20)(      

11. E&CE 380 Frequency Response Analysis Constructing Bode Plots: Asymptotic Approximations The angle of G(j  ) may be written asThe angle of G(j  ) may be written as Thus it is clear that for both magnitude in dB and the angle, the total transfer function may be written in terms of the sum of its componentsThus it is clear that for both magnitude in dB and the angle, the total transfer function may be written in terms of the sum of its components    1 1)( k p j j            22 1 nn j          2 1 1 1 )( / B z j z j Kj G 

12. E&CE 380 Frequency Response Analysis Individual Components: Poles and Zeros at the Origin Poles at the origin:Poles at the origin: Plot Plot a constant slope of – k  20 dB per decade of frequency. a constant slope of – k  20 dB per decade of frequency. The angle is constant at – k  90  The angle is constant at – k  90  Zeroes at the origin: the same, but with the opposite signZeroes at the origin: the same, but with the opposite sign k j)( 1     10 log vs.log20 log vs. )( 1 log20  k j k k j)(  Frequency (rad/sec) Phase (deg) Magnitude (dB) Bode Diagrams -40 -20 0 20 40 10 10 0 1 2 -50 0 50 j 1  j  -90  Slope = -20 dB/dec.

13. E&CE 380 Frequency Response Analysis Individual Components: First-Order Poles –for  << p –for  = p –for  >> p pj/1 1  2 10dB 1log20 /1 1 gain          ppj dB    0 angle, 0)1(log20gain 10dB , 3)2(log20gain 10dB  dB  90 angle, log20gain 10dB         p        p  1 tan angle   45(1)tan angle 1-

14. E&CE 380 Frequency Response Analysis Individual Components: First-Order Poles and Zeros Plot these on a scaled frequency axis  /p.Plot these on a scaled frequency axis  /p. Zeroes are the same but with the opposite sign.Zeroes are the same but with the opposite sign. Asymptotic approximations fit the high frequency and low frequency characteristics.Asymptotic approximations fit the high frequency and low frequency characteristics.  = p is the corner frequency of the pole/zero.  = p is the corner frequency of the pole/zero. Frequency (rad/sec) Phase (deg) Magnitude (dB) First-Order Pole and Zero -20 -10 0 10 20 10 10 0 1 -50 0 50 /p/p pj/1 1  zj/1  Slope = -20 dB/dec. -3dB -45 

15. E&CE 380 Frequency Response Analysis Individual Components: Second-Order Poles Individual Components: Second-Order Poles for  <<  nfor  <<  n for  =  nfor  =  n for  >>  nfor  >>  n 2 21 1        nn j      001     0angle,01log20gain 10dB dB     90angle, 2 1 121 1 log20gain 10dB dB j  n nn j        10 2 10dB log40 21log20gain          A slope of  40dB per decade.

16. E&CE 380 Frequency Response Analysis Individual Components: Second-Order Poles  =  n is the corner frequency and the gain at this point is 1/2  and the angle is  90 . Note that for  = 1, the gain is  6dB and for  = 0, the gain is . Plot the characteristics on a normalized frequency scale.           180 / 2 tan- angle 1- n   2 21 1        nn j     

17. E&CE 380 Frequency Response Analysis Individual Components: Second-Order Poles

18. E&CE 380 Frequency Response Analysis Individual Components: Non-Minimum Phase Non-minimum phase refers to a system with a final phase more negative than expected from the order of denominator and numerator. A zero in the RHS, (  s + z) has the same magnitude characteristics as the zero in the LHS, but has the opposite angle. The final angle is  90  rather than +90 . The final phase of a system with a RHS zero will be 180  more negative than with a LHS zero

19. E&CE 380 Frequency Response Analysis Individual Components: Non-Minimum Phase A second non-minimum phase component is a time delay or transportation lag which is represented by e  s  , where  is the delay time in seconds. The frequency characteristic is e  j   . The magnitude is 1.0 (0dB) for all frequencies and so it does not affect the magnitude plot. The phase angle is . When plotted against log 10 , this results in an exponential decrease in phase.  s e 

20. E&CE 380 Frequency Response Analysis Bode Plot Construction: Example 1 Bode Plot Construction: Example 1 Bode Form:Bode Form: Plot the asymptotic approximations for each term separately, for both magnitude and angle.Plot the asymptotic approximations for each term separately, for both magnitude and angle. Then add them together to get the system asymptotic approximation.Then add them together to get the system asymptotic approximation. Sketch in the Bode plot curve.Sketch in the Bode plot curve. )2)(5( )10(10 )(    sss s s G )2/1)(5/1( )10/1(10 )2)(5( )10(10 )(  jjj j jjj j j G  

21. E&CE 380 Frequency Response Analysis )2)(5( )10(10 )(  sss s sG Example Example 1

22. E&CE 380 Frequency Response Analysis )2)(5( )10(–10)(  sssssG Non-Minimum Phase: Example 1A

23. E&CE 380 Frequency Response Analysis Bode Plot Construction: Example 2 Bode Plot Construction: Example 2 Bode Form:Bode Form: The damping ratio for the second-order term is  = 0.1 and the natural frequency is 10 rad./s.The damping ratio for the second-order term is  = 0.1 and the natural frequency is 10 rad./s. 1002 100 1 5 )( 2     sss sG   2 2 10/10/2.01 1 1/1 5 100210015)(              j j jj j jG

24. E&CE 380 Frequency Response Analysis Example 2 100210015)( 2     sss sG

25. E&CE 380 Frequency Response Analysis Transfer Function Identification Frequency response characteristics can be be obtained experimentally by applying sinusoidal inputs of various frequencies, and measuring the gain and phase relationships between input and output. By fitting asymptotic approximations to the frequency response characteristics obtain from experimental measurements, an approximate transfer function model of the system can be obtained.

26. E&CE 380 Frequency Response Analysis Transfer Function Identification: Major Steps 1. Determine the initial slope  order of poles at the origin. 2. Determine the final slope  difference in order between the denominator and numerator (n  m). 3. Determine the initial and final angle  confirm the results from above or detect the presence of a non-minimum phase system (delays or zeroes in the RHS). 4. Determine the low frequency gain (Bode gain).

27. E&CE 380 Frequency Response Analysis Transfer Function Identification: Major Steps 5. Detect the number and approximate location of corner frequencies and fit asymptotes. –possibly subtract well defined components. –examine expected -3dB points. –try to separate second-order terms and use the standard responses to estimate damping. 6. Sketch the phase plot for the identified transfer function as a check of accuracy.

28. E&CE 380 Frequency Response Analysis Transfer Function Identification: Major Steps 7. Use the phase plot to check for non-minimum phase terms and to calculate the time delay value if one is present. 8. Calculate the frequency response for the identified model and check against the experimental data (MatLab or a few points by hand calculation). 9. Iterate and refine the pole/zero locations and damping of second-order terms.

29. E&CE 380 Frequency Response Analysis

30. Transfer Function Model Identification: Example 1 Initial slope =  20dB/dec.  a 1/j  term.Initial slope =  20dB/dec.  a 1/j  term. Final slope =  40dB/dec.  (n-m) = 2.Final slope =  40dB/dec.  (n-m) = 2. Initial angle =  90  and the final angle is  180  which checks with the gain curve.Initial angle =  90  and the final angle is  180  which checks with the gain curve. Low frequency gain is found to be |K B /  | dB = 35dB, where  = 0.1  K B = 5.62.Low frequency gain is found to be |K B /  | dB = 35dB, where  = 0.1  K B = 5.62. Through asymptotic fitting there are two poles found at  c = 4 and 25, and one zero at  c = 70.Through asymptotic fitting there are two poles found at  c = 4 and 25, and one zero at  c = 70.

31. E&CE 380 Frequency Response Analysis +45  /dec.

32. E&CE 380 Frequency Response Analysis Transfer Function Model Identification: Example 1 The estimated transfer function in Bode form isThe estimated transfer function in Bode form is The final form isThe final form is Note: the asymptotic approximation of the phase, based on this transfer function is quite consistent with the data.Note: the asymptotic approximation of the phase, based on this transfer function is quite consistent with the data.  )25)(4( )70(8 )(  sss s sG  )25/1)(4/1()70/1(62.5)(   jjjjjG  

33. E&CE 380 Frequency Response Analysis

34. Transfer Function Model Identification: Example 2 Initial slope = 0dB/dec.  no pole at the origin.Initial slope = 0dB/dec.  no pole at the origin. Final slope =  40dB/dec.  (n-m) = 2.Final slope =  40dB/dec.  (n-m) = 2. Initial angle = 0  and the final angle is  180  which checks with the gain curve.Initial angle = 0  and the final angle is  180  which checks with the gain curve. Low frequency gain is found to be |K B | dB = 20dB  K B = 10.Low frequency gain is found to be |K B | dB = 20dB  K B = 10. Through asymptotic fitting a simple pole is found at  c = 0.2, a simple zero at  c = 1.0 and a complex pole at  n = 5.0.Through asymptotic fitting a simple pole is found at  c = 0.2, a simple zero at  c = 1.0 and a complex pole at  n = 5.0.

35. E&CE 380 Frequency Response Analysis

36. Transfer Function Model Identification: Example 2 The peak at  n = 5.0 is  15dB which corresponds to a damping ratio of  0.1.The peak at  n = 5.0 is  15dB which corresponds to a damping ratio of  0.1. The estimated transfer function in Bode form isThe estimated transfer function in Bode form is The final form isThe final form is )25)(2.0( )1(50 )( 2    sss s sG )25/5/2.01)(2.0/1( )1/1(10 )( 2    jj j jG   

37. E&CE 380 Frequency Response Analysis

38. Transfer Function Model Identification: Example 2A The magnitude plot is the same as the previous example.The magnitude plot is the same as the previous example. The angle plot continues to go more negative and does not asymptotically approach  180  as expected.The angle plot continues to go more negative and does not asymptotically approach  180  as expected. This is a non-minimum phase system with a time delay, e –s .This is a non-minimum phase system with a time delay, e –s . How do we determine the time delay,  ?How do we determine the time delay,  ?

39. E&CE 380 Frequency Response Analysis Transfer Function Model Identification: Example 2A Plot the phase of the transfer function identified from the magnitude plot and find the phase difference between this plot and the phase data. e –j  term.This difference represents the e –j  term. At  = 20 rad/sec, –   –  /3    0.052 sAt  = 20 rad/sec, –   –  /3    0.052 s (– 60  ) (– 60  ) At  = 50 rad/sec, –   – 5  /6    0.052 sAt  = 50 rad/sec, –   – 5  /6    0.052 s (– 150  ) (– 150  )

40. E&CE 380 Frequency Response Analysis )25)(2.0( )1( 50 )( 2    sss s sG e –0.052s Transfer Function with Delay

41. E&CE 380 Frequency Response Analysis Bode Plots and Stability Analysis In the Nyquist analysis, it became clear that Section II of the plot was the most critical in determining the stability of the closed-loop system.In the Nyquist analysis, it became clear that Section II of the plot was the most critical in determining the stability of the closed-loop system. The Bode plot of the loop transfer function, GH(j  ) provides the same magnitude and angle information as Section II of the Nyquist plot.The Bode plot of the loop transfer function, GH(j  ) provides the same magnitude and angle information as Section II of the Nyquist plot. Therefore, the Bode plot of GH(j  ) can be used to evaluate the stability of the closed-loop system.Therefore, the Bode plot of GH(j  ) can be used to evaluate the stability of the closed-loop system.

42. E&CE 380 Frequency Response Analysis Bode Plots and Stability Analysis Consider the definitions of the gain and phase margins in relation to the Bode plot of GH(j  ).Consider the definitions of the gain and phase margins in relation to the Bode plot of GH(j  ). –Gain Margin: the additional gain required to make | GH(j  ) | = 1 when /GH(j  ) =  180 . On the Bode plot this is the distance, in dB, from the magnitude curve up to 0dB when the angle curve crosses  180 . –Phase Margin: the additional phase lag required to make /GH(j  ) =  180  when | GH(j  ) | = 1. On the Bode plot this is the distance in degrees from the phase curve to  180  when the gain curve crosses 0dB.

43. E&CE 380 Frequency Response Analysis )25)(4( )70(8 )(    sss s sG

44. E&CE 380 Frequency Response Analysis Closed-loop Frequency Response Assuming the closed-loop system is stable, the frequency response of the closed-loop system will directly give the bandwidth of the system, i.e. 3dB down from the steady-state gain.Assuming the closed-loop system is stable, the frequency response of the closed-loop system will directly give the bandwidth of the system, i.e. 3dB down from the steady-state gain. This will be related to the zero crossing frequency of the loop transfer function plot.This will be related to the zero crossing frequency of the loop transfer function plot. Information about the overshoot of the step response can also be obtained from the peak of the magnitude curve in the closed-loop Bode plot.Information about the overshoot of the step response can also be obtained from the peak of the magnitude curve in the closed-loop Bode plot.

45. E&CE 380 Frequency Response Analysis 56010829 )70(8 )(1 )( 23     sss s sG sG