CHAPTER 7 : BODE PLOTS AND GAIN ADJUSTMENTS COMPENSATION

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1 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. Use the bode plots to find a suitable gain to meet the stability specifications. Design the gain adjustment compensator to meet the frequency response specifications. 6.1 INTRODUCTION Frequency response methods, discovered by Nyquist and Bode in the 1930s. Frequency response yields a new vantage point from which to view feedback control system. This technique has distinct advantage in the following situations: 1. When modeling transfer function from physical data. 2. When finding the stability of nonlinear systems. 3. In settling ambiguities when sketching a root locus. -1-

2 6.2 PLOTTING FREQUENCY RESPONSE In this section, we learn a method to draw the frequency response using the bode plot technique. G j M G j G can be plotted in several ways; two of them are 1. as a function of frequency, with separate magnitude and phase plots. The magnitude can be plotted in decibels (db) vs log, where db = 20 log M. The phase curve is plotted as phase angle vs log. The motivation for these plots is shown in next section. 2. as a polar plot, where the phasor length is the magnitude and the phasor angle is the phase. Based on the s-plane concept Magnitude response at particular freq is the product of the vector length from the zeroes of G(s) divided by the product of vector lengths from the poles of G(s) drawn to points on the imaginary axis. Phase response is the sum of angles from the zeroes of G(s) minus the sum of the angles from the poles of G(s) drawn to points on the imaginary axis. -2-

3 Ex 1: Demonstrates how to obtain analytical expression for frequency response and make a plot of the result. Problem: Find the analytical expression for the magnitude freq response and the phase freq response for a system G(s) = 1/(s+2). Also, plot both the separate magnitude and phase diagrams. Solution: Substitute s=j in the system and obtaining G(j )=1/(j +2) G(j )=1/(j +2) = (2- j )/( 2+4) The magnitude of this complex number, G j M 1 / 2 4 The phase angle, G(j )= tan 1 / 2 Refer to Figure 1 for the plot (actual Plot): Figure 1-3-

4 7.3 Asymptotic Approximations: Bode Plots The log-magnitude and phase freq response curves as functions of log are called Bode plots or Bode diagrams. Sketching the bode plots can be simplified because they can be approximated as sequence of straight lines. Consider the following transfer function: G s K s z1 s z 2... s z n s m s p1 s p 2... s p n Then, the magnitude frequency response is the product of the magnitude freq response of each term, or G ( j ) K ( s z1 ) s z 2... s z k s m (s p1 ) ( s p 2 )... ( s p n ) s j Thus, if we know the magnitude response of each pole and zero term, we can find the total magnitude response. The process can be simplified by working with the logarithm of the magnitude since the zero terms magnitude responses would be added and the pole terms magnitude responses subtracted, rather than, respectively, multiplied or divided, to yield the logarithm of the total magnitude response. Converting the magnitude response into db, we obtain 20 log G ( j ) 20 log K 20 log ( s z1 ) 20 log s z log s m 20 log s p1... s j Thus, if we knew the response of each term, the algebraic sum would yield the total response in db. Further, if we could make an approximation of each term that would consist only of straight lines, graphic addition of terms would be greatly simplified. -4-

5 Bode Plots for G(s) = (s+a) Consider a function, G(s) = (s+a), for which we want to sketch separate logarithmic magnitude and phase response plots. Letting s = jω, we have G ( j ) j a a j 1 a At low frequencies when ω approaches zero, G j a The magnitude response in db is 20 log M 20 log a where and is a constant. Equation above is shown plotted in M G j Figure 2 (a) from ω = 0.01a to a. At high frequencies where ω >>a, ω = 0.01a to a becomes j 0 0 G j a a a a The magnitude response in db is 20 log M 20 log a 20 log where a. 20 log a Notice from the middle term that the high-frequency approximation is equal to the low frequency approximation when a, and increases for If we plot db, 20 log M 20 log M 20 log a 20 log a., against log ω, equation 20 log a becomes a straight line: y 20 x where y 20 log M, and x log. The line has a slope of 20 when plotted as db vs. log ω. -5-

6 Figure 2 (a) & (b) Bode Plots for G(s) = 1/(s+a) Let us find the Bode plots for the transfer function G s 1 1 s a a s 1 a This function has a low-frequency asymptote of 20log(1/a), which is found by letting the frequency, s, approach zero. The Bode plot is constant until the break frequency, a rad/s, is reached. The plot is then approximated by the high-frequency asymptote found by letting s approach. Thus, at high frequencies G j 1 s a a s j a j a a a -6-

7 or, in db, 20 log M 20 log 1 20 log 20 log a a Figure 3-7-

8 Notice from the middle term that the high-frequency approximation equals the low-frequency approximation when a, and decreases for a. This result is similar to equation 20 log M 20 log a 20 log 20 log, a except the slope is negative rather than positive. The Bode logmagnitude diagram will decrease at a rate of 20 db/decade rather than increase at a rate of 20 db/decade after the break frequency. The phase plot is the negative of the previous example sine the function is the inverse. The phase begins at 00 and reaches -900 at high frequencies, going through -450 at the break frequency. Both the Bode normalized and scaled log-magnitude and phase plot are shown in Figure 4 (d). Bode Plots for G(s) =s Our next function, G(s) = s, has only a high-frequency asymptote. Letting s=jω, the magnitude is 20 log ω, which is the same as equation 20 log M 20 log a 20 log 20 log. a Hence, the Bode magnitude lot is a straight line drawn with a +20 db/decade slope passing through zero db when ω = 1. The phase plot, which is a constant +900, is shown with the magnitude plot in Figure 4(a). Bode Plots for G(s) = 1/s The frequency response of the inverse of the preceding function, G(s) = 1/s, is shown in Figure 3(b) and is a straight line with a

9 db/decade slope passing through the zero db at ω = 1. The Bode phase plot is equal to a constant Figure 4-9-

10 Ex 2: Bode plots for ratio of first-order factors. Problem: Draw the Bode plots for the system shown in figure below, where G s K s 3 / s s 1 s 2 Solution: Bode plot for open loop system The bode plot is the sum of each first order system Use the normalized plot in order to determine the cut-off frequency easier. The normalized TF is: 3 s K G s s s s So, the cut-off frequencies are at 1, 2, and 3. The magnitude plot should begin a decade below ( 0.1 ) the lowest freq break and extend a decade above the highest break freq ( 100 ). K is chosen at 1 easy to denormalized later for any value of K The plot is shown in Figure

11 Figure 5 (magnitude)

12 Figure 5 (Phase)

13 7.4 Bode Plots For Second Order For G s s 2 2 n s n 2 General equation of second order is given by; G s s 2 2 n s n 2 = s2 s n n n At low frequency: 2 2 G s n n 0 0 At high frequency: G s s The log-magnitude: 20 log 2 40 log Magnitude-phase plots; Figure

14 7.4.2 For G s 1 / s 2 2 n s n 2 Magnitude-phase plots: Reverse of the plots in section Ex 3: (Bode plots for second order and first order system) Draw the Bode log-magnitude and phase plots of G s s 3 s 2 s 2 2s 25 Solution: Convert G(s) to normalized value, getting, s G s s 2 s s Then, Bode log-magnitude shown in Figure 7. Figure

15 7.5 Stability, Gain Margin, and Phase Margin via Bode Plots Determining Stability The specification below should be get in order to ensure the stability of the system using Bode Plots The closed loop system will be stable if the frequency response has a gain less than unity when phase is Ex 4: Use Bode Plots to determine the range of K within which the unity feedback system is stable. Let G s K s 2 s 4 s 5 Solution: Convert G(s) to normalized value, yields, G s K 1 40 s s s Choose K = 40 in order to start the plots at 0db Break frequency at 2, 4, and 5 Lowest frequency, 0.01 ; Highest frequency, 100 Plots,

16 Figure 8 (Ex 4) From the graph, at freq = 7rad/sec, phase is At this point magnitude plot is at -20db. Then the system is stable. The magnitude can move until +20db in order ensure the stability of the system. 20db means gain equal to 10. Then, at this point the gain K = 40*10=400 So, range of K that make the system stable is K 400 Evaluating Gain Margin And Phase Margin

17 Refer Figure 9 for gain margin and phase margin concept; Figure 9: Gain Margin and Phase Margin EX 5: If K=200 in Ex 4, find gain margin and phase margin. Solution: Plots the Bode plots for magnitude and phase using the same procedure as Ex 4 K = 200, means fives times greater than 40 (chosen k value previously) Then, 20 log 5=13.98db. Means bode plots in Figure 8 should increase 13.98db in magnitude

18 Therefore, to find gain margin look at phase plot and the freq when phase is 180o. At this freq, determine from the magnitude plot how much the gain can be increased before reaching 0dB. From Figure 8, the phase angle is 180o at approximately 7 rad/s. On the magnitude plot, the gain is -20 db db = db. Thus, the gain margin is 6.02 db. For phase margin, refer freq at magnitude plot where the gain is 0 db. At this freq, look on the phase plot to find the difference between the phase and 180o. The difference is phase margin. From Figure 8, remember that the magnitude plot is db lower than actual plot, the 0 db crossing ( db for the normalized plot shown in Figure 8) occurs at 5.5 rad/s. At this freq, the phase angle is -165o. Thus, the phase margin is -165-(-180) = 15o Relation Between Closed-loop Transient and Closed-Loop Frequency Responses. A relationship exists between the peak value of the closed-loop magnitude response and the damping ratio is given by equation below, Mp This condition happens at a frequency p n 1 2 p, of 2 Representive log-magnitude plots,

19 7.5.4 Relation Between Closed-loop Transient and Open-loop Frequency Responses Relation between phase margin and damping ratio is given by, 2 M tan Design Via Frequency Response: Transient Response Via Gain Adjustment Introduction Frequency response design methods, unlike root locus methods, can be implemented conveniently without a computer or other tool except for testing the design

20 The bode plots can be designed easily using asymptotic approximations and gain can be read from there Gain Adjustment From previous discussion the phase margin is relates to the damping ratio (equivalently percent overshoot). Thus, any varying in phase margin will result on varying the percent overshoot. From Figure 10, if we desire a phase margin, M, represented by CD, we would have to raise the magnitude curve by AB. Therefore, a simple gain adjustment can be used to design phase margin and, hence, percent overshoot. Procedure involve in designing the gain adjustment compensator: a) Draw the bode magnitude and phase plots for a convenient value of gain. b) Using equations below to determine the required phase margin from the percent overshoot. %OS ln 100 %OS 2 ln c) Find the frequency, M tan M, on the bode phase diagram that yields the desired phase margin, as shown in Figure 10. d) Change the gain by the amount of AB to force the magnitude curve to go through 0 db at M. The amount of gain

21 adjustment is the additional gain needed to produce the required phase margin. Figure 10 EX 6: (Transient response design via gain adjustment) For the position control system shown in Figure 11, find the value of preamplifier gain, K, to yield a 9.5% overshoot in the transient response for a step input. Use only frequency response methods. Figure

22 Solution: Follow the procedure stated previously, Choose K=3.6 to start the magnitude plot at 0 db at 0.1 in Figure 12. Using equation above, a 9.5% overshoot implies 0.6 for the closed-loop dominant poles. From equation for phase margin yields a 59.2o phase margin for a damping ratio of 0.6. Locate on the phase plot the frequency that yields a 59.2o phase margin. This frequency is found where the phase angle is the difference between -180o and 59.2o, or o. The value of the phase margin frequency is 14.8 rad/s. At a freq of 14.8 rad/s on the magnitude plot, the gain is found to be db. This magnitude has to be raised to 0 db to yield the required phase margin. Since the log-magnitude plot was drawn for K=3.6, a 44.2 db increase, or K=3.6*162.2=583.9, would yield the required phase margin for a 9.48% overshoot. The gain-adjusted open-loop transfer function is G s s s 36 s

23 Figure