# Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types

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1 Lecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types Venkata Sonti Department of Mechanical Engineering Indian Institute of Science Bangalore, India, 562 This draft: March 2, 28 Introduction:Frequency domain design With lecture 4, we have completed what is considered the time domain method of system design. The reason being that time domain parameters, rise time, settling time, overshoot and steady state error were directly addressed by the PID controller which differentiates and integrates the time domain error. PID modifies effectively the pole positions and we can use the Root Locus method or Ziegler Nichols for finding the PID gains. The paramaters under the engineer s control are Kp, Ki and Kd, which using the method of Root Locus modify the pole positions and influence the final objective, viz., t r,t s,m p andsse. Let us say we model a system and then build it. We may find that the model differs quite a bit from reality. And so we have to upgrade the model and say we converge. Then we measure the t r,t s,m p etc. and we are not satisfied. Then we specify what we want and proceed with PID design. It seems simple enough. But, it is difficult to measure time domain characteristics. We need to give a step input which can be light and no energy input occurs. Or the input can be strong, then the system can become nonlinear. If the system is stiff, then the response can be very quick and die away. Measuring it is difficult. On the other hand the system may oscillate with a lot of low frequency dynamics. For every kind of system we cannot design a step inputu system. Locally, step inputs create nonlinear responses and deformations. On the other hand, the frequency response of a system can be very easily measured in the lab. Either a hammer impact is given or a sinusoid/random input is given and this method has been standardized. Now, does such a frequency response which captures sustained oscillation behavior capture information about transient response is the question. We will see that the frequency response is characterized by the resonance peak Mp, resonance frequency ω n, BW, Phase and gain margin. Indirectly, these indicate the time domain response parameters t r, etc. Here, the final requirement is still the time response. The tools here are compensators which use bode diagram and nyquist diagram methods.

2 Nyquist Criterion Nyquist Criterion: For a closed loop system to be stable, the Nyquist plot of G(s)H(s) must encircle the (,j) point as many times as the number of poles of G(s)H(s) that are in the right half of s-plane, and the encirclement, if any, must be made in the closkwise direction. (Read from Benjamin Kuo) 2 Nyquist or polar plots The characteristic equation is + or so that when the system is just unstable amplitude of and phase = 8 The Nyquist plot is the polar (amplitude phase) plot of a GH transfer function on the complex plane as s = jω, and ω takes values from to. The length of the arrow from the origin to the plot indicates the amplitude of GH and the angle of the arrow w.r.t the real axis indicates the phase of GH. The left figure (figure ) is for The right figure in figure is for s + s 2 + 3s + 2 s 3 + 4s 2 + 5s + 2 s(s + ) K s(s 2 + 3s + 2) () (2) where K takes values,3,6. At K = 6, the nyquist plot just passes through the point s = +j. Take for example the left figure and look at /(s 2 +3s+2). At ω = the amplitude is.5 and the phase is degrees. At ω = the amplitude is and the phase is dominated by /s 2 = /ω 2 and hence 8 degrees. This can be seen in the plot. Similarly in the right figure of figure the second order transfer function has phase 8 at high frequencies and the third order transfer function has -27 degrees. And as the gain in the third order transfer function is raised it gets closer to the point -+j and finally crosses it going unstable. Thus, a polar plot of GH gives the relative stability of the closed loop system. If we know that at a given gain, the system is stable, then by increasing the gain and seeing which way the polar plot moves we can find out the relative stability of the system. How far is the system from going unstable. Note that we use the polar plot of GH in order to figure the relative stability of the closed loop system. How far the amplitude is from - and how far the phase is from 8 degrees are indicators of stability. These are called gain and phase margins respectively and we will elaborate more on these two in a little while. 2

3 Nyquist Diagrams Nyquist Diagrams Imaginary Axis [ 4 5 2] θ [ 3 2] Imaginary Axis [ 3 2 ] [ ] [ ] Real Axis Real Axis Figure : Nyquist plots 3 Bode Diagrams The same Nyquist information can be plotted with the amplitude and phase on separate plots. The amplitude is plotted after conversion to a db scale and the x axis on a log scale. Such plots are called Bode plots. Bode plot although very simple carries a lot of information such as: relative stability (gain and phase margins), steady state errors, damping coefficient etc. We begin with a simple example, 3. Example : /s the amplitude and phase are given by 2log ω s = jω = 2log ω and φ = jω = 9o The amplitude and phase are plotted in figure 9. Lets look at the amplitude plot. The x axis is on a log scale,.,.,,, etc. The amplitude goes through zero at ω = and has a negative drop of 2 db for every times increase in frequency, i.e., -2 db/decade. The phase plot is relatively simple. 3.2 Example 2: /(s + ) Amplitude is given by s + jω + ω 2 + 2log ω 2 + 3

4 2 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 2: Bode plot for /s. Amplitude and phase. Low frequency ω =, the plot approaches asymptotically. Hence amplitude= is the low frequency asymptote. On the log scale as ω, the plot approaches db. High frequency ω, the plot approaches ω asymptotically. Hence amplitude=/ω is the high frequency asymptote. On the log scale as ω, the amplitude approaches 2log ω. At ω =, this asymptote goes through db and hence intersects the low frequency asymptote. The actual value of the curve at ω = is 2log + = 3dB. The curve lies 3dB below the db asymptote. Phase is given by jω+ as ω, the phase goes to as ω, the phase approaches 9 See figure Example 3: /(s +.) s+. jω+. s +. Magnitude ω 2 +. ω, reaches constant= asymptote ω, ω is the asymptote. db = 2log ω

5 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 3: Bode plot for /s +. Note the asymptotes in the magnitude plot. ω, 2log. = 2dB ω, 2log ω 2 db/decade Phase as before. The meeting point of the asymptotes is the 2 db line at ω =. (note s+.). The actual value of the curve at ω =. is 2log = This value is 3 db below the 2 db asymptote. 3.4 Example 4: /(s + ) s+ jω+ s + Magnitude ω 2 + ω, asymptote. ω, ω asymptote. Following the earlier examples, the low frequency asymptote reaches the value -2 db. The high frequency asymptote goes as -2dB/decade. The two meet at ω =. The actual curve is 3 db below the -2 db asymptote at ω =. See figure 5. 5

6 2 Bode Diagram Phase (deg) Magnitude (db) Asymptote 9 2 Frequency (rad/sec) Figure 4: Bode plot for /s +.. Note the asymptotes in the magnitude plot meet at ω =.. -2 Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 5: Bode plot for /s +. Note the asymptotes in the magnitude plot. 6

7 3.5 Example 4: /((s +.)(s + )(s + )) We examine a complicated case. There are three terms and a (s +.)(s + )(s + ) ω,. = asymptote. ω, asymptote. ω 3 Following the earlier examples, the low frequency asymptote reaches the value db. The high frequency asymptote goes as -6dB/decade. The curve will follow /ω asymptote at ω =. with a slope of 2dB/decade. Then, the curve will follow /ω 2 asymptote at ω = with a slope of 4dB/decade. Then, the curve will follow /ω 3 asymptote at ω = with a slope of 6dB/decade. The actual curve is 3 db below the meeting points of asymptotes at the corresponding ω. See figure 6. Phase goes to at low frequencies and to 27 o at high frequencies. Bode Diagrams Phase (deg); Magnitude (db) w=. w=. w= Frequency (rad/sec) Figure 6: Bode plot for /((s +.)(s + )(s + )). 4 Second order system: ω n 2 s 2 +2ξω n s+ω n 2 Lets look at the closed loop Tr. Fn. M(s) = ω n 2 s 2 + 2ξω n s + ω n 2 + (2ξ/ω n )s + /ω n 2 s 2 7

8 We are interested in values of ξ <<. Substituting s = jω gives M(jω) = [ (ω/ω n ) 2 ] + 2jξω/ω n As can be seen, at low frequencies GH reaches the db asymptote. At high frequencies we get the -4 db/decade asymptote. These two meet at ω = ω n. The phase is degrees at low frequencies and at ω = ω n it is -9 and for higher frequencies it slowly approaches -8. The actual plots around ω = ω n are quite complicated depending on the actual value of ξ. Bode Diagram Phase (deg) Magnitude (db) Frequency (rad/sec) Figure 7: Bode plot of a second order transfer function. 4. M p, ω p and bandwidth of a second order system: ω n 2 M(s) = s 2 + 2ξω n s + ω 2 n { } { M(s) L = y(t) = e σt cosω d t + σ } sinω d t s ω d ω d = ω n ξ 2, σ = ξω n π y(t) =, ω d t p = π, t peak = ω n ξ 2 y(t p ) = + M p M p = e πξ = ξ.6. ξ.6. 8 ξ 2. ξ.

9 For a % settling time we know Looking at M(s) in the frequenc domain: M(jω) = C(jω) R(jω) = t s = 4.6 ξω n ω n t r =.8 ω 2 n (jω) 2 + 2ξω n (jω) + ω 2 n = ( ω ω n ) 2 + j2ξ ω ω n M(ju) = u 2 + j2uξ M(ju) = [( u 2 ) 2 + (2ξu) 2 ] 2 2ξu M(ju) = φ m (u) = tan u 2 In order to find out the frequency at which the peak transfer function value occurs in frequency: dm(u) du = 4u 3 4u + 8uξ 2 = u = u p = u p = 2ξ 2 So M p is purely a function of ξ. ω p = ω n 2ξ 2 valid for ξ <.77 M p = 2ξ ξ 2 For ξ.77,ω p =,M p =. Bandwidth of a second order system is frequency at which M(ω) drops to 7.7% of its value from its low frequency asymptote. as ξ, bandwidth and M p. M(u) = ( u 2 ) 2 + (2ξu) 2 =.77 u 2 = ( 2ξ 2 ) ± 4ξ 4 4ξ Bw = ω n [( 2ξ 2 ) + 4ξ 4 4ξ 2 + 2] 2. Maximum overshoot of unit step response depends on ξ only. 2. Resonance peak of closed loop frequency response M p depends only on ξ. 9

10 M P ω P ω n.77 ζ ζ ζ =.77 Figure 8: 3. Bandwidth is directly proportional to ω n. 4. High bandwidth = largerm p. 5. ω n =.8 t r. 6. t s = f(ξω n ) = 4.6 ξω n for % settling time. 7. Also, the following relations are relevant between bandwidth, ω n, t s, t r and ξ: ω Bw = ω Bw = ω n [( 2ξ 2 ) + 4ξ 4 4ξ 2 + 2] 2 ω n = 4.6 ξt s ω Bw = 4.6 ξt s [( 2ξ 2 ) + π ω n = t p ξ 2 π t p ξ 2 [( 2ξ2 ) + 8. Open loop system must be stable for Bode plot design. 4ξ 4 4ξ 2 + 2] 2 4ξ 4 4ξ 2 + 2] 2 9. For 2 nd order system closed loop damping ratio is approximately equal to PM, if phase margin is between and 6.. bandwidth ω n. We use these relations quite freely even for non-second-order systems...with caution. 5 System Types The type of the open loop system one is dealing with decides its steady state error nature. Hence, it is important to know. Also, the steady state error depends not only on the form of the transfer function but also on the input, like step or ramp. Here we will look at system types with regard to only a step input. More is available in Benjamin Kuo.

11 r(t) R(s) + - e(t) E(s) G(s) c(t) C(s) b(t) B(s) H(s) Figure 9: The error E(s) is given by E(s) = R(s) C(s)H(s) (3) C(s) = E(s)G(s) E(s) = R(s) E(s)G(s)H(s) R(s) E(s) = + G(s)H(s) E(s), the LT of e(t) is given by E(s) = = [ e st e(t) ] s = e() s + s se(s) = e() + e(t)e st dt. e st s de(t).dt dt e stde(t).dt dt e stde(t).dt dt lim se(s) = e() + de = e( ) e() = e( ) s Thus steady state error, i.e., e( ) is given by the above formula. Step response For step response R(s) = R s and let us denote lim k p, then s and if the steady state error then E(s) = R s( + k p ) e( ) = lim s se(s) = lim se(s) = s or k p =. If the loop transfer function has the form R ( + k p ) = s j F(s)

12 then the system is of type j. If j=, then lim F() = Kp s is a finite value. Thus, Kp is finite for type systems and hence the steady state error is not zero. For type and higher systems Kp is infinity. Hence, for step inputs steady state error is not for type systems and is for type and higher systems. A similar set of rules can be defined for ramp inputs. (Read Kuo). 2

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### H(s) = s. a 2. H eq (z) = z z. G(s) a 2. G(s) A B. s 2 s(s + a) 2 s(s a) G(s) 1 a 1 a. } = (z s 1)( z. e ) ) (z. (z 1)(z e at )(z e at ) .7 Quiz Solutions Problem : a H(s) = s a a) Calculate the zero order hold equivalent H eq (z). H eq (z) = z z G(s) Z{ } s G(s) a Z{ } = Z{ s s(s a ) } G(s) A B Z{ } = Z{ + } s s(s + a) s(s a) G(s) a a Collocated versus non-collocated control [H04Q7] Jan Swevers September 2008 0-0 Contents Some concepts of structural dynamics Collocated versus non-collocated control Summary This lecture is based on parts NADAR SARASWATHI COLLEGE OF ENGINEERING AND TECHNOLOGY Vadapudupatti, Theni-625531 Question Bank for the Units I to V SE05 BR05 SU02 5 th Semester B.E. / B.Tech. Electrical & Electronics engineering IC6501