Part IB Paper 6: Information Engineering LINEAR SYSTEMS AND CONTROL. Glenn Vinnicombe HANDOUT 4. The Frequency ResponseG(jω)

Transcription

1 Part IB Paper 6: Information Engineering LINEAR SYSEMS AND CONROL Glenn Vinnicombe HANDOU 4 he Frequency ResponseG(jω)

2 x(t) 2π ω y(t) G(jω) argg(jω) ω t t Asymptotically stable LI systemg(s) x(t)=cos(ωt) y(t)= G(jω) cos ( ωt+ argg(jω) ) +starting transient

3 Summary If a pure sinusoid is input to an asymptotically stable LI system, then the output will also settle down, eventually, to a pure sinusoid. his steady-state output will have the same frequency as the input but be at a different amplitude and phase. he dependence of this amplitude and phase on the frequency of the input is called the frequency response of the system. In this handout we shall: Show how the frequency response can be derived from the transfer function. (by substitutingjω fors) Study how the frequency response can be represented graphically. (using the Bode diagram)

4 Contents 4 he Frequency ResponseG(jω) 4. What is the Frequency Response? Derivation of gain and phase shift Plotting the frequency response Sketching Bode Diagrams Powers ofs:(s) k First Order erms:(+s) Second order terms:(+2ζs+s 2 2 ) Examples Key Points

5 4. What is the Frequency Response? x(t) 2π ω y(t) A φ ω t t Asymptotically stable LI system x(t)=cos(ωt) y(t)= A cos ( ωt+φ ) +starting transient

6 If the input to an asymptotically stable LI system is a pure sinusoid then the steady state output will also be a pure sinusoid, of the same frequency as the input but at a different amplitude and phase. How do we finda, the gain, andφ, the phase shift?

7 How does this relate to Part IA? Consider again a system with input u and outputy, If d 2 y dt 2 +αdy dt +βy=adu dt +bu. then we can use the usual trick, letting u=e jωt so that R(u(t))=cos(ωt). We will find the response to the input cos(ωt) by taking the real part ofy, the response tou=e jωt. o find the solutiony, we assume it takes the form y(t)=ye jωt (4.) for some complex numbery = Y e j argy, so that R(y(t))=R( Y e j argy e jωt )= Y cos(ωt+ argy).

8 Substituting (4.) into the differential equation, and noting that dy dt =[jω]yejωt, d 2 y dt 2 =[jω]2 Ye jωt, etc we obtain Y[jω] 2 e jωt +αy[jω]e jωt +βye jωt =a[jω]e jωt +be jωt or Y = a[jω]+b [jω] 2 +α[jω]+β Note that the transfer function fromū(s) toȳ(s) is given by G(s)= as+b s 2 +αs+β So, it would appear thaty =G(jω), suggesting: Answer: If system has the transfer functiong(s), then A= G(jω), φ=argg(jω)

9 You ve done this in Linear Circuits, Maths and Mechanical Vibrations in Part IA, so I m assuming that you are familiar with these arguments! he notation above is as used in Maths and Mechanical Vibrations. In Linear Circuitsỹ was used to represent a complex phasor, rather than they above. Example:he Capacitor is described by the differential equation i=c dv dt soī(s)=cs v(s) in the absence of initial conditions giving the transfer function v(s) ī(s) = sc. So, the frequency response of a capacitor (from current to voltage) is jωc, which equals its impedance,ṽ/ĩ. he advantage of transfer functions is that they define the response to all possible inputs, whereas the impedence is only valid for sinusoidal (ac) signals.

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11 4.. Derivation of gain and phase shift he previous argument is not the whole story though - we have assumed that the inputu=cos(ωt) has been present since the beginning of time, and have only shown thaty=a cos(ωt+φ) is a possible response. What if the system is at rest untilt= 0, and then a sinusoid is applied? ake an asymptotically stable system with input ū(s) outputȳ(s) and rational transfer functiong(s)= n(s) d(s), so Letu(t)=e jωt, soū(s)= s=jω, ȳ(s)=g(s) ȳ(s)=g(s)ū(s). s jω then, sinceg(s) can t have a pole at s jω = λ + λ λ n + G(jω) s p s p 2 s p n s jω y(t)=λ e p t +λ e p 2 t + +λ n e p nt +G(jω)e }{{}}{{ jωt } y tr (= CF) y ss (= PI)

12 (with the obvious modifications if any poles are repeated). Since all the p k, the poles ofg(s), have a negative real part, theny tr (t) 0 as t leaving the steady-state responsey ss. he steady-state response to the inputu(t)=cos(ωt) is given by the real part of this: ( R(y ss (t))=r e jωt ) G(jω) =R ( G(jω) e j( ) ) ωt+argg(jω) asz= z e j argz = G(jω) cos ( ) ωt+ argg(jω) }{{}}{{} A φ

13 Example: Consider a system with transfer function and an input G(s)= s s+ 2 x(t) = cos(0.5t) (a sinusoid at 0.5rad/s, or 0.5/(2π)= Hz ) So,ω=0.5 and G(jω) = (0.5j) (0.5j)+2 = j+ 2 = j =.7507e j = 0.57e j = he following figures show the impulse response of this system, the inputx(t)=cos(0.5t), and the response to this input.

14 Note how the output initially contains a component near the resonant frequency of the system ω n = 2 (0.225 Hz) but that this quickly decays leaving only a sinusoid at the frequency of the input ω=0.5 (0.08 Hz) (and at an amplitude of 0.57). (he phase lag of.64 is too small to be seen on this diagram) G(s)=/(s s+ 2) g(t) Input:x(t)=cos(0.5t) Response to the inputx(t)=cos(0.5t) y(t)=g(t) x(t) t (sec)

15 4.2 Plotting the frequency response he frequency responseg(jω) is a complex-valued function of frequencyω. At each frequencyω, the complex numberg(jω) can be represented either in terms of its real and imaginary parts, or in terms of its gain (magnitude) and phase (argument). here are a number of common ways of representing this information graphically: he Bode Diagram: wo separate graphs, one of G(jω) vsω(on log-log axes), and one of G(jω) (lin axis) vsω(log axis). he Nyquist Diagram: One single parametric plot, ofr ( G(jω)) againsti ( G(jω)) (on linear axes) asωvaries. he Nichols Diagram: One single parametric plot, of G(jω) (log axis) against (G(jω)) (lin axis) asωvaries.

16 Example: G(s) = 20 Bode Gain Plot s(s s+2) Nyquist Diagram G(jω) (db) ω (rad/s) Bode Phase Plot I ( G(jω) ) R ( G(jω) ) Nichols Chart G(jω) (deg) ω (rad/s) G(jω) (db) G(jω) (deg)

17 Each has its use: he Bode diagram is relatively straightforward to sketch to a high degree of accuracy, is compact and gives an indication of the frequency ranges in which different levels of performance are achieved. he Nyquist diagram provides a rigorous way of determining the stability of a feedback system. he Nichols diagram combines some of the advantages of both of these (although is not quite as good in either specific application) and is widely used in industry. We shall not study the Nichols diagram in this course(!), but the ideas behind its construction will be readily grasped once the two fundamental diagrams that we do study are understood.

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19 Consider G(jω) for ω = 2 G(jω) = ω=2 2j((2j) (2j)+2) = 2j( j) Hence and G(j2) = = argg(j2)=arg arg 2j arg( 2+0.4j) = 0 π/ }{{} π atan(.4/2) = 4.550

20 because R I X So, G(j2)=0.245e 4.550j = j Also, 20log 0 G(j2) = 2.2dB, G(jω)= he corresponding point has been marked with a cross on each of the previous plots.

21 4.3 Sketching Bode Diagrams he bode diagram ofg(s) consists of two curves. Gain Plot: Gain G(jω) (log) vs freqω(log) 2. Phase Plot: Phase G(jω) (lin) vsω(log) It is straightforward to sketch, and gives a lot of insight. Basic idea: polynomials e.g. Clearly Consider a transfer function written as a ratio of factorized G(s)= a (s)a 2 (s) b (s)b 2 (s). log 0 G(jω) =log 0 a (jω) +log 0 a 2 (jω) log 0 b (jω) log 0 b 2 (jω), so we can compute the gain curve by simply adding and subtracting gains corresponding to terms in the numerator and denominator. Similarly G(jω)= a (jω)+ a 2 (jω) b (jω) b 2 (jω) and so the phase curve can be determined in an analogous fashion.

22 Since a polynomial can always be written as a product of terms of the type K, s, +s, +2ζs+s 2 2 (for ζ <, i.e. complex roots) it is suffices to be able to sketch Bode diagrams for these terms. he Bode plot of a complex system is then obtained by adding the gains and phases of the individual terms. Note: Always rewrite the transfer function in terms of these building blocks before starting to sketch a Bode diagram you will find it much easier. For example, if the transfer function has a term(s+a) first rewrite this as a (+s/a) and then collect together all the constants that have been pulled out. he transfer functions in Question 6 on Examples Paper 2 are already given in the right form, but you will need to rewrite the transfer function in Question 7. Example: We would rewriteg(s)= ( ) 0 G(s)= s 000(s+ ) s(s 2 + 5s+ 00) as (+s) (+0.05s+s 2 /00) and begin by considering each term individually.

23 4.3. Powers ofs:(s) k he simplest term in a transfer function is a power ofs, which it is convenient to write in the form a(s)=(s) k wherek>0 if the term appears in the numerator andk<0 if the term is in the denominator. he magnitude and phase of the term are given by log 0 a(jω) =log 0 (ω) k =klog 0 (ω), G(jω)=90k he gain curve is thus a straight line with slopekdecades/decade, or 20k db/decade, intersecting the 0dB line atω =. he phase curve is a constant at 90 k. For =, the case whenk= corresponds to a differentiator, and has a slope 20dB/decade and phase 90. he case whenk= corresponds to an integrator and has a slope 20dB/decade and phase 90.

24 40dB (=00) 20dB (=0) 0dB (=) 20dB (=.) 40dB (=.0)

25 4.3.2 First Order erms:( + s) Bode plot of G(s)=(+s) (for > 0)... replaces byjω to get G(jω)=(+jω) Asymptotes: = G(jω) = +jω }{{} +ω 2 2 G(jω) = (+jω) }{{} atan(ω) ω 0: (i.e. ω / ) +jω jω 20log 0 G(jω) 20log 0 =0 G(jω) =0

26 ω : (i.e. ω / ) 20log 0 G(jω) 20log 0 jω = 20log 0 ω 20log 0 / (which is a straight line with slope=20db/decade andx-axis (i.e. 0db) intercept at ω = /) G(jω) jω = 90 Atω=/, we get 20log 0 G(jω) =20log 0 +j = 20log 0 2 (3dB) G(jω)= (+j)=45

27 Bode diagram ofg(s)=(+s): 40dB (=00) Gain high freq asymptote ( slope= 20 db/decade) 20dB (=0) low freq asymptote 20log 0 G(jω) 0dB (=) 3dB Frequency (rad/s) 0 00

28 90 G(jω) 0 high freq asymptote Phase (Degrees) 45 low freq asymptote Frequency (rad/s) 0 00

29 4.3.3 Second order terms:(+2ζs+s 2 2 ) Bode plot of G(s)= +2ζs+s 2 2 (for > 0, 0 ζ )... replaces byjω to getg(jω)= +2ζjω ω 2 2 ω = 20log 0 G(jω) = 20log ζjω ( G(jω)= ω 2 2 ) + 2ζjω Asymptotes: ω 0: (i.e. ω / ) 20log 0 G(jω) 20log 0 =0 G(jω) =0

30 ω : (i.e. ω / ) 20log 0 G(jω) 20log 0 ω 2 2 = 40log 0 ω = 40log 0 / 40log 0 ω (which is a straight line with slope= 40dB/decade andx-axis (i.e. 0db) intercept atω=/) Atω=/, we get G(jω) ω 2 2 = 80 20log 0 G(jω) = 20log 0 2ζj = 20log 0 2ζ G(jω)= (2ζj)= 90

31 20dB 8dB= 2.5 (/2ζ) ζ= 0.2 0dB Gain (db) 20dB 6dB ζ= 40dB 60dB

32 0 ζ= 0.2 Phase (Degrees) ζ=

33 4.3.4 Examples Example : G(s) = 5 +0s (K= 5, / = 0.) 20log 0 G(jω) =20log log 0 +jω/0. G(jω) = 5 (+jω/0.) We now produce a sketch of the Bode diagram in two stages (this is for clarity all these constructions would normally appear on one pair of graphs ). First we plot the asymptotes and approximations to the true curves for the individual terms, using the 3dB error at the cornerand 0/, /0 approximations. (he exact values are used for the plots here.)

34 Frequency (rad/s) 20 Gain (db) 0 20log 0 +0jω 20log Phase (Degrees) 0 +0jω

35 In the next pair of diagrams, the contributions from the individual terms (now shown as dashed lines) have been added to give the Bode diagram ofg(s) (this has been done for both the asymptotes and the true gain and phase).

36 Frequency (rad/s) 20 Gain (db) 0 20log 0 G(jω) Phase (Degrees) 0 G(jω)

37 he next example has both a pole and a zero it is an example of a phase-lead compensator. Example 2: G(s)= s +s So,G(jω)= jω +jω, 20log 0 G(jω) =20log Note: 0.05= 26dB + 20log 0 +0jω 20log 0 +jω and G(jω) = 0.05 }{{} + (+0jω) (+jω) 0

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39 First we draw the individual terms: Gain (db) log 0 +0jω 20log 0 +jω 40 20log

40 Phase (Degrees) jω jω Frequency (rad/s)

41 and then we sum them (Note how the phase terms sum to produce a maximum phase advance of only about 55.) 20 Gain (db) log 0 G(jω)

42 Phase (Degrees) 90 0 G(jω) Frequency (rad/s)

43 Example 3: G(s)= s = 0.05 s s 20 20log 0 +2jω (+2s) Gain (db) log 0 jω = 20log log 0 ω

44 90 Phase (Degrees) jω 0.05 jω = j Frequency (rad/s)

45 20 Gain (db) log 0 G(jω)

46 Phase (Degrees) 90 0 G(jω) Frequency (rad/s)

47 RHP poles and zeros ( jω)= (+jω) 20log 0 ( jω) = + 20log0 (+jω) hat is, if we have a term( s) instead of(+s) then the gain plot is unchanged but the term s contribution to the phase plot is reversed in sign (so the contribution of a RHP zero to the overall phase diagram is the same as that of a LHP pole at the same location).

48 Comments on Bode sketching A alternative technique for Bode diagrams would be to ignore the asymptotes, and just calculate and plot the true gain and phase over a grid of frequencies. his is not recommended for a number of reasons. Firstly, a lot more points are required to get the same accuracy (particularly when the diagram is to be used for control system analysis and design, as only a small region is required accurately in this case). Secondly, the structure of the diagram is then lost. A practising control engineer will often prefer a good sketch, with the asymptotes shown, to an accurate computer generated diagram since this gives a better idea of how things can be changed to improve the behaviour of the controlled system. When a question asks you to draw a Bode diagram (as in the questions on Examples Paper 2) it s really asking you to produce a drawing showing the straight line asymptotes and approximations and a rough approximation to the true gain and phase by rounding the corners appropriately.

49 4.4 Key Points he frequency response is obtained from the transfer function by replacings withjω. At each frequencyω,g(jω) is a complex number whose magnitude gives the gain of the system at that frequency and whose argument gives the phase shift of the system at that frequency. he gain and the phase shift are conveniently shown on the Bode diagram.