FREQUENCYRESPONSE ANALYSIS


 Bathsheba Patterson
 4 years ago
 Views:
Transcription
1 ECE450/550: Feedback Control Systems. 8 FREQUENCYRESPONSE ANALYSIS 8.: Motivation to study frequencyresponse methods Advantages and disadvantages to rootlocus design approach: ADVANTAGES: Good indicator of transient response. Explicitly shows location of closedloop poles. Tradeoffs are clear. DISADVANTAGES: Requires transfer function of plant be known. Difficult to infer all performance values. Hard to extract steadystate response (sinusoidal inputs). Frequencyresponse methods can be used to supplement root locus: Can infer performance and stability from same plot. Can use measured data when no model is available. Design process is independent of system order (# poles). Time delays handled correctly (e sτ ). Graphical techniques (analysis/synthesis) are quite simple. What is a frequency response? We want to know how a linear system responds to sinusoidal input, in steady state.
2 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 2 Consider system Y (s) = G(s)U(s) with input u(t) = u0 cos(ωt), so s U(s) = u 0 s 2 + ω 2. With zero initial conditions, s Y (s) = u 0 G(s) s 2 + ω 2. Do a partialfraction expansion (assume distinct roots) Y (s) = α s a + α 2 s a α n s a n + α 0 s jω + α 0 s + jω y(t) = α e at + α 2 e a2t + +α n e a nt }{{} +α 0 e jωt + α0 e jωt. If stable, these decay to zero. y ss (t) = α 0 e jωt + α 0 e jωt. Let α0 = Ae jφ.then, y ss = Ae jφ e jωt + Ae jφ e jωt = A ( e j (ωt+φ) + e j (ωt+φ)) = 2A cos (ωt + φ). We find α 0 via standard partialfractionexpansion means: α 0 = [ (s jω)y (s) s= jω [ u0 sg(s) = (s + jω) s= jω = u 0( jω)g( jω) (2 jω) Substituting into our prior result = u 0G( jω). 2 y ss = u 0 G( jω) cos (ωt + G( jω)).
3 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 3 Important LTIsystem fact: If the input to an LTI system is a sinusoid, the steadystate output is a sinusoid of the same frequency but different amplitude and phase. FORESHADOWING: Transfer function at s = jω tells us response to a sinusoid...but also about stability as jωaxis is stability boundary! EXAMPLE: Suppose that we have a system with transfer function G(s) = s. Then, the system s frequency response is G( jω) = s = s= jω jω. The magnitude response is A( jω) = jω = jω = 2 = (3 + jω)(3 jω) The phase response is ( ) 2 φ(jω) = 3 + jω = (2) (3 + jω) = 0 tan (ω/3). θ ω 2. Now that we know the amplitude and phase response, we can find the amplitude gain and phase change caused by the system for any specific frequency. For example, if ω = 3rads, A( j3) = = 2 φ(j3) = tan (3/3) = π/4. 3 ω
4 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Plotting a frequency response There are two common ways to plot a frequency response the magnitude and phase for all frequencies. EXAMPLE: U(s) R C Y (s) G(s) = + RCs Frequency response G( jω) = = = + jωrc + jω + ω 2 (let RC = ) tan (ω). We will need to separate magnitude and phase information from rational polynomials in jω. Magnitude = magnitude of numerator / magnitude of denominator R(num)2 + I(num) 2 R(den)2 + I(den) 2. Phase = phase of numerator phase of denominator ( ) ( ) I(num) I(den) tan tan. R(num) R(den) Plot method #: Polar plot in complex plane Evaluate G( jω) at each frequency for 0 ω<. Result will be a complex number at each frequency: a + jb or Ae jφ.
5 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 5 Plot each point on the complex plane at (a + jb) or Ae jφ for each frequencyresponse value. Result = polar plot. We will later call this a Nyquist plot. ω G( jω) The polar plot is parametric in ω, soitishardtoreadthe frequencyresponse for a specific frequency from the plot. We will see later that the polar plot will help us determine stability properties of the plant and closedloop system. Plot method #2: Magnitude and phase plots We can replot the data by separating the plots for magnitude and phase making two plots versus frequency.
6 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 6 G( jω) Frequency, (rads/sec.) G( jω) Frequency, (rads/sec.) The above plots are in a natural scale, but usually a loglog plot is made This is called a Bode plot or Bode diagram. Reason for using a logarithmic scale Simplest way to display the frequency response of a rationalpolynomial transfer function is to use a Bode Plot. Logarithmic G( jω) versus logarithmic ω, andlogarithmic versus ω. REASON: G( jω) ( ) ab log 0 = log cd 0 a + log 0 b log 0 c log 0 d. The polynomial factors that contribute to the transfer function can be split up and evaluated separately. (s + ) G(s) = (s/0 + ) G( jω) = ( jω + ) ( jω/0 + ) jω + G( jω) = jω/0 + ( ω ) 2. log 0 G( jω) =log 0 + ω2 log 0 + 0
7 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 7 Consider: For ω ωn, ( ) ω 2 log 0 +. ( ω log 0 + ) 2 log 0 () = 0. For ω ωn, ( ) ω 2 ( ) ω log 0 + log 0. KEY POINT: Two straight lines on a loglog plot; intersect at ω =. Typically plot 20 log0 G( jω) ; thatis,indb. 20dB Exact Approximation 0. 0 Atransferfunctionismadeupoffirstorderzerosandpoles,complex zeros and poles, constant gains and delays. We will see how to make straightline (magnitude and phaseplot) approximations forallthese, and combine them to form the appropriate Bode diagram.
8 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Bode magnitude diagrams (a) The log0 ( ) operator lets us break a transfer function up into pieces. If we know how to plot the Bode plot of each piece, then we simply add all the pieces together when we re done. Bode magnitude: Constant gain db = 20 log0 K. Not a function of frequency. Horizontal straight line. If K <, thennegative,else positive. db K > 0. 0 K < Bode magnitude: Zero or pole at origin For a zero at the origin, G(s) = s db = 20 log 0 G( jω) = 20 log 0 jω db. For a pole at the origin, G(s) = s db = 20 log 0 G( jω) = 20 log 0 jω db. 20 db 20 db 20 db 20 db 20 db per decade db per decade 0. 0
9 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 9 Both are straight lines, slope =±20 db per decade of frequency. Line intersects ωaxis at ω =. For an nthorder pole or zero at the origin, Still straight lines. Still intersect ωaxis at ω =. db =±20 log 0 ( jω) n =±20 log 0 =±20n log 0 ω. But, slope =±20n db per decade. Bode magnitude: Zero or pole on real axis, but not at origin For a zero on the real axis, (LHP or RHP), the standard Bode form is ( ) s G(s) = ±, which ensures unity dcgain. If you start out with something like G(s) = (s + ), then factor as ( ) s G(s) = +. Draw the gain term ( )separatelyfromthezeroterm(s/ + ). In general, a LHP or RHP zero has standard Bode form ( ) s G(s) = ±
10 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 0 ( ) ω G( jω) =± + j ( ω 20 log 0 G( jω) =20 log 0 + ( ) ω 2 For ω ωn, 20 log log 0 = 0. ( ) ω 2 ( ω For ω ωn, 20 log log 0 Two straight lines on a log scale which intersect at ω = ωn. For a pole on the real axis, (LHP or RHP) standard Bode form is ) 2 ). G(s) = ( s ± ) ( ) ω 2 20 log 0 G( jω) = 20 log 0 +. This is the same except for a minus sign db 20 db per decade db 20 db per decade
11 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 8.4: Bode magnitude diagrams (b) Bode magnitude: Complex zero pair or complex pole pair For a complexzero pair (LHP or RHP) standard Bode form is ( ) s 2 ( ) s ± 2ζ +, which has unity dcgain. If you start out with something like s 2 ± 2ζ s + ω 2 n, which we have seen before as a standard form, the dcgain is ω 2 n. Convert forms by factoring out ω 2 n [ ( s s 2 ± 2ζ s + ωn 2 = ω2 n ) 2 ± 2ζ ( ) ] s +. Complex zeros do not lend themselves very well to straightline approximation. ( s 2 If ζ =, thenthisis ± ). Double real zero at ωn slope of 40 db/decade. For ζ =, therewillbeovershootorundershootatω ωn. For other values of ζ : 0 Dip amount for 0 < ζ < Dip frequency: ω d = 2ζ 2 Value of H( jω d ) is: 20 log 0 (2ζ ( ζ 2 )). Note: There is no dip unless 0 <ζ </ ζ Dip (db)
12 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 2 We write complex poles (LHP or RHP) as [ ( ) s 2 ( s G(s) = ± 2ζ ) + ]. The resonant peak frequency is ω r = 2ζ 2 Value of H( jω r ) is 20 log 0 (2ζ ( ζ 2 )). Same graph as for dip for complexconjugate zeros. Note that there is no peak unless 0 <ζ </ For ω,magnitude 0dB. For ω,magnitudeslope= 40 db/decade. 40 db 20 db 0 db 20 db Bode Mag: Complex zeros ζ = Bode magnitude: Time delay 20 db 0 db 20 db Bode Mag: Complex poles ζ = ζ = db 0. 0 G(s) = e sτ... G( jω) =. 20 log0 = 0dB. Does not change magnitude response. EXAMPLE: Sketch the Bode magnitude plot for G(s) = 2000(s + 0.5) s(s + 0)(s + 50).
13 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 3 The first step is to convert the terms of the transfer function into Bode standard form. G(s) = G( jω) = (s + 0.5) s(s + 0)(s + 50) = 0 50 ( ) 2 jω jω ( jω 0 + )( jω 50 + ). ( s ) s ( s 0 + )( s 50 + ) We can see that the components of the transfer function are: DC gain of 20 log 0 2 6dB; Pole at origin; One real zero not at origin, and Two real poles not at origin
14 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Bode phase diagrams (a) Bode diagrams consist of the magnitude plots we have seen so far, BUT, also phase plots. These are just as easy to draw. BUT, they differ depending on whether the dynamics are RHP or LHP. Finding the phase of a complex number Plot the location of the number as a vector in the complex plane. Use trigonometry to find the phase. For numbers with positive real part, ( ) I(#) (#) = tan. R(#) For numbers with negative real part, ( ) I(#) (#) = 80 tan. R(#) p 2 (p 2 ) If you are lucky enough to have the atan2(y, x) function, then I R p (p ) (#) = atan2(i(#), R(#)) for any complex number. Also note, ( ) ab cd = (a) + (b) (c) (d). Finding the phase of a complex function of ω This is the same as finding the phase of a complex number, if specific values of ω are substituted into the function.
15 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 5 Bode phase: Constant gain G(s) = K. { 0, K 0; (K ) = 80, K < 0.. Constant phase of 0 or 80. Bode phase: Zero or pole at origin Zero: G(s) = s,... G( jω) = jω = ω 90. Pole: G(s) =,... G( jω) = s jω = j ω = 90 ω. Constant phase of ±90. Bode phase: Real LHP zero or pole ( ) s Zero: G(s) = +. G( jω) = ( j ω ) + ( ) ω = tan. Pole: G(s) = ( ), s + ( G( jω) = () j ω ) + ( ) ω = tan per decade per decade
16 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 6 Bode phase: Real RHP zero or pole ( ) s Zero: G(s) =. G( jω) = ( j ω ) ( ) ω = 80 tan per decade 90 Pole: G(s) = ( ), s ( G( jω) = () j ω ) ( )) = (80 tan ωωn ( ) ω = 80 + tan per decade
17 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Bode phase diagrams (b) Bode phase: Complex LHP zero pair or pole pair Complex LHP zeros cause phase to go from 0 to 80. Complex LHP poles cause phase to go from 80 to 0. Transition happens in about ±ζ decades, centered at ωn. Bode Phase: Complex LHP zeros 80 ζ = Bode Phase: Complex LHP poles 0 ζ = Bode phase: Complex RHP zero pair or pole pair 0. 0 Complex RHP zeros cause phase to go from 360 to 80. Complex RHP poles cause phase to go from 360 to Bode Phase: Complex RHP zeros Bode Phase: Complex RHP poles
18 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 8 Bode phase: Time delay G(s) = e sτ, G( jω) = e jωτ = ωτ 0.ω i ω i 0ω i G( jω) = ωτ in radians. = 56.3ωτ in degrees. Note: Line curve in log scale. EXAMPLE: Sketch the Bode phase plot for 2000(s + 0.5) G(s) = or G( jω) = s(s + 0)(s + 50) 2 ( jω/0.5 + ) jω ( jω/0 + )(jω/50 + ), where we converted to Bode standard form in a prior example. Constant: K =+2. Zerophasecontribution. Pole at origin: Phase contribution of 90. Two real LHP poles: Phase from 0 to 90,each. One real LHP zero: Phase from 0 to
19 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 9 EXAMPLE: Sketch the Bode magnitude and phase plots for G(s) = 200 (s + 3) s (s + 2)(s + 50). First, we convert to Bode standard form, which gives 200 (3) ( ) + s 3 G(s) = s (2)(50) ( )( ) + s 2 + s 50 ( ) 6 + jω 3 G( jω) = ( )( ). jω + jω + jω 2 50 Positive gain, one real LHP zero, one pole at origin, two real LHP poles
20 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Some observations based on Bode plots Nonminimumphase systems Asystemiscalledanonminimumphaseifithaspole(s)orzero(s) in the RHP. Consider G (s) = 0 s + } s + 0 G 2 (s) = 0 s } s + 0 zero at pole at 0 zero at + pole at 0 } } minimum phase nonminimum phase The magnitude responses of these two systems are: jω + G ( jω) =0 jω + 0 = 0 ω2 + ω jω G 2 ( jω) =0 jω + 0 = 0 ω2 + ω which are the same! The phase responses are very different: G( jω), G2( jω) BodeMagnitude Plot Frequency, (rads/sec.) (G( jω)), (G2( jω)) Minimum Phase, G BodePhase Plot Nonminimum Phase, G Frequency, (rads/sec.) Note that the change in phase of G is much smaller than change of phase in G 2.HenceG is minimum phase and G 2 is nonminimumphase
21 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 2 Nonminimum phase usually associated with delay. G 2 (s) = G (s) s s + }{{} Delay s Note: is very similar to a firstorder Padé approximation to a s + delay. It is the same when evaluated at s = jω. Consider using feedback to control a nonminimumphase system. What do the rootlocus plotting techniques tell us? Consequently, nonminimumphase systems are harder to design controllers for; step response often tends to go the wrong way, at least initially. Steadystate errors from Bode magnitude plot Recall our discussion of steadystate errors to step/ramp/parabolic inputs versus system type (summarized on pg. 4 24) Consider a unityfeedback system. If the openloop plant transfer function has N poles at s = 0 then the system is type N K p is error constant for type 0. K v is error constant for type. K a is error constant for type 2... For a unityfeedback system, K p = lim s 0 G(s). At low frequency, atype0systemwillhaveg(s) K p. We can read this off the Bodemagnitude plot directly!
22 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 22 Horizontal yintercept at low frequency = K p. e ss = + K p for step input. Kv = lim s 0 sg(s), andisnonzeroforatypesystem. At low frequency, atypesystemwillhaveg(s) K v s. At low frequency, G( jω) K v.slopeof 20 db/decade. ω Use the above approximation to extend the lowfrequency asymptote to ω =. Theasymptote(NOT THE ORIGINAL G( jω) ) evaluatedatω = is K v. e ss = K v for ramp input. Ka = lim s 0 s 2 G(s), andisnonzeroforatype2system. At low frequency, atype2systemwillhaveg(s) K a s 2. At low frequency, G( jω) K a.slopeof 40 db/decade. ω2 Again, use approximation to extend lowfrequency asymptote to ω =. Theasymptoteevaluatedatω = is K a. Similar for higherorder systems. e ss = K a for parabolic input.
23 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS Example 60 Example Magnitude Magnitude Frequency, (rads/sec.) EXAMPLE : Frequency, (rads/sec.) EXAMPLE 2: Horizontal as ω 0, soweknowthisistype0. Intercept = 6dB...K p = 6dB= 2 [linear units]. Slope = 20 db/decade as ω 0, soweknowthisistype. Extend slope at low frequency to ω =. Intercept = 20 db... K v = 20 db = 0 [linear units].
24 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Stability revisited If we know the closedloop transfer function of a system in rationalpolynomial form, we can use Routh to find stable ranges for K. Motivation: Whatifweonlyhaveopenloopfrequency response? Asimpleexample Consider, for now, that we know the transferfunction of the system, and can plot the rootlocus. EXAMPLE: I(s) r(t) K s(s + ) 2 y(t) K = 2 R(s) We see neutral stability at K = 2. ThesystemisstableforK < 2 and unstable for K > 2. Recall that a point is on the root locus if KG(s) = and G(s) = 80. If system is neutrally stable, jωaxis will have a point (points) where KG( jω) = and G( jω) = 80.
25 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 25 Consider the Bode plot of KG(s)... Aneutralstabilityconditionfrom Bode plot is: KG( jω o ) = AND KG( jω o ) = 80 at the same frequency ω o K = 0. K = 2 K = In this case, increasing K instability KG( jω) < at KG( jω) = 80 =stability. In some cases, decreasing K instability KG( jω) > at KG( jω) = 80 =stability KEY POINT: We can find neutral stability point on Bode plot, but don t (yet) have a way of determining if the system is stable or not. Nyquist found a frequencydomain method to do so. Nyquist stability Poles of closedloop transfer function in RHP the system is unstable. Nyquist found way to count closedloop poles in RHP. If count is greater than zero, system is unstable. Idea: First, find a way to count closedloop poles inside a contour. Second, make the contour equal to the RHP. Counting is related to complex functional mapping.
26 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Interlude: Complex functional mapping Nyquist technique is a graphical method to determine system stability, regions of stability and MARGINS of stability. Involves graphing complex functions of s as a polar plot. EXAMPLE: Plotting f (x), arealfunctionofarealvariablex. f (x) This can be done. EXAMPLE: PlottingF(s), acomplexfunctionofacomplexvariables. F(s)? s x NO! This is wrong! Must draw mapping of points or lines from splane to F(s)plane. jω s I(F) F(s 0 ) F(s) s 0 mapping σ R(F) EXAMPLE: F(s) = 2s +... map the four points: A, B, C, D D j A I(s) R(s) C j B
27 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 27 EXAMPLE: Mapasquarecontour(closedpath)byF(s) = D j A j I(s) F(D) s s + 2. F(A) F(B) R(s) C j B j F(C) FORESHADOWING: By drawing maps of a specific contour, using a mapping function related to the plant openloop frequencyresponse, we will be able to determine closedloop stability of systems. Mapping function: Poles of the function When we map a contour containing (encircling) poles and zeros of the mapping function, this map will give us information about how many poles and zeros are encircled by the contour. Practice drawing maps when we know poles and zeros. Evaluate G(s) s=so = G(s o ) = v e jα α = (zeros) (poles). EXAMPLE: I(s) F(c ) c R(s) α
28 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 28 In this example, there are no zeros or poles inside the contour. The phase α increases and decreases, but never undergoes a net change of 360 (does not encircle the origin). EXAMPLE: I(s) c 2 F(c 2 ) R(s) α One pole inside contour. Resulting map undergoes 360 net phase change. (Encircles the origin). EXAMPLE: I(s) c 3 F(c 3 ) R(s) In this example, there are two poles inside the contour, and the map encircles the origin twice.
29 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Cauchy s theorem and Nyquist s rule These examples give heuristic evidence of the general rule: Cauchy s theorem Let F(s) be the ratio of two polynomials in s. Lettheclosed curve C in the splane be mapped into the complex plane through the mapping F(s). IfthecurveC does not pass through any zeros or poles of F(s) as it is traversed in the CW direction, the corresponding map in the F(s)plane encircles the origin N = Z P times in the CW direction, where Z = # of zeros of F(s) in C, P = # of poles of F(s) in C. Consider the following feedback system: r(t) D(s) H(s) G(s) y(t) T (s) = D(s)G(s) + D(s)G(s)H(s). For closedloop stability, no poles of T (s) I(s) in RHP. R No zeros of + D(s)G(s)H(s) in RHP. Let F(s) = + D(s)G(s)H(s). R(s) Count zeros in RHP using Cauchy theorem! (Contour=entire RHP). The Nyquist criterion simplifies Cauchy s criterion for feedback systems of the above form.
30 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 30 Cauchy: F(s) = + D(s)G(s)H(s). N = # of encirclements of origin. Nyquist: F(s) = D(s)G(s)H(s). N = # of encirclements of. I(s) Cauchy I(s) Nyquist R(s) R(s) Simple? YES!!! Think of Nyquist path as four parts: I(s) I. Origin. Sometimes a special case (later examples). II III II. + jωaxis. FREQUENCYresponse of O.L. system! Just plot it as a polar plot. III. For physical systems=0. I IV R(s) IV. Complex conjugate of II. So, for most physical systems, the Nyquist plot, used to determine CLOSEDLOOP stability, is merely a polar plot of LOOP frequency response D( jω)g( jω)h( jω). We don t even need a mathematical model of the system. Measured data of G( jω) combined with our known D( jω) and H( jω) are enough to determine closedloop stability.
31 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 3 THE TEST: N = # encirclements of point when F(s) = D(s)G(s)H(s). P = # poles of + F(s) in RHP= # of openloop unstable poles. (assuming that H(s) is stable reasonable). Z = # of zeros of + F(s) in RHP= # of closedloop unstable poles. Z = N + P The system is stable iff Z = 0. Be careful counting encirclements! Draw line from in any direction. Count # crossings of line and diagram. N = #CW crossings #CCW crossings. Changing the gain K of F(s) MAGNIFIES the entire plot. ENHANCED TEST: Loop transfer function is KD(s)G(s)H(s). N = # encirclements of /K point when F(s) = D(s)G(s)H(s). Rest of test is the same. Gives ranges of K for stability.
32 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Nyquist test example EXAMPLE: D(s) = H(s) =. G(s) = or, G( jω) = I: At s = 0, G(s) = 5. II: At s = jω, G( jω) = 5 (s + ) 2 III: At s =, G(s) = 0. 5 ( jω + ) 2 5 ( + jω) 2. IV: At s = jω, G(s) = 5 ( jω) 2. I(s) ω R(G( jω)) I(G( jω)) R(s) No encirclements of, N = 0. No openloop unstable poles P = 0. Z = N + P = 0. Closedloopsystemisstable. No encirclements of /K for any K > 0. So, system is stable for any K > 0.
33 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 33 Confirm by checking Routh array. Routh array: a(s) = + KG(s) = s 2 + 2s + + 5K. s 2 + 5K s 2 s 0 + 5K Stable for any K > EXAMPLE: G(s) = (s + ) 2 (s + 0). I: G(0) = 50/0 = II: G( jω) = ( jω + ) 2 ( jω + 0). III: G( ) = 0. IV: G( jω) = G( jω). Note loop to left of origin. System is NOT stable for all K > 0. I(s) ω R(G( jω)) I(G( jω)) Zoom R(s)
34 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Nyquist test example with pole on jωaxis EXAMPLE: Pole(s) at origin. G(s) = s(τs + ). WARNING! Wecannotblindlyfollowprocedure! Nyquist path goes through pole at zero! (Remember from Cauchy s theorem that the path cannot pass directly through a pole or zero.) Remember: We want to count closedloop poles inside a box that encompasses the RHP. So, we use a slightlymodified Nyquist path. I(s) I(s) Zoom II III ρ 0 I R(s) θ R(s) IV The bump at the origin makes a detour around the offending pole. Bump defined by curve: s = lim ρ 0 ρe jθ, 0 θ 90. From above, G(s) s=ρe jθ = ρe jθ (τρe jθ + ), 0 θ 90 Consider magnitude as ρ 0 lim G(s) s=ρe ρ 0 jθ = ρ τρe jθ + ρ.
35 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 35 Consider phase as ρ 0 So, lim G(s) s=ρe jθ = θ (τρe jθ + ). ρ 0 lim G(ρe jθ ) = lim ρ 0 ρ 0 ρ θ + This is an arc of infinite radius, sweeping from 0 to 90 + (a little more than 90 because of contribution from (τs + ) term). WE CANNOT DRAW THIS TO SCALE! Z = N + P. N = # encirclements of. N = 0. P = # Loop transfer function poles inside MODIFIED contour. P = 0. Z = 0. Closedloopsystemisstable. EXAMPLE: G(s) = Use modified Nyquist path again s 2 (s + ) I: Near origin G(s) s=ρe jθ = Magnitude: lim ρ 0 G(ρe jθ ) = ρ 2 e j2θ ( + ρe jθ ). ρ 2 + ρe jθ ρ 2. Phase: lim ρ 0 G(ρe jθ ) = 0 [2θ + ( + ρe jθ )] 2θ +.So, lim G(ρe jθ ) = lim ρ 0 ρ 0 ρ 2 2θ + 0 θ 90.
36 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 36 Infinite arc from 0 to 80 + (a little more than 80 because of + s term.) Z = N + P = = 2. Unstablefor K =. In fact, unstable for any K > 0! Matlab for above G(s) = num=[0 0 0 ]; den=[ 0 0]; nyquist(num,den); s 3 + s 2 + 0s + 0 axis([xmin xmax ymin ymax]); nyquist.m is available on course web site. It repairs the standard Matlab nyquist.m program, which doesn t work when poles are on imaginary axis. nyquist2.m is also available. It draws contours around poles on the imaginary axis in the opposite way to nyquist.m. Counting is different.
37 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Stability (gain and phase) margins Alargefractionofsystemstobecontrolledarestableforsmall gain but become unstable if gain is increased beyond a certain point. The distance between the current (stable) system and an unstable system is called a stability margin. Can have a gain margin and a phase margin. GAIN MARGIN: Factor by which the gain is less than the neutral stability value. Gain margin measures How much can we increase the gain of the loop transfer function L(s) = D(s)G(s)H(s) and still have a stable system? Many Nyquist plots are like this one. Increasing loop gain magnifies the plot. GM GM =/(distance between origin and place where Nyquist map crosses real axis). PM If we increase gain, Nyquist map stretches and we may encircle. For a stable system, GM > (linear units) or GM > 0dB. PHASE MARGIN: Phase factor by which phase is greater than neutral stability value. Phase margin measures How much delay can we add to the loop transfer function and still have a stable system?
38 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 38 PM = Angle to rotate Nyquist plot to achieve neutral stability = intersection of Nyquist with circle of radius. If we increase openloop delay, Nyquist map rotates and we may encircle. For a stable system, PM > 0. IRONY: This is usually easiest to check on Bode plot, even though derived on Nyquist plot! Define gain crossover as frequency where Bode magnitude is 0 db. Define phase crossover as frequency where Bode phase is 80. GM = /(Bode gain at phasecrossover frequency) if Bode gain is measured in linear units. GM = ( Bode gain at phasecrossover frequency) [db] if Bode gain measured in db. PM = Bode phase at gaincrossover ( 80 ). Magnitude Phase Frequency, (rads/sec.) Frequency, (rads/sec.) We can also determine stability as K changes. Instead of defining gain crossover where G( jω) =, usethefrequencywhere KG( jω) =.
39 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 39 You need to be careful using this test. It works if you apply it blindly and the system is minimumphase. You need to think harder if the system is nonminimumphase. Nyquist is the safest bet. PM and performance AbonusofcomputingPM from the openloop frequency response graph is that it can help us predict closedloop system performance. PM is related to damping. Consider openloop 2ndorder system with unity feedback, G(s) = ω 2 n s(s + 2ζ ) T (s) = ω 2 n. s 2 + 2ζ + ωn 2 The relationship between PM and ζ is: (for this system) PM = tan 2ζ + 4ζ 4 2ζ 2 For PM 60, ζ PM 00,socanalsoinferM p from PM. Damping ratio ζ Damping ratio versus PM Phase margin M p, overshoot Overshoot fraction versus PM Phase margin
40 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS : Preparing for control using frequencyresponse methods Bode s gainphase relationship For any stable minimumphase system (that is, one with no RHP zeros or poles), the phase of G( jω) is uniquely related to the magnitude of G( jω) Relationship: G( jω o ) = π M = ln G( jω) ( ) ω u = ln ω o ( dm du ) W (u) du dm du slope n of logmag curve at ω = ω o W (u) = weighting function = ln(coth u /2) π 2 W (u) δ(u). Usingthisrelationship, G( jω) n 90 2 (in radians) if slope of Bode magnitudeplot is constant in the decadeneighborhood of ω So, if G( jω) 90 if n =. W (u) Normalized freq. u So, if G( jω) 80 if n = 2. KEY POINT: Want crossover G( jω) = at a slope of about for good PM. Wewillsoonseehowtodothis(design!).
41 ECE450/ECE550, FREQUENCYRESPONSE ANALYSIS 8 4 Closedloop frequency response Most of the notes in this section have used the openloop frequency response to predict closedloop behavior. How about closedloop frequency response? T (s) = KD(s)G(s) + KD(s)G(s). General approximations are simple to make. If, KD( jω)g( jω) and KD( jω)g( jω) for ω ω c for ω ω c where ω c is the cutoff frequency where openloop magnitude response crosses magnitude=. { T ( jω) = KD( jω)g( jω) + KD( jω)g( jω), ω ω c ; KD( jω)g( jω), ω ω c. Note: ωc ω bw 2ω c.
STABILITY ANALYSIS TECHNIQUES
ECE4540/5540: Digital Control Systems 4 1 STABILITY ANALYSIS TECHNIQUES 41: Bilinear transformation Three main aspects to controlsystem design: 1 Stability, 2 Steadystate response, 3 Transient response
More informationFrequency methods for the analysis of feedback systems. Lecture 6. Loop analysis of feedback systems. Nyquist approach to study stability
Lecture 6. Loop analysis of feedback systems 1. Motivation 2. Graphical representation of frequency response: Bode and Nyquist curves 3. Nyquist stability theorem 4. Stability margins Frequency methods
More information1 (20 pts) Nyquist Exercise
EE C128 / ME134 Problem Set 6 Solution Fall 2011 1 (20 pts) Nyquist Exercise Consider a close loop system with unity feedback. For each G(s), hand sketch the Nyquist diagram, determine Z = P N, algebraically
More informationROOT LOCUS. Consider the system. Root locus presents the poles of the closedloop system when the gain K changes from 0 to. H(s) H ( s) = ( s)
C1 ROOT LOCUS Consider the system R(s) E(s) C(s) + K G(s)  H(s) C(s) R(s) = K G(s) 1 + K G(s) H(s) Root locus presents the poles of the closedloop system when the gain K changes from 0 to 1+ K G ( s)
More informationFrequency Response Techniques
4th Edition T E N Frequency Response Techniques SOLUTION TO CASE STUDY CHALLENGE Antenna Control: Stability Design and Transient Performance First find the forward transfer function, G(s). Pot: K 1 = 10
More informationINTRODUCTION TO DIGITAL CONTROL
ECE4540/5540: Digital Control Systems INTRODUCTION TO DIGITAL CONTROL.: Introduction In ECE450/ECE550 Feedback Control Systems, welearnedhow to make an analog controller D(s) to control a lineartimeinvariant
More informationFREQUENCYRESPONSE DESIGN
ECE45/55: Feedback Control Systems. 9 FREQUENCYRESPONSE DESIGN 9.: PD and lead compensation networks The frequencyresponse methods we have seen so far largely tell us about stability and stability margins
More informationCHAPTER 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.
More informationDESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD
206 Spring Semester ELEC733 Digital Control System LECTURE 7: DESIGN USING TRANSFORMATION TECHNIQUE CLASSICAL METHOD For a unit ramp input Tz Ez ( ) 2 ( z ) D( z) G( z) Tz e( ) lim( z) z 2 ( z ) D( z)
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using
More informationHomework 7  Solutions
Homework 7  Solutions Note: This homework is worth a total of 48 points. 1. Compensators (9 points) For a unity feedback system given below, with G(s) = K s(s + 5)(s + 11) do the following: (c) Find the
More informationThe FrequencyResponse
6 The FrequencyResponse Design Method A Perspective on the FrequencyResponse Design Method The design of feedback control systems in industry is probably accomplished using frequencyresponse methods
More informationModule 3F2: Systems and Control EXAMPLES PAPER 2 ROOTLOCUS. Solutions
Cambridge University Engineering Dept. Third Year Module 3F: Systems and Control EXAMPLES PAPER ROOTLOCUS Solutions. (a) For the system L(s) = (s + a)(s + b) (a, b both real) show that the rootlocus
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 21: Stability Margins and Closing the Loop Overview In this Lecture, you will learn: Closing the Loop Effect on Bode Plot Effect
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 23: Drawing The Nyquist Plot Overview In this Lecture, you will learn: Review of Nyquist Drawing the Nyquist Plot Using the
More informationTopic # Feedback Control
Topic #4 16.31 Feedback Control Stability in the Frequency Domain Nyquist Stability Theorem Examples Appendix (details) This is the basis of future robustness tests. Fall 2007 16.31 4 2 Frequency Stability
More informationControl Systems. Frequency Method Nyquist Analysis.
Frequency Method Nyquist Analysis chibum@seoultech.ac.kr Outline Polar plots Nyquist plots Factors of polar plots PolarNyquist Plots Polar plot: he locus of the magnitude of ω vs. the phase of ω on polar
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition adings: Guzzella, Chapter 9.4 6 Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Emilio Frazzoli Institute
More informationMAS107 Control Theory Exam Solutions 2008
MAS07 CONTROL THEORY. HOVLAND: EXAM SOLUTION 2008 MAS07 Control Theory Exam Solutions 2008 Geir Hovland, Mechatronics Group, Grimstad, Norway June 30, 2008 C. Repeat question B, but plot the phase curve
More informationFrequency domain analysis
Automatic Control 2 Frequency domain analysis Prof. Alberto Bemporad University of Trento Academic year 20102011 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 20102011
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : Steadystate error Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling Analysis Design Laplace
More informationRadar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.
Radar Dish ME 304 CONTROL SYSTEMS Mechanical Engineering Department, Middle East Technical University Armature controlled dc motor Outside θ D output Inside θ r input r θ m Gearbox Control Transmitter
More informationSTABILITY ANALYSIS. Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated using cones: Stable Neutral Unstable
ECE4510/5510: Feedback Control Systems. 5 1 STABILITY ANALYSIS 5.1: Boundedinput boundedoutput (BIBO) stability Asystemmaybe stable, neutrallyormarginallystable, or unstable. This can be illustrated
More informationControl Systems I. Lecture 9: The Nyquist condition
Control Systems I Lecture 9: The Nyquist condition Readings: Åstrom and Murray, Chapter 9.1 4 www.cds.caltech.edu/~murray/amwiki/index.php/first_edition Jacopo Tani Institute for Dynamic Systems and Control
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 22: The Nyquist Criterion Overview In this Lecture, you will learn: Complex Analysis The Argument Principle The Contour
More informationCourse Outline. Closed Loop Stability. Stability. Amme 3500 : System Dynamics & Control. Nyquist Stability. Dr. Dunant Halim
Amme 3 : System Dynamics & Control Nyquist Stability Dr. Dunant Halim Course Outline Week Date Content Assignment Notes 1 5 Mar Introduction 2 12 Mar Frequency Domain Modelling 3 19 Mar System Response
More informationDynamic circuits: Frequency domain analysis
Electronic Circuits 1 Dynamic circuits: Contents Free oscillation and natural frequency Transfer functions Frequency response Bode plots 1 System behaviour: overview 2 System behaviour : review solution
More informationr +  FINAL June 12, 2012 MAE 143B Linear Control Prof. M. Krstic
MAE 43B Linear Control Prof. M. Krstic FINAL June, One sheet of handwritten notes (two pages). Present your reasoning and calculations clearly. Inconsistent etchings will not be graded. Write answers
More informationRaktim Bhattacharya. . AERO 422: Active Controls for Aerospace Vehicles. Frequency ResponseDesign Method
.. AERO 422: Active Controls for Aerospace Vehicles Frequency Response Method Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. ... Response to
More informationTransient response via gain adjustment. Consider a unity feedback system, where G(s) = 2. The closed loop transfer function is. s 2 + 2ζωs + ω 2 n
Design via frequency response Transient response via gain adjustment Consider a unity feedback system, where G(s) = ωn 2. The closed loop transfer function is s(s+2ζω n ) T(s) = ω 2 n s 2 + 2ζωs + ω 2
More informationControl Systems I Lecture 10: System Specifications
Control Systems I Lecture 10: System Specifications Readings: Guzzella, Chapter 10 Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich November 24, 2017 E. Frazzoli (ETH) Lecture
More informationIf you need more room, use the backs of the pages and indicate that you have done so.
EE 343 Exam II Ahmad F. Taha Spring 206 Your Name: Your Signature: Exam duration: hour and 30 minutes. This exam is closed book, closed notes, closed laptops, closed phones, closed tablets, closed pretty
More informationCompensator Design to Improve Transient Performance Using Root Locus
1 Compensator Design to Improve Transient Performance Using Root Locus Prof. Guy Beale Electrical and Computer Engineering Department George Mason University Fairfax, Virginia Correspondence concerning
More informationSoftware Engineering 3DX3. Slides 8: Root Locus Techniques
Software Engineering 3DX3 Slides 8: Root Locus Techniques Dr. Ryan Leduc Department of Computing and Software McMaster University Material based on Control Systems Engineering by N. Nise. c 2006, 2007
More informationECEN 605 LINEAR SYSTEMS. Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability 1/27
1/27 ECEN 605 LINEAR SYSTEMS Lecture 20 Characteristics of Feedback Control Systems II Feedback and Stability Feedback System Consider the feedback system u + G ol (s) y Figure 1: A unity feedback system
More informationPlan of the Lecture. Goal: wrap up lead and lag control; start looking at frequency response as an alternative methodology for control systems design.
Plan of the Lecture Review: design using Root Locus; dynamic compensation; PD and lead control Today s topic: PI and lag control; introduction to frequencyresponse design method Goal: wrap up lead and
More informationMEM 355 Performance Enhancement of Dynamical Systems
MEM 355 Performance Enhancement of Dynamical Systems Frequency Domain Design Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University 5/8/25 Outline Closed Loop Transfer Functions
More informationProcess Control & Instrumentation (CH 3040)
Firstorder systems Process Control & Instrumentation (CH 3040) Arun K. Tangirala Department of Chemical Engineering, IIT Madras January  April 010 Lectures: Mon, Tue, Wed, Fri Extra class: Thu A firstorder
More informationIntro to Frequency Domain Design
Intro to Frequency Domain Design MEM 355 Performance Enhancement of Dynamical Systems Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University Outline Closed Loop Transfer Functions
More informationMAE 143B  Homework 9
MAE 143B  Homework 9 7.1 a) We have stable firstorder poles at p 1 = 1 and p 2 = 1. For small values of ω, we recover the DC gain K = lim ω G(jω) = 1 1 = 2dB. Having this finite limit, our straightline
More informationLecture 6 Classical Control Overview IV. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 6 Classical Control Overview IV Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore Lead Lag Compensator Design Dr. Radhakant Padhi Asst.
More informationECE 486 Control Systems
ECE 486 Control Systems Spring 208 Midterm #2 Information Issued: April 5, 208 Updated: April 8, 208 ˆ This document is an info sheet about the second exam of ECE 486, Spring 208. ˆ Please read the following
More informationThe Frequencyresponse Design Method
Chapter 6 The Frequencyresponse Design Method Problems and Solutions for Section 6.. (a) Show that α 0 in Eq. (6.2) is given by α 0 = G(s) U 0ω = U 0 G( jω) s jω s= jω 2j and α 0 = G(s) U 0ω = U 0 G(jω)
More informationDelhi Noida Bhopal Hyderabad Jaipur Lucknow Indore Pune Bhubaneswar Kolkata Patna Web: Ph:
Serial : 0. LS_D_ECIN_Control Systems_30078 Delhi Noida Bhopal Hyderabad Jaipur Lucnow Indore Pune Bhubaneswar Kolata Patna Web: Email: info@madeeasy.in Ph: 04546 CLASS TEST 089 ELECTRONICS ENGINEERING
More informationAnalysis of SISO Control Loops
Chapter 5 Analysis of SISO Control Loops Topics to be covered For a given controller and plant connected in feedback we ask and answer the following questions: Is the loop stable? What are the sensitivities
More information1 (s + 3)(s + 2)(s + a) G(s) = C(s) = K P + K I
MAE 43B Linear Control Prof. M. Krstic FINAL June 9, Problem. ( points) Consider a plant in feedback with the PI controller G(s) = (s + 3)(s + )(s + a) C(s) = K P + K I s. (a) (4 points) For a given constant
More informationSTABILITY OF CLOSEDLOOP CONTOL SYSTEMS
CHBE320 LECTURE X STABILITY OF CLOSEDLOOP CONTOL SYSTEMS Professor Dae Ryook Yang Spring 2018 Dept. of Chemical and Biological Engineering 101 Road Map of the Lecture X Stability of closedloop control
More informationOverview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros)
Overview of Bode Plots Transfer function review Piecewise linear approximations Firstorder terms Secondorder terms (complex poles & zeros) J. McNames Portland State University ECE 222 Bode Plots Ver.
More informationRobust Control 3 The Closed Loop
Robust Control 3 The Closed Loop Harry G. Kwatny Department of Mechanical Engineering & Mechanics Drexel University /2/2002 Outline Closed Loop Transfer Functions Traditional Performance Measures Time
More informationRecitation 11: Time delays
Recitation : Time delays Emilio Frazzoli Laboratory for Information and Decision Systems Massachusetts Institute of Technology November, 00. Introduction and motivation. Delays are incurred when the controller
More informationControl Systems. University Questions
University Questions UNIT1 1. Distinguish between open loop and closed loop control system. Describe two examples for each. (10 Marks), Jan 2009, June 12, Dec 11,July 08, July 2009, Dec 2010 2. Write
More informationLecture 5: Frequency domain analysis: Nyquist, Bode Diagrams, second order systems, system types
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
More informationDigital Control Systems
Digital Control Systems Lecture Summary #4 This summary discussed some graphical methods their use to determine the stability the stability margins of closed loop systems. A. Nyquist criterion Nyquist
More information(b) A unity feedback system is characterized by the transfer function. Design a suitable compensator to meet the following specifications:
1. (a) The open loop transfer function of a unity feedback control system is given by G(S) = K/S(1+0.1S)(1+S) (i) Determine the value of K so that the resonance peak M r of the system is equal to 1.4.
More informationClass 13 Frequency domain analysis
Class 13 Frequency domain analysis The frequency response is the output of the system in steady state when the input of the system is sinusoidal Methods of system analysis by the frequency response, as
More informationECE317 : Feedback and Control
ECE317 : Feedback and Control Lecture : RouthHurwitz stability criterion Examples Dr. Richard Tymerski Dept. of Electrical and Computer Engineering Portland State University 1 Course roadmap Modeling
More informationDynamic Response. Assoc. Prof. Enver Tatlicioglu. Department of Electrical & Electronics Engineering Izmir Institute of Technology.
Dynamic Response Assoc. Prof. Enver Tatlicioglu Department of Electrical & Electronics Engineering Izmir Institute of Technology Chapter 3 Assoc. Prof. Enver Tatlicioglu (EEE@IYTE) EE362 Feedback Control
More informationControl Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho Tel: Fax:
Control Systems Engineering ( Chapter 8. Root Locus Techniques ) Prof. KwangChun Ho kwangho@hansung.ac.kr Tel: 027604253 Fax:027604435 Introduction In this lesson, you will learn the following : The
More informationChapter 7. Digital Control Systems
Chapter 7 Digital Control Systems 1 1 Introduction In this chapter, we introduce analysis and design of stability, steadystate error, and transient response for computercontrolled systems. Transfer functions,
More informationIntroduction. Performance and Robustness (Chapter 1) Advanced Control Systems Spring / 31
Introduction Classical Control Robust Control u(t) y(t) G u(t) G + y(t) G : nominal model G = G + : plant uncertainty Uncertainty sources : Structured : parametric uncertainty, multimodel uncertainty Unstructured
More informationK(s +2) s +20 K (s + 10)(s +1) 2. (c) KG(s) = K(s + 10)(s +1) (s + 100)(s +5) 3. Solution : (a) KG(s) = s +20 = K s s
321 16. Determine the range of K for which each of the following systems is stable by making a Bode plot for K = 1 and imagining the magnitude plot sliding up or down until instability results. Verify
More informationAMME3500: System Dynamics & Control
Stefan B. Williams May, 211 AMME35: System Dynamics & Control Assignment 4 Note: This assignment contributes 15% towards your final mark. This assignment is due at 4pm on Monday, May 3 th during Week 13
More informationMAE143a: Signals & Systems (& Control) Final Exam (2011) solutions
MAE143a: Signals & Systems (& Control) Final Exam (2011) solutions Question 1. SIGNALS: Design of a noisecancelling headphone system. 1a. Based on the lowpass filter given, design a highpass filter,
More informationControl Systems I. Lecture 5: Transfer Functions. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 5: Transfer Functions Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 20, 2017 E. Frazzoli (ETH) Lecture 5: Control Systems I 20/10/2017
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (HS 25) Solution Exercise Set Loop Shaping Noele Norris, 9th December 26 Exercise (A Nonminimum Phase System) To increase the rise time of the system, we
More informationExercises for lectures 13 Design using frequency methods
Exercises for lectures 13 Design using frequency methods Michael Šebek Automatic control 2016 31317 Setting of the closed loop bandwidth At the transition frequency in the open loop is (from definition)
More informationStability of Feedback Control Systems: Absolute and Relative
Stability of Feedback Control Systems: Absolute and Relative Dr. Kevin Craig Greenheck Chair in Engineering Design & Professor of Mechanical Engineering Marquette University Stability: Absolute and Relative
More informationControls Problems for Qualifying Exam  Spring 2014
Controls Problems for Qualifying Exam  Spring 2014 Problem 1 Consider the system block diagram given in Figure 1. Find the overall transfer function T(s) = C(s)/R(s). Note that this transfer function
More informationVALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur
VALLIAMMAI ENGINEERING COLLEGE SRM Nagar, Kattankulathur 603 203. DEPARTMENT OF ELECTRONICS & COMMUNICATION ENGINEERING SUBJECT QUESTION BANK : EC6405 CONTROL SYSTEM ENGINEERING SEM / YEAR: IV / II year
More informationELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 2010/2011 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques
CAIRO UNIVERSITY FACULTY OF ENGINEERING ELECTRONICS & COMMUNICATIONS DEP. 3rd YEAR, 00/0 CONTROL ENGINEERING SHEET 5 LeadLag Compensation Techniques [] For the following system, Design a compensator such
More informationReview of Linear TimeInvariant Network Analysis
D1 APPENDIX D Review of Linear TimeInvariant Network Analysis Consider a network with input x(t) and output y(t) as shown in Figure D1. If an input x 1 (t) produces an output y 1 (t), and an input x
More informationBASIC PROPERTIES OF FEEDBACK
ECE450/550: Feedback Control Systems. 4 BASIC PROPERTIES OF FEEDBACK 4.: Setting up an example to benchmark controllers There are two basic types/categories of control systems: OPEN LOOP: Disturbance r(t)
More informationCourse roadmap. Step response for 2ndorder system. Step response for 2ndorder system
ME45: Control Systems Lecture Time response of ndorder systems Prof. Clar Radcliffe and Prof. Jongeun Choi Department of Mechanical Engineering Michigan State University Modeling Laplace transform Transfer
More informationIC6501 CONTROL SYSTEMS
DHANALAKSHMI COLLEGE OF ENGINEERING CHENNAI DEPARTMENT OF ELECTRICAL AND ELECTRONICS ENGINEERING YEAR/SEMESTER: II/IV IC6501 CONTROL SYSTEMS UNIT I SYSTEMS AND THEIR REPRESENTATION 1. What is the mathematical
More informationEE C128 / ME C134 Fall 2014 HW 8  Solutions. HW 8  Solutions
EE C28 / ME C34 Fall 24 HW 8  Solutions HW 8  Solutions. Transient Response Design via Gain Adjustment For a transfer function G(s) = in negative feedback, find the gain to yield a 5% s(s+2)(s+85) overshoot
More informationClassify a transfer function to see which order or ramp it can follow and with which expected error.
Dr. J. Tani, Prof. Dr. E. Frazzoli 505900 Control Systems I (Autumn 208) Exercise Set 0 Topic: Specifications for Feedback Systems Discussion: 30.. 208 Learning objectives: The student can grizzi@ethz.ch,
More informationRoot Locus Techniques
4th Edition E I G H T Root Locus Techniques SOLUTIONS TO CASE STUDIES CHALLENGES Antenna Control: Transient Design via Gain a. From the Chapter 5 Case Study Challenge: 76.39K G(s) = s(s+50)(s+.32) Since
More informationH(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
More information(Continued on next page)
(Continued on next page) 18.2 Roots of Stability Nyquist Criterion 87 e(s) 1 S(s) = =, r(s) 1 + P (s)c(s) where P (s) represents the plant transfer function, and C(s) the compensator. The closedloop characteristic
More informationOutline. Classical Control. Lecture 1
Outline Outline Outline 1 Introduction 2 Prerequisites Block diagram for system modeling Modeling Mechanical Electrical Outline Introduction Background Basic Systems Models/Transfers functions 1 Introduction
More informationExercise 1 (A Nonminimum Phase System)
Prof. Dr. E. Frazzoli 559 Control Systems I (Autumn 27) Solution Exercise Set 2 Loop Shaping clruch@ethz.ch, 8th December 27 Exercise (A Nonminimum Phase System) To decrease the rise time of the system,
More informationSTABILITY. Have looked at modeling dynamic systems using differential equations. and used the Laplace transform to help find step and impulse
SIGNALS AND SYSTEMS: PAPER 3C1 HANDOUT 4. Dr David Corrigan 1. Electronic and Electrical Engineering Dept. corrigad@tcd.ie www.sigmedia.tv STABILITY Have looked at modeling dynamic systems using differential
More informationDue Wednesday, February 6th EE/MFS 599 HW #5
Due Wednesday, February 6th EE/MFS 599 HW #5 You may use Matlab/Simulink wherever applicable. Consider the standard, unityfeedback closed loop control system shown below where G(s) = /[s q (s+)(s+9)]
More informationFrequency Response Analysis
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
More informationEC CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING
EC 2255  CONTROL SYSTEM UNIT I CONTROL SYSTEM MODELING 1. What is meant by a system? It is an arrangement of physical components related in such a manner as to form an entire unit. 2. List the two types
More informationKINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING
KINGS COLLEGE OF ENGINEERING DEPARTMENT OF ELECTRONICS AND COMMUNICATION ENGINEERING QUESTION BANK SUB.NAME : CONTROL SYSTEMS BRANCH : ECE YEAR : II SEMESTER: IV 1. What is control system? 2. Define open
More information100 (s + 10) (s + 100) e 0.5s. s 100 (s + 10) (s + 100). G(s) =
1 AME 3315; Spring 215; Midterm 2 Review (not graded) Problems: 9.3 9.8 9.9 9.12 except parts 5 and 6. 9.13 except parts 4 and 5 9.28 9.34 You are given the transfer function: G(s) = 1) Plot the bode plot
More informationExample on Root Locus Sketching and Control Design
Example on Root Locus Sketching and Control Design MCE44  Spring 5 Dr. Richter April 25, 25 The following figure represents the system used for controlling the robotic manipulator of a Mars Rover. We
More informationOutline. Classical Control. Lecture 5
Outline Outline Outline 1 What is 2 Outline What is Why use? Sketching a 1 What is Why use? Sketching a 2 Gain Controller Lead Compensation Lag Compensation What is Properties of a General System Why use?
More informationLinear Control Systems Lecture #3  Frequency Domain Analysis. Guillaume Drion Academic year
Linear Control Systems Lecture #3  Frequency Domain Analysis Guillaume Drion Academic year 20182019 1 Goal and Outline Goal: To be able to analyze the stability and robustness of a closedloop system
More informationChapter 8: Converter Transfer Functions
Chapter 8. Converter Transfer Functions 8.1. Review of Bode plots 8.1.1. Single pole response 8.1.2. Single zero response 8.1.3. Right halfplane zero 8.1.4. Frequency inversion 8.1.5. Combinations 8.1.6.
More informationEECS C128/ ME C134 Final Thu. May 14, pm. Closed book. One page, 2 sides of formula sheets. No calculators.
Name: SID: EECS C28/ ME C34 Final Thu. May 4, 25 58 pm Closed book. One page, 2 sides of formula sheets. No calculators. There are 8 problems worth points total. Problem Points Score 4 2 4 3 6 4 8 5 3
More informationControl Systems I. Lecture 6: Poles and Zeros. Readings: Emilio Frazzoli. Institute for Dynamic Systems and Control DMAVT ETH Zürich
Control Systems I Lecture 6: Poles and Zeros Readings: Emilio Frazzoli Institute for Dynamic Systems and Control DMAVT ETH Zürich October 27, 2017 E. Frazzoli (ETH) Lecture 6: Control Systems I 27/10/2017
More informationSchool of Mechanical Engineering Purdue University. DC Motor Position Control The block diagram for position control of the servo table is given by:
Root Locus Motivation Sketching Root Locus Examples ME375 Root Locus  1 Servo Table Example DC Motor Position Control The block diagram for position control of the servo table is given by: θ D 0.09 See
More informationLecture 1 Root Locus
Root Locus ELEC304Alper Erdogan 1 1 Lecture 1 Root Locus What is RootLocus? : A graphical representation of closed loop poles as a system parameter varied. Based on RootLocus graph we can choose the
More informationTime Response Analysis (Part II)
Time Response Analysis (Part II). A critically damped, continuoustime, second order system, when sampled, will have (in Z domain) (a) A simple pole (b) Double pole on real axis (c) Double pole on imaginary
More informationR10 JNTUWORLD B 1 M 1 K 2 M 2. f(t) Figure 1
Code No: R06 R0 SET  II B. Tech II Semester Regular Examinations April/May 03 CONTROL SYSTEMS (Com. to EEE, ECE, EIE, ECC, AE) Time: 3 hours Max. Marks: 75 Answer any FIVE Questions All Questions carry
More informationRobust Performance Example #1
Robust Performance Example # The transfer function for a nominal system (plant) is given, along with the transfer function for one extreme system. These two transfer functions define a family of plants
More informationLecture 5 Classical Control Overview III. Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore
Lecture 5 Classical Control Overview III Dr. Radhakant Padhi Asst. Professor Dept. of Aerospace Engineering Indian Institute of Science  Bangalore A Fundamental Problem in Control Systems Poles of open
More information7.4 STEP BY STEP PROCEDURE TO DRAW THE ROOT LOCUS DIAGRAM
ROOT LOCUS TECHNIQUE. Values of on the root loci The value of at any point s on the root loci is determined from the following equation G( s) H( s) Product of lengths of vectors from poles of G( s)h( s)
More information