Radar Dish. Armature controlled dc motor. Inside. θ r input. Outside. θ D output. θ m. Gearbox. Control Transmitter. Control. θ D.


 Tamsyn McKinney
 1 years ago
 Views:
Transcription
1 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 θ D dc amplifier Control Transformer Prof. Dr. Y. Samim Ünlüsoy Prof. Dr. Y. Samim Ünlüsoy 1
2 COURSE OUTLINE CH IX I. INTRODUCTION & BASIC CONCEPTS II. MODELING DYNAMIC SYSTEMS III. CONTROL SYSTEM COMPONENTS IV. STABILITY V. TRANSIENT RESPONSE VI. STEADY STATE RESPONSE VII. DISTURBANCE REJECTION VIII. BASIC CONTROL ACTIONS & CONTROLLERS IX. FREQUENCY RESPONSE ANALYSIS X. SENSITIVITY ANALYSIS XI. ROOT LOCUS ANALYSIS Prof. Dr. Y. Samim Ünlüsoy 2
3 FREQUENCY RESPONSE  OBJECTIVES In this chapter : A short introduction to the steady state response of control systems to sinusoidal inputs will be given. Frequency domain specifications for a control system will be examined. Bode plots and their construction using asymptotic approximations will be presented. Prof. Dr. Y. Samim Ünlüsoy 3
4 FREQUENCY RESPONSE INTRODUCTION Nise Ch. 10 In frequency response analysis of control systems, the steady state response of the system to sinusoidal input is of interest. The frequency response analyses are carried out in the frequency domain, rather than the time domain. It is to be noted that, time domain properties of a control system can be predicted from its frequency domain characteristics. Prof. Dr. Y. Samim Ünlüsoy 4
5 FREQUENCY RESPONSE  INTRODUCTION For an LTI system the Laplace transforms of the input and output t are related to each other by the transfer function, T(s). Laplace Domain Input R(s) T(s) C(s) Output In the frequency response analysis, the system is excited by a sinusoidal input of fixed amplitude and varying frequency. Prof. Dr. Y. Samim Ünlüsoy 5
6 FREQUENCY RESPONSE  INTRODUCTION Let us subject a stable LTI system to a sinusoidal input of amplitude R and frequency ω in time domain. r(t)=rsin(ωt) The steady state output of the system will be again a sinusoidal signal of the same frequency, but probably with a different amplitude and phase. c(t)=csin(ωt+ t+φ) Prof. Dr. Y. Samim Ünlüsoy 6
7 FREQUENCY RESPONSE  INTRODUCTION Prof. Dr. Y. Samim Ünlüsoy 7
8 FREQUENCY RESPONSE  INTRODUCTION To carry out the same process in the frequency domain for sinusoidal steady state analysis, one replaces the Laplace variable s with ih s=jω in the input output relation C(s)=T(s)R(s) with the result C(jω)=T(jω)R(j )R(jω)) Prof. Dr. Y. Samim Ünlüsoy 8
9 FREQUENCY RESPONSE  INTRODUCTION The input, output, and the transfer function have now become complex and thus they can be represented by their magnitudes and phases. Input : R(jω)= R(jω) R(jω) Input : Output : C(jω)= C(jω) C(jω) Output : Transfer Function : T(jω)= T(jω) Τ(jω) ) Prof. Dr. Y. Samim Ünlüsoy 9
10 FREQUENCY RESPONSE  INTRODUCTION With similar expressions for the input and the transfer function, the input output relation in the frequency domain consists of the magnitude and phase expressions : C(jω)=T(j )=T(jω)R(j )R(jω) C(jω) =T(jω) R(jω) C(jω)= T(jω)+ R ( jω) Prof. Dr. Y. Samim Ünlüsoy 10
11 FREQUENCY RESPONSE  INTRODUCTION For the input and output described by r(t)=rsin(ωt) c(t)=csin(ωt+ t+φ) the amplitude and the phase of the output can now be written as C=R T(jω) φ = T(jω) Prof. Dr. Y. Samim Ünlüsoy 11
12 FREQUENCY RESPONSE Consider the transfer function for the general closed loop system. C(s) G(s) T(s)= = R(s) 1+G(s)H(s) For the steady state behaviour, insert s=jω. C(j ω) G(j ω) T(j ω)= = R(j ω) 1+G(j ω)h(j ω) T(jω) is called the Frequency Response Function (FRF) or Sinusoidal Transfer Function. Prof. Dr. Y. Samim Ünlüsoy 12
13 FREQUENCY RESPONSE The frequency response function can be written in terms of its magnitude and phase. T(j ω )= T(j ω ) T(j ω ) Since this function is complex, it can also be written in terms of its real and imaginary parts. T(j ω)=re[ T(j ω) ] + jim[ T(j ω) ] Prof. Dr. Y. Samim Ünlüsoy 13
14 FREQUENCY RESPONSE Remember that t for a complex number be expressed in its real and imaginary parts : the magnitude is given by : z=a+bj ( )( ) 2 2 z= a+bj abj = a +b the phase is given by : 1 b z z = tan a Prof. Dr. Y. Samim Ünlüsoy 14
15 FREQUENCY RESPONSE The magnitude and phase of the frequency response function are given by : G(j ) G(j ) T(j ω ω ω ) = = 1+G(j ω )H(j ω ) 1+G(j ω )H(j ω ) [ ] T(j ω )= G(j ω ) 1+G(j ω )H(j ω ) These are called the gain and phase characteristics. Prof. Dr. Y. Samim Ünlüsoy 15
16 FREQUENCY RESPONSE Example 1a For a system described by the differential equation x+2x=y(t) determine the steady state response x ss (t) for a pure sine wave input y(t)= 3sin(0.5t) Prof. Dr. Y. Samim Ünlüsoy 16
17 FREQUENCY RESPONSE Example 1b The transfer function is given by X(s) 1 x+2x=y(t) T(s)= = Y(s) s( s +2) Insert s=jω to get : For ω=0.5 0 [rad/s]: 1 T(jω)= jω jω+2 ( ) 1 1 T(0.5j)= = 0.5j 0.5j j ( ) Prof. Dr. Y. Samim Ünlüsoy 17
18 FREQUENCY RESPONSE Example 1c Multiply and divide by the complex conjugate j j T(0.5j)= = j j T(0.5j)= j Determine the magnitude and the angle. cos  cos ( ) ( ) T(0.5j) = = T(0.5j)= tan = sin + sin + 1 o cos  sin  cos + sin  76 o 104 o Prof. Dr. Y. Samim Ünlüsoy 18
19 FREQUENCY RESPONSE Example 1d The steady state response is then given by : o ( ) ( ) x (t)=3097sin05t t ss ( o ) =2.91sin 0.5t 104 Prof. Dr. Y. Samim Ünlüsoy 19
20 FREQUENCY RESPONSE Example 2a Express the transfer function (input : F, output : y) in terms of its magnitude and phase. my +cy +ky =F k F m c y G(s)= ms 2 1 +cs+k Prof. Dr. Y. Samim Ünlüsoy 20
21 FREQUENCY RESPONSE Example 2b Insert s=jω in the transfer function to obtain the frequency response function. G(s)= ms 2 1 +cs+k 1 1 T(jω)= = m ωj j +c ωj j+k kmω +cωj 2 ( ) ( ) ( 2 ) Write the FRF in a+bj form. Prof. Dr. Y. Samim Ünlüsoy 21
22 FREQUENCY RESPONSE Example 2c Multiply and divide the FRF expression with the complex conjugate of its denominator. ( ) ( ) ( ) ( ) kmω cωj kmω cωj T(jω)= = kmω +cωj kmω cωj kmω +cω ( 2 ) ( ) ( kmω 2 ) cω T(jω)= ( ) ( ) ( 2) 2 j 2 kmω +cω kmω +cω ( ) T(jω)=Re[ T(jω) ] +ImT(jω) [ ] j Prof. Dr. Y. Samim Ünlüsoy 22
23 FREQUENCY RESPONSE Example 2d Obtain the magnitude and phase of the frequency response function. 2 2 z= a +b T(jω) = ( 2 kmω ) + ( cω) 2 2 = ( 2 ) ( ) ( ) kmω +cω kmω +cω ( ) 1 z=tan 1 b a T(jω)=tan 1 cω ( k mω 2 ) Prof. Dr. Y. Samim Ünlüsoy 23
24 FREQUENCY RESPONSE Example 3a The open loop transfer function of a control system is given as : ( ) 300( s +100) ( )( ) Gs = s s+10 s+40 Determine an expression for the phase angle of G(jw) in terms of the angles of its basic factors. Calculate its value at a frequency of 28.3 rad/s. Determine the expression for the magnitude of G(jw) in terms of the magnitudes of its basic factors. Find its value in db at a frequency of 28.3 rad/s. Prof. Dr. Y. Samim Ünlüsoy 24
25 ( ) ( ) 300 s +100 Gs = FREQUENCY RESPONSE s ( s+10 )( s+40 ) Example 3b G(jω) = 300+ G(jω +100) G(jω) G(jω + 10)  G(jω +40) 1 ω 1 ω 1 ω 1 ω = 0 + tan  tan  tan  tan o 1 ω o 1 ω 1 ω =0 +tan 90 tan tan o o G(28.3j) = 0 + tan tan  tan o o o o o o = = 180 Prof. Dr. Y. Samim Ünlüsoy 25
26 ( ) ( ) 300 s +100 Gs = FREQUENCY RESPONSE s ( s+10 )( s+40 ) Example 3c 300 jω G( jω ) = jω jω +10 jω +40 = ω ω ω +10 ω +40 ( ) G 28.3j = ( 300)( 103.9) ( 28.3)( 30.0)( 49.0) = =0.749 Prof. Dr. Y. Samim Ünlüsoy 26
27 FREQUENCY RESPONSE Typical gain and phase characteristics of a closed loop system. T(jw) T ( jω ) M r ω ω r BW ω Prof. Dr. Y. Samim Ünlüsoy 27
28 FREQUENCY DOMAIN SPECIFICATIONS Similar to transient response specifications in time domain, frequency response specifications are defined.  Resonant peak, M r,  Resonant frequency, ω r,  Bandwidth, BW, Cutoff Rate. Prof. Dr. Y. Samim Ünlüsoy 28
29 FREQUENCY DOMAIN SPECIFICATIONS Resonant peak, M r : This is the maximum value of the transfer function magnitude T(jω). T(jω) M r 1 M r depends on the damping ratio ξ only and indicates the relative stability of a stable closed loop system. A large M r results in a large overshoot of the step response. As a rule of thumb, M r between 1.1 and 1.5. should be M = r ω r 1 2ξ 1ξ 2 ω Prof. Dr. Y. Samim Ünlüsoy 29
30 FREQUENCY DOMAIN SPECIFICATIONS Resonant T(jω) frequency, ω r : M r This is the frequency at which the resonant peak is obtained. ω 2 r =ω n 12ξ 1 ω r ω Note that resonant frequency is different than both the undamped and damped natural frequencies! Prof. Dr. Y. Samim Ünlüsoy 30
31 FREQUENCY DOMAIN SPECIFICATIONS Bandwidth, BW : This is the frequency at which the magnitude of the frequency response function, T(jω), drops to of its zero frequency value. T(jω) M r ω r BW ω BW is directly proportional to ω n and gives an indication of the transient response characteristics of a control system. The larger the bandwidth is, the faster the system responds. Prof. Dr. Y. Samim Ünlüsoy 31
32 FREQUENCY DOMAIN SPECIFICATIONS C S Bandwidth, BW : It is also an indicator of robustness and noise filtering characteristics of a control system. T(jω) M r ω r BW ω ( ) ω = ω 1 2 ξ + 4 ξ 4 ξ + 2 BW n Prof. Dr. Y. Samim Ünlüsoy 32
33 FREQUENCY DOMAIN SPECIFICATIONS Cutoff Rate : This is the slope of the magnitude of the frequency response function, T(jω), at higher (above resonant) frequencies. It indicates the ability of a system to distinguish i signals from noise. T(jω) M r ω r BW Two systems having the same bandwidth can have different cutoff rates. ω Prof. Dr. Y. Samim Ünlüsoy 33
34 BODE PLOT Dorf & Bishop Ch. 8, Ogata Ch. 8 The Bode plot of a transfer function is a useful graphical tool for the analysis and design of linear control systems in the frequency domain. The Bode plot has the advantages that  it can be sketched approximately using straightline segments without using a computer.  relative stability characteristics are easily determined, and  effects of adding controllers and their parameters are easily visualized. Prof. Dr. Y. Samim Ünlüsoy 34
35 BODE PLOT Prof. Dr. Y. Samim Ünlüsoy 35
36 BODE PLOT Nise Section 10.2 The Bode plot consists of two plots drawn on semilogarithmic paper. 1. Magnitude of the frequency response function in decibels, i.e., 20 log T(jω) on a linear scale versus frequency on a logarithmic scale. 2. Phase of the frequency response function on a linear scale versus frequency on a logarithmic scale. Prof. Dr. Y. Samim Ünlüsoy 36
37 BODE PLOT Prof. Dr. Y. Samim Ünlüsoy 37
38 BODE PLOT It is possible to construct the Bode plots of the open loop transfer functions, but the closed loop frequency response is not so easy to plot. It is also possible, however, to obtain the closed loop frequency response from the open loop frequency response. Thus, it is usual to draw the Bode plots of the open loop transfer functions. Then the closed loop frequency response can be evaluated from the open loop Bode plots. Prof. Dr. Y. Samim Ünlüsoy 38
39 BODE PLOT It is possible to construct the Bode plots by adding the contributions of the basic factors of T(jω) by graphical addition. Consider the following general transfer function. T(s)= K P p=1 ( 1+T ) ps M Q 2 N s s s ( ) 1+τms 1+2ξ q + 2 m=1 q=1 ωn q ω nq Prof. Dr. Y. Samim Ünlüsoy 39
40 K P p=1 ( 1+Tp s) T(s)= M Q 2 N s s s ( 1+τm s ) 1+2ξ q + 2 m=1 q=1 ωn q ω nq BODE PLOT The logarithmic magnitude of T(jω) can be obtained by summation of the logarithmic i magnitudes of individual id terms. log T ( jω ) =logk+ P log1+jωτp  p N M Q 2ξq ( ) jω log jω  log1+jωτm  log1+ jω+ m q ω n ω q n q 2 Prof. Dr. Y. Samim Ünlüsoy 40
41 K P p=1 ( 1+Tp s) T(s)= M Q 2 N s s s ( 1+τ m s ) 1+2ξ q + 2 m=1 q=1 ωn q ω nq BODE PLOT Similarly, the phase of T(jω) can be obtained by simple summation of the phases of individual terms. P ( ) ( ) M Q 2ξqωn ω 1 o 11 q φ= T jω = tan ωτp N 90  tan ωτm  tan 2 2 p m q ω  ω nq Prof. Dr. Y. Samim Ünlüsoy 41
42 BODE PLOT Therefore, any transfer function can be constructed from the four basic factors : 1. Gain, K  a constant, 2. Integral, 1/jω, or derivative factor, jω pole or zero at the origin, 3. First order factor simple lag, 1/(1+jωT), or lead 1+jωT (real pole or zero), 4. Quadratic factor quadratic lag or lead ω ω ω ω 1+2ξ j + j or 1+2ξ j + j ωn ωn ωn ωn Prof. Dr. Y. Samim Ünlüsoy 42
43 BODE PLOT Some useful definitions : The magnitude is normally specified in decibels [db]. The value of M in decibels is given by : M[dB]=20logM Frequency ranges may be expressed in terms of decades or octaves. Decade : Frequency band from ω to 10ω. Octave : Frequency band from ω to 2ω. Prof. Dr. Y. Samim Ünlüsoy 43
44 Gain Factor K. BODE PLOT The gain factor multiplies the overall gain by a constant value for all frequencies. It has no effect on phase. G(s)=K G(j ω)=k M=20log 20logG(j ω ) φ =20log(K) [db] = 0 M : magnitude, φ : phase. M[dB] 20logK 0 φ[ o ] 0 ω ω Prof. Dr. Y. Samim Ünlüsoy 44
45 BODE PLOT Integral Factor 1/jω pole at the origin. Magnitude is a straight line with a slope of 20 db/decade becoming zero at ω=1 [rad/s]. Phase is constant at 90 o at all frequencies. G(s)= 1,Gjω ( ) = 1 = 1 j s jω ω 20 0 M=20log G(jω) =20log = 20logω ω φ[ o ] Im o Re φ =90 0 φ 901/ω M[dB] decade ω 20 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 45
46 BODE PLOT Double pole at the origin. 1/ω 2 Simply double the slope of the magnitude and the phase, i.e., 40 db/decade becoming zero at ω=1 ω 1 [rad/s] and 180 o phase. Im φ Re G(s)= 1, G ( jω ) = 1 = 1 s jω ω 2 2 M=20log G(jω) 1 =20log ω φ =180 o 2 ( ) 2 = 40logω φ[ o ] M[dB] decade ω 40 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 46
47 BODE PLOT Derivative Factor jω zero at the origin. Magnitude is a straight line with a slope of 20 db/decade becoming zero at ω=1 [rad/s]. Phase is constant at 90 o at all frequencies. G(s)= s, G( jω ) = ωj M=20log G(jω) ( ) =20log ω o φ = φ[ o ] 90 0 M[dB] decade ω 20 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 47
48 BODE PLOT Double zero at the origin. Simply double the slope of the magnitude and the phase, i.e., 40 db/decade becoming zero at ω=1 ω 1 [rad/s] and 180 o phase. G(s)= s ( ) 2 2, G jω =ω M=20log G(jω) ( ) = 40log ω φ =180 o φ[ o ] M[dB] decade ω 40 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 48
49 BODE PLOT First Order Factor Simple lag (Real pole) 1/(1+jωT). 1 G(s)= 1+Ts 1 1jωT 1 ωt G(jω)= =  j 1+jωT1jωT ω T 1+ω T M=20log G(jω)=20log ω T 2 2 M=20log 1+ω T [db] 11 φ =tan (ωt) =tan ωt Prof. Dr. Y. Samim Ünlüsoy 49
50 BODE PLOT First Order Factor Simple lag (Real Pole) 1/(1+jωT). 2 2 M=20log 1+ω T [db] M[dB] T T T T ω For ω << 1 T 20 M 20log1= 0 [db] 40 For ω >> 1 T M 20logωT [db] Prof. Dr. Y. Samim Ünlüsoy 50
51 BODE PLOT First Order Factor It is clear that the actual magnitude curve can be approximated by two straight lines. M[dB] T T T T ω M 20log1 = 0 [db] M 20logωT [db] 40 For ω<< 1 T For ω >> 1 T Prof. Dr. Y. Samim Ünlüsoy 51
52 BODE PLOT First Order Factor ω c =1/T is called the corner (break) frequency. Maximum error between the linear approximation and the exact value will be at the corner frequency. M[dB] T 1 T 10 T 100 T ω 2 2 M= 20log 1+ω T [db] 1 M ω = = 20log 2 T 3[dB] Prof. Dr. Y. Samim Ünlüsoy 52
53 BODE PLOT First Order Factor ω 0.1ω 0.5ω c c ω c 2ω c 10ω c Error [db] M[dB] T 1 T 10 T 100 T ω Prof. Dr. Y. Samim Ünlüsoy 53
54 BODE PLOT First Order Factor Transfer function G(s)=1/(1+Ts) is a low pass filter. At low frequencies the magnitude ratio is almost one, i.e., the output can follow the input. For higher frequencies, however, the output cannot follow the input because a certain amount of time is required to build up output magnitude (time constant!). Thus, the higher the corner frequency the faster the system response will be. Prof. Dr. Y. Samim Ünlüsoy 54
55 BODE PLOT First Order Factor Simple lag 1/(1+jωT) φ =tan (ωt) =tan ωt 0 φ[ o ] 0.1 T T T T ω For ω<< 0.1 T 45 φ o 0 [ ] For ω>> 10 T 90 φ o 90 [ ] Prof. Dr. Y. Samim Ünlüsoy 55
56 φ BODE PLOT First Order Factor It is clear that the actual phase curve can be approximated by three straight lines. o 0 [ ] φ[ o ] T T T T ω Linear variation in the range ω T T 90 φ o 90 [ ] In this case corner frequencies are : 0.1/T and 10/T Prof. Dr. Y. Samim Ünlüsoy 56
57 BODE PLOT First Order Factor ω 001ω 0.01ω c 01ω 0.1ω c ω c 10ω c 100ω c φ [ o ] Error [ o ] Thus the maximum error of the linear approximation is 5.7 o. Prof. Dr. Y. Samim Ünlüsoy 57
58 BODE PLOT First Order Factor Simple lead (Real zero) 1+jωT. G(s)=1+ Ts G(jω)=1+ωTj M=20log G(jω)=20log ω T 2 2 M=20log 1+ω T [db] ( ) 11 φ =tan ωt =tan ωt Prof. Dr. Y. Samim Ünlüsoy 58
59 BODE PLOT First Order Factor Simple lead (Real zero) 1+jωT. 2 2 M=20log 1+ω T [db] M[dB] For ω<< 1 T 40 M 20log1= 0 [db] 20 1 For ω >> T M 20log ω T [db] ω T 1 T 10 T 100 T Prof. Dr. Y. Samim Ünlüsoy 59
60 BODE PLOT First Order Factor It is clear that the actual magnitude curve can be approximated by two straight lines. M[dB] For ω<< 1 T For ω >> 1 T 40 M 20logω T [db] 20 M 20log1=0 [db] T 1 T 10 T 100 T ω Prof. Dr. Y. Samim Ünlüsoy 60
61 BODE PLOT First Order Factor Simple lead 1+jωT. φ[ o ] ( ) φ =tan ωt φ o 0 [ ] o φ 90 [ ] For For ω<< 0.1 T 10 ω >> T T T T T ω Prof. Dr. Y. Samim Ünlüsoy 61
62 BODE PLOT First Order Factor It is clear that the actual phase curve can be approximated by three straight lines. φ[ o ] 90 φ o 90 [ ] φ o 0 [ ] T 1 10 T T 100 T Linear variation in the range ω T T ω Prof. Dr. Y. Samim Ünlüsoy 62
63 BODE PLOT Quadratic Factors As overdamped systems can be replaced by two first order factors, only underdamped systems are of interest here. G(s)= s 2 ω n 2 +2ξω 2 n s+ω n A set of two complex conjugate poles. G(j ω)= 2 1 ω ω j +2ξ j +1 ω n ω n ω ω M=20log G(jω) =20log 1 ωn + 2ξ ω [db] n Prof. Dr. Y. Samim Ünlüsoy 63
64 BODE PLOT Quadratic Factors ω ω M=20log 20logG(jω) =20log 1 ωn + 2ξ ω [db] n Low frequency asymptote, ω<< <<ω n : ( ) M 20log 1 =0 [db] High frequency asymptote, ω>> >>ω n : 2 ω ω M 20log =40log [db] ωn ωn Low and high frequency asymptotes intersect at ω=ω n, i.e. corner frequency is ω n. Prof. Dr. Y. Samim Ünlüsoy 64
65 BODE PLOT Quadratic Factors Therefore the actual magnitude curve can be approximated by two straight lines. M[dB] LF Asymptote 40dB/decade slope ξ (increasing) HF Asymptote ω n /100 ω n /10 ω n 10ω n 100ω n Frequency Prof. Dr. Y. Samim Ünlüsoy 65
66 BODE PLOT Quadratic Factors φ= G(jω)=tan ω 2ξ ω 1 n ω 1 ω ω n 2 At low frequencies, ω 0 : At ω=ω n : n o φ 90 [ ] At high frequencies, ω : o φ 0[ ] o φ 180 [ ] Prof. Dr. Y. Samim Ünlüsoy 66
67 BODE PLOT Quadratic Factors Thus, the actual phase curve can be approximated by three straight lines. 0 φ[ o ] ξ (increasing) o /decade slope 180 ω n /10 ω n 10ω n 100ω n Frequency Corner frequencies are : ω n /10 and 10ω n. Prof. Dr. Y. Samim Ünlüsoy 67
68 BODE PLOT Quadratic Factors It is observed that, the linear approximations for the magnitude and phase will give more accurate results for damping ratios closer to 1.0. The peak magnitude is given by : The resonant frequency : Μ r = 1 2ξ 1 ξ ω 2 r =ω n 12ξ 2 Prof. Dr. Y. Samim Ünlüsoy 68
69 BODE PLOT Quadratic Factors For ξ=0.707 : M r r =1 (or M=20log1=0 db). Thus, there will be no peak on the magnitude plot. Note the difference that in transient response for step input, there will be no overshoot for critically or overdamped systems, i.e., for ξ 1.0. Prof. Dr. Y. Samim Ünlüsoy 69
70 BODE PLOT Example 1a Sketch the Bode plots for the given open loop transfer function of a control system ( 1+s) T(s)= 2 ( )( ) ss+10s s +14s+1000 First convert to standard form ( 1+ jω) T(jω)= ( )( )( )( ) 2 ω ω jω ωj 1000 j +1.4 j +1 T(jω)= ( 1+ jω) ( )( ) 2 ω ω jω 1+0.1ωj j +1.4 j Prof. Dr. Y. Samim Ünlüsoy 70
71 BODE PLOT Example 1b T(jω)= ( ) jω ω ω ( jω)( 1+0.1ωj1ωj ) j +1.4 j +1 Identify the basic factors and corner frequencies : Constant gain K : K=10, 20log10=20 [db] First order factor (simple lead real zero) : T=1 (ω c1 =1/T=1)  for magnitude plot Integral factor : 1/jω First order factor (simple lag real pole) : T=0.1 (ω c1 =1/T=10)  for magnitude plot Quadratic factor (complex conjugate poles) : ω n =ω c1 =100, ξ=0.7  for magnitude plot Prof. Dr. Y. Samim Ünlüsoy 71
72 BODE PLOT Example 1c T(jω)= ( ) jω ω ω ( jω)( 1+0.1ωj1ωj ) j +1.4 j +1 Identify the basic factors and corner frequencies : Constant gain K : K=10, 20log10=20 [db] First order factor (simple lead real zero) : T=1 (ω c2 =0.1/T=0.1, ω c3 =10/T=10) for phase plot Integral factor : 1/jω First order factor (simple lag real pole) : T=0.1 (ω c2 =0.1/T=1, ω c3 =10/T=100) for phase plot Quadratic factor (complex conjugate poles) : ω n =100 10ω n =1000) for phase plot (ω c2 =ω n /10=10, ω c3 =10 Prof. Dr. Y. Samim Ünlüsoy 72
73 BODE PLOT Example 1d 40 M[dB] 1+jω K=10 ω[rad/s] 20 Quadratic factor 40 1/(1+0.1jω) 60 1/jω Bode (magnitude) plot Prof. Dr. Y. Samim Ünlüsoy 73
74 BODE PLOT Example 1e 90 φ[ o ] 1+jω K=10 ω 1/(1+0.1jω) 1/jω Quadratic factor 270 Bode (phase) plot Prof. Dr. Y. Samim Ünlüsoy 74
75 Matlab plot: 60 full blue lines 40 (just 4 lines to plot!) num=[ ] den=[ ] bode(num,den) grid Approximate plots: dashed lines BODE PLOT Example 1f Magnit tude (db) Phase (deg) Bode Diagram 270 lines Frequency Prof. Dr. Y. Samim Ünlüsoy 75
76 STABILITY ANALYSIS Nise Sect. 10.7, pp Transfer functions which have no poles or zeroes on the right hand side of the complex plane are called minimum phase transfer funtions. Nonminimum phase transfer functions, on the other hand, have zeros and/or poles on the right hand side of the complex plane. The major disadvantage of Bode Plot is that stability of only minimum phase systems can be determined using Bode plot. Prof. Dr. Y. Samim Ünlüsoy 76
77 STABILITY ANALYSIS From the characteristic equation : 1 + G(s)H(s) = 0 or G(s)H(s)= 1 Then the magnitude and phase for the open loop transfer function become : 20log G(jω)H(jω) =20log1=0dB G(jω)H(jω)=180 Thus, when the magnitude and the phase angle of a transfer function are 0 db and 180 o, respectively, then the system is marginally stable. o Prof. Dr. Y. Samim Ünlüsoy 77
78 STABILITY ANALSIS If at the frequency, for which phase becomes equal to 180 o, gain is below 0 db, then the system is stable (unstable otherwise). Further, if at the frequency, for which gain becomes equal to zero, phase is above 180 o, then the system is stable (unstable otherwise). Thus, relative stability of a minimum phase system can be determined according to these observations. Prof. Dr. Y. Samim Ünlüsoy 78
79 GAIN and PHASE MARGINS Nise pp Gain Margin : Additional gain to make the system marginally stable at a frequency for which the phase of the open loop transfer function passes through 180 o. Phase Margin : Additional phase angle to make the system marginally stable at a frequency for which the magnitude of the open loop transfer function is 0 db. Prof. Dr. Y. Samim Ünlüsoy 79
80 GAIN and PHASE MARGINS 50 Mag gnitude (db) Gain Margin Phase (deg) Phase Margin Frequency (rad/sec) Prof. Dr. Y. Samim Ünlüsoy 80
81 BODE PLOT Can you identify the transfer function approximately if the measured Bode diagram is available? Prof. Dr. Y. Samim Ünlüsoy 81
ROOT 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 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 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 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 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 information8.1.6 Quadratic pole response: resonance
8.1.6 Quadratic pole response: resonance Example G(s)= v (s) v 1 (s) = 1 1+s L R + s LC L + Secondorder denominator, of the form 1+a 1 s + a s v 1 (s) + C R Twopole lowpass filter example v (s) with
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 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 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 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 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 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 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 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 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 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 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 informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall 2007
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering.4 Dynamics and Control II Fall 7 Problem Set #9 Solution Posted: Sunday, Dec., 7. The.4 Tower system. The system parameters are
More informationStep Response for the Transfer Function of a Sensor
Step Response f the Transfer Function of a Sens G(s)=Y(s)/X(s) of a sens with X(s) input and Y(s) output A) First Order Instruments a) First der transfer function G(s)=k/(1+Ts), k=gain, T = time constant
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 informationFrequency Response of Linear Time Invariant Systems
ME 328, Spring 203, Prof. Rajamani, University of Minnesota Frequency Response of Linear Time Invariant Systems Complex Numbers: Recall that every complex number has a magnitude and a phase. Example: z
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 informationR. W. Erickson. Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder
R. W. Erickson Department of Electrical, Computer, and Energy Engineering University of Colorado, Boulder 8.1. Review of Bode plots Decibels Table 8.1. Expressing magnitudes in decibels G db = 0 log 10
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 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 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 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 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 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 informationDr Ian R. Manchester Dr Ian R. Manchester AMME 3500 : Review
Week Date Content Notes 1 6 Mar Introduction 2 13 Mar Frequency Domain Modelling 3 20 Mar Transient Performance and the splane 4 27 Mar Block Diagrams Assign 1 Due 5 3 Apr Feedback System Characteristics
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 informationAPPLICATIONS FOR ROBOTICS
Version: 1 CONTROL APPLICATIONS FOR ROBOTICS TEX d: Feb. 17, 214 PREVIEW We show that the transfer function and conditions of stability for linear systems can be studied using Laplace transforms. Table
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 informationECE382/ME482 Spring 2005 Homework 6 Solution April 17, (s/2 + 1) s(2s + 1)[(s/8) 2 + (s/20) + 1]
ECE382/ME482 Spring 25 Homework 6 Solution April 17, 25 1 Solution to HW6 P8.17 We are given a system with open loop transfer function G(s) = 4(s/2 + 1) s(2s + 1)[(s/8) 2 + (s/2) + 1] (1) and unity negative
More informationResponse to a pure sinusoid
Harvard University Division of Engineering and Applied Sciences ES 145/215  INTRODUCTION TO SYSTEMS ANALYSIS WITH PHYSIOLOGICAL APPLICATIONS Fall Lecture 14: The Bode Plot Response to a pure sinusoid
More informationSolutions to SkillAssessment Exercises
Solutions to SkillAssessment Exercises To Accompany Control Systems Engineering 4 th Edition By Norman S. Nise John Wiley & Sons Copyright 2004 by John Wiley & Sons, Inc. All rights reserved. No part
More informationMASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering Dynamics and Control II Fall K(s +1)(s +2) G(s) =.
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering. Dynamics and Control II Fall 7 Problem Set #7 Solution Posted: Friday, Nov., 7. Nise problem 5 from chapter 8, page 76. Answer:
More informationFrequency Response. Re ve jφ e jωt ( ) where v is the amplitude and φ is the phase of the sinusoidal signal v(t). ve jφ
27 Frequency Response Before starting, review phasor analysis, Bode plots... Key concept: smallsignal models for amplifiers are linear and therefore, cosines and sines are solutions of the linear differential
More informationThe requirements of a plant may be expressed in terms of (a) settling time (b) damping ratio (c) peak overshoot  in time domain
Compensators To improve the performance of a given plant or system G f(s) it may be necessary to use a compensator or controller G c(s). Compensator Plant G c (s) G f (s) The requirements of a plant may
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 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 informationPerformance of Feedback Control Systems
Performance of Feedback Control Systems Design of a PID Controller Transient Response of a Closed Loop System Damping Coefficient, Natural frequency, Settling time and Steadystate Error and Type 0, Type
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 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 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 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 informationControl for. Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e
Control for Maarten Steinbuch Dept. Mechanical Engineering Control Systems Technology Group TU/e Motion Systems m F Introduction Timedomain tuning Frequency domain & stability Filters Feedforward Servooriented
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 informationAutomatic Control 2. Loop shaping. Prof. Alberto Bemporad. University of Trento. Academic year
Automatic Control 2 Loop shaping Prof. Alberto Bemporad University of Trento Academic year 21211 Prof. Alberto Bemporad (University of Trento) Automatic Control 2 Academic year 21211 1 / 39 Feedback
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 informationIntroduction to Root Locus. What is root locus?
Introduction to Root Locus What is root locus? A graphical representation of the closed loop poles as a system parameter (Gain K) is varied Method of analysis and design for stability and transient response
More information16.30/31, Fall 2010 Recitation # 2
16.30/31, Fall 2010 Recitation # 2 September 22, 2010 In this recitation, we will consider two problems from Chapter 8 of the Van de Vegte book. R +  E G c (s) G(s) C Figure 1: The standard block diagram
More informationPoles and Zeros and Transfer Functions
Poles and Zeros and Transfer Functions Transfer Function: Considerations: Factorization: A transfer function is defined as the ratio of the Laplace transform of the output to the input with all initial
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 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 informationStep Response Analysis. Frequency Response, Relation Between Model Descriptions
Step Response Analysis. Frequency Response, Relation Between Model Descriptions Automatic Control, Basic Course, Lecture 3 November 9, 27 Lund University, Department of Automatic Control Content. Step
More informationAsymptotic Bode Plot & LeadLag Compensator
Asymptotic Bode Plot & LeadLag Compensator. Introduction Consider a general transfer function Ang Man Shun 20225 G(s = n k=0 a ks k m k=0 b ks k = A n k=0 (s z k m k=0 (s p k m > n When s =, transfer
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 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 informationSinusoidal Forcing of a FirstOrder Process. / τ
Frequency Response Analysis Chapter 3 Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A
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 informationRoot Locus Methods. The root locus procedure
Root Locus Methods Design of a position control system using the root locus method Design of a phase lag compensator using the root locus method The root locus procedure To determine the value of the gain
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 information( ) Frequency Response Analysis. Sinusoidal Forcing of a FirstOrder Process. Chapter 13. ( ) sin ω () (
1 Frequency Response Analysis Sinusoidal Forcing of a FirstOrder Process For a firstorder transfer function with gain K and time constant τ, the response to a general sinusoidal input, xt = A tis: sin
More informationFrequency Response DR. GYURCSEK ISTVÁN
DR. GYURCSEK ISTVÁN Frequency Response Sources and additional materials (recommended) Dr. Gyurcsek Dr. Elmer: Theories in Electric Circuits, GlobeEdit, 2016, ISBN:9783330713413 Ch. Alexander, M. Sadiku:
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 informationStability of CL System
Stability of CL System Consider an open loop stable system that becomes unstable with large gain: At the point of instability, K( j) G( j) = 1 0dB K( j) G( j) K( j) G( j) K( j) G( j) =± 180 o 180 o Closed
More informationSingleTimeConstant (STC) Circuits This lecture is given as a background that will be needed to determine the frequency response of the amplifiers.
SingleTimeConstant (STC) Circuits This lecture is given as a background that will be needed to determine the frequency response of the amplifiers. Objectives To analyze and understand STC circuits with
More informationLab 3: Poles, Zeros, and Time/Frequency Domain Response
ECEN 33 Linear Systems Spring 2 2 P. Mathys Lab 3: Poles, Zeros, and Time/Frequency Domain Response of CT Systems Introduction Systems that are used for signal processing are very often characterized
More information2.010 Fall 2000 Solution of Homework Assignment 1
2. Fall 2 Solution of Homework Assignment. Compact Disk Player. This is essentially a reprise of Problems and 2 from the Fall 999 2.3 Homework Assignment 7. t is included here to encourage you to review
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Illinois Institute of Technology Lecture 2: Drawing Bode Plots, Part 2 Overview In this Lecture, you will learn: Simple Plots Real Zeros Real Poles Complex
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 information(a) Find the transfer function of the amplifier. Ans.: G(s) =
126 INTRDUCTIN T CNTR ENGINEERING 10( s 1) (a) Find the transfer function of the amplifier. Ans.: (. 02s 1)(. 001s 1) (b) Find the expected percent overshoot for a step input for the closedloop system
More informationAppendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2)
Appendix A: Exercise Problems on Classical Feedback Control Theory (Chaps. 1 and 2) For all calculations in this book, you can use the MathCad software or any other mathematical software that you are familiar
More informationEE3CL4: Introduction to Linear Control Systems
1 / 30 EE3CL4: Introduction to Linear Control Systems Section 9: of and using Techniques McMaster University Winter 2017 2 / 30 Outline 1 2 3 4 / 30 domain analysis Analyze closed loop using open loop
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 Deartment, Middle East Technical University Armature controlled dc motor Outside θ D outut Inside θ r inut r θ m Gearbox Control Transmitter θ D
More informationSchool of Mechanical Engineering Purdue University
Case Study ME375 Frequency Response  1 Case Study SUPPORT POWER WIRE DROPPERS Electric train derives power through a pantograph, which contacts the power wire, which is suspended from a catenary. During
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 informationLecture 7:Time Response PoleZero Maps Influence of Poles and Zeros Higher Order Systems and Pole Dominance Criterion
Cleveland State University MCE441: Intr. Linear Control Lecture 7:Time Influence of Poles and Zeros Higher Order and Pole Criterion Prof. Richter 1 / 26 FirstOrder Specs: Step : Pole Real inputs contain
More informationControl of Manufacturing Processes
Control of Manufacturing Processes Subject 2.830 Spring 2004 Lecture #19 Position Control and Root Locus Analysis" April 22, 2004 The Position Servo Problem, reference position NC Control Robots Injection
More informationToday (10/23/01) Today. Reading Assignment: 6.3. Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10
Today Today (10/23/01) Gain/phase margin lead/lag compensator Ref. 6.4, 6.7, 6.10 Reading Assignment: 6.3 Last Time In the last lecture, we discussed control design through shaping of the loop gain GK:
More informationChapter 2. Classical Control System Design. Dutch Institute of Systems and Control
Chapter 2 Classical Control System Design Overview Ch. 2. 2. Classical control system design Introduction Introduction Steadystate Steadystate errors errors Type Type k k systems systems Integral Integral
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 informationINSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad
INSTITUTE OF AERONAUTICAL ENGINEERING (Autonomous) Dundigal, Hyderabad  500 043 Electrical and Electronics Engineering TUTORIAL QUESTION BAN Course Name : CONTROL SYSTEMS Course Code : A502 Class : III
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 informationBoise State University Department of Electrical Engineering ECE461 Control Systems. Control System Design in the Frequency Domain
Boise State University Department of Electrical Engineering ECE6 Control Systems Control System Design in the Frequency Domain Situation: Consider the following block diagram of a type servomechanism:
More informationLearn2Control Laboratory
Learn2Control Laboratory Version 3.2 Summer Term 2014 1 This Script is for use in the scope of the Process Control lab. It is in no way claimed to be in any scientific way complete or unique. Errors should
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 informationStudio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter
Studio Exercise Time Response & Frequency Response 1 st Order Dynamic System RC LowPass Filter i i in R out Assignment: Perform a Complete e in C e Dynamic System Investigation out of the RC LowPass
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 informationagree w/input bond => + sign disagree w/input bond =>  sign
1 ME 344 REVIEW FOR FINAL EXAM LOCATION: CPE 2.204 M. D. BRYANT DATE: Wednesday, May 7, 2008 9noon Finals week office hours: May 6, 47 pm Permitted at final exam: 1 sheet of formulas & calculator I.
More informationChapter 8: Frequency Domain Analysis
Chapter 8: Frequency Domain Analysis Samantha Ramirez Preview Questions 1. What is the steadystate response of a linear system excited by a cyclic or oscillatory input? 2. How does one characterize the
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 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 informationSteady State Frequency Response Using Bode Plots
School of Engineering Department of Electrical and Computer Engineering 332:224 Principles of Electrical Engineering II Laboratory Experiment 3 Steady State Frequency Response Using Bode Plots 1 Introduction
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 informationUsing MATLB for stability analysis in Controls engineering Cyrus Hagigat Ph.D., PE College of Engineering University of Toledo, Toledo, Ohio
Using MATLB for stability analysis in Controls engineering Cyrus Hagigat Ph.D., PE College of Engineering University of Toledo, Toledo, Ohio Abstract Analyses of control systems require solution of differential
More informationNotes on the Periodically Forced Harmonic Oscillator
Notes on the Periodically orced Harmonic Oscillator Warren Weckesser Math 38  Differential Equations 1 The Periodically orced Harmonic Oscillator. By periodically forced harmonic oscillator, we mean the
More informationSystems Analysis and Control
Systems Analysis and Control Matthew M. Peet Arizona State University Lecture 24: Compensation in the Frequency Domain Overview In this Lecture, you will learn: Lead Compensators Performance Specs Altering
More informationRaktim Bhattacharya. . AERO 632: Design of Advance Flight Control System. Preliminaries
. AERO 632: of Advance Flight Control System. Preliminaries Raktim Bhattacharya Laboratory For Uncertainty Quantification Aerospace Engineering, Texas A&M University. Preliminaries Signals & Systems Laplace
More information