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
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
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
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
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
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
FREQUENCY RESPONSE - INTRODUCTION Prof. Dr. Y. Samim Ünlüsoy 7
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
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
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
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
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
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
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 a-bj = a +b the phase is given by : -1 b z z = tan a Prof. Dr. Y. Samim Ünlüsoy 14
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
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
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+2-0.25+ j ( ) Prof. Dr. Y. Samim Ünlüsoy 17
FREQUENCY RESPONSE Example 1c Multiply and divide by the complex conjugate. 1-0.25 - j -0.25 - j T(0.5j)= = -0.25+ j -0.25 - j 1+0.0625 T(0.5j)= -0.235-0.941j Determine the magnitude and the angle. cos - cos + 2 2 ( ) ( ) T(0.5j) = -0.235 + -0.941 = 0.97-0.941 T(0.5j)= tan = -104-0.235 sin + sin + -1 o cos - sin - cos + sin - 76 o -104 o Prof. Dr. Y. Samim Ünlüsoy 18
FREQUENCY RESPONSE Example 1d The steady state response is then given by : o ( ) ( ) x (t)=3097sin05t-104 0.97 0.5t ss ( o ) =2.91sin 0.5t -104 Prof. Dr. Y. Samim Ünlüsoy 19
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
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 k-mω +cωj 2 ( ) ( ) ( 2 ) Write the FRF in a+bj form. Prof. Dr. Y. Samim Ünlüsoy 21
FREQUENCY RESPONSE Example 2c Multiply and divide the FRF expression with the complex conjugate of its denominator. ( ) ( ) ( ) ( ) 2 2 1 k-mω -cωj k-mω -cωj T(jω)= = k-mω +cωj k-mω -cωj k-mω +cω 2 2 2 2 ( 2 ) ( ) ( k-mω 2 ) -cω T(jω)= ( ) 2 + 2 2 ( ) ( 2) 2 j 2 k-mω +cω k-mω +cω ( ) T(jω)=Re[ T(jω) ] +ImT(jω) [ ] j Prof. Dr. Y. Samim Ünlüsoy 22
FREQUENCY RESPONSE Example 2d Obtain the magnitude and phase of the frequency response function. 2 2 z= a +b T(jω) = ( 2 k-mω ) + ( cω) 2 2 = 2 2 2 2 2 2 ( 2 ) ( ) ( ) k-mω +cω k-mω +cω ( ) 1 z=tan -1 b a T(jω)=tan -1 -cω ( k -mω 2 ) Prof. Dr. Y. Samim Ünlüsoy 23
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
( ) ( ) 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 100 0 10 40 o -1 ω o -1 ω -1 ω =0 +tan -90 -tan -tan 100 10 40 o -1 28.3 o -1 28.3-1 28.3 G(28.3j) = 0 + tan - 90 - tan - tan 100 10 40 o o o o o o = 0 + 15.8-90 - 70.5-35.3 = -180 Prof. Dr. Y. Samim Ünlüsoy 25
( ) ( ) 300 s +100 Gs = FREQUENCY RESPONSE s ( s+10 )( s+40 ) Example 3c 300 jω + 100 G( jω ) = jω jω +10 jω +40 = 2 2 300 ω +100 2 2 2 2 ω ω +10 ω +40 ( ) G 28.3j = 2 2 300 28.3 + 100 2 2 2 2 28.3 28.3 + 10 28.3 + 40 ( 300)( 103.9) ( 28.3)( 30.0)( 49.0) = =0.749 Prof. Dr. Y. Samim Ünlüsoy 26
FREQUENCY RESPONSE Typical gain and phase characteristics of a closed loop system. T(jw) T ( jω ) M r 1 0.707 0 ω ω r BW ω Prof. Dr. Y. Samim Ünlüsoy 27
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
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 1-ξ 2 ω Prof. Dr. Y. Samim Ünlüsoy 29
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 1-2ξ 1 ω r ω Note that resonant frequency is different than both the undamped and damped natural frequencies! Prof. Dr. Y. Samim Ünlüsoy 30
FREQUENCY DOMAIN SPECIFICATIONS Bandwidth, BW : This is the frequency at which the magnitude of the frequency response function, T(jω), drops to 0.707 of its zero frequency value. T(jω) M r 1 0.707 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
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 1 0.707 ω r BW ω ( ) 2 4 2 ω = ω 1 2 ξ + 4 ξ 4 ξ + 2 BW n Prof. Dr. Y. Samim Ünlüsoy 32
FREQUENCY DOMAIN SPECIFICATIONS Cut-off 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 1 0.707 ω r BW Two systems having the same bandwidth can have different cutoff rates. ω Prof. Dr. Y. Samim Ünlüsoy 33
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
BODE PLOT Prof. Dr. Y. Samim Ünlüsoy 35
BODE PLOT Nise Section 10.2 The Bode plot consists of two plots drawn on semi-logarithmic 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
BODE PLOT Prof. Dr. Y. Samim Ünlüsoy 37
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
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
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
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 -1-1 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
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. 1 2 2 ω ω ω ω 1+2ξ j + j or 1+2ξ j + j ωn ωn ωn ωn Prof. Dr. Y. Samim Ünlüsoy 42
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
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
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) -20 1 =20log = -20logω ω φ[ o ] Im o Re φ =-90 0 φ -90-1/ω M[dB] decade 0.1 1 10 ω -20 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 45
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 40 0-40 φ[ o ] 0-180 M[dB] decade 0.1 1 10 ω -40 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 46
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 =90 20 0-20 φ[ o ] 90 0 M[dB] decade 0.1 1 10 ω 20 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 47
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 40 0-40 φ[ o ] 180 0 M[dB] decade 0.1 1 10 ω 40 db/decade slope ω Prof. Dr. Y. Samim Ünlüsoy 48
BODE PLOT First Order Factor Simple lag (Real pole) 1/(1+jωT). 1 G(s)= 1+Ts 1 1-jωT 1 ωt G(jω)= = - j 1+jωT1-jωT 2 2 2 2 1+ω T 1+ω T M=20log G(jω)=20log 1 2 2 1+ω T 2 2 M=-20log 1+ω T [db] -1-1 φ =tan (-ωt) =-tan ωt Prof. Dr. Y. Samim Ünlüsoy 49
BODE PLOT First Order Factor Simple lag (Real Pole) 1/(1+jωT). 2 2 M=-20log 1+ω T [db] M[dB] 0 0.1 1 10 100 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
BODE PLOT First Order Factor It is clear that the actual magnitude curve can be approximated by two straight lines. M[dB] 0 0.1 1 10 100 T T T T ω -3-20 M -20log1 = 0 [db] M -20logωT [db] -40 For ω<< 1 T For ω >> 1 T Prof. Dr. Y. Samim Ünlüsoy 51
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] -3-20 -40 0 0.1 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
BODE PLOT First Order Factor ω 0.1ω 0.5ω c c ω c 2ω c 10ω c Error [db] 0.04 1 3 1 0.04 M[dB] 0 0.1 T 1 T 10 T 100 T ω -3-20 -40 Prof. Dr. Y. Samim Ünlüsoy 53
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
BODE PLOT First Order Factor Simple lag 1/(1+jωT). -1-1 φ =tan (-ωt) =-tan ωt 0 φ[ o ] 0.1 T 1 10 100 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
φ BODE PLOT First Order Factor It is clear that the actual phase curve can be approximated by three straight lines. o 0 [ ] -5.7-45 φ[ o ] 0 0.1 T 1 10 100 T T T ω Linear variation in the range 0.1 10 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
BODE PLOT First Order Factor ω 001ω 0.01ω c 01ω 0.1ω c ω c 10ω c 100ω c φ [ o ] -0.57-5.7-45 -84.3-89.4 Error [ o ] 0.6 5.7 0-5.7-0.6 Thus the maximum error of the linear approximation is 5.7 o. Prof. Dr. Y. Samim Ünlüsoy 57
BODE PLOT First Order Factor Simple lead (Real zero) 1+jωT. G(s)=1+ Ts G(jω)=1+ωTj M=20log G(jω)=20log 2 2 1+ω T 2 2 M=20log 1+ω T [db] ( ) -1-1 φ =tan ωt =tan ωt Prof. Dr. Y. Samim Ünlüsoy 58
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 0 0.1 T 1 T 10 T 100 T Prof. Dr. Y. Samim Ünlüsoy 59
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] 0 0.1 T 1 T 10 T 100 T ω Prof. Dr. Y. Samim Ünlüsoy 60
BODE PLOT First Order Factor Simple lead 1+jωT. φ[ o ] 90 45-1 ( ) φ =tan ωt φ o 0 [ ] o φ 90 [ ] For For ω<< 0.1 T 10 ω >> T 0 01 0.1 T 1 10 100 T T T ω Prof. Dr. Y. Samim Ünlüsoy 61
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 45 5.7 o 0 [ ] 0 0.1 T 1 10 T T 100 T Linear variation in the range 0.1 10 ω T T ω Prof. Dr. Y. Samim Ünlüsoy 62
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 2 2 2 ω ω M=20log G(jω) =-20log 1- ωn + 2ξ ω [db] n Prof. Dr. Y. Samim Ünlüsoy 63
BODE PLOT Quadratic Factors 2 2 2 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
BODE PLOT Quadratic Factors Therefore the actual magnitude curve can be approximated by two straight lines. M[dB] 20 0-20 -40-60 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
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
BODE PLOT Quadratic Factors Thus, the actual phase curve can be approximated by three straight lines. 0 φ[ o ] ξ (increasing) -90-90 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
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 1-2ξ 2 Prof. Dr. Y. Samim Ünlüsoy 68
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
BODE PLOT Example 1a Sketch the Bode plots for the given open loop transfer function of a control system. 100000( 1+s) T(s)= 2 ( )( ) ss+10s+10 0.1s +14s+1000 First convert to standard form. 100000( 1+ jω) T(jω)= ( )( )( )( ) 2 ω ω jω 10 1+0.1ωj 1000 j +1.4 j +1 T(jω)= 100 100 10( 1+ jω) ( )( ) 2 ω ω jω 1+0.1ωj j +1.4 j +1 100 100 Prof. Dr. Y. Samim Ünlüsoy 70
BODE PLOT Example 1b T(jω)= ( ) 10 1+ jω ω ω ( jω)( 1+0.1ωj1ωj ) j +1.4 j +1 Identify the basic factors and corner frequencies : 2 100 100 -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
BODE PLOT Example 1c T(jω)= ( ) 10 1+ jω ω ω ( jω)( 1+0.1ωj1ωj ) j +1.4 j +1 Identify the basic factors and corner frequencies : 2 100 100 -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
BODE PLOT Example 1d 40 M[dB] 1+jω 20 0 1 10 100 1000 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
BODE PLOT Example 1e 90 φ[ o ] 1+jω 0-90 -180 1 10 100 1000 K=10 ω 1/(1+0.1jω) 1/jω Quadratic factor -270 Bode (phase) plot Prof. Dr. Y. Samim Ünlüsoy 74
Matlab plot: 60 full blue lines 40 (just 4 lines to plot!) num=[100000 100000] den=[0.1 15 1140 10000 0] bode(num,den) grid Approximate plots: dashed lines BODE PLOT Example 1f Magnit tude (db) Phase (deg) 20 0-20 -40-60 0-90 -180 Bode Diagram -270 lines 10-1 10 0 10 1 10 2 10 3 Frequency Prof. Dr. Y. Samim Ünlüsoy 75
STABILITY ANALYSIS Nise Sect. 10.7, pp.638 638-641 641 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
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
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
GAIN and PHASE MARGINS Nise pp. 638-641 641 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
GAIN and PHASE MARGINS 50 Mag gnitude (db) 0-50 -100 Gain Margin -150-90 Phase (deg) -135-180 -225 Phase Margin -270 10-1 10 0 10 1 10 2 10 3 Frequency (rad/sec) Prof. Dr. Y. Samim Ünlüsoy 80
BODE PLOT Can you identify the transfer function approximately if the measured Bode diagram is available? Prof. Dr. Y. Samim Ünlüsoy 81