ME 304 CONTROL SYSTEMS Radar Dish Mechanical Engineering Department, Middle East Technical University
Armature Inside Outside controlled θr input dc motor θD output
θm
Gearbox Control Transmitter
θD
dc amplifier Control Transformer Prof. Dr . Y . Samim Ünlüsoy
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 1 CH IX COURSE OUTLINE 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
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 2 FREQUENCY RESPONSE -OBJECTIVES
In this chapter :
A short introduction to the steady state resppyonse of control systems to sinusoidal inputs will be given.
Frequency domain specifications for a control system will be examined.
Bode plots and their construction usinggyp asymptotic a pproximations will be presented.
ME 304 CONTROL SYSTEMS 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 freqqyuency 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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 4 FREQUENCY RESPONSE - INTRODUCTION
For an LTI system the Laplace transforms of the input and out put are rel at ed to each other by the transfer function, T(s).
Laplace Domain R(s) C(s) Input T(s) Output
In the frequency response anal ysi s, the system is excited by a sinusoidal input of fixed amplitude and varying frequency.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 5 FREQUENCY RESPONSE - INTRODUCTION
Let us subjjyect a stable LTI system to a sinusoidal input of amplitude R and freqqyuency 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+φ)
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 6 FREQUENCY RESPONSE - INTRODUCTION
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 7 FREQUENCY RESPONSE - INTRODUCTION
To carryyp out the same process in the frequency domain for sinusoidal steady state analysis, one replaces the Laplace varibliable s wihith s=jω in the input output relation C(s)= T(s)R(s) with the result C(jω))T(j=T(jω)R(jω)
ME 304 CONTROL SYSTEMS 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(j∠ ω)
Output : C(jω))C(j= C(jω)C(j∠ ω)
Transfer T(jω))(j=T(jω) ∠Τ(jω) Function :
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 9 FREQUENCY RESPONSE - INTRODUCTION
With similar expressions for the input and the trans fer func tion, the inpu t output relation in the frequency domain consists of the magnitude and phase expressions : C(jω)=T(jω)R(jω)
C(jω))(j=T(jω)(j) R(jω)
∠∠∠(C(jω)= T(jω)+ R jω)
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 10 FREQUENCY RESPONSE - INTRODUCTION
For the input and output described by
r(t)=Rsin(ωt)
c(t)=Csin(ωt+φ) the amplitude and the phase of the output can now be written as
C=R T(jω) φφ∠= ∠T(jω)
ME 304 CONTROL SYSTEMS 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(jω ) G(jω ) T(jω )== R(jω ) 1+G(jωω )H(j )
T(jω) is called the Frequency Response Function (FRF) or Sinusoidal Transfer Function.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 12 FREQUENCY RESPONSE
The frequency response function can be written in terms of its magnitude and phase. T(j)T(j)T(j)T(jω )= T(j)T(j)ω ∠ ω
Since this function is compp,lex, it can also be written in terms of its real and imaginary parts.
T(jω )=Re[ T(jωω )] + jIm[ T(j )]
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 13 FREQUENCY RESPONSE
Remem ber that for a compl ex numb er be expressed in its real and imaginary parts : z=a+bj the magnitude is given by :
z=()() a+bj a-bj= a22 +b
the phase is given by : b ∠zztan= tan-1 a
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 14 FREQUENCY RESPONSE
The magnitude and phase of the frequency response function are given by :
G(jω ) G(jω ) T(jω ) = = 1+G(j)H(j)ω ωωω 1+G(j)H(j)
∠ω∠ω∠T(j )= G(j )- [ 1+G(j1+G(j)H(j) ω )H(j ω )]
These are called the gain and phase characteristics.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 15 FREQUENCY RESPONSE – Example 1a
For a system described by the differen tia l equation x+2x=y(t)
determine the steady state response xss(t) for a pure sine wave input
y(t)=3sin(0.5t)
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 16 FREQUENCY RESPONSE – Example 1b
The transfer function is given by X(s) 1 T(s)= = x+2x=y(t) Y(s) s() s+2
Insert s=jω to get : 1 T(jω)= jω()jω+2 For ω=0[d/]0.5 [rad/s]: 11 T(0.5j)= = 0.5j( 0.5j+2) -0.25+ j
ME 304 CONTROL SYSTEMS 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+ jjj-0.25 - j 1+0.0625 T(0.5j)= -0.235 -0.941j
Determine the magnitude and the angle.
22 T(0.5j) =( -0.235) +( -0.941) =0.97
cos - cos + -0.941 sin + sin + ∠T(0.5j) = t an-1o = -104 cos - cos + -0.235 sin - sin - 76o -104o
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 18 FREQUENCY RESPONSE – Example 1d
The steady state response is then given by :
o xx(t)=30ss (t)=3097sin05t( .97) sin( 0.5t- 104 ) o =2.91sin( 0.5t -104 )
ME 304 CONTROL SYSTEMS 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 c y 1 m G(s)= 2 F ms +cs+k
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 20 FREQUENCY RESPONSE – Example 2b
Insert s=jω in the transfer function to obtain the frequency response function.
1 G(s)= ms2 +cs+k 11 T(jω)= = 2 2 m(ωj+) c(ωj+k) ( k-mω )+cωj
Wr ite the FRF in a+bj form.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 21 FREQUENCY RESPONSE – Example 2c
Multiply and divide the FRF expression with the complex conjugate of its denominator.
22 1 (k-mω )-cωjk-m( ω )-cωj T(jω)= = 2 k-mω22+cωjk-mω -cωj 2 2 ( ) ( ) (k-mω ) +c( ω)
⎡⎤⎡⎤2 ⎢⎥⎢⎥(k-mω ) -cω T(jω)=+j⎢⎥⎢⎥ 2222 ⎢⎥⎢⎥k-mω22+c()ω k-mω +c()ω ⎣⎦⎣⎦⎢⎥⎢⎥()()
T(jω)=Re[ T(jω)+ImT(j] [ ω)j]
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 22 FREQUENCY RESPONSE – Example 2d
Obtain the magnitude and phase of the frequency response function. z= a22 +b
2 2 2 ()k-mω + ()cω 1 T(jω)= = 2 2 2 ⎡⎤2 2 2 2 ⎢⎥()k-mω +c()ω (k-mω ) +c()ω ⎣⎦
-1 -cω -1 b ∠T(jω)=tan ∠z=tan 2 a (kmk -mω )
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 23 FREQUENCY RESPONSE – Example 3a
The open loop transfer 300( s +100) function of a control Gs=() system is given as : ss+10s+40()()
Determine an expression for the phase angle of G(jw ) in terms of the angles of its basic factorsfactors.. 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 val ue in dB at a frequency of 28.3 rad/s.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 24 300( s +100) Gs=( ) FREQUENCY RESPONSE ss+10s+40 ( )( ) – Example 3b
∠∠∠G(jω)= 300+ G(jω +100)- ∠∠ G(jω)- G(jω + 10) - ∠ G(jω +40)
-1⎛⎞ωωωω -1 ⎛⎞ -1 ⎛⎞ -1 ⎛⎞ = 0 + tan⎜⎟ - tan ⎜⎟ - tan ⎜⎟ - tan ⎜⎟ ⎝⎠100 ⎝⎠ 0 ⎝⎠ 10 ⎝⎠ 40
o-1o-1-1⎛⎞ωωω ⎛⎞ ⎛⎞ =0 +tan⎜⎟ -90 -tan ⎜⎟ -tan ⎜⎟ ⎝⎠100 ⎝⎠ 10 ⎝⎠ 40
o-1o-1⎛⎞28.3 ⎛⎞ 28.3 -1 ⎛⎞ 28.3 ∠G(28.3j) = 0 + tan⎜⎟ - 90 - tan ⎜⎟ - tan ⎜⎟ ⎝⎠100 ⎝⎠ 10 ⎝⎠ 40 = 0oooooo + 15.8 - 90 - 70.5 - 35.3 = -180
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 25 300( s +100) Gs=( ) FREQUENCY RESPONSE ss+10s+40 ( )( ) – Example 3c
300 jω + 100 Gj()ω = jω jω +10 jω +40
300 ω22+100 = ωω2222+10 ω +40
300 28. 322 + 100 G28.3j=( ) 28.3 28.322 + 10 28.3 22 + 40 (300)( 103.9) ==0.749 ()()()28.3 30.0 49.0
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 26 FREQUENCY RESPONSE
Typical gain and phase characteristics of a closed loop system. |T(jw)| ∠T(jω) 0 Mr 1 0.707
ω ωr BW ω
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 27 FREQUENCY DOMAIN SPECIFICATIONS
Similar to transient response specifications in time domain, frequency response specifications are defined.
- Resonant peak, Mr,
- Resonant frequency, ωr, - Bandwidth, BW, --CutoCutoff Rate.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 28 FREQUENCY DOMAIN SPECIFICATIONS
Resonant peak, Mr : This is the maximum value of the |T(jω)| transfer function magnitude Mr 1 |T(jω)|.
Mr depends on the damping ratio ξ only and indicates the relative stability of a stable closed loop system.
ωr ω A large Mr results in a large overshoot of the step response. 1 M=r As a rule of thumb, Mr should be 2 between 1.1 and 1.5. 2ξ 1-ξ
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 29 FREQUENCY DOMAIN SPECIFICATIONS
|T(jω)| RtResonant frequency, ω : Mr r 1 This is the frequency at which the resonant peak is obtained.
2 ωrn=ω 1-2ξ
ωr ω
Note that resonant frequency is different than both the undamppped and damped natural fre quencies!
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 30 FREQUENCY DOMAIN SPECIFICATIONS
Bandwidth, BW : |T(jω)| Mr This is the frequency at 1 which the magnitude of 0.707 the frequency response function, |T(jω)|, drops to 0.707 of its zero frequency value. ωr BW ω
BW is directly proportional to ωn and gives an indication of the transient response characteristics of a control system. The larger the bandwidth is, the faster the system responds.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 31 FREQUENC Y D OMAIN SPE CIFI CATI ON S
|T(jω)| M r Bandwidth, BW : 1 0.707 It is also an indicator of robustness and noise filtering characteristics of a control system.
ωr BW ω
242 ωBW= ωξ+ξξ+ω n (12− ξ+) 4 ξ− 4ξ+ 2
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 32 FREQUENCY DOMAIN SPECIFICATIONS
CutCut--offoff Rate : |T(jω)|
This is the slope of the Mr 1 magnitude of the frequency 0.707 response fi|(jfunction, |T(jω)|, at higher (above resonant) frequencies.
It indicates the ability of a ωr BW ω system to diditiihstinguish signals from noise.
Two systems hhiaving th e same bbdidhandwidth can have different cutoff rates.
ME 304 CONTROL SYSTEMS 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 anal ysi s and desi gn 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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 34 BODE PLOT
ME 304 CONTROL SYSTEMS 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 ftifunction in decib iblels, 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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 36 BODE PLOT
ME 304 CONTROL SYSTEMS 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 freqqypuency 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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 38 BODE PLOT
It is pppyossible 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.
P K1+Ts∏ ( p ) T(s)= p=1 ⎛⎞ M Q ss2 s1+N τ s1+2⎜⎟ξ + ∏∏( mq) ⎜⎟2 m=1 q=1 ⎜⎟ωn ω ⎝⎠q nq
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 39 P K1+Ts∏ ()p p=1 T(s)= ⎛⎞ M Q 2 N ⎜⎟ss BODE PLOT s1+∏∏( τ s1+2) ξ + mq⎜⎟ω 2 m=1 q=1 ⎜⎟nq ω ⎝⎠nq
The logarithmic magnitude of T(jω) can be obtained by summation of the lithilogarithmic magnititdudes of iidiidlndividual terms.
P log T() jω =logK+∑ log1+jωτp - p 2 ⎛⎞ N M Q 2ξq jω -log() jω -log1+j∑∑ωτ -log1+ jω+⎜⎟ m ωω⎜⎟ mqnnqq⎝⎠
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 40 P K1+Ts∏ ()p p=1 T(s)= ⎛⎞ M Q ⎜⎟ss2 s1+N ∏∏( τ s1+2) ξ + BODE PLOT mq⎜⎟ω 2 m=1 q=1 ⎜⎟nq ω ⎝⎠nq
Similarly, the phase of T(jω) can be obtained by simple summation of the phases of individual terms.
⎛⎞ Q 2ξω ω PM⎜⎟qnq φ∠=Tj()ω =tan∑∑∑-1ωτ -N 90 o - tan -1ωτ -tan -1 pm() ⎜⎟22 ppqm q ⎜⎟ω - ω ⎝⎠nq
ME 304 CONTROL SYSTEMS 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.
22 ⎡ ⎛⎞⎛⎞ωω⎤⎡ ⎛⎞⎛⎞ ωω ⎤ 1 ⎢1+2ξ ⎜⎟⎜⎟j+jor1+2⎥⎢ξ ⎜⎟⎜⎟j+j ⎥ ⎢ ωω⎥⎢ ωω ⎥ ⎣ ⎝⎠⎝⎠nn⎦⎣ ⎝⎠⎝⎠ nn ⎦
ME 304 CONTROL SYSTEMS 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]=20lo gM
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..
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 43 BODE PLOT
Gain Factor K.
The gain factor multiplies the overall gain by a constant value for all frequencies.
It has no effect on phase. G(()s)=K M[dB] G(jω )=K 20logK MM20logG(j)=20log G(jω ) 0 ω =20log(K) [dB] φ[o] φ = 0 0 ω M : magnitude, φ : phase.
ME 304 CONTROL SYSTEMS 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]. o Phase is constant at --9090 at all frequencies. 111 M[dB] decade G(s)= ,,Gj G( jω)==-j sjωω 20 ω 0 M=20log G(jω) 0.1 1 10 -20 ⎛⎞1 =20log⎜⎟ = -20logω -20 dB/decade slope ⎝⎠ω φ[o] o Im Re ω φ =-90 0 φ -1/ω -90
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 45 -1/ω2 Im BODE PLOT Re Double pole at the origin. φ
Simply double the slope of the magnitude and the phase, i.e., -40-40 dB/decade becoming zero at ω=1 [rad/s] and -180o phase.
M[dB] decade 111 G(s)= , G( jω)==- 40 222 s (jω) ω ω 0 0.1 1 10 M=20log G(jω) -40 -40 dB/decade slope ⎛⎞1 =20log = -40logω φ[o] ⎜⎟2 ⎝⎠ω ω 0 o φ =-180 -180
ME 304 CONTROL SYSTEMS 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]. o Phase is constant at 90 at all frequencies. M[dB] decade G(s)= s , G( jω)= ωj 20 ω 0 0.1 1 10 M20lG(jM=20log G(jω) -20 20 dB/decade slope =20log()ω φ[o] o φ =90 90 ω 0
ME 304 CONTROL SYSTEMS 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 [rad/s] and 180o phase. M[dB] decade G(s)= s22 , G( jω)=-ω 40 ω 0 M=20log G(jω) 0.1 1 10 -40 = 40log()ω 40 dB/decade slope φ[o] o φ =180 180 ω 0
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 48 BODE PLOT – First Order Factor
Simple lag (Real pole) 1/(1+jωT). 1 G(s)= 1+Ts 11-jωT1 ωT G(jω)= = - j 1+jωT1-jωT 1+ω22T1+ω 22T
⎛⎞1 M=20log G(jω)=20log⎜⎟ ⎜⎟22 ⎝⎠1+ω T
M=-20log 1+ω22T[dB]
φ =tan-1( -ωT=-tan) -1ωT
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 49 BODE PLOT – First Order Factor Simple lag (Real Pole) 1/(1+jωT).
M=-20log 1+ω22T[dB]
M[dB] 0.1 1 10 100 T T T T ω 0 1 For ω << T -20 M≅ -20log1= 0 [dB] 1 For ω >> T -40 M-20≅ l[d]logω T [dB]
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 50 BODE PLOT – First Order Factor
It is clear that the actual magnitude curve can be approximated b y two strai ght lines.
M[dB] 0.1 1 10 100 0 T T T T ω -3
M20l10M ≅ -20log1 = 0 [dB] -20 M≅ -20logω T [dB]
-40
1 1 For ω<< For ω >> T T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 51 BODE PLOT – First Order Factor
ωc=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= -20log 1+ω22T[dB] M[dB] 0.1 1 10 100 0 T T T T ω -3 ⎛⎞1 M⎜⎟ω = = -20log 2 -20 ⎝⎠T ≅ -3[dB]
-40
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 52 BODE PLOT – First Order Factor
ω 0.1ωc 0.5ωc ωc 2ωc 10ωc Error 0.04 1133110.04 [dB]
M[dB] 0.1 1 10 100 0 T T T T ω -3
-20
-40
ME 304 CONTROL SYSTEMS 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 out put 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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 54 BODE PLOT – First Order Factor
Simple lag 1/(1+jωT).
φ =tan-1() -ωT=-tan -1ωT φ[o] 0.1 1 10 100 ω T TTT 0 0.1 For ω<< T
φ≅0 [o ] -45 10 For ω>> T
-90 φ≅-90 [o ]
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 55 BODE PLOT – First Order Factor
It is clear that the actual phase curve can be approximated by three straight lines. φ[o] 0.1 1 10 100 0 T T T T ω φ≅0 [o ] -5.7 Linear variation in the range 0.1 10 -45 ≤≤ω TT
-90 φ≅-90 [o ]
In this case corner frequencies are : 0.1/T and 10/T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 56 BODE PLOT – First Order Factor
ω 0010.01ωc 010.1ωc ωc 10ωc 100ωc φ [o] --0.570.57 --5.75.7 --4545-84.3 -84.3 --89.489.4 Error 0.6 5.7 00--5.75.7 --0.60.6 [o]
Thus the maximum error of the linear approximation is 5.7o.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 57 BODE PLOT – First Order Factor
Simple lead (Real zero) 1+jωT.
G(s)=1+Ts
G(jω)1)=1+ωTj
⎛⎞22 M=20log G(jω)=20log⎜⎟ 1+ω T ⎝⎠
M=20log 1+ω22T[dB]
φ =tan-1(ωT=tan) -1ωT
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 58 BODE PLOT – First Order Factor Simple lead (Real zero) 1+jωT.
M=20log 1+ω22T[dB] M[dB] 1 For ω<< T 40 M≅ 20log1= 0 [dB] 1 For ω >> T 20 M20logT≅ ω [dB]
ω 0 0.1 1 10 100 T T T T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 59 BODE PLOT – First Order Factor
It is clear that the actual magnitude curve can be approximated b y two strai ght lines. 1 1 For ω<< For ω >> M[dB] T T
40 M≅ 20logω T [dB]
20
M20log1=0≅ [dB] ω 0 0.1 1 10 100 T T T T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 60 BODE PLOT – First Order Factor
Simple lead 1+jωT.
-1 φ[o] φ =tan ()ωT 0.1 For ω<< T 90 φ≅0 [o ] 10 For ω >> T 45 φ≅90 [o ] ω 0 010.1 1 10 100 T T T T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 61 BODE PLOT – First Order Factor
It is clear that the actual phase curve can be approximated b y three strai ght lines.
φ[o]
90 φ≅90 [o ]
Linear var iati on 45 in the range 0.1 10 ≤≤ω TT 5.7 φ ≅ 0 [o ] 0 0.1 1 10 100 ω T T T T
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 62 BODE PLOT – Quadratic Factors
As overdamped systems can be replaced by two first order factors,,y only underdam ped systems are of interest here.
ω2 A set of two complex G(s)= n 22 s+2ξωnns+ω conjugate poles. 1 G(jω )= 2 ⎛⎞ωω ⎛⎞ ⎜⎟j+2ξ ⎜⎟j+1 ⎝⎠ωωnn ⎝⎠
222 ⎡⎤⎛⎞ωω ⎛ ⎞ M=20logG(jω)=- 20log⎢⎥ 1- ⎜⎟+2 + ⎜ 2ξ ⎟[dB] ⎢⎥ωω ⎣⎦⎝⎠nn ⎝ ⎠
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 63 BODE PLOT – Quadratic Factors
222 ⎡⎤⎛⎞ωω ⎛ ⎞ MM20logG(j=20log G(jω))20log1=-20log⎢ 1- ⎜⎟⎥ +2 ⎜ξ ⎟[dB] ⎢⎥ωω ⎣⎦⎝⎠nn ⎝ ⎠
Low frequency asymptote, ω<<ωn : M20log1=0[dB]≅− ( )
High frequency asymptote, ω>>ωn :
2 ⎛⎞ωω ⎛⎞ M-20log≅ ⎜⎟ =-40log ⎜⎟ [dB] ⎝⎠ωωnn ⎝⎠
Low and high frequency asymptotes intersect at
ω=ωn, i.e. corner frequency is ωn.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 64 BODE PLOT – Quadratic Factors Therefore the actual magnitude curve can be approximated by two straight lines.
M[dB]
LF 20 Asymptote ξ (increasing) 0 HF -20 Asymptote
-40 -40dB/decade -60 slope
ωn/100 ωn/10 ωn 10ωn 100ωn Frequency
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 65 BODE PLOT – Quadratic Factors
ω 2ξ ω φ=∠G(jω)=-tan-1 n 2 ⎛⎞ω 1-⎜⎟ ⎝⎠ωn
At low frequencies, ω→0 : φ≅0[o ] o At ω=ωn : φ ≅−90 [ ] o At high frequencies, ω→ : φ≅-180 [ ]
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 66 BODE PLOT – Quadratic Factors
Thus, the actual phase curve can be approximated by three straight lines.
φ[o] 0 ξ (increasing)
-90 -90o/decade slope
-180 ωn/10 ωn 10ωn 100ωn Frequency
Corner frequencies are : ωn/10 and 10ωn.
ME 304 CONTROL SYSTEMS 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 dampi ng rati os cl oser to 110.0.
The peak magnitude is given by : 1 Μr = 2ξξξ1−ξ2
The resonant frequency :
2 ωrn=ω 1-2ξ
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 68 BODE PLOT – Quadratic Factors
For ξ=0.707 :
Mr=1 ((g)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 overdamppyed systems ,,, i.e., for ξ ≥ 1.0.
ME 304 CONTROL SYSTEMS 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 sss+100( s+10)( 0. 1s +14s+1000)
First convert to standard form. 100000( 1+ jω) T(jω)= 2 ⎡⎤⎛⎞ωω ⎛⎞ (jω)(10)( 1+0.1ωj)( 1000)⎢⎥ j +1.4 j +1 ⎜⎟100 ⎜⎟ 100 ⎣⎦⎢⎥⎝⎠ ⎝⎠
10( 1+ jω) T(jω)= 2 ⎡⎤⎛⎞ωω ⎛⎞ ()(jω 1+0.1ωjj⎢⎥ ) +1.4j +1 ⎜⎟100 ⎜⎟ 100 ⎣⎦⎢⎥⎝⎠ ⎝⎠
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 70 BODE PLOT – Example 1b
10( 1+ jω) T(jω)= 2 ⎡⎛⎞ωω ⎛⎞⎤ (jω)(1+0. 1ωjjj)⎢⎥ j +1.4j 4 j +1 ⎜⎟100 ⎜⎟ 100 ⎣⎢⎝⎠ ⎝⎠⎦⎥
Identify the basic factors and corner frequencies : -Constant gain K : K=10, 20log10=20 [dB] -First order factor (simple lead – real zero) : T=1 ( ωc1=1/T=1) - for magnitude plot -Integral factor : 1/jω -First order factor (simple lag –real pole) : T=0.1 (ωc1=1/T=10) - for magnitude plot -Quadratic factor (complex conjugate poles) : ωn=ωc1=100, ξ=0.7 - for magnitude plot
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 71 BODE PLOT – Example 1c
10( 1+ jω) T(jω)= 2 ⎡⎛⎞ωω ⎛⎞⎤ (jω)(1+0. 1ωjjj)⎢⎥ j +1.4j 4 j +1 ⎜⎟100 ⎜⎟ 100 ⎣⎢⎝⎠ ⎝⎠⎦⎥
Identify the basic factors and corner frequencies : -Constant gain K : K=10, 20log10=20 [dB] -First order factor (simple lead –real zero) : T=1 (ωc2=0.1/T=0.1, ωc3=10/T=10) –for phase plot -Integral factor : 1/jω -First order factor (simple lag –real pole) : T=0.1 (ωc2=0.1/T=1, ωc3=10/T=100) –for phase plot -Quadratic factor (complex conjugate poles) : ωn=100 (ωc2=ωn/10=10, ωc3=10ωn=1000) –for phase plot
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 72 BODE PLOT – Example 1d
M[dB] 1+jω 40
20 K=10
1 10 100 1000 0 ω[rad/s]
-20 Quadratic factor -40 1/(1+0.1jω)
-60 1/jω Bode (magnitude) plot
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 73 BODE PLOT – Example 1e
φ[o] 1+jω 90
1 10 100 1000 K=10 0 ω
1/(1+0.1jω) -90 1/jω
-180 Quadratic factor
-270 Bode (phase) plot
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 74 BODE PLOT – Example 1f
Bode Diagram Matlab plot: 60
full blue lines 40 (just 4 lines 20 0 ude (dB) tt to plot !) -20
Magni -40 num=[100000 100000] den=[0. 1 15 1140 10000 0] -60 0 bode(num,den) grid -90 deg) ((
-180 Approximate Phase plots: dashed -270 -1 0 1 2 3 lines 10 10 10 10 10 Frequency
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 75 STABILITY ANALYSIS Nise Sect. 10. 7, pp. 638-641
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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 76 STABILITY ANALYSIS
From the characteristic equation : 1 + G(s)H(s) = 0 or G(s)H(s)= --11 Then the magnitude and phase for the open loop transfer function become : 20log G(jω)H(jω) =20log1=0dB ∠G(jω)H(jω)=-180o Thus, when the magnitude and the phase angle of a transfer function are 0 dB and --1180o, respectilively, th en t he system is marginally stable.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 77 STABILITY ANALSIS
If at the frequency, for which phase becomes equal to -180-180o, gain is below 0 dB, then the system is stable (unstable otherwise).
Further, if at the frequency, for which gain becomes equal to zero, ph ase is above --180180o, then the system is stable (unstable otherwise).
Thus, relative stability of a minimum phase system can be determined according to these observations.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 78 GAIN and PHASE MARGINS Nise pp. 638 -641
Gain Margin : Additional gain to make the system marginally stable at a freqqyuency for which the phase of the open loop transfer function passes through -180-180o.
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.
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 79 GAIN and PHASE MARGINS
50 )) 0 GiGain Margin -50 nitude (dB gg
Ma -100
-150 -90
-135 Phase deg) (( -180 MiMargin Phase -225
-270 -1 0 1 2 3 10 10 10 10 10 Frequency (rad/sec)
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 80 BODE PLOT
Can you identify the transfer function approximately if the measured Bode diagram is available ?
ME 304 CONTROL SYSTEMS Prof. Dr. Y. Samim Ünlüsoy 81