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By the term frequency response, we mean the steady-state response of a system to asinusoidal input r(t)=Xsint where “” is measured in rad/sec.
)(tyss
the transient response
(t0 t tf).
the steady state response
(tf < t ).
)(ty
0
)(tyt
t0 ts=tf t
To obtain and study frequency-response of the system G(j), the input signalfrequency is varied over a certain range.The information obtained from such analysis is different from root-locus analysis.In fact, the frequency response and root-locus approaches complement each other.The frequency-response methods are most powerful in conventional (classical)control theory.
Frequency-response methods were developed in 1930s and 1940s by Nyquist,Bode, Nichols, and many others.The Nyquist stability criterion enables us to investigate both the absolute andrelative stabilities of linear closed-loop systems from a knowledge of their open-loop frequency response characteristics, G(j)H(j).The introduction of logarithmic plots, often called Bode plots, simplifies thedetermination of the graphical portrayal of the frequency response.
The steady state outputs to sinusoidal input:The steady-state response of system G(s) can be obtained directly from thesinusoidal input (r(t)=Xsint) transfer function G(j ), that is, the transferfunction in which s=j is replaced by j, where is frequency.
The steady state outputs to sinusoidal Input:Suppose that the transfer function G(s)
))...()((
)(
)(
)()(
)(
)(
21 nssssss
sZ
sP
sZsG
sR
sY
G(s)
R(s) Y(s)
polessystem
n
n
Input
ss
b
ss
b
ss
b
js
a
js
a
s
XsGsRsGsY
2
2
1
122
)()()()(
The steady-state response of a stable, linear, time-invariant system to a sinusoidalinput does not depend on the initial conditions.
where a and the b, (where i = 1,2, . . . , n) are constants and ā is the complexconjugate of a. The inverse Laplace transform of system output Y(s) gives
polessystem
tsn
tsts
Input
jj nebebebeaaety 21
21)(
For a stable system, the terms contain e-st approach zero as t approaches infinity.
The steady state outputs to sinusoidal Input:Similarly, we obtain the following expression for G(-j):
Then,
j
j
ejG
ejGjG
)(
)()(
j
ejGXa
j
ejGXa
jj
2
|)(|,
2
|)(|
The steady state response is
)sin()sin(|)(|
2|)(|)(
)()(
tYtjGX
j
eejGXeaaety
tjtjjj
ss
G(s)R(s) Y(s)
y(t)r(t) = Xsint
ts t
x(t)
yss(t)
The steady state response
Results: A stable, linear, time-invariant system subjected to a sinusoidal input will,at steady state, have a sinusoidal output of the same frequency as the input. But theamplitude and phase of the output will, in general, be different from those of theinput.
)()(
)(
jG
jR
jY
The function G(j) is called the sinusoidal transfer function.
Example 9.1: The steady state response of an RC circuits, r(t)=Xsint:
0
1
111
452
1
)1(tan2
1
1,|)(|
)(
)(
1
RC
jGjR
jY
RC
From yss(t), it can be seen that for 0, the amplitude of the steady-stateoutput yss(t) is almost equal to X. The phase shift of the output is small for small.The magnitude of output is X|G(j1)|=X/2 and phase (1)=450 at=1=1/=1/RC.For , the amplitude of the output is small and almost inverselyproportional to . The phase shift approaches -900 and yss(t)=0 as . Thisis a phase-lag network.
Example 9.1: The steady state response of an RC circuits:
0
1
111
452
1
)1(tan2
1
1,|)(|
)(
)(
1
RC
jGjR
jY
RC
Results: The magnitude of the transfer function |G(j)| is almost equal to onefor the range of frequency of 0 1=1/. However it rapidly degreased tozero for 1=1/ < ∞. Thus the frequency 1=1/ is called corner frequencyfort the simple poles/zeros.
Presenting Frequency-Response Characteristics in Graphical Forms:The transfer function G(j) is characterized by its magnitude and phase angle,with frequency as the parameter.
)(
)()()()(
)(
jG
jGjGjGjR
jY
G(j)R(j) Y(j)
j
0s=+j plane
0
There are three common methods are used study the frequency response of thesystems:1. Bode diagram or logarithmic plot2. Nyquist diagram or polar plot3. Log-magnitude-versus-phase plot (Nichols plots)In this lecture Bode diagram of feedback systems will be subject t be studied.
Results: An interval of two frequencies with a ratio equal to 10 is called a decade, so that the range of frequencies from 1 to 2, where 2 = 101, is called a decade. The magnitude curve has two asymptotes one has 0 dB/decade gain for <<1/ the other has -20 dB/decade gain for >> 1/.
Bode diagram of feedback systems is obtained from the knowledge of their looptransfer (open-loop) function frequency response characteristics, G(j)H(j) toenables us to investigate both their stability and frequency performances.
Frequency response characteristics of the characteristic equation;
A Bode diagram gives the frequency performance of the closed loop system bystudying the loop transfer function G(j)H(j) only.
A Bode diagram consists of two graphs:a) Magnitudes graph; A plot of the logarithm of the magnitude of a sinusoidalloop transfer function G(j)H(j) in the decibel, usually abbreviated dB;20log10|G(j)H(j)| = 0 (dB)b) Angle graph; A plot of the phase angle;G(j)H(j) = -1800
Basic Factors of G(j)H(j). The basic factors that very frequently occur in anarbitrary loop transfer function G(j)H(j) are:1. Gain K2. Integral and derivative factors (j) 1
3. First-order factors (1+jT) 1
4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) 1
Once the logarithmic plots of these basic factors are well understood, it ispossible to utilize them in constructing a composite logarithmic plot for anygeneral form of loop transfer function G(j)H(j) by sketching the curves foreach factor and adding individual curves graphically, because adding thelogarithms of the gains corresponds to multiplying them together.
3. First-order factors (1+j)-n , Simple pole with negative real axes. (Phase Lag)
Magnitudes graph;
For low frequencies, such that << 1/ (corner frequency), the log magnitude may be approximated by
1
1)()(,1,1
ssHsGn
)(tan1)(
1()()()()()( 1
2
jHjGjHjGjHjG
dBjHjG 2101010 1log201log20)()(log20
dBdBjHjG 01log20)(1log20)()( 210
For high frequencies, such that >> 1/ (corner frequency),
dBdBjHjG )log(20)(1log20)()( 210
The slope is 0 dB/decade
The slope is -20 dB/decade
Results: At = 1/ , the log magnitude equals 0 dB; at = 10/, the logmagnitude is -20 dB. Thus, the value of -20 log dB decreases by 20 dB forevery decade of . For >>1/, the log-magnitude curve is thus a straight linewith a slope of -20 dB/decade (or -6 dB/octave).
3. First-order factors (1+j)n , Simple pole with negative real axes. (Phase Lead)
Magnitudes graph;
For low frequencies, such that << 1/ (corner frequency), the log magnitude may be approximated by
1)()(,1,1 jjHjGn
dBdBjHjG 01log20)(1log20)()(log20 21010
For high frequencies, such that >> 1/ (corner frequency),
dBdBjHjG )log(20)(1log20)()(log20 21010
The slope is 0 dB/decade
The slope is 20 dB/decade
Results: At = 1/ , the log magnitude equals 0 dB; at = 10/ , the logmagnitude is 20 dB. Thus, the value of 20 log dB increases by 20 dB forevery decade of . For >>1/, the log-magnitude curve is thus a straight linewith a slope of 20 dB/decade (or 6 dB/octave).
The two asymptotes just derived are independent of the value of damping ratio .Near the frequency =n, a resonant peak occurs. The damping ratio determines the magnitude of this resonant peak.Errors obviously exist in the approximation by straight-line asymptotes. Themagnitude of the error depends on the value of . It is large for small values of .
4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 )-1, (Phase Lag) for a range of 0 < < 1.
Note that the optimum value of = 0.707.
Figure shows exact log-magnitude curves, together with the straight-lineasymptotes and the exact phase-angle curves for the quadratic factor withseveral values of .
4. Quadratic factors (1+ j 2 (j/n) + (j/n)2 ) -1, (Phase Lag) for a range of 0 < < 1.
The phase angle is a function of both o and . At = 0, the phase angle equals 00. At the corner frequency = n, the phase angle is -90" regardless of , since
At = , the phase angle becomes -1800. The phase-angle curve is skew symmetricabout the inflection point-the point where = -90°. There are no simple ways tosketch such phase curves. We need to refer to the phase-angle curves shown inabove Figure.
1) First rewrite the sinusoidal transfer function G(j)H(j) as a product of basicfactors discussed above.
2) Then identify the corner frequencies for first order factors of zeros/poles andnatural frequencies for quadratic factors of zeros/poles.
3) Finally, draw the asymptotic log-magnitude curves, |G(j)H(j)| with proper slopes between the corner frequencies and natural frequencies with the damping demping ratio. The exact curve, which lies close to the asymptotic curve, can be obtained by adding proper corrections.
4) The phase-angle curve of G(j)H(j) can be drawn by adding the phase-angle curves of individual factors.
The use of Bode diagrams employing asymptotic approximations requires muchless time than other methods that may be used for computing the frequencyresponse of a transfer function.The ease of plotting the frequency-response curves for a given transfer functionand the ease of modification of the frequency-response curve as compensation isadded are the main reasons why Bode diagrams are very frequently used inpractice.
Example 9.4: Plot the Bode diagram of the following system.
Solution 9.4; The loop transfer function;
The zero of G(j) is =20 rad/secThe corner frequency of the simple pole of G(s) is = 2 rad/sec andthe natural frequency of the second order pole is n 13 rad/sec
bode(sys,{wmin,wmax}) draws the Bode plot for frequencies between WMIN and WMAX (in radians/second).
bode(sys1,sys2,...,w) graphs the Bode response of multiple LTI models SYS1,SYS2,... on a single plot.
bode(sys1,'r',sys2,'y--',sys3,'gx')
[mag,phase] = bode(sys,w) and [mag,phase,w] = bode(sys) return the response magnitudes and phases in degrees (along with the frequency vector W if unspecified). No plot is drawn on the screen. mag(:,:,k) and phase(:,:,k) determine the response at the frequency w(k).
Plotting Bode Diagram: MATLABExample 9.5: The block diagram of a disk drive head position control, including effect of flexure head mount where the blocks dynamics are given in equations (1). Use Matlab to plot Bode diagram of the system for the control gain K=400.
The main advantage of using the Bode diagram is that multiplication factors innumerator of G(j)H(j) are converted to addition and in denominator ofG(j)H(j) are converted to subtraction factors.
A simple method based on asymptotic approximations for sketching anapproximate magnitudes graph (log-magnitude) curve is available. Suchapproximation by straight line asymptotes is sufficient if only rough information onthe frequency-response characteristics is needed. Should the exact curve be desired,corrections can be made easily to these basic asymptotic plots.
Expanding the low frequency range by use of a logarithmic scale for thefrequency is highly advantageous since characteristics at low frequencies are mostimportant in practical systems. Although it is not possible to plot the curves rightdown to zero frequency because of the logarithmic frequency (log 0 = -), thisdoes not create a serious problem.
The experimental determination of a transfer function can be made simple iffrequency response data are presented in the form of a Bode diagram.
Minimum-Phase Systems and Nonminimum-Phase Systems.
Transfer functions having neither poles nor zeros in the right-half s= jcomplex plane are minimum-phase transfer functions, whereas those having polesand/or zeros in the right-half s plane are nonminimum-phase transfer functions.
Systems with minimum-phase transfer functions are called minimum-phase systems,whereas those with nonminimum-phase transfer functions are called nonminirnum-phase systems.
For systems with the same magnitude characteristic, the range in phase angle of the
minimum-phase transfer function is minimum among all such systems, while therange in phase angle of any nonminimum-phase transfer function is greater than thisminimum.
It is noted that for a minimum-phase system, the transfer function can be uniquelydetermined from the magnitude curve alone. For a nonminimum-phase system, thisis not the case.
Multiplying any transfer function by all-pass filters does not alter the magnitudecurve, but the phase curve is changed.
Bode Diagram of Minimum and Nonminimum Phase Systems
Minimum-Phase Systems and Nonminimum-Phase Systems.Nonminimum-phase situations may arise in two different ways. One is simplywhen a system includes a nonminimum-phase element or elements. The othersituation may arise in the case where a minor loop is unstable.
For a minimum-phase system, the phase angle at = becomes -90°(n- m), wherem and n are the degrees of the numerator and denominator polynomials of thetransfer function, respectively.
Minimum-Phase Systems
The magnitude graph satisfies 20log10G(j) = -20(n-m) dB/decade for
The angle graph satisfies G(j) = -900(n-m) for
Bode Diagram of Minimum and Nonminimum Phase Systems
Nonminimum Phase Systems: Nonminimum-phase situations may arise in twodifferent ways. One is simply when a system includes a nonminimum-phaseelement or elements. The other situation may arise in the case where a minor loopis unstable.
For a nonminimum-phase system, the slope of the log-magnitude curve at = isequal to -20(n–m) dB/d.But the phase angle at = differs from -90°(n-m).
Nonminimum Phase Systems
The magnitude graph satisfies 20log10G(j) = -20(n-m) dB/decade for
The angle graph does not satisfy G(j) = -900(n-m) for
Results: To detect whether the system is minimum phase is necessary to examineboth the slope of the high-frequency asymptote of the log-magnitude curve and thephase angle at = .If the slope of the log-magnitude curve as approaches infinity is -20(n - m) dB/dand the phase angle at = is equal to -90°(n - m), then the system is minimumphase.
In designing a control system, we require that the system be stable. Furthermore, itis necessary that the system have adequate relative stability.One of the important problems in analyzing a control system is to find all closed-loop poles or at least those closest to the j axis (or the dominant pair of closed-loop poles).If the open-loop frequency-response characteristics of a system are known, it maybe possible to estimate the closed-loop poles closest to j axis.Although the stability of closed loop systems are usually studied with Nyquiststability criterion, however, Bode diagram can effectively be used.
Phase and Gain Margins.In general, the closer the Magnitude and angle graphs of the loop transfer functionG(j)H(j) to 1 (0 dB) and -1800 respectively, the more oscillatory is the systemresponse. The closeness of the Magnitude graph of the loop transfer function,|G(j)H(j)| to the reference point 1 (0 dB) and the angle graph =G(j)H(j) tothe reference point -1800 can be used as a measure of the margin of the stability.
The gain crossover frequency, gc, is the frequency at which |G(j)H(j)|, themagnitude of the open loop transfer function, is unity.The phase margin PM is 1800 plus the phase angle = G(j)H(j) of the open-loop transfer function at the gain crossover frequency, gc, or
Phase Margin: The phase margin is defined as the change in open loop phase shift required to make a closed loop system unstable. The phase margin is that amount of additional phase lag at the gain crossover frequency required to bring the system to the verge of instability.The phase margin is the difference in phase between the phase curve and -180 deg at the point corresponding to the frequency that gives us a gain of 0dB (the gain cross over frequency, gc).
Phase Margin:Likewise, the gain margin is the difference between the magnitude curve and 0dBat the point corresponding to the frequency that gives us a phase of -180 deg (the phase cross over frequency, gc).
Figures illustrate the phase margin of both a stable system and an unstable systemin Bode diagrams.The phase margin is positive for PM >0 and negative for PM<0. For a minimumphase system to be stable, the phase margin must be positive.In the Bode diagram, the critical point, j axis, in the complex plane correspondsto the 0 dB and -180" lines.
Gain Margin: The gain margin is defined as the change in open loop gainrequired to make the system unstable. Systems with greater gain margins canwithstand greater changes in system parameters before becoming unstable inclosed loop.Keep in mind that unity gain in magnitude is equal to a gain of zero in dB.
Gain margin: The gain margin is the reciprocal of the magnitude |G(j)H(j)| at thefrequency, pc, at which the phase angle is -1800. Defining the phase crossover frequency pc, to be the frequency at which the phase angle of the open-loop transfer function equals -1800 gives the gain margin Kg:
The gain margin expressed in decibels is positive if Kg is greater than unity andnegative if Kg is smaller than unity. Thus, a positive gain margin (in decibels)PM>0 means that the system is stable, and a negative gain margin (in decibels)PM<0 means that the system is unstable. The gain margin is shown in Figuresillustrate the phase margin of both a stable system and an unstable system in Bodediagrams.
For a stable minimum-phase system, the gain margin indicates how much the gaincan be increased before the system becomes unstable.For an unstable system, the gain margin is indicative of how much the gain mustbe decreased to make the system stable.The gain margin of a first or second order system is infinite since the Bodediagrams for such systems do not cross the reference points 0 dB and -1800
degrees.Thus, theoretically, first- or second order systems cannot be unstable.
Comments:
The phase and gain margins of a control system are a measure of the closeness ofthe Bode diagrams to the reference point 0 dB and -1800.
Therefore, these margins may be used as design criteria. It should be noted thateither the gain margin alone or the phase margin alone does not give a sufficientindication of the relative stability. Both should be given in the determination ofrelative stability.
For a minimum-phase system, both the phase and gain margins must be positivefor the system to be stable. Negative margins indicate instability.
Proper phase and gain margins ensure us against variations in the systemcomponents and are specified for definite positive values. The two values boundthe behavior of the closed-loop system near the resonant frequency.
For satisfactory performance, the phase margin should be between 300 and 60°,300<PM< 60°, and the gain margin should be greater than 6 dB, GM>6 dB.
With these values, a minimum-phase system has guaranteed stability, even if theopen loop gain and time constants of the components vary to a certain extent.
Although the phase and gain margins give only rough estimates of the effectivedamping ratio of the closed-loop system, they do offer a convenient means fordesigning control systems or adjusting the gain constants of systems.
For minimum-phase systems, the magnitude and phase characteristics of the openloop transfer function are definitely related.
The requirement that the phase margin be 300<PM< 60° means that in a Bodediagram the slope of the log magnitude curve at the gain crossover frequency, gc
should be more gradual than -40 dB/decade.
In most practical cases, a slope of -20 dB/decade is desirable at the gain crossoverfrequency, gc, for stability. If it is -40 dB/decade the systems could be eitherstable or unstable. (Even if the system is stable, however, the phase margin issmall.) If the slope at the gain crossover frequency is -60 or steeper, the system ismost likely unstable.
Nonminimum phase system:For nonminimum phase systems, the correct interpretation of stability marginsrequires careful study. The best way to determine the stability of nonminimumphase systems is to use the Nyquist diagram approach rather than Bodediagram approach.
Bode DiagramsRelative Stability of Nonminimum phase system
Example 9.12; a) Obtain phase crossover frequency pc and gain margin GM ofthe given system given in (1).
b) Obtain gain crossover frequency gc and phase margin PM of the given system.c) Draw the Bode diagram of the system and show pc, GM, gc, and PM on theBode diagram.
)1()50)(5(
2500)()(
ssssHsG
Solution 9.12; The system transfer function and characteristic equation are;
Method 1: i) The gain crossover frequency gc can approximately be readfromthe Bode diagram directly.ii) The gain crossover frequency gc can approximately be obtained from theslope log magnitude graph of Bode diagram.First, the slope between two corner frequencies f, and l is determined, wherethe positive sign of log magnitude equation at f, changes to negative at l.
Solution 9.12; b) The gain crossover frequency gc and phase margin PM:The gain crossover frequency gc can be obtained approximately from method1,based on slope of the log magnitude at the gain crossover frequency gc, or frommethod 2, based on analytically solution of the log magnitude equation;
dBjHjGjHjGgcgc
0|)()(|log201|)()(|)1( 10
))50(log205log20log20(2500log20)()(log20 2210
2210101010 jHjG
Mag .(dB) Phase
0.1 39.99 -91.26
5 3 -140.71
50 -37.03 -219.28
As seen from the table that the positive log magnitudeequation at f = 5 rad/s changed to negative at l = 50rad/s and;
a) Draw the Bode diagram of the system for K=10 and determine its stability.b) Draw the Bode diagram of the system for K=100 and determine its stability.
)1()5)(1(
)()(
sss
KsHsG
Solution 8.13; a) For K=10 phase crossover frequency, pc and gain margin GM;
Homework 9.4: Consider the system shown in Figure where system blockstransfer function are;
1)(,)50)(10)(1(
1)(
)(
sHsss
sG
KsG
p
cc
a) Draw the Bode diagram of the loop transfer function,b) Determine the gain Kc such that the phase margin is PM=50°.What is the gain margin PM in this case?
Gp(s)R(s)
H(s)
E(s) Y(s)Gc(s)
Homework 9.3: Consider the system shown in Figure where system blockstransfer function are;
1)(,)2(
1)(,)(
2
sH
sssGKsG pcc
a) Plot the frequency response for this system when Kc = 4.b) Calculate the phase and magnitude at = 0.5,1,2, 4, and .
In addition to the phase margin PM, gain margin GM, resonant peak Mr, and resonantfrequency r, there are other frequency-domain quantities commonly used in performancespecifications. They are the cutoff frequency, bandwidth BW, and the cutoff rate.
Cutoff Frequency and Bandwidth: Both cutoff Frequency and Bandwidth are related to the closed loop system Bode diagram such that the frequency b, at which the magnitude of the closed-loop frequency response is 3 dB below its zero-frequency value is called the cutoff frequency. Thus
G(s)R(s) Y(s)
H(s)
bfordBjR
jY
jHjG
jG
jR
jY
,3
)0(
)0(
)()(1
)(
)(
)(
For systems in which |Y(j0)/R(j0)|= 0 dB, bfordBjHjG
jG
jR
jY
,3
)()(1
)(
)(
)(
The closed-loop system filters out the signal components whose frequencies are greaterthan the cutoff frequency and transmits those signal components with frequencies lowerthan the cutoff frequency.
Closed Loop Frequency Performances: Cutoff Frequency and Bandwidth
Bandwidth: The frequency range 0≤≤ b in which the magnitude of the closed loop doesnot drop -3 dB is called the bandwidth of the system. The bandwidth indicates the frequencywhere the gain starts to fall off from its low-frequency value. Thus, the bandwidth indicateshow well the system will track an input sinusoid. Note that for a given natural fequency n
the rise time increases with increasing damping ratio . On the other hand, the bandwidthdecreases with the increase in . Therefore, the rise time and the bandwidth are inverselyproportional to each other.
Bandwidth: The specification of the bandwidth may be determined by the following factors:1. The ability to reproduce the input signal. A large bandwidth corresponds to a small rise time, or fast response. Roughly speaking, we can say that the bandwidth is proportional to the speed of response. 2. The necessary filtering characteristics for high-frequency noise.For the system to follow arbitrary inputs accurately, it must have a large bandwidth. Fromthe viewpoint of noise, however, the bandwidth should not be too 1arge.Thus there areconflicting requirements on the bandwidth, and a compromise is usually necessary for gooddesign. Note that a system with large bandwidth requires high-performance components, sothe cost of components usually increases with the bandwidth.
Cutoff Rate. The cutoff rate is the slope of the log-magnitude curve near the cutofffrequency. The cutoff rate indicates the ability of a system to distinguish the signal fromnoise.
G(s)R(s) Y(s)
H(s)
bfordBjR
jY
jR
jY
jHjG
jG
jR
jY
,3)0(
)0(
)(
)(
)()(1
)(
)(
)(
It is noted that a closed-loop frequency response curve with a steep cutoff characteristicmay have a large resonant peak magnitude, which implies that the system has a relativelysmall stability margin.
The resonant peak is the value of the maximum magnitude (in decibels) of the closed-loop frequency response. The resonant frequency is the frequency that yields the maximum magnitude.
G(s)R(s) Y(s)
H(s)
bfordBjR
jY
den
num
jHjG
jG
jR
jY
,3
)0(
)0(
)()(1
)(
)(
)(
dB
-3
Bandwidth
b in log scale
% M-File 1 for obtaining the resonant peak and resonant frequency [mag,phase,w] = bode(num,den,w); or [mag,phase,w]= bode(sys,w);[Mp,k]= max(mag);resonant-peak = 20*log10(Mp);resonant-frequency = w(k)% M-File 2 The bandwidth can be obtained by entering the following lines in the program:n = l ;while 20*log10(mag(n)) > = -3; n = n + 1;endbandwidth = w(n)
Cutoff Rate.
Closed Loop Frequency Performances: Cutoff Frequency and Bandwidth
Example 9.14; Consider the system shown in Figure . Using MATLAB, obtain a Bode diagram for the closed loop transfer function. Obtain also the resonant peak, resonant frequency, and bandwidth.
G(s)R(s) Y(s)
H(s)
1)(,)1)(15.0(
1)(
sH
ssssG
Solution 9.14;
15.15.0
1
1)1)(15.0(
1
)()(1
)(
)(
)(23
sssssssHsG
sG
sR
sY
bfordBjR
jY
jR
jY
den
num
jR
jY
jHjG
jG
jR
jY
,3)0(
)0(
)(
)(
)(
)(
)()(1
)(
)(
)( dB
-3
Bandwidth
b in log scale
Cutoff Rate.
The closed loop system transfer function is
Resonant Peak, Resonant Frequency, and Bandwidth: MATLAB
Remark 1; The frequency performances such Phase Margin PM, Gain margin GM andstability are related to the loop (oplen loop) frequency responce. That is open loopsystem Bode diagram.
G(s)R(s) Y(s)
H(s)
0)()(1)(,)()(1
)(
)(
)(
sHsGsP
sHsG
sG
sR
sY
0
10
180|)()()2(
0|)()(|log20
1|)()(|)1(
pc
gc
gc
jHjG
dBjHjG
jHjG
Remark 2; Resonant Peak, Resonant Frequency, Cutoff Frequency and Bandwidth arerelated to the closed loop system Bode diagram.
The static position, velocity, and acceleration error constants describe the low-frequency behaviour of type 0, type 1, and type 2 systems, respectively.
The static position, velocity, and acceleration error constants describe the low-frequency behaviour (region I) of type 0, type 1, and type 2 systems, respectively.For a given system, only one of the static error constants is finite and significant.(The larger the value of the finite static error constant, the higher the loop gain is as approaches zero.)The type of the system determines the slope of the log-magnitude curve at lowfrequencies (region I). Thus, information concerning the existence and magnitudeof the steady state error of a control system to a given input can be determined fromthe observation of the low-frequency region (region I) of the log-magnitude curve.
Figure 2 shows an example ofthe log-magnitude plot of a type0 system. In such a system, themagnitude of G(j)H(j) equalsKp at low frequencies, or
Determination of static position error constants Kp from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by
)1)...(1)(1(
)1)...(1)(1()()(
21
sTsTsTs
sTsTsTKsHsG
pN
mba G(s)R(s) Y(s)
H(s)
log 0
|G(j
)H(j
)|[d
B]
gc
-20 dB/d20 log10Kp
-40 dB/d
0 dB/decade
Figure 2:
pKjHjG
)()(lim0
From Bode diagram20log10|G(j)H(j)|= 20log10Kp pKjHjG
Determination of static velocity error constants Kv from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by
)1)...(1)(1(
)1)...(1)(1()()(
21
sTsTsTs
sTsTsTKsHsG
pN
mbaG(s)
R(s) Y(s)
H(s)
log 0
|G(j
)H(j
)|[d
B]
gc
-20 dB/d
-40 dB/dFigure 2:2 =1
vK10log20
1
Figure 2 shows an example of the log magnitude plot of a type 1 system. Theintersection of the initial -20 dB/d segment (or its extension) with the line 1 = 1has the magnitude 20 log Kv.This may be seen as follows: In a type 1 system
Determination of static velocity error constants Kv from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by
)1)...(1)(1(
)1)...(1)(1()()(
21
sTsTsTs
sTsTsTKsHsG
pN
mbaG(s)
R(s) Y(s)
H(s)
.,1|||)()(| 1
1
11
vv Kor
j
KjHjG
The intersection of the initial -20dB/dsegment (or its extension) with the 0dB line has a frequency numericallyequal to Kv. To see this, define thefrequency at this intersection to be 1;then
Determination of static acceleration error constants Ka from Bode diagram.Consider the feedback control system shown in Figure. Assume that the loop(open-loop) transfer function is given by
)1)...(1)(1(
)1)...(1)(1()()(
21
sTsTsTs
sTsTsTKsHsG
pN
mbaG(s)
R(s) Y(s)
H(s)
Figure 2 shows an example of the log-magnitude plot of a type 2 system. The intersection of the initial -40-dB/decade segment (or its extension) with the = 1line has the magnitude of 20 log Ka. Since at low frequencies
Determination of static acceleration error constants Ka from Bode diagram.For the given feedback control system the loop (open-loop) transfer function is;
)1)...(1)(1(
)1)...(1)(1()()(
21
sTsTsTs
sTsTsTKsHsG
pN
mbaG(s)
R(s) Y(s)
H(s)
The frequency a at the intersection of the initial -40 dB/d segment (or itsextension) with the 0 dB line gives the square root of Ka numerically. This can beseen from the following:
Determination of static acceleration error constants Ka from Bode diagram.For the given feedback control system the loop (open-loop) transfer function is;
Example 9.15; Consider the system given in (1). Determine the finite andsignificant static error constants for a given system. Draw the magnitude graph ofthe Bode diagram of the system. Obtain the finite static error constants of thesystem from the driven bode diagram.
Solution 9.15; Since the system is type 1, it has finite static velocity error constant, Kv. The system Bode diagram is;
a) Plot the Bode Diagram the system.b) Obtain Phase Margin, PM, and Gain Margin, PM. c) Find the finite static error constant.
Gp(s)R(s)
H(s)
E(s) Y(s)Gc(s)
Figure 1: Satellite repair.
Homework 9.6: The space shuttle has been used to repair satellites and the Hubble telescope. Figure 1 illustrates how a crew member, with his feet strapped to the platform on the end of the shuttle's robotic arm, used his arms to stop the satellite's spin.
Figure 2: Satellite repair.
The control system of the robotic arm has a closed-loop transfer function;
Homework 9.7: Consider the system shown in Figure where system blockstransfer function are;
10
10)(
14.1
1)(,)(
2
ssH
sssGKsG pcc
a) Sketch the Bode plot of the loop transfer function for Kc=2.b) Obtain Phase Margin, PM, and Gain Margin, PM. c) Find the finite static error constant.d) Obtain the closed loop system response y(t) to a unit step input, R(s) = 1/s.e) Determine the bandwidth of the system.
Gp(s)R(s)
H(s)
E(s) Y(s)Gc(s)
Bode Diagrams: Homework
9/17/2001
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Electrical & Electronics Engineering
Chapter 09: Control Systems Design by the Root Locus Method