Frequency Response Analysis and Design

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Frequency Response Analysis & Design K. Craig 1

Frequency Response

Analysis & Design

Dr. Kevin Craig

Professor of Mechanical Engineering

Rensselaer Polytechnic Institute




Frequency Response Analysis & Design

• In conventional control-system analysis there are two

basic methods for predicting and adjusting a system’sperformance without resorting to the solution of the

system’s differential equation. They are:

– Root-Locus Method – Frequency-Response Method

• For the comprehensive study of a system by

conventional methods it is necessary to use bothmethods of analysis.




• Root-Locus Method

– Precise root locations are known and actual timeresponse is easily obtained by means of the

inverse Laplace Transform.

• Frequency-Response Method

– Frequency response is the steady-state response

of a system to a sinusoidal input. In frequency-

response methods, we vary the frequency of the

input signal over a certain range and study the

resulting steady-state response.

– The design of feedback control systems in industry

is probably accomplished using frequency-

response methods more often than any other,

primarily because it provides good designs in the

face of uncertainty in the plant model.




– Many times performance requirements are given in

terms of frequency response and/or time response.

– Noise, which is always present in any system, can

result in poor overall performance. Frequency

response permits analysis with respect to this.

– When the transfer function for a component is

unknown, the frequency response can be determined

experimentally and an approximate expression for the

transfer function can be obtained from the graph of theexperimental data.

– The Nyquist stability criterion enables one to

investigate both the absolute and relative stabilities oflinear, time-invariant closed-loop systems from a

knowledge of their open-loop frequency-response

characteristics.




– Frequency-response tests are, in general, simple

and can be made accurately by readily-availableequipment, e.g., dynamic signal analyzer.

– Correlation between frequency and transient

responses is indirect, except for 2nd-ordersystems.

– In designing a closed-loop system, we adjust the

frequency-response characteristic of the open-loop transfer function by using several design

criteria in order to obtain acceptable transient-

response characteristics for the system. – It is the easiest method to use for designing

compensation.




• Linearity and Time Invariance (LTI)

– A frequency-domain transfer function is limited todescribing elements that are linear and time

invariant. These are severe restrictions and virtually

no real-world system fully meets them.

– Homogeneity: If the input to a system r(t) generates

an output c(t), then an input kr(t) generates an output

kc(t) for any k.

– Superposition: If the input r 1(t) generates an output

c1(t), and the input r 2(t) generates an output c2(t),

then the input r 1(t) + r 2(t) generates the output c1(t) +

c2(t).

– Time Invariance: If the input r(t) generates an output

c(t), then the input r(t-τ) generates the output c(t-τ) for

all τ > 0.




– Homogeneity and Superposition are attributes for

linearity.

– No real-world system is completely LTI, however,

for most control systems, components are

designed to be close enough to being LTI that thenon-LTI behavior can be ignored or avoided.

– In practice, most control systems are designed to

minimize non-LTI behavior.




• For a stable, linear, time-invariant system, the

mathematical model is the linear ODE with constant

coefficients:

• qo is the output (response) variable of the system

• qi is the input (excitation) variable of the system

• an and bm are the physical parameters of the system

n n 1

o o on n 1 1 0 on n 1

m m 1

i i im m 1 1 0 im m 1

d q d q dqa a a a qdt dt dt

d q d q dq b b b b q

dt dt dt

−

− −

−

− −

+ + + + =

+ + + +




• If the input to this system is a sine wave, the steady-state

output (after the transients have died out) is also a sinewave with the same frequency, but with a different

amplitude and phase angle.

• System Input:

• System Steady-State Output:

• Both amplitude ratio, Qo/Qi , and phase angle, φ, change

with frequency, ω.• The frequency response can be determined analytically

from the Laplace transfer function:

i i

q Q sin( t)= ω

o oq Q sin( t )= ω + φ

G(s) s = iω Sinusoidal

Transfer FunctionM( ) ( )ω ∠φ ω




• A negative phase angle is called phase lag, and apositive phase angle is called phase lead.

• If the system being excited were a nonlinear or time-

varying system, the output might contain frequencies

other than the input frequency and the output-input

ratio might be dependent on the input magnitude.

• Any real-world device or process will only need to

function properly for a certain range of frequencies;

outside this range we don’t care what happens.




System

Frequency

Response




• When one has the frequency-response curves for any

system and is given a specific sinusoidal input, it is aneasy calculation to get the sinusoidal output.

• What is not obvious, but extremely important, is that

the frequency-response curves are really a completedescription of the system’s dynamic behavior and

allow one to compute the response for any input, not

just sine waves.• Every dynamic signal has a frequency spectrum and if

we can compute this spectrum and properly combine it

with the system’s frequency response, we can

calculate the system time response.




• The details of this procedure depend on the nature ofthe input signal; is it periodic, transient, or random?

• For periodic signals (those that repeat themselves

over and over in a definite cycle), Fourier Series isthe mathematical tool needed to solve the response

problem.

• Although a single sine wave is an adequate model ofsome real-world input signals, the generic periodic

signal fits many more practical situations.

• A periodic function qi(t) can be represented by aninfinite series of terms called a Fourier Series.




( )

( )

( )

0i n n

n 1

T2

n i

T

2T

2

n i

T

2

a 2 2 n 2 nq t a cos t b sin t

T T T T

2 na q t cos t dt

T

2 n b q t sin t dt

T

∞

=

−

−

⎡ ⎤π π⎛ ⎞ ⎛ ⎞= + +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

π⎛ ⎞= ⎜ ⎟⎝ ⎠

π⎛ ⎞= ⎜ ⎟⎝ ⎠

∑

∫

∫

Fourier

Series




t

qi(t )

+0.01-0.01

1.5

-0.5

Consider the Square Wave

( )

( )

0 0.01

0 0.01 0

0 0.01

n

0.01 0

0 0.01

n

0.01 0

i

0.5dt 1.5dta

0.5 average value

T 0.022 n 2 n

a 0.5cos t dt 1.5cos t dt 00.02 0.02

1 cos n2 n 2 n b 0.5sin t dt 1.5sin t dt0.02 0.02 50n

4q t 0.5 s

−

−

−

− +

= = =

π π⎛ ⎞ ⎛ ⎞= − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− ππ π⎛ ⎞ ⎛ ⎞= − + =⎜ ⎟ ⎜ ⎟ π⎝ ⎠ ⎝ ⎠

= +

π

∫ ∫

∫ ∫

∫ ∫

( ) ( )4

in 100 t sin 300 t3

π + π +

π




• The term for n = 1 is called the fundamental or firstharmonic and always has the same frequency as the

repetition rate of the original periodic wave form (50

Hz in this example); whereas n = 2, 3, … gives the

second, third, and so forth harmonic frequencies as

integer multiples of the first.

• The square wave has only the first, third, fifth, and so

forth harmonics. The more terms used in the series,

the better the fit. An infinite number gives a “perfect”

fit.




-0.01 -0.008 -0.006 -0.004 -0.002 0 0.002 0.004 0.006 0.008 0.01

-1

-0.5

0

0.5

1

1.5

2

time (sec)

a m p l i t u d e

Plot of the

Fourier

Series for

the square

wave

through

the thirdharmonic

( ) ( ) ( )i

4 4q t 0.5 sin 100 t sin 300 t

3= + π + π

π π




• For a signal of arbitrary periodic shape (rather than

the simple and symmetrical square wave), theFourier Series will generally include all the harmonics

and both sine and cosine terms.

• We can combine the sine and cosine terms using:

• Thus

( ) ( ) ( )2 2

1

A cos t Bsin t Csin t

C A B

Atan

B

−

ω + ω = ω + α

= +

α =

( ) ( ) ( )i i0 i1 1 1 i2 1 2q t A A sin t A sin 2 t= + ω + α + ω + α +




• Example: Frequency Response of a Lead

Compensator ( ) Ts 1D s K 1

Ts 1+= α <α +

( )

( ) ( )( )

( ) ( ) ( )

( ) ( )

2

2

1 1

Tj 1D j K Tj 1

T 1M K T 1

Tj 1 Tj 1

tan T tan T− −

ω+ω =α ω +

ω +ω =αω +

φ ω = ∠ ω+ − ∠ α ω+

= ω − α ω

0.1 T 1 K 1α = = =




• What about transient inputs?

– The best path to understanding the meaning of M

and Φ is to relate the frequency response G(jω) to

the transient responses calculated by the Laplace

transform.

( )2

n

2 2

n n

G ss 2 s

ω=

+ ζω + ω

Unit Step Response




Magnitude

Phase

Frequency Response Plots

( )2

n

2 2

n n

G ss 2 s

ω=+ ζω + ω




– What features of the frequency response

correspond to the transient-response

characteristics?

• Damping of the system can be determined from

the transient response overshoot or from the

peak in the magnitude of the frequencyresponse.

• Rise time can be estimated from bandwidth.

• Peak overshoot in the step response can be

estimated from the peak overshoot in the

frequency response.

– Thus we see that essentially the same informationis contained in the frequency-response curve as is

found in the transient-response curve.




Review of Frequency-Response

Performance Specifications

• Let V be a sine wave (U = 0) and wait for transients

to die out.

• Every signal will be a sine wave of the same

frequency. We can then speak of amplitude ratios

and phase angles between various pairs of signals.

1 2

1 2

C AG G (i )(i )V 1 G G H(i )

ωω =+ ω




• The most important pair involves V and C. Ideally(C/V)(iw) = 1.0 for all frequencies.

• Amplitude ratio and phase angle will approximate the

ideal values of 1.0 and 0 degrees for some range oflow frequencies, but will deviate at higher

frequencies.




Typical Closed-LoopFrequency-Response

Curves

As noise is generally in a band of frequencies above

the dominant frequency

band of the true signal,

feedback control systemsare designed to have a

definite passband in order

to reproduce the true

signal and attenuate noise.




• The frequency at which a resonant peak occurs, ωr , is a

speed-of-response criterion. The higher ωr , the faster the

system response.• The peak amplitude ratio, Mp, is a relative-stability

criterion. The higher the peak, the poorer the relative

stability. If no specific requirements are pushing thedesigner in one direction or the other, Mp = 1.3 is often

used as a compromise between speed and stability.

• For systems that exhibit no peak, the bandwidth is usedfor a speed of response specification. The bandwidth is

the frequency at which the amplitude ratio has dropped to

0.707 times its zero-frequency value. It can of course bespecified even if there is a peak. It is the maximum

frequency at which the output of a system will

satisfactorily track an input sinusoid.




• If we set V = 0 and let U be a sine wave, we can

measure or calculate (C/U)(iω) which should ideally

be 0 for all frequencies. A real system cannot

achieve this perfection but will behave typically as

shown.

Closed-Loop Frequency Response to a Disturbance Input




• Gain margin (GM) and phase margin (PM) are in the

nature of safety factors such that (B/E)(iω) stays far

enough away from 1 ∠ -180° on the stable side.

• Gain margin is the multiplying factor by which the

steady-state gain of (B/E)(iω) could be increased

(nothing else in (B/E)(iω) being changed) so as to put

the system on the edge of instability, i.e., (B/E)(iω))

passes exactly through the -1 point. This is called

marginal stability.

• Phase margin is the number of degrees of additional

phase lag (nothing else being changed) required to

create marginal stability.• Both a good gain margin and a good phase margin

are needed; neither is sufficient by itself.




A system must have adequate stability margins.

Both a good gain margin and a good phase margin

are needed.

Useful lower bounds:

GM > 2.5 PM > 30°

Open-Loop Performance Criteria:

Gain Margin and Phase Margin




Bode Plot View of





• It is important to realize that, because of model

uncertainties, it is not merely sufficient for a system to be

stable, but rather it must have adequate stabilitymargins.

• Stable systems with low stability margins work only on

paper; when implemented in real time, they arefrequently unstable.

• The way uncertainty has been quantified in classical

control is to assume that either gain changes or phasechanges occur. Typically, systems are destabilized

when either gain exceeds certain limits or if there is too

much phase lag (i.e., negative phase associated withunmodeled poles or time delays).

• As we have seen these tolerances of gain or phase

uncertainty are the gain margin and phase margin.




Frequency-Response Curves

• The sinusoidal transfer function, a complex function

of the frequency ω, is characterized by its magnitude

and phase angle, with frequency as the parameter.• There are three commonly used representations of

sinusoidal transfer functions:

– Bode diagram or logarithmic plot: magnitude ofoutput-input ratio vs. frequency and phase angle

vs. frequency

– Nyquist plot or polar plot: output-input ratio plottedin polar coordinates with frequency as the

parameter

– Log-magnitude vs. phase plot (Nichols Diagram)




Bode Diagrams

• Advantages of Logarithmic Plots:

– Rapid manual graphing is possible.

– Wide ranges of amplitude ratio and frequency, bothlow and high, are conveniently displayed.

– Amplitude ratio exhibits straight-line asymptote

regions of definite slope. These are helpful inidentifying model type from experimental data.

– Complex transfer functions are easily plotted and

understood as graphical sums of simple (zero-order,1st-order, 2nd-order) basic systems since the dB

(logarithmic) technique changes multiplication into

addition and division into subtraction.




IEEE

Control

Systems

Magazine

June 2007




• A sinusoidal transfer function may be represented by two

separate plots:

– Magnitude (dB) vs. frequency (log10)

– Phase angle (degrees) vs. frequency (log10)

• The log magnitude (Lm) of a transfer function in dB

(decibel) is:

• Frequency Bands:

– Octave

• An octave is a frequency band from f 1 to f 2 where

f 2/f 1 = 2.

– Decade• A decade is a frequency band from f 1 to f 2 where

f 2/f 1 = 10.

( )1020log G iω

x2

1

f 2 where x = # of octaves

f

=

x2

1

f 10 where x = # of decades

f

=




– As a number doubles, the dB value increases by 6

dB.

– As a number increases by a factor of 10, the dBvalue increases by 20 dB.

• Generalized Form of the Sinusoidal TransferFunction:

0.01 40 dB

0.1 20 dB0.5 6 dB

1.0 0 dB

2.0 6 dB10.0 20 dB

100.0 40 dB

= −

= −= −

=

==

=

( ) ( )( )

( ) ( ) ( )

r

1 2

m 2

3 2

n n

K 1 i T 1 i TG i

2 1i 1 i T 1 i i

+ ω + ωω =

⎡ ⎤⎛ ⎞ ⎛ ⎞ζω + ω + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟

ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦

Note that, when expressed in dB,

the reciprocal of a number differs

from its value only in sign.




• The log magnitude (Lm) of G(iω) is given by:

• The phase angle is given by:

• Both the log magnitude and angle are functions offrequency.

( ) [ ] [ ] ( ) [ ]

( ) [ ] [ ] ( )

1 2

2

3 2

n n

Lm G i Lm K Lm 1 i T r Lm 1 i T

2 1m Lm i Lm 1 i T Lm 1 i i

⎡ ⎤ω = + + ω + + ω⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞ζ− ω − + ω − + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟

ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1 2

2

3 2

n n

G i K 1 i T r 1 i T

2 1m i 1 i T 1 i i

∠ ω = ∠ + ∠ + ω + ∠ + ω⎡ ⎤⎛ ⎞ ⎛ ⎞ζ

− ∠ ω − ∠ + ω − ∠ + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦




• When the log magnitude and angle are plotted as

functions of log10(ω), the resulting curves are referred

to as Bode Plots.• These two curves can be combined into a single

curve of log magnitude vs. angle with frequency as

the parameter. This curve is called the NicholsDiagram.

• Drawing Bode Plots

– The generalized form of a transfer function showsthat the numerator and denominator have 4 basic

types of factors:

( ) ( ) ( ) p

m r 2

2

n n

2 1K i 1+i T 1 i i

±± ± ⎡ ⎤⎛ ⎞ ⎛ ⎞ζ

ω ω + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦




– The curves of log magnitude and angle vs. log10(ω)

can be drawn for each factor. Then these curves can

be added together graphically to get the curves forthe complete transfer function. Asymptotic

approximations to these curves are normally used.

– Gain K (positive)

– Integral and derivative factors (iω)± m

• The log magnitude curve is a straight line with a

slope ± m(20) dB/decade = ± m(6) dB/octave when

plotted against log(ω). It goes through the point 0

dB at ω = 1.

[ ] ( )10Lm K 20log K constantK 0

= =∠ = °

( ) ( )

( ) ( )

m10 10

m

Lm i m20log i m20log

i m 90 constant

±

±

⎡ ⎤ω = ± ω = ± ω⎣ ⎦

⎡ ⎤∠ ω = ± ° =⎣ ⎦




K(jω)n is the only class of term that affects the

slope at the lowest frequency because all other

terms are constant in that region.

t 1




– 1st-Order Factors (1 + iωT)± 1

• ω < 1/T: straight-line asymptote with zero slope

• ω > 1/T: straight-line asymptote with ± 20 dB/decade slope• ω = 1/T: exact value is ± 3.01 dB

• ωcf = corner frequency = 1/T = frequency at which the

asymptotes to the log magnitude curve intersect

• Phase angle straight-line asymptotes: 0° at ω < 0.1ωcf , ± 45° atω = ωcf , ± 90° at ω > 10ωcf

• Angle curve is symmetrical about ωcf when plotted vs. log10(ω)

( )

( )

1

10

2 2

10

10

Lm 1 i T 20log 1 i T

20log 1 T

0 dB for T << 1

20log T for T >> 1

±⎡ ⎤+ ω = ± + ω⎣ ⎦

= ± + ω

≈ ω

≈ ± ω ω( ) ( )

1 11 i T tan T± −⎡ ⎤∠ + ω = ± ω

⎣ ⎦




G(jω) = 10jω + 1




dB = 20 log10 (amplitude ratio)

decade = 10 to 1 frequency change

octave = 2 to 1 frequency change

Bode Plotting of1st-Order

Frequency

Response

Note that varying the

time constant shifts the

corner frequency to the

left or to the right, but

the shapes of the curves

remain the same.




• For the case where the exponent of the first-order term is ± r,

the corner frequency is unchanged, and the asymptotes are

still straight lines: the low-frequency asymptote is a horizontal

line at 0 dB, while the high-frequency asymptote has a slope

of ± (20)r dB/decade. The error involved in the asymptotic

expressions is r times that for (1 + iωT)

±1

. The phase angle isr times that of (1 + iωT)±1 at each frequency point.

– 2nd-Order Factors

• For ζ > 1, the quadratic can be factored into two 1st-order

factors with real poles which can be plotted as described for

a 1st-order factor.• For 0 < ζ < 1, the quadratic is plotted without factoring, as it is

the product of two complex-conjugate factors.

( )

1

2

2n n

2 11 i i

±⎡ ⎤⎛ ⎞ ⎛ ⎞ζ

+ ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦




• For ω << ωn

: the log magnitude ≈ 0 dB

• For ω >> ωn: the log magnitude ≈ ± 40 log10 (ω/ωn) dB

• The low-frequency asymptote is a horizontal line at 0 dB.

• The high-frequency asymptote is a straight line with a slope of

± 40 dB/decade.

• The asymptotes, which are independent of ζ, cross at ωcf = ωn.

These are not accurate for a factor with low values of ζ.

• Phase angle: 0° at ω = 0, ± 90° at ω = ωn, ± 180° at ω = ∞

( )

11 2 2 22

2

102 2n n n n

2 1 2Lm 1 i i 20log 1

± ⎡ ⎤⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞ζ ω ζω+ ω + ω = ± − +⎢ ⎥

⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟ω ω ω ω⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎣ ⎦ ⎣ ⎦

( )

12

2 1n 22

n n2

n

2

2 11 i i tan

1

±

−

ζω⎡ ⎤⎛ ⎞ ωζ

∠ + ω + ω = ±⎢ ⎥⎜ ⎟ ωω ω⎝ ⎠⎣ ⎦ −ω




Frequency Response

of a2nd-Order System

Note: The plots shown arefor a 2nd-order term with an

exponent of –1. For a 2nd-

order term with an exponent

of +1, the magnitudes of thelog magnitude and phase

angle are the same except

with a sign change.




• Some Observations on 1st-Order Factors

– Time Constant τ• Time it takes the step response to reach 63%

of the steady-state value

– Rise Time Tr = 2.2 τ• Time it takes the step response to go from 10%

to 90% of the steady-state value

– Delay Time Td = 0.69 τ• Time it takes the step response to reach 50%

of the steady-state value




• Some Observations on 2nd-Order Factors

– When a physical system exhibits a naturaloscillatory behavior, a 1st-order model (or even a

cascade of several 1st-order models) cannot

provide the desired response. The simplest model

that does possess that possibility is the 2nd-order

dynamic system model.

– This system is very important in control design.

– System specifications are often given assuming

that the system is 2nd order.

– For higher-order systems, we can often usedominant pole techniques to approximate the

system with a 2nd-order transfer function.



d




– Thus, when 2nd-order components are used in

feedback system design, large values of ωn (small

lags) are desirable since they allow the use oflarger loop gain before stability limits are

encountered.

– For frequency response, a resonant peak occursfor ζ < 0.707. The peak frequency is ωp and the

peak amplitude ratio depends only on ζ.

– At ω = ωn, the amplitude ratio is 1/2ζ and the phase

is -90°. – The phase angle at the frequency where the

resonant peak occurs is given by:

2 p n

2K 1 2 peak amplitude ratio

2 1ω = ω − ζ =

ζ − ζ

21

p

1 2tan− − ζ

φ = ±ζ

B d id h




– Bandwidth

• The bandwidth is the frequency where the amplitude ratio

drops by a factor of 0.707 = -3dB of its gain at zero or low-frequency.

• For a 1st-order system, the bandwidth is equal to 1/τ.

• The larger (smaller) the bandwidth, the faster (slower) the

step response.

• Bandwidth is a direct measure of system susceptibility to

noise, as well as an indicator of the system speed of

response.• For a 2nd-order system:

• As ζ varies from 0 to 1, BW varies from 1.55ωn to 0.64ωn.

For a value of ζ = 0.707, BW = ωn. For most design

considerations, we assume that the bandwidth of a 2nd-

order all pole system can be approximated by ωn.

2 2 4

nBW 1 2 2 4 4= ω − ζ + − ζ + ζ




( )2

Gω




( ) ( )

( )

2

n

n

2

n

2 2

n n

1

4 2

G s s s 2

T s

s 2 s

2PM tan

1 4 2

−

ω

= + ζω

ω=

+ ζω + ω⎡ ⎤

ζ⎢ ⎥=⎢ ⎥

+ ζ − ζ⎣ ⎦

PM

100

ζ ≈( )

( )

( )

n

n

2n

2 2

n n

G ss s 2

T ss 2 s

ω=

+ ζω

ω=+ ζω + ω

Note: In practice, Mr is rarely used; most designers

prefer to use PM to specify the damping because

imperfections that make systems nonlinear or cause

delays usually erode phase more significantly than

magnitude.

Closed-Loop Bandwidth with respect to PM




Closed Loop Bandwidth with respect to PM

for a 2nd-Order System

( ) ( )

( ) c c

KG jT j 1 , KG

1 KG j

ωω = ≅ ω << ω ω >> ω

+ ω

B d Pl tti P d f C it C




• Bode Plotting Procedure for Composite Curves:

– Rewrite the sinusoidal transfer function as a product of

the four basic factors.

– Determine the value of 20log10(K) = Lm(K) dB

– Plot the low-frequency magnitude asymptote through

the point Lm(K) at ω = 1 with a slope 20(m) dB per

decade.

– Complete the composite magnitude asymptotes

• Extend the low-frequency asymptote until the first frequency

break point, then step the slope by ± r(20) or ± p(40),depending on whether the break point is from a 1st-order or

2nd-order term in the numerator or denominator. Continue

through all break points in ascending order.

( ) ( ) ( )

p

m r 2

2

n n

2 1K i 1+i T 1 i i

±

± ± ⎡ ⎤⎛ ⎞ ⎛ ⎞ζω ω + ω + ω⎢ ⎥⎜ ⎟ ⎜ ⎟ω ω⎝ ⎠ ⎝ ⎠⎣ ⎦

Sk t h i th i t it d




– Sketch in the approximate magnitude curve:

Increase the asymptote value by a factor of +3 dB

at 1st-order numerator break points, and decreaseit by a factor of -3 dB at 1st-order denominator

break points. At 2nd-order break points, sketch in

the resonant peak (or valley) using the relationthat at the break point ω = ωn:

– Plot the low-frequency asymptote of the phase

curve, φ = ± m(90°). – As a guide, sketch in the approximate phase curve

by changing the phase by ± 90° or ± 180° at each

break point in ascending order.

( ) ( )

1

2

102

n n

2 1

Lm 1 i i 20log 2

±⎡ ⎤⎛ ⎞ζ

+ ω + ω = ± ζ⎢ ⎥⎜ ⎟ω ω⎝ ⎠⎣ ⎦




– Locate the asymptotes for each individual phase

curve so that their phase change corresponds to

the steps in the phase toward or away from the

approximate curve. Sketch in each individual

phase curve as indicated by the detailed phase

plots for the individual terms. – Graphically add each phase curve. Use grids if an

accuracy of about ± 5° is desired. If less accuracy

is acceptable, the composite curve can be doneby eye. Keep in mind that the curve will start at

the lowest-frequency asymptote and end on the

highest-frequency asymptote and will approachthe intermediate asymptotes to an extent that is

determined by how close the break points are to

each other.




( )

( ) ( )

s i2 1 2 12000 s 0.5 0.5 0.5

s s i is s 10 s 50s 1 1 i 1 1

10 50 10 50

ω⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟+ ⎝ ⎠ ⎝ ⎠= =ω ω+ + ⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞+ + ω + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠




( )

( )( )

s2 12000 s 0.5 0.5

s ss s 10 s 50s 1 1

10 50

⎛ ⎞+⎜ ⎟+ ⎝ ⎠=+ + ⎛ ⎞⎛ ⎞+ +

⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠




( ) 22

10 2.5

s 0.2s s 0.4s 4s s 1

4 2

=⎛ ⎞+ +

+ +⎜ ⎟⎝ ⎠




( )2

22

0.01 s 0.01s 1

s 0.02s s 1

4 2

+ +

⎛ ⎞+ +⎜ ⎟

⎝ ⎠

More Bode ( )( )12 s s

1 12 6

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠




More Bode

Plotting

Examples

( ) ( )( )

( )( )( ) ( )

s 2 s 6 50 2 6L s

s ss s 1 s 5 s 10s s 1 1 1

5 10

⎜ ⎟⎜ ⎟+ + ⎝ ⎠⎝ ⎠= =+ + + ⎛ ⎞⎛ ⎞+ + +⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

( )

2

2

8 s s1

10 42 8

⎡ ⎤⎛ ⎞+ +⎢ ⎥⎜ ⎟

⎝ ⎠⎢ ⎥




( ) ( )

( )

2

22

10 4s 2s 8 2 2L s

s s 2s 10 s s

s 1510

⎢ ⎥⎜ ⎟+ + ⎝ ⎠⎢ ⎥⎣ ⎦= =⎡ ⎤+ + ⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

( )3 s

1s 3 8 3

⎛ ⎞+⎜ ⎟+ ⎝ ⎠




( ) ( )

( )22

s 3 8 3L s

ss s 8s 1

8

+ ⎝ ⎠= =+ ⎛ ⎞+⎜ ⎟

⎝ ⎠

( )

2

24 s

12

⎛ ⎞+⎜ ⎟⎝ ⎠




( ) ( )

( )( ) 22 2

2

s 2 250 2L s

s s 10 s 6s 25 s s 6s 1 s 1

10 5 25

⎜ ⎟+ ⎝ ⎠= =⎡ ⎤+ + + ⎛ ⎞ ⎛ ⎞+ + +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦



( )( )285 s 1 s 2s 43 25+ + +




( ) ( )( )

( )( )( )( )

( )( )

2 2 2

2

85 s 1 s 2s 43.25G s

s s 2s 82 s 2s 101

85 s 1 s 1 6.5j

s s 1 9j s 1 10j

+ + +=

+ + + +

+ + ±=

+ ± + ±

Nyquist Plot

Bode Plots




• Advantages of Working with Frequency Response in

terms of Bode Plots: – Bode plots of systems in series simply add, which

is quite convenient.

– Bode’s important phase-gain relationship is givenin terms of logarithms of phase and gain.

– A much wider range of system behavior – from

low- to high-frequency behavior – can bedisplayed.

– Bode plots can be determined experimentally.

– Dynamic compensator design can be basedentirely on Bode plots.




• Why is it important for an engineer to know how to

hand-plot frequency responses? – Allows engineer to deal with simple problems but

also to check computer results for more

complicated cases. – Often approximations can be used to quickly

sketch the frequency response and deduce

stability as well as determine the form of theneeded dynamic compensations.

– Hand plotting is useful in interpreting frequency-

response data that have been generatedexperimentally.

• Minimum Phase and Nonminimum Phase Systems




• Minimum-Phase and Nonminimum Phase Systems

– Transfer functions having neither poles nor zeros in

the RHP are minimum-phase transfer functions.

– Transfer functions having either poles or zeros in the

RHP are nonminimum-phase transfer functions.

– 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 the range in phase angle of anynonminimum-phase transfer function is greater than

this minimum.

– For a minimum-phase system, the transfer functioncan be uniquely determined from the magnitude

curve alone. For a nonminimum-phase system, this

is not the case.

– Consider as an example the following two systems:




Frequency (rad/sec)

P h a s e ( d e g ) ; M a g n

i t u d e ( d B )

Bode Diagrams

-6

-4

-2

0From: U(1)

10-2 10-1 100-200

-150

-100

-50

0

T o :

Y ( 1 )

Consider as an example the following two systems:

( ) ( )1 1

1 2 1 22 2

1 T s 1 T s

G s G s 0 T T1 T s 1 T s

+ −

= = < <+ +

G1(s)

G2(s)

A small amountof change in

magnitude

produces a small

amount ofchange in the

phase of G1(s)

but a much

larger change inthe phase of

G2(s).

T1 = 5, T2 = 10

– These two systems have the same magnitude




y g

characteristics, but they have different phase-angle

characteristics. – The two systems differ from each other by the factor:

– This factor has a magnitude of unity and a phase angle

that varies from 0° to -180° as ω is increased from 0 to

∞. – For the stable minimum-phase system, the magnitude

and phase-angle characteristics are uniquely related.

This means that if the magnitude curve is specified overthe entire frequency range from zero to infinity, then the

phase-angle curve is uniquely determined, and vice

versa. This is called Bode’s Gain-Phase relationship.

1

1

1 T sG(s)

1 T s

−=

+

• Bode’s Gain-Phase Relationship




When the slope of the magnitude

vs. ω on a log-log scale persists at

a constant value for approximately

a decade of frequency, the

relationship is particularly simpleand is given by the relationship

where n is the slope of themagnitude curve in units of decade

of amplitude per decade of

frequency.

Bode s Gain Phase Relationship

– For any minimum-phase system (i.e., one with no

RHP zeros or poles), the phase of G(jω) isuniquely related to the magnitude of G(jω).

( )G j n 90∠ ω ≅ × °

– For stability we want the angle of G(jω) > -180° for a PM




> 0. Therefore, we adjust the magnitude curve so that it

has a slope of -1 at the crossover frequency, ωc, that is,where the magnitude = 1.

– If the slope is -1 for a decade above and below the

crossover frequency, then the PM ≈ 90°. – However, to ensure a reasonable PM, it is usually

necessary only to insist that a slope of -1 (-20 dB per

decade) persist for a decade in frequency that is

centered at the crossover frequency.

– So a design procedure is to adjust the slope of the

magnitude curve so that it crosses over magnitude 1 with

a slope of -1 for a decade around ωc to provide

acceptable PM, and hence adequate damping. Then

adjust the system gain to give a ωc that will yield the

desired bandwidth (and, hence, speed of response).

Simple Design Example

Design Objective: good damping and an

approximate bandwidth of 0 2 rad/s




( ) ( )DKD s K T s 10.01(20s 1)

= += +

Adjust gain K to produce the desired bandwidth and adjust the

breakpoint ω1 = 1/TD to provide the -1 slope at ωc.

ω1 = .05 rad/sωc = .2 rad/s

approximate bandwidth of 0.2 rad/s.

Step

Response



– It is therefore possible to detect whether a system is




p y

minimum phase by examining both the slope of the

high-frequency asymptote of the log-magnitude curveand the phase angle at ω = ∞. If the slope of the log-

magnitude curve as ω → ∞ is –20(q – p) dB/decade

and the phase angle at ω = ∞ is equal to -90°(q – p),then the system is minimum phase.

– Nonminimum-phase systems are slow in response

because of their faulty behavior at the start of the

response.

– In most practical control systems, excessive phase

lag should be carefully avoided. A common example

of a nonminimum-phase element that may be present

in a control system is transport lag:dts

dte 1−τ = ∠ − ωτ

– Dead-time approximation comparison




10-1

100

101

102

103

104

105

-400

-350

-300

-250

-200

-150

-100

-50

0

frequency (rad/sec)

p h a s e a n g l e ( d e g r e s s )

Dead-Time Phase-Angle Approximation Comparison

pp p

( )o dt

i dt

Q 2 ss

Q 2 s

− τ=

+ τ

( )

( )

( )

2

dt

dto

2

i dt

dt

s2 s

Q 8sQ s

2 s8

τ− τ +

=τ

+ τ +

dts

dte 1−τ = ∠ − ωτ

τdt = 0.01

Step ResponseUnit Step Responses




Time (sec.)

A m p l i t u d e

Step Response

0 2 4 6 8 10 12-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

From: U(1)

T o : Y ( 1 )

Unit Step Responses

21

s s 1+ +

2

s 1

s s 1

++ +

2

s 1

s s 1

− +

+ +





Time (sec.)

A m p l i t u d e

Step Response

0 2 4 6 8 10 12-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

From: U(1)

T o : Y ( 1 )

21

s s 1+ +

2s

s s 1+ +

2s 1

s s 1+

+ +

Unit Step Responses





Time (sec.)

A m p l i t u d e

0 2 4 6 8 10 12-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

From: U(1)

T o : Y ( 1 )

Unit Step Responses

2 s 1s s 1− ++ +

2

s

s s 1

−

+ +

2

1

s s 1+ +




( )

( )

1

2

s 1G s 10

s 10

s 1G s 10

s 10

+=

+−=

+

Another Example:

Minimum-Phase

&

Nonminimum-Phase

Systems

• Review of Steady-State Error




Review of Steady State Error

– Assume H(s) = 1 and D(s) = 0. The error is then

E(s) which equals R(s) – C(s).

+-

C(s)R(s)G

c(s ) G(s)

H(s)

E(s)

B(s)

D(s)

Σ+

+

c

E(s) 1

R(s) 1 G (s)G(s)= + c

R(s)

E(s) 1 G (s)G(s)= +

sss 0 s 0

c

R(s)e (t) limsE(s) lims

1 G (s)G(s)→ →

= =

+

– Step Input: R(s) = 1/s




p p ( )

– Ramp Input: R(s) = 1/s2

sss 0 s 0

c c p

1

1 1se (t) lims lim1 G (s)G(s) 1 G (s)G(s) 1 K → →

= = =+ + +

p cs 0K limG (s)G(s)→≡

2

sss 0 s 0

c c

s 0c v

11se (t) lims lim

1 G (s)G(s) s sG (s)G(s)

1 1limsG (s)G(s) K

→ →

→

= =+ +

= =v c

s 0K limsG (s)G(s)

→≡

Static Error Constants:

K p

and K v

• Steady-State Errors: System Type and Gain as




y y yp

Related to Log Magnitude Curves

– Consider a unity-feedback control system.

– The steady-state error of this closed-loop system

depends on the system type and the gain. The

system error coefficients are determined by thesetwo characteristics. For any given log magnitude

curve the system type and gain can be

determined. – The steady-state step, ramp, and parabolic error

coefficients describe the low-frequency behavior

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 ωapproaches zero.

– The type of the system determines the slope of the

log-magnitude curve at low frequencies. – Information concerning the existence and

magnitude of the steady-state error of a control

system to a given input can be determined fromthe observation of the low-frequency region of the

log-magnitude curve.

– Type 0 System( ) 0K




yp y

• The slope at low frequencies is zero.

• The magnitude at low frequencies is 20log10K0.

• The gain K0 is the steady-state step error coefficient.

– Type 1 System• The slope at low frequencies is –20 dB/decade.

• The intercept of the low-frequency slope of –20 dB/decade (or

its extension) with the 0 dB axis occurs at the frequency ω = K1.

• The value of the low-frequency slope of –20 dB/decade (or its

extension) at the frequency ω = 1 is equal to 20log10K1.

• The gain K1 is the steady-state ramp error coefficient.

( )( )

1K G s

s Ts 1=

+

( ) 0K G s

Ts 1=

+




– Type 2 System

• The slope at low frequencies is –40 dB/decade.• The intercept of the low-frequency slope of –40

dB/decade (or its extension) with the 0 dB axis occurs at

a frequency ω = (K2)1/2.

• The value on the low-frequency slope of –40 dB/decade

(or its extension) at the frequency ω = 1 is equal to

20log10K2.

• The gain K2 is the steady-state parabolic error coefficient.

( )( )

2

2

K G s

s Ts 1=

+

N t l St bilit




Neutral Stability

• If the closed-loop transfer function of a system is

known, which is usually not the case, one can

determine the stability of the system by simplyinspecting the denominator in factored form to

observe whether the real parts are positive or

negative.• One can determine closed-loop stability by evaluating

the frequency response of the open-loop transfer

function and then performing a test on that response.This is the Nyquist Stability Test.




• All points on the root locus have the property:




• At the point of neutral stability, the root locus

conditions hold for s = jω, so:

• Thus a Bode plot of a system that is neutrally stable

(that is, with K defined such that a closed-loop root

falls on the imaginary axis) will satisfy these

conditions.

• The Bode magnitude response corresponding to

neutral stability passes through 1 (0 dB) at the same

frequency at which the phase passes through 180°.

( ) ( )KG s 1 G s 180

°

= ∠ =

( ) ( )KG j 1 G j 180°ω = ∠ ω =



Neutral Stability Example ( )( )( ) ( )

22

K 0.1K KG s

ss 10 s 1 1 s 1

= =⎛ ⎞+ + + +⎜ ⎟




( )( ) ( )1 s 110

+ +⎜ ⎟⎝ ⎠

K = 242




K = 32




K = 900

Nyquist Stability Criterion





• The advantages of the Nyquist stability criterion over

the Routh criterion are: – It uses the open-loop transfer function, i.e.,

(B/E)(s), to determine the number, not the

numerical values, of the unstable roots of theclosed-loop system characteristic equation. The

Routh criterion requires the closed-loop system

characteristic equation to determine the same

information.



– In addition to answering the question of absolute




stability, Nyquist also gives some useful results on

relative stability, i.e., gain margin and phasemargin. Furthermore, the graphical plot used,

keeps the effects of individual pieces of hardware

more apparent (Routh tends to "scramble themup") making needed design changes more

obvious.

– The Nyquist Stability Criterion will handle stabilityanalysis of complex systems with one or more

resonances, with multiple magnitude-curve

crossings of 1.0, and with multiple phase-curve

crossings of 180°. It handles open-loop unstable

systems, nonminimum-phase systems, and

systems with pure time delay.

• The Argument Principle




g p

– Consider the transfer function H1

(s) with poles and

zeros shown below.

– The evaluation of H1(s) on the contour C1 is also

shown.

( )1 2 1 2α = θ + θ − φ + φ

Since there are no poles or zeros within C1, α undergoes

no net change of 360° and H1(s) will not encircle the origin.

– Consider the transfer function H2(s) with poles and

zeros shown below Note the pole within C




zeros shown below. Note the pole within C1.

– The evaluation of H2(s) on the contour C1 is alsoshown.

Since there is a pole within C1, α undergoes a net change

of -360° after a full traverse of C1. Therefore H2(s) will

encircle the origin in the CCW direction.

( )1 2 1 2α = θ + θ − φ + φ

– The Argument Principle can be stated as follows:




• A contour map of a complex function will

encircle the origin Z – P times, where Z is the

number of zeros and P is the number of poles

of the function inside the contour.

– How is the Argument Principle applied in controldesign?

Consider a contour C1 encirclingthe entire RHP, where a pole

would cause an unstable system.

The resulting evaluation of H(s)will encircle the origin only if H(s)

has a RHP pole or zero.

– What makes all this contour behavior useful is that

t l ti f l KG( ) b




a contour evaluation of an open-loop KG(s) can be

used to determine stability of the closed-loopsystem.

– The closed-loop roots are the solutions to

1 + KG(s) = 0. Apply the principle of the argumentto the function 1 + KG(s). If the evaluation contour

of this function enclosing the entire RHP contains

a pole or zero of 1 + KG(s), then the evaluatedcontour will encircle the origin.

– Note that 1 + KG(s) is simply KG(s) shifted to the

right one unit.

( )

( ) ( )

( )

( )

Y s KG s

T sR s 1 KG s= = +

– If the plot of 1 + KG(S) encircles the origin, the plot




of KG(s) will encircle -1 on the real axis.

– Therefore, we can plot the contour evaluation of

the open-loop KG(s), examine its encirclements of

-1, and draw conclusions about the origin

encirclements of the closed-loop function 1 +KG(s). This is called a Nyquist plot, or polar plot,

because we plot the magnitude of KG(s) versus

the angle of KG(s).

Nyquist Plots

Evaluation of

KG(s)&

1 + KG(s)

– To determine whether an encirclement is due to a




pole or zero, we write 1 + KG(s) in terms of poles

and zeros of KG(s):

– The poles of 1 + KG(s) are also the poles of G(s).

It is safe to assume that the poles of G(s) (i.e.,

factors of a(s)) are known, the rare existence of

any of these poles in the RHP can be accountedfor. Assuming that there are no poles of G(s) in

the RHP, an encirclement of -1 by KG(s) indicates

a zero of 1 + KG(s) in the RHP, and thus anunstable root of the closed-loop system.

( ) b(s) a(s) Kb(s)

1 KG s 1 K a(s) a(s)

++ = + =

A clockwise contour C enclosing a zero of 1 + KG(s)




– A clockwise contour C1 enclosing a zero of 1 + KG(s),

i.e., a closed-loop root, will result in KG(s) encircling-1 in a CW direction.

– If C1 encloses a pole of 1 + KG(s), i.e., an unstable

open-loop pole, there will be a CCW KG(s)encirclement of -1.

– The net number of CW encirclements, N, equals the

number of zeros (closed-loop system roots) in theRHP, Z, minus the number of open-loop poles in the

RHP, P.

– N = Z – P. This is the key concept of the Nyquist

Stability Criterion.

– Any KG(s) that represents a physical system will

h t i fi it f i h




have zero response at infinite frequency, i.e., has

more poles than zeros. This means that the bigarc of C1 corresponding to s at infinity results in

KG(s) being an infinitesimally small point near the

origin for that portion of C1.

• Procedure for Plotting the Nyquist Plot

Plot KG(s) for jω ≤ s ≤ +jω




– Plot KG(s) for –jω ≤ s ≤ +jω.

• Evaluate KG(jω) for ω = 0 to ωh, where ωh is so largethat the magnitude of KG(jω) is negligibly small for ω >

ωh, then reflect the image about the real axis, adding it

to the preceding image. The magnitude of KG(jω) willbe small at high frequencies for any physical system.

The Nyquist plot will always be symmetrical with

respect to the real axis. – Evaluate the number of CW encirclements of -1, and call

that number N. N is negative for CCW encirclements.

– Determine the number of unstable (RHP) poles of G(s),and call that number P.

– Z = N + P, which is the number of unstable closed-loop

roots. Z = 0 for stability.




Feedback Control SystemBlock Diagram

Plausibility Demonstration for the Nyquist Stability Criterion

– Consider a sinusoidal input to the open-loop




– Consider a sinusoidal input to the open-loop

configuration. Suppose that at some frequency,(B/E)(iω) = -1 = 1 ∠ 180°. If we would then close

the loop, the signal -B would now be exactly the

same as the original excitation sine wave E and

an external source for E would no longer be

required. The closed-loop system would maintain

a steady self-excited oscillation of fixed amplitude,

i.e., marginal stability.

– It thus appears that if the open-loop curve

(B/E)(iω) for any system passes through the -1

point, then the closed-loop system will bemarginally stable.

– However, the plausibility argument does not make

clear what happens if curve does not go exactly




clear what happens if curve does not go exactly

through -1. The complete answer requires a rigorousproof and results in a criterion that gives exactly the

same type of answer as the Routh Criterion, i.e., the

number of unstable closed-loop roots.

– Instead, we state a step-by-step procedure for the

Nyquist criterion.

1. Make a polar plot of (B/E)(iω) for 0 ≤ ω < ∞ , eitheranalytically or by experimental test for an existing

system. Although negative ω's have no physical

meaning, the mathematical criterion requires that we

plot (B/E)(-iω) on the same graph. Fortunately this iseasy since (B/E)(-iω) is just a reflection about the real

(horizontal) axis of (B/E)(+iω).

Polar Plot of Open-Loop( ) ( )1 2

Bi G G H i

Eω = ω




Frequency Response

Simplified Version of


E




Examples

of Polar Plots

2. If (B/E)(iω) has no terms (iω)k, i.e., integrators,

as multiplying factors in its denominator the plot




as multiplying factors in its denominator, the plot

of (B/E)(iω) for -∞ < ω < ∞ results in a closedcurve. If (B/E)(iω) has (iω)k as a multiplying

factor in its denominator, the plots for +ω and -ωwill go off the paper as ω → 0 and we will not get

a single closed curve. The rule for closing such

plots says to connect the "tail" of the curve at ω→ 0− to the tail at ω → 0+ by drawing k clockwise

semicircles of "infinite" radius. Application of thisrule will always result in a single closed curve so

that one can start at the ω = -∞ point and trace

completely around the curve toward ω = 0-

and ω= 0+ and finally to ω = +∞, which will always be

the same point (the origin) at which we started

with ω = -∞.



3. We must next find the number Np of poles of

G1G2H(s) that are in the right half of the complex




G1G2 (s) t at a e t e g t a o t e co p e

plane. This will almost always be zero since thesepoles are the roots of the characteristic equation of

the open-loop system and open-loop systems are

rarely unstable. If the open-loop poles are not

already factored and thus apparent, one can apply

the Routh criterion to find out how many unstable

ones there are, if any. If G1G2H(iω) is not known

analytically but rather by experimentalmeasurements on an existing open-loop system,

then it must have zero unstable roots or else we

would never have been able to run the necessaryexperiments because the system would have been

unstable. We thus generally have little trouble

finding Np

and it is usually zero.

4 We now return to our plot (B/E)(iω) which has




4. We now return to our plot (B/E)(iω), which has

already been reflected and closed in earliersteps. Draw a vector whose tail is bound to the -

1 point and whose head lies at the origin, where

ω = -∞. Now let the head of this vector tracecompletely around the closed curve in the

direction from w = -∞ to 0- to 0+ to +∞, returning

to the starting point. Keep careful track of the

total number of net rotations of this test vector

about the -1 point, calling this Np-z and making it

positive for counter-clockwise rotations and

negative for clockwise rotations.



• Nyquist Plot Example



Frequency Response Analysis & Design K. Craig 124N = 0, P = 0, therefore Z = 0.

• Often the control systems engineer is more interested




y g

in determining a range of gains K for which thesystem is stable than in testing for stability at a

specific value of K.

– To accomplish this, scale KG(s) by K and examineG(s) to determine stability for a range of gains K.

This is possible because an encirclement of -1 by

KG(s) is equivalent to an encirclement of -1/K by

G(s).

– Therefore, instead of having to deal with KG(s),

we need only consider G(s), and count the

number of encirclements of the point -1/K.




• In this example, the open-loop pole at s = 0 creates

i fi it it d f G( ) t 0 T tl




an infinite magnitude of G(s) at ω = 0. To correctly

determine the number of -1/K encirclements, we mustdraw this arc in the proper half plane. Should it cross

the positive real axis or the negative one? It is also

necessary to assess whether the arc should sweep180°, 360°, or 540°.

Modify the C1 contour, as shown.

Because the phase of G(s) is thenegative of the sum of the angles

from all the poles, the evaluation

results in a Nyquist plot moving

from +90° for s just below thepole at s = 0, across the positive

real axis to -90° for s just above

the pole.

• Nyquist Plot Example




Nyquist Stability Criterion Examples ( ) 2

1G s

s=




( ) 2

1G s

s=




Contour Nyquist Plot

s

0+

ω =

0

−

ω =

ω = +∞

ω = −∞

( ) 2 2

0

1G s

s=

+ ω




0

0 2ω =




( ) ( )

( )

K s 2KG s

s 10

+=

+

Closed-Loop System

is stable for any K > 0




( )

( )( )

2

K KG s

s 10 s 2=

+ +

Closed-Loop System is stable for 0 < K < 576;

K > 576 Closed-Loop System has two unstable roots




( ) ( )( )

( )( )3

K s 10 s 1KG s

s 100 s 2

+ +=

+ +

Closed-Loop System

is stable for any K > 0

( ) ( )

( )

K s 1KG s

s s 3

+=

+

Range of K for stability is -1/K < 0

N = 0, P = 0, Z = N + P = 0

Closed Loop System is Stable for K > 0




( )Closed-Loop System is Stable for K > 0

Range of K for stability is -1/K < 0

N = -1, P = 1, Z = N + P = 0

Closed-Loop System is Stable for K > 0




Closed Loop System is Stable for K > 0




( )( )( )2

1G s

s 1 s 2s 2=

+ + +



s 1+




( )s 1

G s s 10

+

= +

For any K > 0, N = 0, P = 0, and Z = N + P = 0

Closed-Loop System is Stable




( )s 1

G ss 10

−= −

+




( )s 1

G ss 10

+=

−




Nyquist Stability

Analysis

of aSystem

with

Dead Time

– The Nyquist criterion treats without approximation

systems with dead times. Since the frequency




response of a dead time element τdt is given bythe expression 1∠-ωτdt, the (B/E)(iω) for the

system of Figure (a) spirals unendingly into the

origin. With low loop gain, the closed-loop system

is stable, i.e., Np = 0 and Np-z = 0.

– Raising the gain, Figure (b), expands the spirals

sufficiently to cause the test vector to experience

two net rotations, i.e., Np-z = -2, causing closed-

loop instability. Further gain increases expand

more and more of these spirals out to the region

beyond the -1 point, causing Np-z to increase,indicating the presence of more and more

unstable closed-loop roots.

Stability Margins




• Two open-loop performance criteria in common use

to specify relative stability are gain margin and phase

margin.• The open-loop frequency response is defined as

(B/E)(iω). One could open the loop by removing the

summing junction at R, B, E and just input a sinewave at E and measure the response at B. This is

valid since (B/E)(iω) = G1G2H(iω).

• The utility of open-loop frequency-response rests onthe Nyquist stability criterion.

• Gain margin (GM) and phase margin (PM) are in the

nature of safety factors such that (B/E)(iω) stays far




enough away from 1 ∠ -180° on the stable side.• Gain margin is the multiplying factor by which the

steady-state gain of (B/E)(iω) could be increased

(nothing else in (B/E)(iω) being changed) so as to putthe system on the edge of instability, i.e., (B/E)(iω))

passes exactly through the -1 point. This is called

marginal stability.• Phase margin is the number of degrees of additional

phase lag (nothing else being changed) required to

create marginal stability.

• Both a good gain margin and a good phase margin

are needed; neither is sufficient by itself.

Open-Loop Performance Criteria:





A system must have adequate stability margins. Both a good gain margin and a good phase margin

are needed.

Useful lower bounds:GM > 2.5 PM > 30°

Bode Plot View of




The phase margin can be

determined for any value of K.




y

Indicate on the figure where

for selected trial values of K.

( )KG j 1ω =

If we wish a certain PM, we

simply read the value of

corresponding to the

frequency that would create

the desired PM and note thatthe magnitude at this

frequency is 1/K.

( )G jω

• It is important to realize that, because of model

uncertainties, it is not merely sufficient for a system to be




stable, but rather it must have adequate stabilitymargins.

• Stable systems with low stability margins work only on

paper; when implemented in real time, they arefrequently unstable.

• The way uncertainty has been quantified in classical

control is to assume that either gain changes or phasechanges occur. Typically, systems are destabilized

when either gain exceeds certain limits or if there is too

much phase lag (i.e., negative phase associated with

unmodeled poles or time delays).

• As we have seen these tolerances of gain or phase

uncertainty are the gain margin and phase margin.

• The PM is more commonly used to specify control system

performance because it is most closely related to the




damping ratio of the system.

• One can also relate transient-response overshoot (Mp)

and frequency-response resonant peak (Mr ) to phase

margin (PM) for a second-order closed-loop system.

( )( )

( )

2

n

n

2

n

2 2

n n

1

4 2

G ss s 2

T ss 2 s

2PM tan

1 4 2

−

ω=

+ ζω

ω=

+ ζω + ω

⎡ ⎤ζ⎢ ⎥=⎢ ⎥+ ζ − ζ⎣ ⎦

PM

100ζ ≈ ( )

2

nG ss s 2

ω=

+ ζω




( )( )

( )

2

n

n

2

n

2 2

n n

1

4 2

G s

s s 2

T ss 2 s

2PM tan

1 4 2

−

ω=

+ ζωω

=+ ζω + ω

⎡ ⎤ζ⎢ ⎥=

⎢ ⎥+ ζ − ζ⎣ ⎦

100

( )( )

n2

n

2 2

n n

s s 2

T ss 2 s

+ ζω

ω=

+ ζω + ω

( ) ( )( )

( )( )

2

2 2 2

85 s 1 s 2s 43.25G s

s s 2s 82 s 2s 101

+ + +=

+ + + +




( )( )( )( )2

85 s 1 s 1 6.5j

s s 1 9j s 1 10j

+ + ±=+ ± + ±

Nyquist Plot

Bode Plots

The actual stability margin can beassessed only be examining the

Nyquist plot to determine its closest

approach to the -1 point.

Vector MarginVector Margin is

the distance to




the -1 point fromthe closest

approach of the

Nyquist plot.

This is a single-margin

parameter and it

removes allambiguities in

assessing

stability thatcome from using

GM and PM in

combination.

Conditionally-Stable System

2

Here an increase in gain can

make the system stable.The Bode plot yields a PM = +10° and a GM

= 0.7 for K = 7. These are conflicting. For




Root-Locus Plot Nyquist Plot (K = 7)

( ) ( )2

3

K s 10KG s

s+=

g

systems like this, use the root-locus and/orNyquist plot to determine stability.



( )

s100 1

10G s

s s

⎡ ⎤+⎢ ⎥⎣ ⎦=⎡ ⎤ ⎡ ⎤

Note: The RHP pole at s = 1 causes -

180º shift from the -90º that you would

normally expect from a normal system




s 1 11 100− +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ with all the singularities in the LHP.

K = 1

K = 0.1112

Marginal

Stability

Nyquist Plot




K = 1

( )

( )

3 2

2

s 2L s

s s 2

K s 2s 2

+=+ −

+ +




K = 0.5

( )( )

3 2T s s s Ks 2K 2= + + + −




K = 0.5




K = 1




K = 1.5




K = 2




K = 2.5




K = 2.5

( ) ( )

( )( )2

3.2 s 1G s

s s 2 s 0.2s 16

+=

+ + +

Note: Even though the PM is

92.8º, the GM is only 1 dB.




( )( ) ( )

( )2

1 1G s H s

s 1s 2 s 4= =

++ +



K = 5.9




ζ = 0.707

K = 5.9




K = 31ζ = 0.262ζ = 0.944




K = 31




Same

Root-Locus

Plots

Same

Closed-Loop Poles

Different Zeros




P = 0

N = 0 for 0 < K < 4Therefore Z = 0

P = 0N = 2 for 4 < K < ∞

Therefore Z = 2

( ) ( )

( )22

K s 1KG s

s s 10

+=

+




K = 50.9

Closed-Loop Bode Plot




Frequency-Response Design




• Here we consider the design and compensation of

SISO, linear, time-invariant control systems by the

frequency-response approach.

• In this approach, transient-response performance,

which is usually most important, is specified in an

indirect manner.

• Phase margin, gain margin, and resonant peak

magnitude give a rough estimate of system damping.

• Gain crossover frequency resonant frequency and




• Gain crossover frequency, resonant frequency, andbandwidth give a rough estimate of the speed of

transient response.

• Static error constants give the steady-state accuracy.• Although the correlation between the transient

response and frequency response is indirect, the

frequency domain specifications can be convenientlymet in the Bode diagram approach.

• Design the open loop by the frequency response

method, determine the closed-loop poles and zeros,

and check that the transient response specifications




and check that the transient-response specificationshave been met. If not, then iterate.

• This method can be used for systems or components

whose dynamic characteristics are given in the formof frequency-response data.

• When dealing with high-frequency noises, this

approach is more convenient.• Two approaches in frequency-domain design:

– Polar-plot approach

– Bode-diagram approach

• Polar-Plot Approach

– When a compensator is added, the polar plot does

not retain the original shape and therefore we need




not retain the original shape, and, therefore, we needto draw a new polar plot, which is time consuming

and inconvenient.

• Bode-Diagram Approach

– The compensator can simply be added to the original

Bode diagram, and thus plotting the complete Bode

diagram is a simple matter. – If the open-loop gain is varied, the magnitude curve is

shifted up or down without changing the slope of the

curve, and the phase curve remains the same. – For design purposes, the Bode diagram is preferred.

• A common approach to the Bode Diagram is:

– First adjust the open-loop gain so that the




– First adjust the open-loop gain so that therequirement on the steady-state accuracy is met.

– Then plot the magnitude and phase curves of

the uncompensated, but gain-adjusted, open-loop system.

– Reshape the open-loop transfer function with the

addition of a suitable compensator to meet gainmargin and phase margin specifications.

– Try to meet other specifications, if any.



• The gain in the low-frequency region should be large

enough for steady-state error and disturbance




enough for steady-state error and disturbancerejection properties.

• Near the gain crossover frequency, chosen for speed

of response requirements, the slope of the log-magnitude curve should be -20 dB/decade and

should extend over a sufficiently wide frequency band

to assure a proper phase margin.

• In the high-frequency region, the gain should be

attenuated as rapidly as possible to minimize noise

effects.

• Consider the following design problem:

– Given a plant transfer function G2

(s), find a

compensator transfer function G (s) which yields




compensator transfer function G1(s) which yields

the following:

• stable closed-loop system

• good command following

• good disturbance rejection

• insensitivity of command following to modelingerrors (performance robustness)

• stability robustness with unmodeled dynamics

• sensor noise rejection

• Without closed-loop stability, a discussion of performance

is meaningless. It is critically important to realize that the

compensator G1(s) is actually designed to stabilize ai l l l t U f t t l th tG ( )∗




compensator G (s) is actually designed to stabilize anominal open-loop plant . Unfortunately, the true

plant is different from the nominal plant due to

unavoidable modeling errors, denoted by δG2(s). Thus

the true plant may be represented by:

• Knowledge of δG2

(s) should influence the design of G1

(s).

We assume here that the actual closed-loop system,

represented by the true closed-loop transfer function, is

absolutely stable.

2G (s)∗

2 2 2G (s) G (s) G (s)∗= + δ

1 2 2

1 2 2

G (s) G (s) G (s)

1 G (s) G (s) G (s)

∗

∗

⎡ ⎤+ δ⎣ ⎦⎡ ⎤+ + δ⎣ ⎦

(unity feedback assumed)

Design a Good Single-Input, Single-Output Control Loop

• stable closed-loop system

• good command following




Smooth transition from the low to high-frequency range, i.e., -20 dB/decade

slope near the gain crossover frequency

Frequencies for good

command following,

disturbance reduction,

sensitivity reduction

Sensor noise,

unmodeled high-

frequency dynamics

are significant here.

Gain below this level

at high frequencies

Gain above this level

at low frequencies

good command following

• good disturbance rejection

• insensitivity of command

following to modeling

errors (performancerobustness)

• stability robustness with

unmodeled dynamics

• sensor noise rejection

Lead Compensation

• Lead compensation approximates PD control A PDt t f f ti h th f ( )G K K+




Lead compensation approximates PD control. A PDcompensator transfer function has the form:

• The PD compensator was shown to have a stabilizing

effect on the root-locus of a second-order system.• The Bode diagram shows the stabilizing influence in the

increase in phase at frequencies above the break point.

We use this compensation by locating the break point so

that the increased phase occurs in the vicinity of crossover

(where the log magnitude = 0 dB), thus increasing the

phase margin.

• The magnitude of this compensation continues to grow

with increasing frequency (impossible with physical

elements) thus amplifying high-frequency noise.

( )c p dG s K K s= +

B )

Bode Diagrams

30

40

50

From: U(1)Bode Diagram:

PD Controller




Frequency (rad/sec)

P h a s e ( d e g ) ; M a g n i t u d e ( d B

0

10

20

10-2 10-1 100 101 1020

20

40

60

80

100

T o : Y ( 1 )

( )c p d

d p

p

G s K K s

K K 1 s

K

= +⎛ ⎞

= +⎜ ⎟⎜ ⎟⎝ ⎠

K p =1 K d = 1

PD Controller

D(s) = T s + 1




D(s) = TDs + 1

• In order to alleviate the high-frequency amplification

of the PD compensation, a first-order pole is added in

the denominator at frequencies higher then thebreakpoint of the PD compensator The phase




the denominator at frequencies higher then thebreakpoint of the PD compensator. The phase

increase (or lead) still occurs, but the amplification at

high frequencies is limited.

• Lead Compensation Characteristics

– improves stability margins; adds damping to the system

– yields a small change in steady-state accuracy

– yields a higher gain crossover frequency which means

higher bandwidth which means a reduction in settling time

– is more susceptible to high-frequency noise because of

increase in high-frequency gain due to the increase inbandwidth

– raises the order of the system by one

• Lead Compensator: A high-pass filter

( )c c c

1s

Ts 1 TG s K K (0 < < 1)1

++= α = α




• The minimum value of α (usually 0.05) is limited by

the physical construction of the compensator, either

analog or digital. Therefore the maximum phase lead

that may be produced by a lead compensator is

about 65°.• Lead Compensator Polar Plot:

( )c c cTs 1 TG s K K (0 1)1Ts 1

sT

+= α = αα + +α

c j T 1K (0 1) j T 1ω +α < α <ωα +

m

m

m

112sin ( )

1 1

2

1 sin

1 sin

− α− α

φ = =+ α + α

− φα =

+ φ( ) ( )1 1tan T tan T− −φ = ω − α ω

Maximum Phase Increase for Lead Compensation

While oneld i




could increase

the phase lead

up to 90° using

higher values ofthe lead ratio

1/α, this

produces

higheramplifications

at higher

frequencies.

Select a value of 1/α that is a good compromise between an acceptable

phase margin and an acceptable noise sensitivity at high frequencies.

65° is the usual maximum. Use a double lead compensator if needed.

φm




12

1( )+ α

ω = 0 ω = ∞

φm

1

21( )− α

α

ω m

Polar Plot of Lead Compensator

with K c = 1

Bode Diagrams

0

Bode Diagram

c

( j T 1)

K ( j T 1)

ω +

α ωα +




Frequency (rad/sec)

P h

a s e ( d e g ) ; M a g n i t u d e (

d B )

-20

-15

-10

-5

10-1

100

101

102

10

20

30

40

50

c

( j T 1)

0.1

T 1K 1

α ωα +α =

==

m1 1

T T

1

T

− −⎛ ⎞⎛ ⎞ω = ⎜ ⎟⎜ ⎟α⎝ ⎠⎝ ⎠

= α

c

1Mag K = α

α

ωm occurs midway between the 2 break-point frequencies

on a log scale.

Φm

( )c c c

1s

s zTG s K K 1 s ps

+ += =

++




( ) psT

++α

m

1 1

T T

z p

− −⎛ ⎞⎛ ⎞ω = ⎜ ⎟⎜ ⎟α⎝ ⎠⎝ ⎠

=

• The amount of phase lead at the midpoint depends

only on α. Our task is to select a value of α that is a

good compromise between an acceptable phase

margin and an acceptable noise sensitivity at high




g p y g

frequencies.

• If a phase lead greater than 65° is required, then a

double lead compensator would be required.

• In lead-network designs, there are 3 primary design

parameters:

– Gain crossover frequency ωgc, which determines

bandwidth, rise time, and settling time.

– Phase margin, which determines the damping

coefficient and the overshoot.

– Low-frequency gain, which determines the steady-

state error characteristics.

Lead Compensation Design Procedure

P i f ti i t h th f




• Primary function is to reshape the frequency-

response curve to provide sufficient phase-lead angle

to offset the excessive phase lag associated with thecomponents of the fixed system.

• Assume performance specifications are given in

terms of phase margin, gain margin, static error

constant, and so on.

• Assume the following lead compensator:

( )c c c

1

sTs 1 TG s K K (0 < < 1)1Ts 1

sT

++= α = αα + +

α

• Let Kcα = K. Determine gain K to satisfy steady-state error

requirements or bandwidth requirements.

– To meet error requirement, pick K to meet ess

– To meet bandwidth requirement pick K so that ω is a




To meet bandwidth requirement, pick K so that ωgc is a

factor of two below the desired closed-loop bandwidth

• Draw the Bode diagram of the gain-adjusted, but

uncompensated, system, i.e., KG(s). Evaluate the phase

margin.

• Determine the necessary phase lead angle φ to be added

to the system. Add 5° - 10° to this value to compensate

for the shift in ωgc.

• Determine the attenuation factor α by using

mm

m

1 1 sin( )sin( )

1 1 sin( )

− α − φφ = ⇒ α =

+ α + φ

• Determine the frequency where the magnitude of

KG(s) is equal to

10

1

20log

⎛ ⎞

− ⎜ ⎟α⎝ ⎠




Select this frequency as the new ωgc. This

corresponds to

and the maximum phase shift φm occurs at this

frequency.

• Determine pole and zero frequencies of the lead

compensator:

g ⎜ ⎟α⎝ ⎠

m

1

Tω = α

zero pole

1 1

T Tω = ω =

α

• Calculate Kc = K/α.

• Draw the compensated frequency response and check the

PM.

• Check the gain margin to be sure it is satisfactory If not




• Check the gain margin to be sure it is satisfactory. If not,

modify the pole-zero location of the compensator.

• Summary – The design problem is to find the best values for the

parameters given the requirements. For a lead

controller, if the low frequency gain is kept the same,the crossover frequency will increase. If the crossover

frequency is kept the same, the low frequency gain will

decrease. So assume a fixed value of one of the three:

PM, low-frequency gain, or ωgc, and then adjust theother two iteratively to meet specifications.

Lead Compensation for a DC Motor ( )( )

1G ss s 1

=+

Performance Specifications:

ess < 0.1 for a unit ramp input and Mp < 25% (PM > 45°)1 1




p p ( )

ss

v cs 0

1 1e

K limsG (s)G(s)→

= ≡ K = 10

α = 0.2

Lead Compensation for a Temperature Control System

( )

( )

K KG s

s s

1 s 1 10.5 2

=⎛ ⎞ ⎛ ⎞

+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Kp

= 9 and PM > 25°




( )⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

ss

p

p cs 0

1e

1 K

K limG (s)G(s)→

=+

≡ K = 9

PM = 7°, ωgc

= 1.9

( ) ( )

( )1

s 1D s 3

s 3

+=

+PM = 16°

( ) ( )( )2s 1.5D s 10s 15+=+

PM = 38°



Lag Compensation

• Lag Compensation Characteristics




• Lag Compensation Characteristics

– reduces the system gain at higher frequencies

without reducing the system gain at low frequencies – reduces the system bandwidth and so the system

has a slower response speed

– has improved steady-state accuracy since the totalsystem gain and hence, low-frequency gain, can be

increased because of the reduced high-frequency

gain – is less susceptible to high-frequency noise since the

high-frequency gain is reduced

• Lag Compensation approximates a Proportional-

Integral (PI) Compensator.

( ) K 1D s s⎛ ⎞= +⎜ ⎟

PI




( )I

D s ss T

⎛ ⎞= +⎜ ⎟⎝ ⎠

PI Compensation has infinitegain at zero frequency, which

reduces steady-state errors, but

has phase decrease atfrequencies lower than the

break point at ω = 1/TI. The

break point is located at a

frequency substantially lowerthan ωgc so the system’s PM is

not affected significantly.

Compensation

( )c c Ts 1G s K Ts 1+= β β +

Lag Compensator: A low-pass filter




( )

c

Ts 1

1s

TK ( 1)1

sT

β β +

+

= β >+

β

Frequency Response

of a Lag Compensation

with α = 10.

Note that α and β are used

interchangeably.

Polar Plot of Lag Compensator

c j T 1K ( 1)ω +β β >




ω = 0ω = ∞

K c K cβ

c ( ) j T 1

β βωβ +

Bode Diagrams

20

Bode Diagram: Lag Compensation




Frequency (rad/sec)

P h a s e ( d e g ) ; M a g n i t u d e ( d B )

0

5

10

15

10

-2

10

-1

10

0

10

1

-50

-40

-30

-20

-10

c

c

( j T 1)K ( j T 1)

10

T 1

K 1

ω +βωβ +

β =

==

• A lag compensator is essentially a low-pass filter. It

permits a high gain at low frequencies whichimproves steady state performance and reduces gain




improves steady-state performance and reduces gain

in the higher range of frequencies so as to improve

the phase margin. The phase-lag characteristic is ofno consequence for compensation purposes.

• The exact location of the lag compensator pole and

zero is not critical provided they are close to theorigin (but not too close) and their ratio is that

required to meet steady-state error requirements.

• The closed-loop pole created by the lag compensator

will adversely affect the transient response (settlingtime) to both a command and a disturbance




time) to both a command and a disturbance.

• The attenuation due to the lag compensator will shift

ωgc to a lower frequency point where the phasemargin is acceptable. The bandwidth of the system

will be reduced and this will result in a slower

transient response.• A lag-compensated system tends to be less stable as

it acts approximately as PI controller. To avoid this, T

should be made sufficiently larger than the largest

time constant of the system.

• Conditional stability may occur when a system having

saturation or limiting, which reduces the effective loop




gain, is compensated by use of a lag compensator.

To avoid this, the system must be designed so that

the effect of lag compensation becomes significantonly when the amplitude of the input to the saturating

element is small. This can be done by means of

minor feedback-loop compensation.

Lag Compensation Design Procedure

• Lag Compensator: 1




• Lag Compensator:

• Define Kcβ = K. Determine gain K to satisfy steady-

state error requirements.

• Draw the Bode diagram of the gain-adjusted, but

uncompensated, system, i.e., KG(s). Evaluate the

phase margin and gain margin.

( )c c c

1s

Ts 1 TG s K K ( 1)

1Ts 1 sT

++= β = β >

β + + β

• If the specifications on gain margin and phase margin

are not satisfied, find the frequency where the phaseangle of KG(s) is -180° plus the required phase




angle of KG(s) is 180 plus the required phase

margin plus 5° to 12° (to compensate for the phase

lag of the lag compensator). Choose this frequency

as the new ωgc.

• The pole and zero of the lag compensator must be

located substantially lower than the new ωgc to

prevent detrimental effects of phase lag due to the

lag compensator. Choose the zero location 1 octave

to 1 decade below the new ωgc.

• Determine the attenuation necessary to bring the

magnitude curve down to 0 dB at the new ωgc. Since

this attenuation is -20 log10β, determine β. The pole

of the compensator is now determined.




• Calculate Kc = K/β.

• Example:


PM > 40° GM > 10 dBunit-ramp-input steady-state error < 0.2

Compensator:

1G(s) s(s 1)(0.5s 1)= + +

c10s 1G (s) 5100s 1

+=+

Lag-Compensation Design for a Temperature Control System

( )

( )

K KG s

s s1 s 1 10.5 2

=

⎛ ⎞ ⎛ ⎞+ + +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠


Kp = 9 and PM > 40°




0.5 2⎝ ⎠ ⎝ ⎠

( )5s 1

D s 3

15s 1

+=

+

Note: A lead or a lag compensator

could meet these specifications.The bandwidth of the lead design

is higher by a factor of 3 than that

of the lag design.

Lag-Compensation Design for a DC Motor


ess

< 0.1 for a unit ramp input and Mp < 25% (PM > 45°)

( )( )

1G ss s 1

=+




( )

10s 1

D s 10100s 1

+

= +

PID Compensation

• For problems that need




For problems that need

phase-margin improvement

at ωgc

and low-frequency gain

improvement, it is effective to

use both derivative and

integral control.

• The PID Control Transfer

Function is given by:

( ) ( )D

I

K 1D s T s 1 s

s T

⎡ ⎤⎛ ⎞= + +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

PID Compensation Design for Spacecraft Attitude Control




Step Response

Step-Disturbance

Response

Lead-Lag Compensation




• Lead-Lag Compensator:

• We often choose γ = β in designing a lead-lag

compensator, although this is not necessary.

1 2

c c

1 2

1 1s sT T

G (s) K ( > 1, > 1)1

s s

T T

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠= γ β

⎛ ⎞⎛ ⎞γ+ +⎜ ⎟⎜ ⎟

β⎝ ⎠⎝ ⎠

Im Polar Plot of Lead-Lag Compensator

K c = 1 and γ = β

1




Re

0

ω = 0

ω = ∞

ω ω= 1

1

Lag Compensator

Lead Compensator

1

1 2

1

T Tω =

Bode Diagrams

-5

0

Lead-Lag Compensator

K c = 1, γ = β = 10, T2 = 10T1, T1 = 1




Frequency (rad/sec)

P h

a s e ( d e g ) ; M a g n i t u d

e ( d B )

-15

-10

10-3

10-2

10-1

100

101

102

-50

0

50

1

1 2

1

T T

ω =



• Example:

P f S ifi ti

K G(s)

s(s 1)(s 2)=

+ +





GM > 10 dB PM > 50°unity-ramp-input steady-state error < 0.1

Compensator:

c

(s 0.7) (s 0.15)G (s) K 20

(s 7) (s 0.015)

+ += =

+ +

Comparison:

Lead, Lag, Lead-Lag Compensators




• Lead compensation achieves the desired result

through the merits of its phase-lead contribution.

• Lag compensation accomplishes its result through

the merits of its attenuation property at high

frequencies.

• In some design problems both lag compensation and

lead compensation may satisfy the specifications.

• Lead Compensation:

– improves stability margins

– yields a higher gain crossover frequency whichmeans higher bandwidth which means a reduction




means higher bandwidth which means a reduction

in settling time

– is more susceptible to high-frequency noisebecause of increase in high-frequency gain due to

the increase in bandwidth

– requires a larger gain than a lag network to offsetthe attenuation inherent in the lead network. This

means larger space, greater weight, and higher

cost.

• Lag Compensation:

– reduces the system gain at higher frequencies

without reducing the system gain at lowfrequencies




frequencies

– reduces the system bandwidth and so the system

has a slower response speed – has improved steady-state accuracy since the

total system gain and hence, low-frequency gain,

can be increased because of the reduced high-frequency gain

– is less susceptible to high-frequency noise since

the high-frequency gain is reduced

• Lead - Lag Compensators:

– can result in both fast response and good static

accuracy– can result in an increase in low-frequency gain




– can result in an increase in low-frequency gain

(which improves steady-state accuracy) while at

the same time the system bandwidth and stabilitymargins can be increased.




Frequency-ResponseDesign Problems




Problem # 1

( )( )( )

5G s

s s 1 0.2s 1=

+ +( )D s K 1= =




Closed-Loop Bandwidth: 7.08 rad/sec




Problem # 2 ( )( )( )

1G ss 0.2s 1 0.02s 1

=+ +( )D s K 1= =



( )( )( )

100G ss 0.2s 1 0.02s 1

= + +( ) s 0.2D s 0.05s 0.01

+=+



Problem # 3 ( )( )( )

1G s

s 0.2s 1 0.05s 1=

+ +( )D s K 100= =




0.2 rad/s

200 rad/s

( )( )( )

1G s

s 0.2s 1 0.05s 1=

+ +( )s s1 12 4D s K 100s s

1 1

0.2 50

⎛ ⎞⎛ ⎞+ +⎜ ⎟⎜ ⎟= = ⎜ ⎟⎜ ⎟

⎜ ⎟⎜ ⎟+ +

⎝ ⎠⎝ ⎠




More Frequency-Response Design Problems

( )

( )( )

50000G s

s s 1 s 50

=

+ +

Design a lead compensator so that the

PM ≥ 50º and ωBW ≥ 20 rad/s.




( ) c c

1s

Ts 1 TD s K K (0 < < 1)1Ts 1

s T

++= α = α

α ++ α




mm

m

1 1 sin( )sin( )1 1 sin( )− α − φφ = ⇒ α =+ α + φ m 60 0.07φ = ° ⇒ α =

m

1

Tω =

αm

1 120rad/s 5.3 75.6T T

ω = ⇒ = =α

( ) s 5.3D s 1.85

s 75.6+=

+




BW

36rad/sω =

( ) 2

1G s

s 1=

−

The transfer function is similar to the transferfunction for an inverted pendulum.

Design a lead compensator to achieve a PM =

30º and ωc = 1 rad/s . Correlate your designwith a root-locus plot.

1




( ) c c

1s

Ts 1TD s K K (0 < < 1)1Ts 1

sT

++= α = αα + +

α

mm

m

1 1 sin( )sin( )1 1 sin( )− α − φφ = ⇒ α =+ α + φ m 30 0.33φ = ° ⇒ α =

m1

Tω =

αm

1 11rad/s 0.57 1.74T T

ω = ⇒ = =α

( ) s 0.57D s 3.48

s 1.73+=+




( )

( )( )

K G s

s 0.2s 1 0.005s 1=

+ +

Design a lead compensator sothat the steady-state error to a

unit-ramp reference input is <

0.01. For the dominant closed-loop poles, the damping ratio ζ

should be ≥ 0 4




should be ≥ 0.4.

( ) ( ) rampv SSs 0v

1

K limsD s G s K e 0.01 K > 100K →= = = <

( ) Ts 1D s (0 < < 1)Ts 1

+= αα +

PMPM 40

100ζ ≈ ⇒ > °

( )

s1

s 1010D s 10s s 100

1100

+ += = ++

K = 121




( ) ( )( )2 20.05 s 25G s

s s 0.1s 4+=

+ +

For the satellite attitude-control system, amplitude-stabilizethe system using lead compensation so that the GM is ≥ 2

(6 dB) and the PM ≥ 45º, keeping the bandwidth as high as

possible with a single lead compensator.




( )

s 0.06

D s 0.958 s 6

+

= +




K = 0.958




( )10

G ss s

s 1 11.4 3

=⎛ ⎞⎛ ⎞

+ +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

For the open-loop system, shown,design a lag compensator with

unity DC gain so that the PM ≥ 40º.

What is the bandwidth of theclosed-loop system?




K = 1

K = 0.111




The lag compensator needs to lower the gain by a factor of

about 10 at the crossover frequency of 0.89 rad/s.

( )s 1Ts 1 0.04D ssTs 1

1

0.004

++= =

α + +

Frequency Response Analysis and Design

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