chapter 2 數學基礎 -...

自動控制

Chapter 2 數學基礎

NTU-BIME-自控講義_Joe-Air Jiang

Automatic Control Systems_F. Golnaraghi & B. C. Kuo

本章目的本章目的


1.1. 介紹複數。介紹複數。2.2. 介紹頻域分析以及頻率圖。介紹頻域分析以及頻率圖。3.3. 介紹微分方程式和狀態空間系統。介紹微分方程式和狀態空間系統。4.4. 介紹介紹拉氏拉氏轉換的基本理論。轉換的基本理論。5.5. 舉例說明應用舉例說明應用拉氏拉氏轉換求解線性常微分方程式的方法。轉換求解線性常微分方程式的方法。6.6. 介紹轉移函數觀念，及如何應用來建置線性非時變系統的模型。介紹轉移函數觀念，及如何應用來建置線性非時變系統的模型。7.7. 論述非時變系統之穩定性和羅斯赫維茲準則。論述非時變系統之穩定性和羅斯赫維茲準則。

8.8. 以專案研究方式示範以專案研究方式示範 MATLAB MATLAB 工具的用法。工具的用法。

本章大部份習題的求解均可用本章大部份習題的求解均可用MATLAB MATLAB 工具工具

傳統控制理論所需數學工具包括複變數理論、微分和差分方程式、拉氏轉換、z 轉換等。。

現代控制理論需要更多精深的數學背景，諸如矩陣理論、集合理論、線性現代控制理論需要更多精深的數學背景，諸如矩陣理論、集合理論、線性

代數和轉換、變分法、數學規劃、機率理論及其它更高深的數學。代數和轉換、變分法、數學規劃、機率理論及其它更高深的數學。

2



Chapter 2 數學基礎複數 Rectangular form : (2-1) , where 1 z x jy j

圖圖22--1 Complex number z representation 1 Complex number z representation in rectangular and polar forms. in rectangular and polar forms.

(2-2)

2 2

1tan

R x y

θ yx

(2-3)

cos sincos sin

jθe θ j θz R θ jR θ

(2-4)(2-5) Euler formula

RRez j

*z x jy

* cos sin jθz R θ R θ Re 2 2 2*zz R x y

(2-6)

(2-7)

(2-8)

(2-9)

cossin

x R θy R θ

3




4




複變數

圖圖22--2 2 複數平面複數平面

►► 例題例題 22--11--1 1 FindFind jj 33 and and jj 44..

2

3

33 2 2

4 3 2

1 cos sin2 2

1 1 1 1

1

πj

π πj j

π πj j e

j j

j e ej j j j

►► 例題例題 22--11--2 2 FindFind zznn using Eq. (2using Eq. (2--6).6). nn n jnθ njθz R e R nθRe

Figure 2-2 illustrates the complex s-plane, in which any arbitrary point s = s1is defined by the coordinates = 1, and = 1, or simply s1 = 1 + j1.

5



Chapter 2 數學基礎複變函數

)]lm[G()]([R)( sjsGesG (2-11)

圖圖22--3 Single3 Single--valued mapping from the svalued mapping from the s--plane to the plane to the GG((ss))--plane.plane.

The function G(s) is said to be a function of the complex variable s if for every value of s, there is one or more corresponding values of G(s).

The function G(s) is also represented by

where Re[G(s)] denotes the real part of G(s), and Im[G(s)] represents the imaginary part of G(s).

6



Chapter 2 數學基礎複變函數

)1(1)(

ss

sG (2-12)

If for every value of s there is only one corresponding value of G(s) in the G(s)-plane, G(s) is said to be a single-valued function, and the mapping from points in the s-plane onto points in the G(s)-plane is described as single-valued (Fig. 2-3).

If the mapping from the G(s)-plane to the s-plane is also single-valued, the mapping is called one-to-one. However, there are many functions for which the mapping from the function plane to the complex-variable plane is not single valued.

For instance, given the function

it is apparent that for each value of s, there is only one unique corresponding value for G(s). However, the inverse mapping is not true; for instance, the point G(s) = is mapped onto two points, s = 0 and s = –1, in the s-plane.

7



Chapter 2 數學基礎解析函數(Analytic Function)

)1(1)(

ss

sG (2-12)

A function G(s) of the complex variable s is called an analytic function in a region of the s-plane if the function and all its derivatives exist in the region.

Ex.

The function given in Eq. (2-2) is analytic at every point in the s-plane except at the point s = 0 and s = –1.

Ex. The function G(s) = s + 2 is analytic at every point in the entire s-plane.

Singularities and Poles of a FunctionThe singularities of a function are the points in the s-plane at which the function or its derivatives does not exist. A pole is the most common type of singularity and plays a very important role in the studies of classical control theory.

8




Definition of Pole:If a function G(s) is analytic and single-valued in the neighborhood of si, it is said to have a pole of order r at s = si if the limit

has a finite, nonzero value. In other words, the denominator of G(s) must include the factor (s – si)r, so when s = si, the function becomes infinite. If r = 1, the pole at s = si is called a simple pole.

Ex. 2)3)(1()2(10)(

sssssG (2-13)

The function in Eq. (2-13) has a pole of order 2 at s = – 3 and simple poles at s = 0 and s = – 1.

9



Chapter 2 數學基礎Zeros of a Function

Definition of Zero:If a function G(s) is analytic and single-valued in the neighborhood of si, it is said to have a zero of order r at s = si if the limit

has a finite, nonzero value. Or, simply, G(s) has a zero of order r at s = si if 1/G(s) has an rth-order pole at s = si.

Ex. The function in Eq. (2-13) has a simple zero at s = – 2.

The total numbers of poles and zeros of a rational function is the same, counting the ones at infinity.

Ex. The function in Eq. (2-13) has four finite poles at s = 0, – 1, – 3, and – 3; there is one finite zero at s = – 2, but there are three zeros at infinity, since

310lim ( ) lim 0

s sG s

s (2-14)

10




圖2-4 Graphical representation of in the s-plane: x poles and O zeros.

2)3)(1()2(10)(

sssssG

11




12




13



Chapter 2 數學基礎Polar Representation

)1(1)(

ss

sG (2-15)

2

2

1

2 2

1 2 1

2 1 5

tan (2 1.107 )63.441

πjjθ

jθ

s j Re e

s j Re

R

θ

(2-16)

11 tantan22

12 121 1 1 1(2 )2 (2 1) 2 5 2 5

ππ jj jG j e e e

j j

(2-17)

To find the polar representation of G(s) in Eq. (2-12) at s = 2j, we look at individual components. That is

圖圖22--55 Graphical representation of Graphical representation of ss11 = 2= 2jj + 1 in the + 1 in the ss--plane.plane.

1tan 21

14



Chapter 2 數學基礎Ex. 2-1-3Find the polar representation of G(s) given below for s = jω, where ω is a constant varying from zero to infinity.

)8)(2(16

161016)(

2

sssssG (2-18)

To evaluate Eq.(2-18) at s = jω, we look at individual components. Thus,

12 2

1 1

1 1

2 21

1 11

1

2 2 (2-19)sin (2-20)

2 cos (2-21)

2 (2-22)

tan (2-23)2

jjω ω eω R

R

R ωω R

R

15



Chapter 2 數學基礎Ex. 2-1-3 (continued)

圖圖22--6 6 Graphical representation of components of 16/[(j + 2) /[(j + 8)].

See Fig.2-6 for a graphical representation of components of 16/[(j + 2)/[(j + 8)]. Hence,

1

2

1 1 1

1

2 2

1 22

20

2 ( sin cos ) (2-24)

2 (2-25)

8 8 (2-26)

tan (2-27)8

16 16

j

j

jω R j

jω R e

jω ω eω R

R

e

(2-28)

1

2

2 2

2 2

1 12 2 (2-29)

1 18 8

j

j

jω ω e

jω ω e

16




As a result, G(s = jω) becomes:Ex. 2-1-3 (continued)

Table 2-2 describes different R and values as changes. As shown, the magnitude describes as the frequency increases. The phase goes from 0 to – 180.

1 2( )

2 2 2 2

2 2 2 2

11 2

16( ) (2 - 30)( )2 8

16where ( ) ( ) (2 31)2 8

similarly,we can defineIm ( )tan ( ) (2 - 32)Re ( )

j jG jω e eG jωω ω

R G ω G jωω ω

G jω G s jωG jω

17




18



Chapter 2 數學基礎Ex. 2-1-3 (continued)Alternative Approach:Alternative Approach: If we multiply both numerator and denominator of Eq. (2.18) by the complex conjugate of the denominator, i.e.,

( 2)( 8) 1( 2)( 8)

jω jωjω jω

we get2 2 2 2

22 2

2 2

2 2

2 2

16( 2)( 8)( )( 2 )( 8 )

16 [(16 ) 10 ]( 4)( 64)Real Imaginary

16 (16 ) (10 )( 4)( 64)

16( 4)( 64)

j

j j

jω jωG jωω ω

ω j ωω ω

ω ω eω ω

e Reω ω

(2-33)

12

1

10 /where tan(16 ) /

Im( ( ))tanRe( ( ))

ω Rω R

G jωG jω

19



圖 2.7 Graphical representation of 16/[(j + 2) /[(j + 8)] for a fixed value of .


See Fig. 2-7 for a graphical representation of 16/[(j + 2) /[(j + 8)] for a fixed value of ω.

Ex. 2-1-3 (continued)

20



Chapter 2 數學基礎Ex. Consider the following second order system:

))(()(

21 pspsKsG

(2(2--34)34)

where (– p1) and (– p2) are poles of the function G(s). By definition, if s = j, G(j) is the frequency response function of G(s), because has a unit of frequency (rad/s):

))(()(

21 pjpjKsG

(2-35)

21

)(pjpj

KjGR

(2-36)

The magnitude of G(j)

(2-37)

And the phase angle of G(j) is

2121)( pjpjKjG

21



Chapter 2 數學基礎 For the general case, where

1

1

( )( )

( )

m

kkn

ii

s zG s K

s p

(2-38)

The magnitude and phase of G(s) are as follows

1

1

1 1

...( )

...( ) ( ... ) ( ... )

m

n

m n

jω z jω zR G jω K

jω p jω pG jω ψ ψ

(2-39)

22




23




24



Chapter 2 數學基礎頻域圖 Let G(s) be the forward-path transfer function of a linear control system with

unity feedback. The frequency-domain analysis of the closed-loop system can be conducted from the frequency-domain plots of G(s) with s replaced by j.

The function G(j) is generally a complex function of the frequency , and can be written as

)()()( jGjGjG (2-40)where G(j) denotes the magnitude of G(j), and G(j) is the phase of G(j).

The following frequency-domain plots of G(j) versus are often used in the analysis and design of linear control systems in the frequency domain.

25




COMPUTER-AIDED CONSTRUCTION OF THE FREQUENCY-DOMAIN PLOTS

The MATLAB and ACSYS computer programs can be used for this purpose.

Polar plots The polar plot of a function of the complex variable s, G(s), is a plot of the

magnitude of G(j) versus the phase of G(j) on polar coordinates as is varied from zero to infinity.

From a mathematical viewpoint, the process can be regarded as the mapping of the positive half of the imaginary axis of the s-plane onto the G(j)-plane.

A simple example of this mapping is shown in Fig. 2-8. For any frequency = 1, the magnitude and phase of G(j1) are represented by a vector in the G(j)-plane. In measuring the phase, counterclockwise is referred to as positive, and clockwise is negative.

26




圖圖 22--8 8 Polar plot shown as a mapping of the positive half of the j -axis in the s-plane onto the G(j)-plane.

27



Chapter 2 數學基礎Ex. 2-2-1To illustrate the construction of the polar plot of a function G(s), consider the function

TssG

11)( (2-41)

where T is a positive constant. Setting s = jω, we have

TjjG

11)( (2-42)

In terms of magnitude and phase, Eq.(2-42) is rewritten as

TT

jG

1

22tan

11)(

(2-43)

When is zero, the magnitude of G(j) is unity, and the phase of G(j) is at 0. Thus, at = 0, G(j) is represented by a vector of unit length directed in the 0direction.

As increases, the magnitude of G(j) decreases, and the phase becomes more negative. As increases, the length of the vector in the polar coordinates decreases, and the vector rotates in the clockwise (negative) direction.

28




When approaches infinity, the magnitude of G(j) becomes zero, and the phase reaches – 90. This is presented by a vector with an infinitesimally small length directed along the – 90-axis in the G(j)-plane.

By substituting other finite values of into Eq. (2-43), the exact plot of G(j) turns out to be a semicircle, as shown in Fig. 2-9.

圖圖22--99

29



Chapter 2 數學基礎Ex2-2-2As a second illustrative example, consider the function

2

1

1( )1

jωTG jωjωT

(2-44)

where T1 and T2 are positive real constants. Eq.(2-44) is re-written as

(2-45)2 2

1 122 12 2

1

1( ) (tan tan )1ω TG jω ωT ωTω T

The polar plot of G(j), in this case, depends on the relative magnitudes of T1and T2.

If T2 is greater than T1, the magnitude of G(j) is always greater than unity as is varied from zero to infinity, and the phase of G(j) is always positive.

If T2 is less than T1, the magnitude of G(j) is always less than unity, and the phase is always negative.

The polar plots of G(j) of Eq. (2-45) that correspond to these two conditions are shown in Fig. 2-10.

30




圖圖22--1010

The general shape of the polar plot of a function G(j) can be determined from The following information.

1. The behavior of the magnitude and phase of G(j) at = 0 and = ∞.2. The intersections of the polar plot with the real and imaginary axes,

and the values of at these intersections.

31




32




33



Chapter 2 數學基礎Ex. 2-2-3In frequency-domain analyses of control systems, often we have to determine the basic properties of a polar plot. Consider the following transfer function:

)1(10)(

ss

sG (2-46)

By substituting s = jω in Eq.(2-46),the magnitude and phase of G(jω) at ω = 0 and ω = ∞ are computed as follows:

0 0

0 0

2

2

10lim ( ) lim (2-47)

lim ( ) lim 10 / 90 (2-48)

10lim ( ) lim 0 (2-49)

lim ( ) lim 10 / 180 (2- 0( 5 ))

ω ω

ω ω

ω ω

ω ω

G jωω

G jω jω

G jωω

G ω jωj

34




Thus, the properties of the polar plot of G(jω) at ω = 0 and ω = ∞ are ascertained. Next, we determine the intersections, if any, of the polar plot with the two axes of the G(j)-plane. If the polar plot of G(j) intersects the real axis, at the point of intersection, the imaginary part of G(j) is zero; that is,

Im[ ( )] 0 (2-51)G jω

To express G(j) as the sum of its real and imaginary parts, we must rationalizeG(j) by multiplying its numerator and denominator by the complex conjugate of its denominator. Therefore, G(j) is written

2

4 2 4 2

10( )( 1)( )( 1)( )( 1)

10 10

Re[ ( )] Im[ ( )] (2-52)

jω jωG jωjω jω jω jω

ω ωjω ω ω ω

G jω j G jω

35




When we set Im[G(j)] to zero, we get = , meaning that the G(j) plot intersects only with the real axis of the G(j)-plane at the origin.

Similarly, the intersection of G(j) with the imaginary axis is found by setting Re[G(j)] of Eq. (2-52) to zero.

The only real solution for is also = , which corresponds to the origin of the G(j)-plane. The conclusion is that the polar plot of G(j) does not intersect any one of the axes at any finite nonzero frequency.

Under certain conditions, we are interested in the properties of the G(j) at infinity, which corresponds to = 0 in this case. From Eq. (2-52), we see that Im[G(j)] = and Re[G(j)] = – 10 at = 0.

Based on this information, as well as knowledge of the angles of G(j) at = 0 and = , the polar plot of G(j) is easily sketched without actual plotting, as shown in Fig. 2-11.

36




Figure 2-11Polar plot of G(s) = 10/[s(s + 1)].

37



Chapter 2 數學基礎Ex. 2-2-4Given the transfer function

We want to make a rough sketch of the polar plot of G(jω).The following calculations are made for the properties of the magnitude and phase of G(jω) at ω = 0 and ω = ∞:

0 0

0 0

3

5lim ( ) lim (2-54)

lim ( ) lim 5 / 90 (2-55)

10lim ( ) lim 0 (2-56)

ω ω

ω ω

ω ω

G jωω

G jω jω

G jωω

)2)(1(10)(

ssssG (2-53)

(2-56-a)

38




To find the intersections of the G(j) plot on the real and imaginary axes of the G(j)-plane, we rationalize G(j) to give

10( )( 1)( 2)( ) (2-57)( 1)( 2)( )( 1)( 2)

jω jω jωG jωjω jω jω jω jω jω

After simplification, the last equation is written

Setting Re[G(j)] to zero, we have = , and G(j) = 0, which means that the G(j) plot intersects the imaginary axis only at the origin.

Setting Im[G(j)] to zero, we have = 2 rad/sec. This gives the point of intersection on the real axis at

( 2) 5 / 3 (2-59)G j

2

2 2 2 2 2 230 10(2 )( ) Re[ ( )] Im[ ( )] (2-58)

9 (2 ) 9 (2 )j ωG jω G jω j G jω

ω ω ω ω

39




Figure 2-12 Polar plot of G(s) = 10/[s(s + 1)(s + 2)].

40




41



Chapter 2 數學基礎Bode plots The Bode plot of the function G(j) is composed of two plots, one with the

amplitude of G(j) in decibels (dB) versus log10 or , and the other with the phase of G(j) in degrees as a function of log10 or .

A Bode plot is also known as a corner plot or an asymptotic plot of G(j). In simple terms, the Bode plot has the following features:

Consider the function:

1 2

1 2

( )( )...( )( ) (2-60)( )( )...( )

dT smj

n

K s z s z s zG s es s p s p s p

42




where K and Td are real constants, and the z’s and the p’s may be real or complex (in conjugate pairs) numbers.

In Chapter 7, Eq. (2-60) is the preferred form for root-locus construction, since the poles and zeros of G(s) are easily identified.

For constructing the Bode plot manually, G(s) is preferably written in the following form:

1 1 2(1 )(1 )...(1 )( ) (2-61)(1 )(1 )...(1 )

dT smj

a b n

K T s T s T sG s es T s T s T s

where K1 is a real constant, the T ’s may be real or complex (in conjugate pairs) numbers, and Td is the real time delay.

If the Bode plot is to be constructed with a computer program, then either forms of Eq. (2-60) or Eq. (2-61) can be used.

Because practically all the terms in Eq. (2-61) are of the same form, then without loss of generality, we can use the following transfer function to illustrate the construction of the Bode diagram.

43




10dB

10 10 1 10 2

2 210 10 10

( ) 20log ( )

20log 20log 1 20log 1

20log 20log 1 20log 1 2 /a a n

G jω G jω

K jωT jωT

jω jω T j ξω ω ω

(2-62)

(2-63)

1 22 2

(1 )(1 )( )(1 )(1 2 / / )

dT s

a n n

K T s T sG s es T s ξs ω s ω

where K, Td, T1, T2, Ta, , and n are real constants. It is assumed that the second-order polynomial in the denominator has complex-conjugate zeros.

The magnitude of G(j) in dB is obtained by multiplying the logarithm (base 10) of G(j) by 20; we have

The phase of G(j) is

1 22 2

( ) (1 ) (1 ) (1 )

(1 2 / / ) rada

n n d

G jω K jωT jωT jω jωT

ξω ω ω ω ωT

(2-64)

44




In general, the function G(j) may be of higher order than that of Eq. (2-62) and have many more factored terms. However, Eqs. (2-63) and (2-64) indicate that additional terms in G(j) would simply produce more similar terms in the magnitude and phase expressions, so the basic method of construction of the Bode plot would be the same.

We have also indicated that, in general, G(j) can contain just five simple types of factors:

45




One of the unique characteristics of the Bode plot in that each of the five types of factors listed can be considered as a separate plot; the individual plots are then added or subtracted accordingly to yield the total magnitude in dB and the phase plot of G(j).

The curves can be plotted on semilog graph paper or linear rectangular-coordinate graph paper, depending on whether or log10 is used as the abscissa.

We shall now investigate sketching the Bode plot of different types of factors.

Real constant KdB 1020log constant (2-65)

0 0 (2-66)

180 0

K KK

KK

The Bode plot of the real constant K is shown in Fig. 2-13 in semilog coordinates.

46




47




48




圖圖22--1313 Bode plot of constant Bode plot of constant KK..

49




圖圖22--13 Bode plot of constant 13 Bode plot of constant KK..

50



Chapter 2 數學基礎Poles and Zeros at the Origin, (jω)±p

1010

( 20 log ) 20 dB/decade (2-68)log

d p ω pd ω

The magnitude of (j)p in dB is given by

10 1020log ( ) 20 log dB (2-67)pjω p ω

for 0. The last expression for a given p represents a straight line in either semilog or

rectangular coordinates. The slopes of these lines are determined by taking the derivative of Eq. (2-67) with respect to log10; that is,

These lines pass through the 0-dB axis at = 1. Thus, a unit change in log10corresponds to a change of 20p dB in magnitude. Furthermore, a unit change in log10 in the rectangular coordinates is equivalent to one decade of variation in , that is, from 1 to 10, 10 to 100, and so on, in the semilog coordinates. Thus, the slopes of the straight lines described by Eq. (2-68) are said to be 20p dB/decadeof frequency.

51




Instead of decades, sometimes octaves are used to represent the separation of two frequencies. The frequencies 1 and 2 are separated by one octave if 2/1 = 2. The number of decades between any two frequencies 1 and 2 is given by

number of octaves 1/ 0.301 decades 3.32 decades (2-71)

10 2 1 210

10 1

log ( / )number of decades log (2-69)log 10ω ω ω

ω

Similarly, the number of octaves between 1 and 2 is

10 2 1 210

10 1

log ( / ) 1number of octaves log (2-70)log 2 0.301ω ω ω

ω

Thus, the relation between octaves and decades is

52



Chapter 2 數學基礎Substituting Eq. (2-71) into Eq. (2-67), we have

( ) 90 (2-73)pjω p

20 dB/decade 20 0.301 6 dB/octave (2-72)p p p

For the function G(s) = 1/s, which has a simple pole at s = 0, the magnitude of G(j) is a straight line with a slope of – 20 dB/decade, and passes through the 0-dB axis at = 1 rad/sec.

The phase of (j)p is written

The magnitude and phase curves of the function (j)p are shown in Fig. 2-14for several values of p.

53




Figure 2.14 Bode plot of (j)p.

54




Figure 2.14 Bode plot of (j)p.

55



Chapter 2 數學基礎Simple zero,1+jωT Consider the function

( ) 1 (2-74)G jω jωT

2 210 10( ) 20log ( ) 20log 1 (2-75)

dBG jω G jω ω T

10( ) 20log 1 0 dB (2-76)dB

G jω

where T is a positive real constant. The magnitude of G(j) in dB is

To obtain asymptotic approximations of G(j) dB, we consider both very large and very small values of . At very low frequencies, T << 1, Eq. (2-75) is approximated by

since 2T2 is neglected when compared with 1. At very high frequencies, T >> 1, we can approximate 1 + 2T2 by 2T2; then

Eq. (2-75) becomes

56




2 210 10( ) 20log 20log (2-77)

dBG jω ω T ωT

Equation (2-76) represents a straight line with a slope of 20 dB/decade of frequency. The intersect of these two lines is found by equating Eq. (2-76) to Eq. (2-77), which gives

1/ (2-78)ω T

57




This frequency is also the intersect of high-frequency approximate plot and the low frequency approximate plot, which is the 0-dB axis.

The frequency given in Eq. (2-78) is also known as the corner frequency of the Bode plot of Eq. (2-74), since the asymptotic plot forms the shape of a corner at this frequency, as shown in Fig. 2-15.

The actual plot G(j) dB of Eq. (2-74) is a smooth curve, and deviates only slightly from the straight line approximation.

The actual values and the straight-line approximation of 1 + jT dB as functions of T are tabulated in Table 2-3.

The error between the actual magnitude curve and the straight-line asymptotes is symmetrical with respect to the corner frequency = 1/T. It is useful to remember that the error is 3 dB at the corner frequency, and 1 dB at 1 octave above ( = 2/T) and 1 octave below ( = 1/2T) the corner frequency. At 1 decade above and below the corner frequency, the error is dropped to approximately 0.3 dB.

58




Based on these facts, the procedure of drawing 1 + jT) dB is as follows:

59




Figure 2.15 Bode plot of G(s) = 1 + Ts and G(s) = 1/(1 + Ts).

60




Figure 2.15 Bode plot of G(s) = 1 + Ts and G(s) = 1/(1 + Ts).

61




Table 2-3 Values of (1 + jT) versus T.

62



Chapter 2 數學基礎Simple pole, 1/(1+jωT) For the function

1( ) (2-80)1

G jωjωT

dB1 ( ) 0 dB (2-81)ωT G jω

10dB1 ( ) 20log (2-82)ωT G jω ωT

the magnitude, G(j) in dB, is given by the negative of the right side of Eq.(2-75), and the phase G(j) is the negative of the angle in Eq. (2-79).

Therefore, it is simple to extend all the analysis for the case of the simple zero to the Bode plot of Eq. (2-80). The asymptotic approximations of G(j) dB at low and high frequencies are

The corner frequency of the Bode plot of Eq. (2-80) is still at = 1/T, except that at high frequencies the slope of the straight-line approximation is – 20 dB/decade.

63




The phase of G(j) is 0 degrees at = 0, and – 90 when = . The magnitude in dB and phase of the Bode plot of Eq. (2-80) are shown in

Fig. 2-15. The data in Table 2-3 are still useful for the simple-pole case if appropriate

sign changes are made to the numbers. For instance, the numbers in 1 + jTdB, the the straight-line approximation of 1 + jTdB, the error (db), and the (1 + jT) columns should all be negative.

At the corner frequency, the error between the straight-line approximation and the actual magnitude curve is – 3 dB.

64



Chapter 2 數學基礎Quadratic Poles and Zeros

2

2 2 2 21( ) (2-83)

2 1 (2 / ) (1 / )n

n n n n

ωG ss ξω s ω ξ ω s ω s

2

1( ) (2-84)1 ( / ) 2 ( / )n n

G jωω ω j ξ ω ω

Consider the second-order transfer function

We are interested only in the case when ≤ 1, because otherwise G(s) would have two unequal real poles, and the Bode plot can be obtained by considering G(s) as the product of two transfer functions with simple poles.

By letting s = j, Eq. (2-83) becomes

The magnitude of G(j) in dB is22 2 2

10 10dB( ) 20log ( ) 20log 1 ( / ) 4 ( / ) (2 -85)n nG jω G jω ω ω ξ ω ω

65




10 10dB( ) 20log ( ) 20log 1 0 dB (2 -86)G jω G jω

At very low frequencies /n << 1, Eq. (2-85) can be approximated as

The low-frequency asymptote of the magnitude plot of Eq. (2-83) is a straight line that lies on the 0-dB axis. At very high frequencies /n >> 1, the magnitude in dB of G(j) in Eq. (2-83) becomes

410 10dB

( ) 20log ( / ) 40log ( / ) dB (2 - 87)n nG jω ω ω ω ω

This equation represents a straight line with a slope of – 40 dB/decade in the Bode plot coordinates. The intersection of the two asymptotes is found by equating Eq. (2-86) to Eq. (2-87), yielding the corner frequency at = n.

The actual magnitude curve of G(j) in this case may differ strikingly from the asymptotic curve. The reason for this is that the amplitude and phase curves of the second-order G(j) depend not only on the corner frequency n, but also on the damping ratio , which does not enter the asymptotic curve.

The actual and the asymptotic curves of G(j) dB are shown in Fig. 2-16 for several values of .

66




The errors between the two sets of curves are shown in Fig. 2-17 for the same set of values of .

The standard procedure of constructing the second-order G(j) dB is to first locate the corner frequency n, and – 40-dB/decade line to the right of n.

The actual curve is obtained by making corrections to the asymptotes by using either the data from the error curves of Fig. 2-17 or the curves in Fig. 2-16 for the corresponding .

The phase of G(j) is given by

21 2( ) tan 1 (2-88)

n n

ξω ωG jωω ω

and is plotted as shown in Fig. 2-16 for various values of .

The analysis of the Bode plot of the second-order transfer function of Eq. (2-83) can be applied to the second-order transfer function with two complex zeros.

67




22

2 1( ) 1 (2-89)n n

ξG s s sω ω

For the function

the magnitude and phase curves are obtained by inverting those in Fig. 2-16. The errors between the actual and the asymptotic curves in Fig. 2-17 are also inverted.

68




69




70




Figure 2-17 Errors in magnitude curves of Bode plots of

71




72




73




Pure Time Delay, exp(– jωTd)

1( ) ( ) (2 - 91)djωTG jω G jω e

The magnitude of the pure time delay term is equal to unity for all values of . The phase of the pure time delay term is

(2-90)djωTde ωT

which decreases linearly as a function of . Thus, for the transfer function

the magnitude plot G(j) dB is identical to that of G1(j) dB . The phase plot G(j) is obtained by subtracting Td radians from the

phase curve of G1(j) at various .

74



Chapter 2 數學基礎Ex. 2-2-5As an illustrative example on the manual construction of Bode plot, consider the function

10(1 0.1 )) (2-93)(1 0.5 )(1 0.2 )

j ωG(jωjω j ω j ω

10( 10)( ) (2-92)( 2)( 5)

sG ss s s

The first step is to express G(s) in the form of of Eq. (2-92) and set s = j

(keeping in mind that for computer plotting, this step is unnecessary); we have

Equation (2-93) shows that G(j) has corner frequencies at = 2, 5, and 10 rad/sec. The pole at s = 0 gives a magnitude curve that is a straight line with slope of – 20 dB/decade, passing through the = 1 rad/sec point on the 0-dB axis.

The complete Bode plot of the magnitude and phase of G(j) is obtained by adding the component curves together, point by point, as shown in Fig. 2-18. The actual curves can be obtained by a computer program and are shown in Fig. 2-18.

75




Figure 2-18 Bode plot of G(s) = 10(s + 10)/[s(s + 2)(s + 5)]

76




Figure 2-18 Bode plot of G(s) = 10(s + 10)/[s(s + 2)(s + 5)]

77




78



Chapter 2 數學基礎96年臺大生機所入學考Consider a minimum-phase system whose asymptotic amplitude frequency response is depicted in Fig. C.【計分：10分】 (1) Determine the transfer function G(s) of the system. (2) Determine the two gain crossover frequencies g1 and g2.【計分：10分】<Some useful values: 101.5 = 31.623, log1.5 = 0.1761, log0.2 = 0.699, and log5 = 0.699.>

20 dB/decade 20 dB/decade

40 dB/decade

30

51 20 g2g1

Fig. CdB

0

79



Chapter 2 數學基礎<Sol.>

(1) The low-frequency asymptote has a slope of 20 dB/decade. Ks is a factor of the transfer function with 20logK = 30. This gives

K =31.623

The transfer function is the obtained as

31.623( )1 1(1 ) 1 15 20

sG ss s s

(2) The two gain crossover frequencies g1 and g2 can be determined as follows:(i) Gain crossover frequency g1:

120log(31.623 ) 0g

This gives 1 0.0316g

80



Chapter 2 數學基礎(ii) Gain crossover frequency g1:

2 2 2 21 120log(31.623 ) 20log 20log 20log 05 20g g g g

This gives

2 56.234g

成功大學九十二年度系統及船舶機電工程研究所入學考Plot the Bode plots of the following cases:

Case (a))1.01)(5.01(

20)(sss

sGA Case (b)

)04.01)(4.01(2)(

2

ssssGB

0.1 rad/sec, 1.0 rad/sec, 2.5 rad/secω ω ω 25 rad/secωCalculate the magnitude (dB) value and phase angle (degree) values at frequencies

and , respectively forfor Case (b). <Sol.>

81




82




83




84



Chapter 2 數學基礎Magnitude-Phase Plot Definition: The magnitude-phase plot of G(j) is a plot of the magnitude of G(j)

in dB versus its phase in degrees, with as a parameter on the curve. One of the most important applications of this type of plot is that when G(j) is

the forward-path transfer function of a unity-feedback control system, the plot can be superposed on the Nichols chart (see Chapter 8) to give information on the relative stability and frequency response of the system.

When the gain factor K of the transfer function varies, the plot is simply raised or lowered vertically according to the value of K in dB.

However, in the construction of the plot, the property of adding the curves of the individual components of the transfer function in the Bode plot does not carry over to this case. Thus, it is best to make the magnitude-phase plot by computer or transfer the data from the Bode plot.

85



Chapter 2 數學基礎GAIN- AND PHASE-CROSSOVER POINTS

Gain-Crossover Point: The gain-crossover point on the frequency-domain plot of G(j) is the point at which G(j) = 1 or G(j) dB = 0 dB. The frequency at the gain-crossover point is called the gain-crossover frequency g.

Phase-Crossover Point: The phase-crossover point on the frequency-domain plot of G(j) is the point at which G(j) = 180. The frequency at the phase-crossover point is called the phase-crossover frequency p.

The gain and phase crossovers are interpreted with respect to three types of plots: Polar Plot: The gain-crossover point (or points) is where |G(j)| = 1. The phase

crossover point (or points) is where G(j) = 180 (see Fig. 2-19). Bode Plot: The gain-crossover point (or points) is where the magnitude curve

G(j) dB crosses the 0-dB axis. The phase-crossover point (or points) is where the phase curve crosses the 180-axis (see Fig. 2-18).

Magnitude-Phase Plot: The gain-crossover point (or points) is where the |G(j)|curve crosses the 0-dB axis. The phase-crossover point (or points) is where the |G(j)| curve crosses the 180-axis (see Fig. 2-20).

Some Definitions:

86



Chapter 2 數學基礎Ex. 2-2-6As an illustrative example, the polar plot and the magnitude-phase plot of Eq. (2-92) are shown in Fig. 2-19and 2-20, respectively. The Bode plot of the function is already shown in Fig. 2-18. The relationships among these three plots are easily identified by comparing the curves in Figs. 2-18, 2-19, and 2-20.

Figure 2-19 Polar plot of G(s) = 10(s + 10)/[s(s + 2)(s + 5)]

10( 10)( ) (2-92)( 2)( 5)

sG ss s s

87



Chapter 2 數學基礎Ex. 2-2-6

Figure 2-20Magnitude-phase plot of G(s) = 10(s + 10)/[s(s + 2)(s + 5)]

88




89




90




91




Minimum-Phase and Nonminimum-Phase Functions

Definitions: The transfer functions that do not have poles or zeros in the right-half s-plane

are called the minimum-phase transfer functions. When a transfer function has either a pole or a zero in the right-half s-plane, it is

called a nonminimum-phase transfer function. The magnitude and phase characteristics of a minimum-phase function are

uniquely related.

Given a minimum phase function G(s), knowing its magnitude characteristics |G(j)| completely defines the phase characteristics, G(j). Conversely, given G(j), |G(j)| is completely defined.

Nonminimum-phase transfer functions do not have the unique magnitude-phase relationships.

92




Additional properties of the minimum-phase transfer functions are as follows:

Ex. Given the function1( ) (2-94)

1G jω

jωT

The magnitude of G(j) is the same whether T is positive (nonminimum phase) or negative (minimum phase). However, the phase of G(j) is different for positive and negative T.

93



Chapter 2 數學基礎Ex 2-2-7

10( 10)( ) (2-95)( 2)( 5)

sG ss s s

As an illustrative example of the properties of the nonminimum-phase transfer function, consider that the zero of the transfer function of Eq. (2-95) is in the right-half s-plane; that is,

The magnitude plot of the Bode diagram of G(j) is identical to that of the minimum-phase transfer function in Eq. (2-92), as shown in Fig. 2-18.

The phase curve of the Bode plot of G(j) of Eq. (2-95) is shown in Fig. 2-21(a), and the polar plot is shown in Fig. 2-21(b).

Notice that the nonminimum-phase function has a net phase shift of 270(from –180 to + 90) as varies from to 0, whereas the minimum-phase transfer function of Eq. (2-92) has a net phase change of only 90 (from –180to – 90) over the same frequency range.

• Do not use the Bode plot and the gain-phase plot of a nonminimum-phase transfer function for stability studies.

94




Figure 2-21

95




Figure 2-21

96



Chapter 2 數學基礎Here are some important notes:

• A Bode plot is also know as a corner plot or an asymptotic plot.

• The magnitude of the pure time delay term is unity for all .

• The magnitude and phase characteristics of a minimum-phase function are uniquely related.

• Do not use the Bode plot and the gain-phase plot of a nonminimum-phase transfer function for stability studies.

97



Chapter 2 數學基礎 INTRODUCTION TO DIFFERENTIAL

EQUATIONS A wide range of systems in engineering

are modeled mathematically by differential equations. These equations generally involve derivatives and integrals of the dependent variables with respect to the independent variable.

For instance, a series electric RLC(resistance-inductance-capacitance) network can be represented by the differential equation

( ) 1( ) ( ) ( ) (2-96)di tRi t L i t dt e tdt C

where R is the resistance; L, the inductance; C, the capacitance; i(t), the current in the network; and e(t), the applied voltage.

98



Chapter 2 數學基礎 Equation (2-96) is referred to as a second-order differential equation,

and we refer to the system as a second-order system.

Linear Ordinary Differential Equations The differential equation of an nth-order system is written

1

1 1 01( ) ( ) ( )... ( ) ( ) (2-97)

n n

nn nd y t d y t dy ta a a y t f t

dt dt dt

which is also known as a linear ordinary differential equation if the coefficients a0, a1, … , an–1 are not functions of y(t).

First-order linear O.D.E.:

Second-order linear O.D.E.:

0( ) ( ) ( ) (2-98)dy t a y t f t

dt

2

1 02( ) ( ) ( ) ( ) (2-99)d y t dy ta a y t f t

dt dt

99



Chapter 2 數學基礎Nonlinear Differential Equations The differential equation that describes the motion of the pendulum shown in Fig.

A is

Figure A Simple pendulum

2

2( ) sin ( ) 0 (2-100)d θ tml mg θ t

dt

Because (t) appears as a sine function, Eq. (2-100) is nonlinear, and the system is called a nonlinear system.

First-Order Differential Equations: State Equations2

In general, an nth-order differential equation can be decomposed into n first-order differential equations. Because, in principle, first-order differential equations are simpler to solve than higher-order ones, first-order differential equations are used in the analytical studies of control systems.

100



Chapter 2 數學基礎Ex. For the differential equation in Eq. (2-96), if we let

1

12

( ) ( ) (2-101)

( )( ) ( ) (2-102)

x t i t dt

dx tx t i tdt

then Eq. (2-96) is decomposed into the following two first-order differential equations:

12

21 2

( ) ( ) (2-103)

( ) 1 1( ) ( ) ( ) (2-104)

dx t x tdt

dx t Rx t x t e tdt LC L L

Ex. For the differential equation in Eq. (2-97), let us define

101



Chapter 2 數學基礎1

2

1

1

( ) ( )( )( )

(2-105)

( )( )n

n n

x t y tdy tx t

dt

d y tx tdt

then the nth-order differential equation is decomposed into n first-order differential equations:

12

23

0 1 1 2 2 1 1

( ) ( )

( ) ( )(2-106)

( ) ( ) ( ) ... ( ) ( ) ( )nn n n n

dx t x tdt

dx t x tdt

dx t a x t a x t a x t a x t f tdt

State Equation;x1, x2, … , xn are called

the state variables.

102



Chapter 2 數學基礎Definition of State Variables The state of a system refers to the past, present, and future conditions of the

system. From a mathematical perspective, it is convenient to define a set of state variables and state equations to model dynamic systems.

In general, there are some basic rules regarding the definition of a state variable and what constitutes a state equation.

The state variables must satisfy the following conditions:

The state variables of a system are defined as a minimal set of variables, x1(t), x2(t), …, xn(t), such that knowledge of these variables at any time t0 and information on the applied input at time t0 are sufficient to determine the state of the system at any time t t0.

103



Chapter 2 數學基礎 Space state form for n variables:

( ) ( ) (2 -107)t t x Ax Bu

1

2

( )( )

( ) (2 -108)

( )n

x tx t

t

x t

x

1

2

( )( )

( ) (2 -109)

( )p

u tu t

t

u t

u

11 12 1

21 22 2

1 2

(2 -110)

n

n

n n nn

a a aa a a

n n

a a a

A ( )

where x(t) and u(t) are state vector having n rows and input vector having p rows, respectively.

The coefficient matrices A and B are defined as:

104




11 12 1

21 22 2

1 2

(2 -111)

p

p

n n np

b b bb b b

n p

b b b

B ( )

The Output Equation 11 12 1

21 22 2

1 2

11 12 1

21 22 2

1 2

(2 -113)

(2-114)

n

n

q q qn

p

p

q q qp

c c cc c c

c c c

d d dd d d

d d d

C

D

1

2

( )( )

( ) ( ) (2 -112)

( )q

y ty t

t t

y t

y Cx Du

An output of a system is a variable that can be measured, but a state variable does not always satisfy this requirement.

105




For instance, in an electric motor, such state variables as the winding current, rotor velocity, and displacement can be measured physically, and these variables all qualify as output variables.

On the other hand, magnetic flux can also be regarded as a state variable in an electric motor, since it represents the past, present, and future states of the motor, but it cannot be measured directly during operation and therefore does not ordinarily qualify as an output variable.

In general, an output variable can be expressed as an algebraic combination of the state variables. For the system described by Eq. (2-97), if y(t) is designated as the output, then the output equation is simply y(t) = x1(t).



Chapter 2 數學基礎拉氏拉氏轉換轉換

1. 1. 齊次方程式的解與微分方程式的特解，可於同一運算中求出。齊次方程式的解與微分方程式的特解，可於同一運算中求出。

2. 2. 拉氏拉氏轉換將轉換將微分微分方程式轉換成含方程式轉換成含 ss 變數的變數的代數代數方程式。方程式。

拉氏拉氏轉換的定義轉換的定義

對某個有限的實數對某個有限的實數，，函數 f (t) 滿足下列的條件：

則則 f f ((tt) ) 的的拉氏拉氏轉換定義為轉換定義為

(2(2--115)115)

(2(2--116)116)

之之拉氏拉氏轉換轉換

因果系統的響應不會發生在訊號輸入之前

(2(2--117)117)

其中其中，s = + j 。

嚴格而言，單邊拉氏轉換應定義成 t = 0 至 t = 。符號 0 代表由 t = 0 的左邊取 t 0 的極限。

單邊單邊拉氏拉氏轉換轉換 (one(one--sided Laplace sided Laplace

transform) transform)

系統稱為因果系統(causal system) 或實際可實現的系統

(physically realizable system)。

107



Chapter 2 數學基礎Ex. 2-4-1

Ex. 2-4-2

假設假設 f f ((tt) ) 為單位步階函數為單位步階函數

(2(2--118) 118)

f f ((tt) ) 之之拉氏拉氏轉換為轉換為

(2(2--119) 119)

條件條件

(2(2--120) 120)

指數函數指數函數

f f ((tt) ) 之之拉氏拉氏轉換為轉換為

其中其中為一實數常數。為一實數常數。

(2(2--121)121)

(2(2--122)122)( )

00

1( )s t

t st eF s e e dts s

108




109



Chapter 2 數學基礎反反拉氏拉氏轉換轉換

拉氏拉氏轉換的重要定理轉換的重要定理

反反拉氏拉氏轉換積分則定義為轉換積分則定義為

★★ ACSYS ACSYS 工具工具 (Tfsym) (Tfsym) 亦可用來解部份分式展開和反亦可用來解部份分式展開和反拉氏拉氏轉換。轉換。

的反拉氏轉換 (2(2--123)123)

(2(2--124)124)

其中，c 是實常數，它大於F(s) 所有奇點的實數部份。

■■ 定理定理 1. 1. 乘上常數乘上常數

( ) ( )f t F s±

■■ 定理定理 2. 2. 和及差和及差

1 1 2 2( ) ( ), ( ) ( )f t F s f t F s ± ±

k k 為一常數為一常數

( ) ( )k f t kF s± (2(2--125)125)

1 2 1 2( ) ( ) ( ) ( )f t f t F s F s ± (2(2--126)126)

拉氏轉換表: 附錄 D

110



Chapter 2 數學基礎拉氏拉氏轉換的重要定理轉換的重要定理

■■ 定理定理 3.3. 微分微分

( ) ( )f t F s±

★★ 對對f f ((tt) ) 的高階微分而言的高階微分而言

其中，其中， f f ((ii)) (0) (0) 代表代表 f f ((tt) ) 對對 tt 之第之第 ii 階微分在階微分在 tt = 0 = 0 時之值。時之值。

■■ 定理定理 4. 4. 積分積分

( ) ( )f t F s±

★★ 對對 nn 階積分而言階積分而言

(2(2--128)128)

0

( ) ( ) lim ( ) ( ) (0)t

df t sF s f t sF s fdt

± (2(2--127)127)

0

( )( )t F sf d

s ± (2(2--129)129)

(2(2--130)130)

111



Chapter 2 數學基礎拉氏拉氏轉換的重要定理轉換的重要定理

若 sF(s) 包含了實數部份是零或正值的任何極點時，終值定理就不適用；亦即在定理中 sF(s) 必須在s-Plane的右半平面是可解析的。

( ) ( )f t F s±

■■ 定理定理 6. 6. 初值定理初值定理

0lim ( ) lim ( )t s

f t sF s

假設時間極限存在假設時間極限存在

(2(2--132)132)

■■ 定理定理 5. 5. 時間的移位時間的移位

( ) ( )f t F s± ( ) ( ) ( )Tssf t T u t T e F s ±

其中，其中，uuss ((t t TT) ) 為向右位移為向右位移TT 時間單位的步階函數。時間單位的步階函數。

(2(2--131)131)

( ) ( )f t F s±

■■ 定理定理 7. 7. 終值定理終值定理

0lim ( ) lim ( )t s

f t sF s

終值定理只有當終值定理只有當 sFsF((ss) ) 在在 jj 軸和軸和s s 平面右半平面沒有極點時才正確平面右半平面沒有極點時才正確

(2(2--133)133)

112



Chapter 2 數學基礎Ex. 2Ex. 2--44--33

Ex. 2Ex. 2--44--44

由於由於 sFsF((ss) ) 在在 ss 平面的右半邊和虛軸上平面的右半邊和虛軸上是可解析的，故可使用終值定理。是可解析的，故可使用終值定理。

由於由於 sFsF((ss) ) 函數有兩個極點位於虛軸上，函數有兩個極點位於虛軸上，故終值定理不能應用於本例中。故終值定理不能應用於本例中。

(2(2--134)134)

(2(2--135)135)

(2(2--136)136)

■■ 定理定理 8. 8. 複數移位複數移位

( ) ( )f t F s± ( ) ( )te f t F s ±

為常數為常數

■■ 定理定理 9. 9. 實數迴旋實數迴旋 ((複數乘法複數乘法) )

1 1 2 2( ) ( ), ( ) ( )f t F s f t F s ± ±

1 2 1 2 1 2 2 10 0( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )

t tF s F s f t f t f f t d f f t d ± ± ±

其中，符號其中，符號＊＊代表代表 tt 領域之迴旋領域之迴旋 (convolution)(convolution)。。

，且，且 tt < 0 < 0 時，時，ff11((tt) = 0) = 0，，ff22((tt) = 0 ) = 0

(2(2--137)137)

(2(2--138)138)

113




★★ s s 領域中兩函數乘積之反領域中兩函數乘積之反拉氏拉氏轉換並不等於轉換並不等於 tt 領域中兩對應函數之乘積，即領域中兩對應函數之乘積，即

11 2 1 2( ) ( ) ( ) ( )F s F s f t f t ±

■■ 定理定理 10. 10. 複數迴旋複數迴旋 ((實數乘法實數乘法) )

實數實數 tt 領域的乘積等於複數領域的乘積等於複數 ss 領域中的迴旋，即領域中的迴旋，即

1 2 1 2( ) ( ) ( ) ( )f t f t F s F s ±

其中其中，＊，＊代表複數迴旋。代表複數迴旋。

★★ 表表 22--4 4 摘錄摘錄拉氏拉氏轉換的重要定理。轉換的重要定理。

(2(2--139)139)

(2(2--140)140)

114



Chapter 2 數學基礎乘上常數

和及差

微分

其中

積分

對時間移位初值定理

終值定理若 sF(s) 在 s 平面的虛軸上及其右邊並無極點

複數移位

實數迴旋

表 2-4 拉氏轉換定理

115



Chapter 2 數學基礎以部份分式展開法求反拉氏轉換

部份分式展開法部份分式展開法

有理函數的反有理函數的反拉氏拉氏轉換運算可使用部份分式展開法和轉換運算可使用部份分式展開法和拉氏拉氏轉換表來求出。轉換表來求出。

其中，P(s) 和 Q(s) 為 s 之多項式

其中， a0，a1，，an1 為實係數

★★ GG((ss) ) 僅有簡單極點時僅有簡單極點時其中，其中， ss11 ss22 ssnn。。

( )( )( )i

i

s is s

Q sK s sP s

其中，其中， (i = 1，2，，n)

(2(2--141)141)

(2(2--142)142)

(2(2--143)143)

(2(2--144)144)

116




Ex. 2Ex. 2--55--11

(2(2--145)145)

例如，為找出係數例如，為找出係數 KKss11 ：：

求算下列函數的部份分式形式：求算下列函數的部份分式形式：

係數 K 1，K 2 和 K 3 求法

(2(2--148)148)

(2(2--147)147)

(2(2--146)146)

(2(2--149)149)

(2(2--150)150)

(2(2--151)151)

117




118




119



Chapter 2 數學基礎極點在 s = si 是 r 階(i 1，2，，n r )

(n r) 個係數 Ks1，Ks2，，Ks(n r)對應 (n r) 個單階極點，這些係數可依 (2-145) 式的方法求之

★★ G(s) G(s) 有多階極點時有多階極點時

GG((ss) ) 可展開成：可展開成：

項項重重複複極極點點

項項單單階階極極點點

多階極點的係數多階極點的係數 AA11，，，，A A rr，則如下求算，則如下求算︰︰

(2(2--152)152)

(2(2--153)153)

(2(2--154)154) (2(2--156)156)

(2(2--155)155) (2(2--157)157)

120




Ex. 2Ex. 2--55--22 求算下列函數的部份分式形式：求算下列函數的部份分式形式：

對應單階極點的係數求法如下：對應單階極點的係數求法如下：

三階極點的係數則為三階極點的係數則為

<Sol.><Sol.>

(2(2--164)164)

(2(2--163)163)

(2(2--162)162)

(2(2--161)161)

(2(2--160)160)

(2(2--159)159)

(2(2--158)158)

121




完整的部份分式展開式(2(2--165)165)(2(2--165)165)

122




123



Chapter 2 數學基礎★★ GG((ss) ) 有簡單共軛複數極點時有簡單共軛複數極點時

其中，P(s) 和 Q(s) 為 s 之多項式

假設假設 (2(2--141) 141) 式的有理函數式的有理函數 GG((ss) ) 包括了一對包括了一對複數極點複數極點︰︰

和

這些極點的對應係數是這些極點的對應係數是

(2(2--141)141)

(2(2--167)167)

(2(2--166)166)

124



Chapter 2 數學基礎Ex. 2Ex. 2--55--33

考慮 Second-order prototype 函數

若假設之值小於 1，試求g (t) = ?

(2-170) (2-171)

其中，

(2(2--168)168)

<Sol.><Sol.>(2(2--169)169)

可解出 (2-169) 式的係數為

或

(2(2--172)172)

(2(2--173)173)

(2(2--174)174)反反拉氏拉氏轉換轉換

(2(2--175)175)

(2(2--176)176)2

2( ) sin 1 0

1n tn

ng t e t t

125




拉氏拉氏轉換在解線性常微分方程式的應用轉換在解線性常微分方程式的應用

用用拉氏拉氏轉換的方法來解線性常微分方程式的步驟如下轉換的方法來解線性常微分方程式的步驟如下︰︰

1. 1. 利用利用拉氏拉氏轉換及轉換表將微分方程式轉換至轉換及轉換表將微分方程式轉換至 ss 領域。領域。

2. 2. 處理轉換過後的代數方程式，解出輸出變數。處理轉換過後的代數方程式，解出輸出變數。

3. 3. 對此代數方程式進行部份分式展開。對此代數方程式進行部份分式展開。

4. 4. 利用利用拉氏拉氏轉換表求解反轉換表求解反拉氏拉氏轉換。轉換。

Linear ordinary differential equations

2

1 02( ) ( ) ( ) ( )d y t dy ta a y t f t

dt dt

First-order linear system :

Second-order linear system :

(2-177)

(2-178)

0( ) ( ) ( )dy t a y t f t

dt

126



Chapter 2 數學基礎First-order prototype system

where, where, is known as the is known as the time constant time constant of the system, which is a measure of howof the system, which is a measure of howfast the system responds to initial conditions of external excifast the system responds to initial conditions of external excitations. tations.

)()(1)( tftydt

tdy

(2(2--179)179)

The first-order prototype form:

Ex. 2-6-1 求解微分方程式求解微分方程式 (2(2--179) 179) 。。<Sol.><Sol.> For a unit step input

0, 0( ) ( )

1, 0 s

tf t u t

t

( ) ( ) ( )su t τ y t y t

1 ( ) ( )sτY s Y ss

Eq. (2-179) becomes

(2-180)

(2-181)

If y(0)=y(0)=0, and (y(t)) = Y(s), we have． 1( ( ))su ts

L L

(2-182)

127




(2(2--183)183)1 1( )

1Y s

s τs

Pole: s = – 1/

由部分分式由部分分式，，(2(2--183)183)式式變為變為1

)( /10

sK

sKsY

(2(2--184)184)

1 and 1 /10 KK

圖2-22 一階一階RCRC電路系統的單電路系統的單位步階響應。位步階響應。

再利用再利用拉氏拉氏轉換轉換，，由由(2(2--184)184)式得出式得出

/( ) 1 t τy t e (2(2--185)185)

where is the time for y(t) to reach 63% of its final value of .1)(lim

tyt

128




129



Chapter 2 數學基礎Second-Order Prototype System The second-order prototype form:

22

2( ) ( )2 ( ) ( ) n n

d y t dy tξω ω y t f tdt dt

(2(2--186)186)

where, where, is known as the is known as the damping ratiodamping ratio, , nn is the is the natural frequencynatural frequency of the of the system. system.

Ex. 2-6-2求解微分方程式：求解微分方程式：

其中，us(t)是單位步階函數。

(2(2--187)187)

(1)0(0) ( ) / 2.ty dy t dt 起始條件為起始條件為 y(0) = 1 及及

<Sol.><Sol.>(2(2--188)188)(2(2--188)188)

(2-187) 式等號兩邊取其拉氏轉換

(2(2--189)189)(2(2--189)189) 初值代入

130



Chapter 2 數學基礎部份分式展開

(2(2--190)190)

Ex. 2-6-3

全解

★★ (2(2--191) 191) 式的第一項為穩態解或特殊積分，而後兩項為暫態解或齊次解。式的第一項為穩態解或特殊積分，而後兩項為暫態解或齊次解。

★★ 應用終值定理應用終值定理 (2(2--133) 133) 式，式，y y ((tt) ) 的穩態解的穩態解

(2(2--192)192)

求解微分方程式：求解微分方程式：

(2(2--193)193)

yy((tt) ) 和和 dydy((tt)/)/dt dt 的初值皆為零。的初值皆為零。

<Sol.><Sol.>其中， = 0.5455 和 n = 31.62。

(2(2--194) 194)

131



Chapter 2 數學基礎查表法

(2(2--195)195)

(2(2--196)196)

(2(2--197)197)

相對應的極點分別為 s = 0， + j 與 j，其中

部份分式展開式部份分式展開式

其中

★★ 部份分式展開法：部份分式展開法：

(2(2--198)198)(2(2--198)198)(2(2--199)199)(2(2--199)199)

(2(2--200)200)

(2(2--201)201)(2(2--201)201)

(2(2--202)202)(2(2--202)202)

132




(2(2--203)203)(2(2--203)203)

角度角度，見，見圖圖 22--23 23

(2(2--204)204)

角度角度，見，見圖圖 22--23 23

(2(2--204)204)

圖圖 22--2323 ss 平面之根位置平面之根位置

133



Chapter 2 數學基礎反反拉氏拉氏轉換轉換

(2(2--204) 204) 式的式的代入代入 (2(2--205)205)式式

(2(2--205)205)

(2(2--206)206)

(2(2--207)207)

134




135




136




線性系統的脈衝響應及轉移函數線性系統的脈衝響應及轉移函數★★ 脈衝響應脈衝響應

Impulse response y (t) = g (t)

LTI SystemLTI System

gg ((tt))

輸入 u (t)

Impulse (t)

輸出 y (t)

圖圖22--2424Graphical representation of an Graphical representation of an impulse function.impulse function.

0ˆ

( )2

0

t τ εuu t τ ε t τ εε

t τ ε

(2(2--208)208)

Definition of impulse function: Fig. 2-24.

A rectangular function u(t) of a very large magnitude û/2ε becomes an impulse function for very small duration as ε 0.

For û = 1, u(t) = (t) is also known as unit impulse or Dirac delta function.

137



Chapter 2 數學基礎 For t = 0 in Eq.(2-208), using Eq.(2-116) and

noting the actual limits of the integral are defined from t = 0– to t = , it is very easy to verify that the Laplace transform of (t) is unity, i.e.,

(2(2--116)116)(2(2--116)116)

[ ( )] 1 as 0t ±

)()(

))(())(()(

sFsY

tutysG

L L

(2(2--209)209)

The response of any system can be characterized by its impulse response g(t), which is defined as the output when the input is a unit-impulse function (t).

Once the impulse response of a linear system is known, the output of the system y(t), with any input, u(t), can be found by using the transfer function.

We define

as the transfer function of the system.

138




Ex. 2Ex. 2--77--11 利用利用 (2(2--186)186) 式之二階系統來解式之二階系統來解 Ex. 2Ex. 2--55--33::

(2(2--210)210)

(2(2--211)211)因此，

0 1sin1

)( 2

2

ttetg n

tn n

(2(2--212)212)

(2(2--213)213)

0

( ) * ( ) ( ) ( )

s

t

s

y t u g tG su g t d

s

L L

L

22 2

2( ) ( )2 ( ) ( )n n n

d y t dy t y t u tdt dt

2

2 2 ( ( ))( ) ( ( )) 2

n

n n

y tG su t s s

LL

Transfer function

Impulse response

For a unit-step input u(t) = us(t), using the convolution properties of Laplace transforms,

From the inverse Laplace transform of Eq.(2-213), the output y(t) is therefore

139




.cos 1 dtgu

t

s 0

)(

或或

0 1sin1

1)( 2

2

ttety n

tn

(2(2--214)214)

140




141



Chapter 2 數學基礎轉移函數轉移函數 ((單輸入單輸出系統單輸入單輸出系統))

假設起始條件為零假設起始條件為零轉移函數轉移函數 GG((ss) ) 的定義的定義(2(2--215)215)

轉移函數轉移函數 GG((ss) ) 與輸入和輸出的與輸入和輸出的拉氏拉氏轉換的關係轉換的關係(2(2--216)216)

線性非時變系統，其輸入線性非時變系統，其輸入--輸出的關係常以微分方程式來描述。一線性非時變系統輸出的關係常以微分方程式來描述。一線性非時變系統的輸入的輸入--輸出關係，為常數實係數輸出關係，為常數實係數 nn 階微分方程式階微分方程式

係數 a0，a1，，an1 和

b0，b1，，

bm 為實數

u(t) 與 y(t) 之間的轉移函數為

(2(2--217)217)

(2(2--218)218)

(2(2--219)219)

142



Chapter 2 數學基礎★★ 轉移函數的性質總結轉移函數的性質總結

1. 1. 轉移函數只定義於線性非時變系統，對非線性系統則無意義。轉移函數只定義於線性非時變系統，對非線性系統則無意義。

2. 2. 系統的輸入變數與輸出變數之間的轉移函數定義為脈衝響應的系統的輸入變數與輸出變數之間的轉移函數定義為脈衝響應的拉氏拉氏轉換；換言轉換；換言

之，是輸出的之，是輸出的拉氏拉氏轉換與輸入的轉換與輸入的拉氏拉氏轉換之比。轉換之比。

3. 3. 系統的所有起始條件均假設為零。系統的所有起始條件均假設為零。

4. 4. 轉移函數與輸入無關。轉移函數與輸入無關。

5. 5. 連續資料系統的轉移函數表示成僅為複變數連續資料系統的轉移函數表示成僅為複變數 ss 的函數，而非實變數、時間或任的函數，而非實變數、時間或任

何其它獨立變數的函數。離散資料系統可用差分方程式來表示，當使用何其它獨立變數的函數。離散資料系統可用差分方程式來表示，當使用 zz 轉換轉換

時，轉移函數就變成複變數時，轉移函數就變成複變數 zz 的函數。的函數。

★★ 適當轉移函數適當轉移函數

如果 (2-219) 式的分母多項式的階數大於分子多項式的階數 (即 n > m)，則稱轉移

函數 (2-219) 式是嚴格適當的 (strictly proper)；若 n = m，則稱轉移函數為適當

的 (proper)，若 m > n，則稱轉移函數為不適當的 (improper)。★★ 特性方程式特性方程式將轉移函數的分母設為零即可得線性系統的特性方程式將轉移函數的分母設為零即可得線性系統的特性方程式

(2(2--220)220)

(2(2--219)219)

143



Chapter 2 數學基礎轉移函數轉移函數 ((多變數系統多變數系統))若一線性系統有若一線性系統有 pp 個輸入和個輸入和 qq 個輸出時，第個輸出時，第 ii 個輸出和第個輸出和第 jj 個輸入之間的轉移函數個輸入之間的轉移函數

可定義為：可定義為：其中，Rk (s) = 0，k = 1，2，，p，k j。

注意 (2-221) 式之定義僅針對第 j 個輸入的影響，假設其它輸入為零。系統的第 i 個輸

出轉換與所有的

輸入轉換之關係

矩陣矩陣--向量的形式向量的形式︰︰

q 1 的轉換輸出向量

p 1 的轉換輸出向量

q p 的轉移函數矩陣

(2(2--222)222)

(2(2--221)221)

(2(2--223)223)

(2(2--224)224)

(2(2--225)225)

(2(2--226)226)

144



Chapter 2 數學基礎STABILITY OF LINEAR CONTROL

SYSTEMS 前前言言

1. 對應於系統暫態響應的齊次解是由特性方程式的根來決定。

2. 線性控制系統的設計可視為如何安排系統轉移函數的極點和零點位置，使系統的表現能依照預先指定的規格。

3. 在多種用於設計的性能規格當中，重要的要求是︰系統為穩定的系統為穩定的。

4. 穩定性分為絕對穩定性(absolute stability)和相對穩定性(relative stability)。絕對穩定性只告訴我們系統是否穩定，其答案為是或非。一旦發現系統是穩定的，我們所感興趣的是來決定它究竟有多穩定，其穩定程度則以相對穩定性來測定。

5. 針對線性非時變系統來定義下列兩種形式的響應︰

當系統同時接受輸入與初始條件時，整體響應可寫成︰

整體響應整體響應 = = 零態響應零態響應 + + 零輸入響應零輸入響應

1) 1) 零態響應零態響應零態響應是當所有系統的初始狀態為零，只由輸入所引發的響應。零態響應是當所有系統的初始狀態為零，只由輸入所引發的響應。

2) 零輸入響應零輸入響應零輸入響應是當所有的輸入為零，只由初始狀態所引發的響應。零輸入響應是當所有的輸入為零，只由初始狀態所引發的響應。

離散與連續資料離散與連續資料系統均適用系統均適用

145




BOUNDED-INPUT, BOUNDED-OUTPUT (BIBO) STABILITY－CONTINUOUS-DATA SYSTEMS

1. 線性非時變系統： u(t) = 輸入、y(t) = 輸出、和 g(t) = 脈衝響應

2. 設初始條件為零，若系統對應於有限輸入 u(t) 的輸出 y(t) 為有限時，則該系統為有有

限輸入有限輸出限輸入有限輸出 ((BIBOBIBO) ) 穩定穩定，或簡稱穩定。

3. u(t)、y(t) 和 g(t) 之間關係的迴旋積分為

(2-227)

(2-228)

(2-229) 若 u(t) 為有限的，

(2-230) 其中 M 為一有限的正數

LTI Systemg(t)

u(t) y(t) = u(t) g(t)

146




(2-231)

若 y(t) 要為有限的，或

(2-232) 其中 N 為一正數

則下列條件必須成立︰ (2-233)

亦即，對於任一有限的正數 Q，

(2-234) (2-234) 式之涵意為︰在 g() 對之曲線下的面積必須是有限的。

RELATIONSHIP BETWEEN CHARACTERISTIC EQUATION ROOTS AND STABILITY

1. 特性方程式的根與條件 (2-234) 式之間的關係：

轉移函數 G(s)

(2-235)

147




(2-236)

2. 當 s 的值為 G(s) 的極點時，G(s) = ，(2-236) 式變為

(2-237)

3. 若一個或多個特性方程式的根在 s 平面的右半邊或 j 軸上，亦即 0；則

(2-238)

由於 est = e t，其中為 s 的實數部份。

上式兩邊取絕對值

(2-237) 式 (2-239)

若要若要BIBOBIBO穩定，特性方穩定，特性方

程式的根必須全部落在程式的根必須全部落在 ss平面的左半部平面的左半部

4. 因此，若要 BIBO 穩定，特性方程式的根或G(s) 的極點，不能落在 s 平面的右半邊或 j 軸上，它們必

須全部落於 s 平面的左半邊。

一系統若是BIBO穩定，則簡稱為穩定，否則即是不穩定

違反了 BIBO 穩定的要求

Ex. 當一系統有根在 j 軸上時，例如在 s = j0 和 s = j0 上，若輸入為弦波

sin 0t，則輸出將會是 t sin 0t 的形式；這是無限的，因此系統不是穩定的。

148




ZERO-INPUT AND ASYMPTOTIC STABILITY OF CONTINUOUS-DATA SYSTEMS

連續資料系統的零輸入穩定性與漸近穩定性連續資料系統的零輸入穩定性與漸近穩定性

1. 零輸入穩定性是指當輸入為零，而系統僅由其初始條件驅動時的穩定狀況。

2. 令一 n 階系統的輸入為零以及由初始條件所引發的輸出為 y(t)，則 y(t) 可表示成

(2-240)

其中(2-241)

gk(t) 表示由 y(k)(t0) 所引發的零輸入響應。

零輸入穩定性也是由特性方程式的根來決定。

3. 定義零輸入穩定如下︰

若由有限初始狀態 y(k)(t0) 所引起的零輸入響應 y(t)，在 t 趨近於無限大時，會達到零，則該系統稱為零輸入穩定或簡稱穩定；否則該系統為不穩定。

149




數學定義數學定義：：

若一線性非時變系統對任一組有限的y(k)(t0)，存在一組由 y(k)(t0) 所決定的 M，使得

且且

則該系統為零輸入穩定零輸入穩定。零輸入穩定性又稱為漸近穩定性。

4. (2-240) 式的兩邊取絕對值：

(2-244)

因為所有的初始條件皆假設為有限值，所以 (2-242) 式要求下列的條件要成立︰

(2-245)

5. 令 n 個特性方程式的根以來表示。

若 n 個根當中的 m 個為單根，而其餘則為多重根，則 y(t) 的形式為

(2-242) (2-243)

由由Condition 2) Condition 2) 所造成所造成

(2-246) 其中 Ki 和 Li 為常數係數。

150




時的響應 y(t) 。所以，為了要滿足

(2-242) 和 (2-243) 式的條件，ssii 的實部必須是負的的實部必須是負的。

6. 當特性方程式有單根落於 j 軸上且沒有根落於右半平面時，我們稱這種情形為臨界穩定或臨界不穩定。

7. 若系統被設計成一積分器 (或速度控制系統)，則該系統會有根 (s) 在 s = 0 且視為穩定。

8. 若系統被設計成一振盪器，則特性方程式會有單根在 j 軸上且視為穩定。

系統特性方程式的根特意放在 j軸上，則定義為穩定

9. 令一連續資料線性非時變之 SISO 系統，其特性方程式的根或 A 的特徵值為 si = i + ji，i = 1，2，，n。若任一根為複數，則其共軛複數必亦為根。

對線性非時變系統而言，BIBO 零輸入穩定和漸近穩定皆要求特性方程式的根落在 s平面的左半邊。因此，若一系統為 BIBO 穩定，則它必定也是零輸入或漸近穩定。

對於一不穩定線性系統，是指其至少有一個特性方程式的根不落在 s 平面的左半邊。

(2-246)

151




表表 22--55：系統可能的穩定狀況與對應之特性方程式的根。

Table 2-5 Stability Conditions of Linear Continuous-Data Time-Invariant SISO Systems

152




Ex. 2-11-1下列有一些閉迴路轉移函數及其相關的穩定性條件。下列有一些閉迴路轉移函數及其相關的穩定性條件。

153




1. 線性非時變 SISO 系統的穩定性，可由檢查系統特性方程式根的位置來得知。

2. 在 s 平面上之穩定與不穩定的區域示於圖 2-25 中。

決定穩定性的方法決定穩定性的方法

3. 若系統的參數皆為已知，特性方程式的根可用 MATLAB 內的各種元件，如本章前面所討論過的各種MATLAB Toolbox 視窗來求解。

4. ACSYS 發展的其它大部份工具同樣也可以用來求取特性方程式之根 (極點)。轉移函數符號工具 (Transfer Function Symbolic Tool, tfsymtfsym)，轉移函數計算器(Transfer Function Calculctor, tfcaltfcal)，及控制器設計工具 (Controller Design Tool, controls)。利用 ACSYSACSYS 元件，如時間響應分析工具 (Time Response Analysis Tool, timetool) 及頻率響應工具 (Freguency Response Tool, freqtool)，以及羅斯羅斯--赫維茲赫維茲穩定度常式(Routh-Hurwitz Stability routine, tfrouth)同樣也能更深入探討穩定性研究。

154




圖 2-25 在 s 平面上的穩定與不穩定區域

155




◆◆ 無須解根即可決定線性連續資料系統穩定性的幾種知名方法：無須解根即可決定線性連續資料系統穩定性的幾種知名方法：

1. 羅斯羅斯--赫維茲赫維茲準則準則此準則是一種代數方法，用來決定具常係數常係數特性方程式之線性非時變系統的絕對穩定性絕對穩定性。它可測試是否有任何特性方程式的根落於 s 平面的右半邊，而落於 j 軸上及右半平面根的個數亦可找出來。

2. 奈氏奈氏準則準則此為一半圖解法，以觀察迴路轉移函數之奈氏圖的行為，來決定在右半s 平面中，閉迴路轉移函數之極點與零點數目的差。

3. 波德波德圖圖此圖為迴路轉移函數 G( j)H( j) 的分貝振幅及角度相位相對於頻率的圖。閉迴路系統的穩定性可由觀察此圖的行為而得知。

156



Chapter 2 數學基礎22--13 13 羅斯羅斯--赫維茲赫維茲準則準則1. 線性非時變 SISO 系統的特性方程式：

(2-247)

其中所有的係數都是實數。

2. 兩個必要的但非充分的條件：

1) 多項式所有的係數都是具有同樣的符號。多項式所有的係數都是具有同樣的符號。2) 沒有缺項。沒有缺項。

3. 使(2-247)式中所有的根沒有一個有正實數部份，則需(2-247) 式的係數有下列關係((代數定律代數定律))︰

一個多項式所有的係數都不為零且均同號，卻仍可能有零點位於 s 平面的右半邊。(非充分的條件 )

因為求根的電腦程式可以輕易地解出多項式的零點，所以羅斯-赫維茲準則的大好處在於可應用到至少有一個未知參數

的方程式。

(2-248)

(2-249)

(2-250)

(2-251)

除非至少有一個根有正實數部份，否則所有的比值必須是正值且非零值。

157



Chapter 2 數學基礎66--55--1 1 羅斯羅斯表表 [1] [1] 1. 赫維茲準則給定 (2-247) 式的所有根落在 s 平面左半面的充分與必要條件。

2. 赫維茲準則規定了方程式的 n 個赫維茲行列式均必須為正才行。

3. 由高階項開始算，第一列由第一，第三，第五，，等係數組成，第二列由第二，第四，第六，，等係數組成，如下表所示︰

Ex. 六階的方程式：

(2-252) 羅斯表或羅斯陣列：見下頁

1. 左邊 s 項所形成的行是做為識別參考之用，有助於計算結果的記錄。

2. 羅斯表的後一列必定是 s0 之列。

3. 若羅斯表中第一行的所有元素都是同號的，則多項式的根全部位於 s 平面的左半

邊。若在第一行中的元素有符號的改變時，則符號改變的次數即為具有正實部份

或在右半 s 平面根的數目。

158




►► 例題例題 2.13.12.13.1 考慮方程式

(2-253) 試判斷根的位置。<Sol.><Sol.>

1. 方程式並無缺項且所有的係數都是同號，正好滿足必要條件，所以沒有根位

於 s 平面的右半邊或虛軸上。 (充分條件仍必須檢查。)

159




2. 羅斯表：

3. 在第一行中有兩次符號改變，所以方程式有兩個根位於 s 平面的右半邊。

4. 解 (2-253) 式的根可得四個根︰s = 1.005 j0.933 和 s = 0.755 j1.444。

160




22--1313--2 2 羅斯羅斯表過早結束的特殊情況表過早結束的特殊情況

★★ 羅斯羅斯表可能發生下列的困難：表可能發生下列的困難：

1) 羅斯表的任何一列的第一個元素是零，而其它的元素卻不是。

2) 羅斯表的一整列元素都是零。

第一種情況：第一種情況：以一個非常小的正數代替第一行中的零，然後繼續羅斯表的進行。

161



Chapter 2 數學基礎►► 例題例題 22--1313--22 考慮一線性系統的特性方程式︰

(2-254) 試判斷根的位置。

<Sol.><Sol.> 因為所有的係數皆不為零且同號，所以我們需要使用羅斯-赫維茲準則。

1. 羅斯表：

2. 以一個非常小的正數來取代 s2 列中的零，所得結果如下︰

3. 在羅斯表的第一行有兩個符號改變，所以 (2-254) 式有兩個根在右半 s平面。

4. 解 (2-254) 式可得根為 s = 0.091 j0.902 和 s = 0.406 j1.293；後面兩個根很明顯地是在右半 s 平面。

如果方程式有純虛根，則上述之法可能會得到不正確的答案 [19，20]。

162




第二種特殊情況：第二種特殊情況：在羅斯表正常完成前其中的一整列元素都是零，則表示有下列的一種或多種情形可能發生︰

1. 方程式至少有一對大小相等、符號相反的實根。2. 方程式有一或多對虛根。3. 方程式有對稱於 s 平面原點的共軛複數根對 (例如 s = 1 j1，s = 1 j1)。◆ 整列都是零的情形，可用輔助方程式 (auxiliary equation) A(s) = 0 來克服。

輔助方程式的係數，是在羅斯表中全列為零的上一列的係數

◆ 輔助方程式必定是偶次多項式，亦即只有 s的偶數次方出現。

輔助方程式的根，必定也滿足原來的方程式輔助方程式的根，必定也滿足原來的方程式

◆ 想繼續羅斯表的進行，請執行下列步驟︰

1. 利用在零那列的上一列的係數來形成輔助方程式 A(s) = 0。2. 取輔助方程式對 s 的微分，可得 dA(s)/ds = 0。3. 以 dA(s)/ds = 0 的係數來取代零的列。4. 以新形成的係數列來取代零的列，並以平常的方式繼續羅斯表的進行。5. 照平常的方式來解釋第一行係數的變號，如果有的話。

163



Chapter 2 數學基礎►► 例題例題 22--1313--33 考慮下列方程式，它可能是某一線性控制系統的特性方程式。


1. 羅斯表：

2. 用 s2 列的係數來形成輔助方程式︰

(2-256) A(s) 對 s 的微分：

(2-257)

3. 以上式的係數 8 和 0 來取代原來表中s1 列的零，則羅斯表的其餘部份為

4. 在整個羅斯表的第一行中並沒有符號的改變，所以 (2-257) 式沒有根在右半 s 平面。

5. 解 (2-256) 式中的輔助方程式，可得兩根 s = j 和 s = j，其亦為 (2-255) 式的根。

系統有兩個根在 j 軸，系統是臨界穩定臨界穩定。

164



Chapter 2 數學基礎►► 例題例題 22--1313--44 考慮一個三階控制系統具有特性方程式


1. 羅斯-赫維茲準則適用於求出使系統穩定之 K 的臨界值，亦即使至少有一根落於j 軸上但沒有根落於右半 s 平面的 K 值。

2. 羅斯表如下︰

3. 為使系統穩定，所有羅斯表第一行的係數必須同號。亦即需

(2-259)

和 (2-260)(2-261)

K 使系統達到穩定的條件

4. 將 K = 273.57 代入輔助方程式 (可由羅斯表中 s2 列的係數求得) 。

165




(2-262) 根在 s = j1097 和 s = j1097 系統的零輸入響應是頻率為 1097.27 rad/sec 的弦波。

(2-263) 試決定出使系統穩定的 K 值範圍。

<Sol.><Sol.>1. 羅斯表如下︰

►► 例題例題 22--1313--55 考慮一閉迴路控制系統的特性方程式

166



Chapter 2 數學基礎2. 穩定性的條件：

K > 0 (2-264)

若要閉迴路系統穩定，K 必須符合

羅斯-赫維茲準則只在特性方程式是代數的，且所有的係數是實數時才適用。

羅斯-赫維茲準則的另一限制為︰它只適用於決定特性方程式的根是在 s平面的左邊或右邊，穩定邊界為 s 平

面的 j 軸。

(2-265)

(2-266)

167



Chapter 2 數學基礎補充範例 (摘自Nise, 4th ed. 2005.)►► 例題例題 11 考慮一閉迴路控制系統，Fig. (a)Fig. (a).

試判斷根的位置。<Sol.><Sol.>1. Equivalent system: Fig. (b).2. Characteristic Equation:

3 210 31 1030 0s s s 3. Routh table:

Number of sign changes = 2.

可以乘上或除以任意的正值倍數!!!

168



Chapter 2 數學基礎►► 例題例題 22 決定下列閉迴路轉移函數的穩定性：

5 4 3 210( )

2 3 6 5 3T s

s s s s s

<Sol.><Sol.>1. Routh table:

169



Chapter 2 數學基礎2. Determining signs in the first column of a Routh table with zero

as first element in a row:

無論 0+ 或 0

均可

Number of sign changes = 2.

170



Chapter 2 數學基礎特性方程式的根取倒數時，其根在 s-plane上的分布區域不變 !!1. Original characteristic equation:

0011

1 asasas n

nn

2. Replace s by 1/d:

011101

1

1

ad

ad

ad

n

n

n

1 1

1 1 0

11 1 0

1 1 1 11

1 1 0

n n n

n

n nn

a a ad d d d

a d a d a dd

►► 例題例題 33 利用反係數方法重做例題例題 22，判斷T(s)的穩定性。

<Sol.><Sol.>

The polynomial with reciprocal roots is a polynomial with the coefficients written in reverse order.

171



Chapter 2 數學基礎1. Characteristic equation with reciprocal coefficients:

5 4 3 2( ) 3 5 6 3 2 1D s s s s s s 2. Routh table:2. Routh table:

Routh table for Example 3

Number of sign changes

= 2.

System unstable!!

172



Chapter 2 數學基礎►► 例題例題 44 決定下列閉迴路轉移函數有幾個右半平面的極點?

<Sol.><Sol.>5684267

10)( 2345

ssssssT

1. Routh table:

Entire row is zero, then it needs to be corrected by the method of auxiliary equation

2. Auxiliary equation:86)( 24 sssP 0124)( 3 ss

dssdP

173



Chapter 2 數學基礎3. Routh table shows that all entries in the first column are

positive. Hence, there are no right-half-plane poles. Remarks1. 當一純偶或純奇次多項式是原特性多項式的因子時，Routh Table將會出現整列

為零的情況。2. 偶次多項式具有對稱於原點的根。3. 此類對稱根有三大類:

1) 對稱根且為實數；2) 對稱根且為虛數；3) 對稱象限內的根(共軛複數根) 。

174



Chapter 2 數學基礎►► 例題例題 55 決定下列閉迴路轉移函數有幾個極點在右半平面、左半平面、及在 j 軸上?

<Sol.><Sol.>8 7 6 5 4 3 2

20( )12 22 39 59 48 38 20

T ss s s s s s s s

1. Routh table:

Entire row is zero, then it needs to be corrected by the method of auxiliary equation

2. Auxiliary equation:4 2( ) 3 2P s s s

3( ) 4 6 0dP s s sds

No sign change

175



Chapter 2 數學基礎3. 由於輔助方程式開始之後的Routh Table並無變號，故其根均位於虛軸上。

4. 整體 Routh Table 的第一行變號二次，故其有二根均位於右半平面內。

5. 結論: 見下表

►► 例題例題 66 決定下圖所示系統有幾個極點在右半平面、左半平面、及在 j 軸上?

<Sol.><Sol.>1. Closed loop transfer function:

176




5 4 3 21( )

2 3 2 3 2 1T s

s s s s s

2. Characteristic equation:

5 4 3 22 3 2 3 2 1 0s s s s s or 5 4 3 22 3 2 3 2 0s s s s s

倒數係數型特性方程式

3. Routh table:

倒數型 Routh table 較易判定正、負號

Two poles in rhp

177



Chapter 2 數學基礎►► 例題例題 77 決定下圖所示系統有幾個極點在右半平面、左半平面、及在 j 軸上?

<Sol.><Sol.>1. Closed loop transfer function:

8 7 6 5 4 3 2128( )

3 10 24 48 96 128 192 128T s

s s s s s s s s

2. Characteristic equation:

8 7 6 5 4 3 23 10 24 48 96 128 192 128 0s s s s s s s s 3. Routh table: see next page.4. Auxiliary equation:

6 4 2( ) 8 32 64P s s s s

5 3( ) 6 32 64 0dP s s s sds

In the 6th line of Routh Table.

178



Chapter 2 數學基礎 Routh table:

Entire row is zero !

5. 結論: 見下表 (see next page).

179




►► 例題例題 8 8 決定下列所示系統有幾個極點在右半平面、左半平面、及在 j 軸上?

<Sol.><Sol.>

0 3 1 102 8 1 010 5 2 0

1 0 0

u

y

x x

x

1. (sI A):

180



Chapter 2 數學基礎0 0 0 3 1 3 1

( ) 0 0 2 8 1 2 8 10 0 10 5 2 10 5 2

s ss s s

s s

I A

2. Characteristic equation det (sI A):3 2det( ) 6 7 52 0s s s s I A

3. Routh table:

4. 結論：Routh Table 第一行變號一次，故有一個極點位於rhp，兩個極點位於lhp.

181




MATLAB MATLAB 工具與個案研究工具與個案研究

轉移函數工具盒的說明與用法轉移函數工具盒的說明與用法

若已進入 MATLAB 符號工具盒，只要在 ACSYS 視窗內按下合適的按鍵式是在 MATLAB 命令視窗

內輸入 tfsym，便可用 ACSYS 轉移函數及符號工

具。此種符號工具視窗如圖圖 22--2626 所示。

點選「初學者說明」(Help for 1st Time User) 鍵來

查閱如何使用此工具盒各項指令。

如圖圖 22--2727 所示圖圖 22--26 26 轉移函數符號視窗轉移函數符號視窗

圖圖 22--27 27 符號協助對話盒符號協助對話盒

182



Chapter 2 數學基礎依指示，按下「轉移函數與反依指示，按下「轉移函數與反拉氏拉氏」」(Transfer Funciton and Inverse (Transfer Funciton and Inverse Laplace) Laplace) 鍵來執行程式。吾人必須在鍵來執行程式。吾人必須在 MATLAB MATLAB 命令視窗內執行此程式命令視窗內執行此程式

才行。輸入如才行。輸入如圖圖22--2828 所示的轉移函數即可求得時間響應。所示的轉移函數即可求得時間響應。

圖圖 22--28 28 在在 MATLAB MATLAB 命令命令視窗內，視窗內，GG((ss) (2) (2--267)267)針對針對脈衝輸入的反脈衝輸入的反拉氏拉氏轉換轉換

183




為了求出為了求出 (2(2--267) 267) 式對不同輸入函數式對不同輸入函數 ((如步階或弦波如步階或弦波) ) 的時間表示式，使用者可結合輸入的時間表示式，使用者可結合輸入轉移函數轉移函數 ((例如，例如，1/1/ss 代表單位步階輸入代表單位步階輸入) ) 與與 tftool tftool 輸入視窗的轉移函數。如此一來，如輸入視窗的轉移函數。如此一來，如欲求取欲求取 (2(2--267) 267) 式對單位步階輸入的時間表示式時，便可使用下列的轉移函數：式對單位步階輸入的時間表示式時，便可使用下列的轉移函數：

並重複做先前所述的步驟即可。並重複做先前所述的步驟即可。(2(2--268)268)

611635

)3)(2)(1(35)( 23

sss

ssss

ssG (2(2--267)267)

)2()10005.34(1000)( 22

2

2nn

n

ssssssY

(2(2--269)269)

利用利用 TFsymTFsym 工具，可知此系統的時間表示式為工具，可知此系統的時間表示式為

Ex. 2Ex. 2--1414--11Find the inverse Laplace transform of the transfer function

184




MATLABMATLAB工具的穩定性工具的穩定性

Ex. 2Ex. 2--1414--2 2 重做重做Ex2Ex2--1313--11

圖圖22--29 Entering characteristic 29 Entering characteristic polynomial for Example 2polynomial for Example 2--1414--2 2 usingusing tfrouth tfrouth module.module.

010532 234 ssss (2(2--270)270)

結果見圖結果見圖22--3030輸入至輸入至MATLAB,MATLAB,如圖如圖22--2929

185




186




Ex. 2Ex. 2--1414--3 3 重做重做Ex2Ex2--1313--220322 234 ssss (2(2--271)271)

結果見圖結果見圖22--3131

187




Ex. 2Ex. 2--1414--4 4 重做重做Ex2Ex2--1313--33

047884 2345 sssss (2(2--272)272)

結果見圖結果見圖22--3232。。

044)( 2 ssA(2(2--273)273)

188




Ex. 2Ex. 2--1414--5 5 Considering the characteristic equation of a Considering the characteristic equation of a closeclose--loop control system loop control system

04)2(3 23 sKKss (2(2--274)274)It is desired to find the range of K so that the system is stable.See Figs.2-33.2-34 and 2-35 for more details.

圖圖22--33 Entering characteristic 33 Entering characteristic polynomial for Example 2polynomial for Example 2--1414--5 using5 usingtfrouth tfrouth modulemodule

189




Ex. 2Ex. 2--1414--5 5

圖圖22--3434

190




Ex. 2Ex. 2--1414--5 5

圖圖22--3535

191



Chapter 2 數學基礎Homework:2-5: (a), (b), and (d)2-7: (e)2-9: (a)2-16: (d) and (e)2-182-192-22: (c)2-26: (c), (g), and (i)2-29: (c) and (f)2-31: (g) and (h)2-35: (d) and (g)2-38: (b) and (e)2-432-45: (a), (b), and (c)2-49: But refer to the figure shown in the next page (即第八版，圖P6-15).2-502-52

192




Figure 2P-49

193




194




195




196




197




198




199

chapter 2 數學基礎 -...

Documents