Chapter 2
Mathematical
Foundation
Automatic Control Systems, 9th Edition
F. Golnaraghi & B. C. Kuo
Section 2- 0, p. 16
Main Objectives of This Chapter
1.
2.
To introduce the fundamentals of complex variables
To introduce frequency domain analysis and frequency
plots
3. To introduce differential equations and state space systems
4. To introduce the fundamentals of Laplace transforms
5. To demonstrate the applications of Laplace transforms to
solve linear ordinary differential equations
6. To introduce the concept of transfer functions and how to
apply them to the modeling of linear time-invariant systems
7. To discuss stability of linear time-invariant systems and the
Routh-Hurwitz criterion
8. To demonstrate the MATLAB tools using case studies
2-1
Section 2- 1, pp. 16-17
2-1 Complex-Variable Concept
• Rectangular form:
z  x  jy
(2-1)
magnitude of z
 j  1
 x  R cos
(2-2)

 y  R sin 
(2-3)
phase of z
z  R cos  jR sin  (2-4)
• Euler formula:
e j  R cos  jR sin  (2-5)
• Polar form:
z  Re j  R (2-6)
2-2
Section 2- 1, p. 17
Conjugate of the Complex Number
• Conjugate
z  x  jy
z *  x  jy
z  R cos  jR sin   e j
z*  R cos  jR sin   e j (2-8)
(2-7)
*
2
2
2
• zz  R  x  y (2-9)
2-3
Section 2- 1, p. 18
Basic Mathematical Properties
2-4
Section 2- 1, p. 18
Examples 2-1-1 & 2-1-2
• Example 2-1-1
Find j3 and j4.
• Example 2-1-2
Find zn using Eq. (2-6).
(2-10)
2-5
Section 2- 1, p. 19
Complex s-plane
   1

  1
2-6
Section 2- 1, p. 19
Function of a Complex Variable
• G(s) = Re[G(s)] + jIm[G(s)]
• Single-valued function:
(2-11)
one-to-one mapping
1
• Not single-valued function: G( s) 
s( s  1)
(2-12)
2-7
Section 2- 1, p. 20
Analytic Function
• 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 its all derivatives exist in the region.
• G(s) = 1/[s(s+1)] is analytic at every point in the s-plane
except at the points s = 0 and s = 1.
• G(s) = (s+2) is analytic at every point in the s-plane.
2-8
Section 2- 1, p. 20
Poles and Zeros of a Function
• A pole of order r at s = pi if the limit lim [( s  pi ) r G( s)]
s  pi
has a finite, nonzero value.
r
• A zero of order r at s = zi if the limit lim [( s  zi ) G( s)]
s  zi
has a finite, nonzero value.
• Example:
(2-13)
2-9
Section 2- 1, p. 22
Polar Representation
Polar representation of G(s) in Eq. (2-12) at s = 2j.
(2-15)
(2-16)
(2-17)
2-10
Section 2- 1, pp. 22-23
Example 2-1-3
16
Polar representation of G ( s) 
for s = j,
( s  2)( s  8)
 = 0  .
j  8  R2 ( j sin 2  cos2 )
2-11
Section 2- 1, p. 23
Example 2-1-3 (cont.)
• Graphical representation of G(j)
2-12
Section 2- 1, p. 24
Example 2-1-3 (cont.)
• The magnitude decreases as the frequency increases.
• The phase goes from 0º to 180º.
2-13
Section 2- 1, p. 24
Example 2-1-3 (cont.)
• Alternative approach
2-14
Section 2- 1, p. 24
Example 2-1-3 (cont.)
2-15
Section 2- 1, p. 26
Frequency Response
• G(j) is the frequency response function of G(s) .
• Second order system:
• General case:
2-16
Section 2- 2, p. 26
2-2 Frequency-Domain Plots
•
magnitude
phase
• Frequency-domain plots of G(j) versus :
– Polar plot
– Bode plot
– Magnitude-phase plot
2-17
Section 2- 2, p. 27
Polar Plots
• A plots of the magnitude of G(j) versus the phase of G(j)
on polar coordinates as  is varied from zero to infinity.
2-18
Section 2- 2, p. 27
Example 2-2-1
Consider the function G ( s ) 
1 , where T > 0.
1  Ts
2-19
Section 2- 2, p. 28
Example 2-2-2
Consider the function G ( j )  1  jT2 , where T1, T2 > 0.
1  jT1
2-20
Section 2- 2, p. 28
General Shape of Polar Plot
• The behavior of the magnitude and phase of G(j)
at  = 0 and  = .
• The intersections of the polar plot with the real
and imaginary axes, and the values of  at the
intersections.
– Intersections with the real axis: Im[G(j)] = 0
– Intersections with the imaginary axis: Re[G(j)] = 0
2-21
Section 2- 2, p. 30
Example 2-2-3
10
Consider the transfer function G ( s) 
s( s  1)
1. The magnitude and phase of G(j) at  = 0 and  = 
2. Determine the intersections, if any.
Re[G(j)] = 0   = , G(j) = 0; Im[G(j)] = 0   = 
2-22
Section 2- 2, p. 30
Example 2-2-3 (cont.)
2-23
Section 2- 2, p. 31
Example 2-2-4
1.
10
s( s  1)( s  2)
The magnitude and phase of G(j) at  = 0 and  = 
2.
Determine the intersections, if any.
Consider the transfer function G ( s) 
Re[G(j)] = 0   = , G(j) = 0
Im[G(j)] = 0   =  2 rad/sec, G( j 2 )   5 3
2-24
Section 2- 2, p. 31
Example 2-2-4 (cont.)
2-25
Section 2- 2, p. 32
Bode Plot
• Bode plot of function G(j) is composed of two plots:
– The amplitude of G(j) in dB versus log10 or 
– The phase of G(j) in degree as a function of log10 or 
• Bode plot  corner plot or asymptotic plot
• For constructing Bode plot manually, G(s) is preferably
written in the form of (2-61).
2-26
Section 2- 2, p. 32-33
Magnitude and Phase of G(j): Example
•
Magnitude of G(j) in dB:
• Phase of G(j):
2-27
Section 2- 2, p. 34
Five Simple Types of Factors
1. Constant factors: K
2. Poles or zeros at the origin of order p: (j)p
3. Poles or zeros at s = 1/T of order q: (1+jT)q
4. Complex poles and zeroes of order r:
(1+j2/n2/2n)r
5. Pure time delay: e  jTd
2-28
Section 2- 2, p. 34
Real Constant K: Magnitude
2-29
Section 2- 2, p. 34
Real Constant K: Phase
2-30
Section 2- 2, p. 34
Poles and Zeros at the Origin: Magnitude
Slope:
2-31
Section 2- 2, p. 36
Poles and Zeros at the Origin: Phase
2-32
Section 2- 2, p. 35
Decade versus Octave
2-33
Section 2- 2, p. 37
Simple Zero, 1+jT
Consider the function G(j) = 1+jT
(2-74)
• Magnitude:
– At very low frequencies, T << 1:
– At very high frequencies, T >> 1:
– Intersection of (2-76) and (2-77):  = 1/T
(2-78)
• Phase:
2-34
Section 2- 2, p. 38
Values of (1+jT) versus T
2-35
Section 2- 2, p. 39
Simple Pole, 1/(1+jT)
For the function G(j) = 1/(1+jT)
• Magnitude:
(2-80)
• Phase: G( j)   tan1 T
2-36
Section 2- 2, p. 38
Bode plots of G(s)=1+Ts & G(s)=1/(1+Ts)
2-37
Section 2- 2, p. 39
Quadratic Poles and Zeros: Magnitude
Consider the second-order transfer function
 1
• Magnitude:
– At very low frequencies, T << 1:
– At very high frequencies, T >> 1:
2-38
Section 2- 2, p. 40
Quadratic Poles and Zeros: Magnitude
2-39
Section 2- 2, p. 40
Quadratic Poles and Zeros: Phase
• Phase:
2-40
Section 2- 2, p. 42
Pure Time Delay
 jTd
e
1
• Magnitude:
• Phase: e  jTd  Td
(2-90)
• For the transfer function
G( j )  G1 ( j )e jTd
(2-91)
G( j )  G1 ( j )
G( j )  G1 ( j )  Td
2-41
Section 2- 2, pp. 42-43
Example 2-2-5
Consider the function
corner frequencies:  = 2, 5, and 10 rad/sec
2-42
Section 2- 2, p. 43
Example 2-2-5 (cont.)
2-43
Section 2- 2, pp. 44-45
Magnitude-Phase Plot
Example 2-2-6:
Polar plot
Magnitude-phase plot
2-44
Section 2- 2, p. 46
Gain- and Phase-Crossover Points
• Gain-crossover point: the point at which G( j )  1
– Gain-crossover frequency g
• Phase-crossover point: the point at which G( j )  180
– Phase-crossover frequency p
2-45
Section 2- 2, p. 47
Minimum- & Nonminimum-Phase
• Minimun-phase transfer function:
no poles or zeros in the right-half s-plane
– Unique magnitude-phase relationship
– For G(s) with m zeros and n poles, excluding the pole
at s = 0, when s = j and  = 0  , the total phase
variation of G(j) is (nm)/2.
– G(s)  0,  if   0.
• Nonminimun-phase transfer function:
has either a pole or a zero in the right-half s-plane
– a more positive phase shift as  =   0
2-46
Section 2- 2, p. 47
Example 2-2-7
90
270
Minimum-phase
transfer function
Nonminimum-phase
transfer function
2-47
Section 2- 3, p. 49
2-3 Introduction to Differential Eqs.
• Linear ordinary differential equation:
– Coefficients a0, a1, …, an1 are not function of y(t)
– First-order linear ODE:
– Second-order linear ODE:
• Nonlinear differential equation:
2-48
Section 2- 3, p. 50
State Equations: Second-Order
• Differential equation of a series electric RLC network:
• State variables:
• State equations:
di (t )
1
R
1

i (t )dt  i (t )  e(t )

dt
LC
L
L
2-49
Section 2- 3, p. 50
State Equations: nth-Order
• The nth-order differential equation:
• State variables
• State equations
2-50
Section 2- 3, p. 51
Space State Form
• Space state form for n state variables:
state vector
coefficient matrix
input vector
2-51
Section 2- 3, p. 52
Output Equation
• General form:
• Output equation of the system described by Eq. (2-97):
y(t )  x1 (t )
2-52
Section 2- 4, p. 52,54
2-4 Laplace Transform
• Condition:
• Definition:
• Inverse Laplace Transform:
2-53
Section 2- 4, p. 53
Examples 2-4-1 & 2-4-2
• Example 2-4-1: f(t) = unit-step function
• Example 2-4-2: f(t) = exponential function
 R
2-54
Section 2- 4, p. 54
Important Theorems
Theorem 1. Multiplication by a Constant
Theorem 2. Sum and Difference
Theorem 3. Differentiation
2-55
Section 2- 4, p. 55
Important Theorems (cont.)
Theorem 4. Integration
Theorem 5. Shift in Time
Theorem 6. Initial-Value Theorem:
Theorem 7. Final-Value Theorem:
2-56
Section 2- 4, pp. 55-56
Examples 2-4-3 & 2-4-4
• Example 2-4-3:
sF(s) is analytic on the imaginary axis and in the righthalf s-plane
• Example 2-4-4:
 f(t)= sint
sF(s) has two poles on the imaginary axis of the s-plane
 the final-value theorem cannot be applied in this case
2-57
Section 2- 4, pp. 56-57
Important Theorems (cont.)
Theorem 8. Complex Shift
Theorem 9. Real Convolution (Complex Multiplication)
2-58
Section 2- 5, pp. 57-58
2-5 Inverse Laplace Transform by
Partial-Fraction Expansion
Partial-Fraction Expansion:
– The order of P(s) is greater than that of Q(s)
• G(s) Has Simple Poles:
s1  s2    sn
2-59
Section 2- 5, p. 58
Example 2-5-1
2-60
Section 2- 5, p. 59
G(s) Has Multiple-Order Poles
i  1,2,..., n  r
2-61
Section 2- 5, p. 60
Example 2-5-2
2-62
Section 2- 5, p. 61
G(s) Has Simple Complex-Conjugate Poles
• G(s) contains a pair of complex poles: s    j
2-63
Section 2- 5, pp. 61-62
Example 2-5-3
• The second-order prototype function:
2-64
Section 2- 6, p. 62
2-6 Application of the Laplace Transform to
the Solution of Linear ODEs
•
Procedure:
1. Transform the differential equation to the s-domain
by Laplace transform using Laplace transform table.
2. Manipulate the transformed algebra equation and
solve for the output variable.
3. Perform partial-fraction expansion to the
transformed algebra equation.
4. Obtain the inverse Laplace transform from the
Laplace transform table.
2-65
Section 2- 6, p. 63
First-Order Prototype System
• First-order prototype form:
: time constant
• Example 2-6-1: Find the solution of Eq. (2-179) for
a unit step function
Step 1
Step 2
Step 3
Step 4
2-66
Section 2- 6, p. 63
Typical Unit-Step Response
2-67
Section 2- 6, pp. 64-65
Second-Order Prototype System
• 2nd-order prototype form:
: damping ratio; n: natural frequency
• Example 2-6-2:
y (0)  1
us(t): unit step function; initial conditions: 
 (1)
 y (0)  2
Step 1
Step 2
Step 3
Step 4
2-68
Section 2- 6, p. 65
Example 2-6-3
The initial value of y(t) and dy(t)/dt are zero.
Method 1: Using Laplace transform table
2-69
Section 2- 6, p. 66
Example 2-6-3 (cont.)
Method 2: Using partial-fraction expansion
2-70
Section 2- 6, p. 66
Example 2-6-3 (cont.)
Method 2: Using partial-fraction expansion (cont.)
2-71
Section 2- 7, p. 68
2-7 Impulse Response &
Transfer Functions of Linear Systems
• Impulse function:
uˆ
  as   0
2
• Dirac delta function: (t)
For uˆ  1, u(t )   (t )
• Laplace transform of (t)
is unity, i.e.,
2-72
Section 2- 7, pp. 68-69
Impulse Response
• Impulse response: the output when the input is a unitimpulse function (t). ( initial condition = 0)
• The response of any system can be characterized by its
impulse response, g(t).
G(s): transfer function
Y ( s)  G( s) R( s)  y(t )  g (t )  r (t )
• Example 2-7-1:
transfer function:
impulse response:
unit-step response:
2-73
Section 2- 7, p. 70
Transfer function (SISO systems)
• The transfer function of a LTI system is defined as the
Laplace transform of the impulse response, with all the
initial conditions set to zero.
• Input-output relation of a LTI system
with zero initial conditions:
• Characteristic Equation:
2-74
Section 2- 7, pp. 71-72
Transfer Function (Multivariable Syst.)
A linear system has p inputs and q outputs:
• The transfer function between jth input and ith output:
Rk (s)  0, k  1,2,..., p, k  j.
• The ith output transform:
• Matrix-vector form:
2-75
Section 2- 8, p. 72
2-8 Stability of Linear Control Systems
• Absolute stability
• Relative stability
• Two types of response for LTI systems:
– Zero-state response
– Zero-input response
• Total response
= zero-state response + zero-input response
2-76
Section 2- 9, p. 73
2-9 BIBO Stability
• With zero initial conditions, the system is said to be BIBO
(bounded-input, bounded output) stable, or simple stable,
if its output y(t) is bounded to a bounded input u(t).
•
• u(t) is bounded:
• y(t) is bounded:
• Condition:
2-77
Section 2- 10, p. 74
2-10 Relationship Between
Characteristic Equation Roots and Stability
• For BIBO stability, the roots of the characteristic equation,
or the poles of G(s), must all lie in the left-half s-plane.
• A system is said to be unstable if it is not BIBO stable.
• When a system has roots on the j-axis, e.g., s =  j0,
if the input is a sinusoid, sin 0t, then the output will be of
the form of tsin 0t, which is unbounded, and the system
is unstable.
2-78
Section 2- 11, p. 75
2-11 Zero-input and Asymptotic Stability
• Zero-input stability: yZI (t )  0 as t  
If the zero-input response y(t), subject to the finite initial
conditions, y(k ) (t0 ) , reaches zero as t approaches infinity.
• Condition: the roots of the characteristic equation must
all be in the left-half s-plane
• Zero-input stability = Asymptotic stability
• Marginally stabile or marginally unstable:
The characteristic equation has simple root on j-axis
and none in the right-half s-plane.
2-79
Section 2- 11, p. 76
Stability Conditions
(BIBO stable)
2-80
Section 2- 11, p. 76
Example 2-11-1
2-81
Section 2- 12, p. 77
2-12 Methods of Determining Stability
Stability of LTI SISO systems:
• Checking the location of the
roots of the characteristic
equation of the system
• Without involving root solving:
– Routh-Hurwitz criterion
– Nyquist criterion
– Bode diagram
2-82
Section 2- 13, p. 78
2-13 Routh-Hurwitz Criterion
• A method of determining the location of zeros of a
polynomial with constant real coefficients, without
actually solving for the zeros.
• Necessary but not sufficient conditions:
– All coefficients of the equation have the same sign
– None of the coefficients vanish
2-83
Section 2- 13, p. 79
Routh’s Tabulation (1/2)
• Step 1: The 1st row consists of the 1st, 3rd, 5th, …, coefficients
The 2nd row consists of the 2nd, 4th, 6th, …, coefficients
• Step 2:
Routh’s tabulation
(Routh’s array)
• Example:
2-84
Section 2- 13, p. 79
Routh’s Tabulation (2/2)
• Step 3: Investigate the signs of the coefficients of the 1st column
• The root of the root all in the left half of
the s-plane if all the elements of the 1st
column of the Routh’s tabulation are of
the same sign.
• The number of changes of signs in the
elements of the 1st column equals
the number of roots with positive real
parts, or those in the right-half s-plane.
2-85
Section 2- 13, p. 80
Example 2-13-1
•
• Routh’s tabulation:
• Two sign changes in the 1st column
 Two roots in the right half of the s-plane
• Four roots of Eq. (2-253): s = 1.005  j0.933 and s = 0.755  j1.444
2-86
Section 2- 13, p. 81
Special Cases when Routh’s Tabulation
Terminates Prematurely
1. The first element in any one row of Routh’s
tabulation is zero, but the others are not.
2. The elements in one row of Routh’s tabulation
are all zero.
– At least one pair of real roots with equal magnitude
but opposite signs.
– One or more pairs of imaginary roots.
– Pairs of complex-conjugate roots forming symmetry
about the origin of the s-plane,
e.g., s = 1  j1, s = 1  j1
2-87
Section 2- 13, p. 81
Special Case 1 & Example 2-13-2
The first element in any one row of Routh’s tabulation is
zero, but the others are not.
 Replace the zero element of the first column by an
arbitrary small positive number , and then proceed with
Routh’s tabulation.
• Example 2-13-2:
• Two sign changes in the 1st column
 Two roots in the right half of the s-plane
•
Four roots of Eq. (2-254): s = 0.091  j0.902 and s = 0.406  j1.293
2-88
Section 2- 13, pp. 81-82
Special 2
The elements in one row of Routh’s tabulation are all zero.
1. Form the auxiliary equation A(s) = 0 by using the
coefficients of the row just preceding the row of zeros.
2. Take the derivative: dA(s) / ds = 0
3. Replace the row of zeros with the coefficients of
dA(s) / ds = 0.
4. Continue with Routh’s tabulation with the newly
formed row of coefficients replacing the row of zeros.
5. Interpret the change of signs, if any, of the coefficients in
the first column of Routh’s tabulation.
2-89
Section 2- 13, p. 82
Example 2-13-3
•
• No sign change in the first column
 No root in the right-half s-plane.
• Roots of A(s): s =  j
which are also twos of the roots of Eq. (2-255)
2-90
Section 2- 13, pp. 82-83
Example 2-13-4
• Determine the critical value of K for stability:
Sol:
2-91
Section 2- 13, pp. 83-84
Example 2-13-5
• Consider the characteristic equation of a close-loop
control system is
Find the range of K so that the system is stable.
Sol:
disregarded
2-92