Review Chapter
Review Chapter
Review Chapter
In linear system, analytic and graphical techniques is easy in control design and
analysis.
- Classical(conventional) Control
: SISO(single-input and single-output) system Control,
main control factors: input, output, error
- Modern Control
: MIMO(Multi-input and Multi-output)system Control,
main control factor: state variable
( → applicable to linear or nonlinear time invariant or
time varying systems)
Linear
u1(t) → y1(t)
System
Linear
a1u1(t)+a2u2(t)→
(t)→ →a1y1(t)+a2y2(t)
System
Linear
u2(t) → → y2(t)
System
superposition
ex)
d 2x dx
Linear: a + b + cx = 0
dt dt
Nonlinear:
2
d 2x dx
a + b + cx = 0
dt dt
d 2 x dx
+ + cx 2 = 0
dt dt
• Laplace Transformation
- Laplace Transformation : mathematical tool used for the solution of
ordinary linear differential equations. Converting many functions into algebraic
functions of a complex variable
Definition
∞
Given f(t) such that ∫ f (t ) ⋅ e −σ t dt < ∞ ,
0
∞
Laplace Transform : F(s)= ∫ f (t )e − st dt or, F ( s ) = L[ f (t )]
0
∞ ∞
− at − st
∞
−( s + a ) t e −( s + a ) t
F ( s ) = ∫ e e dt = ∫ e dt = −
0
0 ( s + a) 0
1
= --- ①
s+a
, where C is real constant and is chosen larger than the real parts of all singular
points of F(s)
Important Theorems of the Laplace Transform
1. Multiplication by a constant
L[ Kf (t )] = KF ( s )
d n f (t )
= S F ( s) − S f (0) − S f ′(0) … − Sf
n −1 n− 2 ( n −2 )
L n
n
(0) − f ( n −1) (0)
dt
f(t) F(s)
unit impulse δ (t ) 1
1
unit step u(t)
s
- 1
t
s2
1
e − at
s+a
5
S 2 X ( s ) + S − 2 + 3SX ( s ) + 3 + 2 X ( s ) =
s
5
→ S 2 X ( s ) + 3SX ( s ) + 2 X ( s ) + S + 1 =
s
5
→ X ( s )( S 2 + 3S + 2) + S + 1 =
s
− S2 − S +5
→ X ( s )( S 2 + 3S + 2) =
S
−S −S +5
2
− S2 − S +5
→ X ( s) = =
S ( S 2 + 3S + 2) S ( S + 1)( S + 2)
A B C A( S + 1)( S + 2) + BS ( S + 2) + CS ( S + 1)
= + + =
S S +1 S + 2 S ( S + 1)( S + 2)
A( S 2 + 3S + 2) + B( S 2 + S ) + C ( S 2 + S )
=
S ( S + 1)( S + 2)
A + B + C = −1
5
3 A + 2 B + C = −1 ∴ A=
2A = 5 2
7
B +C = −
2 3
∴ B = −5, C=
17 2
2B + C = −
2
By partial – fraction expansion ,
5 5 3
X ( s) = − +
2 S S + 1 2( S + 2)
Taking the inverse Laplace transform
5 3
x(t ) = − 5e −t + e − 2 t (t ≥ 0)
2 2
↑ ↑
Steaty-
teaty-state transient period
period
H(s)
H(s)
(GH + 1)
→ R= C
G
G
∴C = R
1 + GH
C 1
E= = R
G 1 + GH
G ( s) R( s)
Now, C ( s) =
→ analytical Solution for C(t)
1 + G( s) H ( s)
Inverse Laplace transform
Characteristic equation
→ ao S n + a1 ⋅ S n −1 + … + an −1 ⋅ S + an =0
S5 a1 a3 a5 0
S 4
a1a2 − ao a3 a1a4 − ao a5 a1a6 − ao ⋅ 0
=A =B = a6 0
a1 a1 a1
S3 A ⋅ a3 − a1 ⋅ B A ⋅ a5 − a1a6 A ⋅ 0 − a1 ⋅ 0
=C =D =0 0
A A A
BC − AD C ⋅ a6 − A ⋅ 0 C ⋅0 − A⋅0
S2 =E = a6 =0 0
C C C
ED − C ⋅ a6
S1 =F 0 0 0
E
F ⋅ a6 − E ⋅ 0
S0 = a6 0 0 0
F
<Conclusion>
i) If all the elements are of the same sign
→ all roots are in the left half of the S plane
(which means that the system is stable)
ii) If change of signs in the elements
→ the number of sign changes = number of roots with positive real
parts. (which means that the system is unstable)
S3 ao =1 a2 =1
S2 a1 = −4 a3 = 6
( − 4) × 1 − 1 × 6
S1 = 2 .5 0
−4
2 .5 × 6 − ( − 4 ) × 0
S0 =6 0
2 .5
∴ two sign changes ⇒ We can say that two roots located in the right half of the
S-plane by Routh tabulation. Therefore, the system is unstable.
In fact we know that the roots are 2, -1, and 3 from equation (S-2)(S+1)(S-3)=0.
-1 2 3
R(s) C(s)
+ G(s)
-
H(s)
- characteristic equation :
1 + G(s)H(s) = 0 or G(S)H(S) =-1
Since G(S)H(S) is a complex quantity , it can be split into angle and magnitude
angle condition :
∠G ( s ) H ( s ) = ±180 °( 2k + 1) , ( k = 0,1,2 ⋯) ①
---①
magnitude condition :
G( s) H (s) = 1 ②
---②
∴ The values of S which satisfy ① and ② are the roots of the characteristic
equation. (or closed loop poles)
R(s) C(s)
+ K
- s ( s + 1)
- loop transfer function
k
G (s) H ( s) =
S ( S + 1)
- closed-loop transfer function
k
C (s) k G (s) S ( S + 1) k
= 2 (← = = 2 )
R( s) S + S + k 1 + G( s) H ( s) 1 + k S +S +k
S ( S + 1)
- characteristic equation : S2 + S + k = 0
1 1 1
ex) Let’s check the point p = − + i ( k = ) on the locus:
2 2 2
k 0 .5
G (s) = =
S ( S + 1) ( −0.5 + 0.5i )(0.5 + 0.5i )
0 .5
= = −1
− 0.25 − 0.25
k
∠ = 180°
S ( S + 1) s = p
∴
k
=1
S ( S + 1) s = p
k
Ex) ∠ = −∠S − ∠S + 1 = ±180°(2n + 1), (n = 0,1,2, ⋯)
S ( S + 1)
1
If we choose p = − + 2 j, what value is k?
2
k 17
=1 ∴k =
1 1 4
(− + 2 j )( + 2 j )
2 2
(If we choose k=17/4, then the root of the characteristic equation can be p )
s ( s + 2)
-
1
s+3
H(s)
- Characteristic equation :
S(S+2)(S+3) + K(S+1) =0
P(s)
ii) rewrite rhe form as 1 + k =0
q(s)
s +1
In this case, 1 + k =0
s ( s + 2)( s + 3)
iii) matlab program
〉p=[1 1];
〉q=[1 5 6 0];
〉rlocus(p,q)
Root Locus
8
2
Imaginary Axis
-2
-4
-6
-8
-3 -2.5 -2 -1.5 -1 -0.5 0
Real Axis
[Home Work1]
d 2 x(t ) dx(t ) 1(t > 0)
[1] + 5 ⋅ + 4 x (t ) = u (t ) ,where u (t ) = ; unit step function
0(t < 0)
s s
dt 2 dt
dx(t )
Initical conditions : x(o) = −1, x′(0) = =2
dt t =0
Solve the equation using Laplace transform
R(s) C(s)
+ K ( s + 2)
- s 2 + 2s + 3