Loading Documentmodern control theory is based on the description of system equations in terms of n first-order differential equations, which may be combined into a first-order vector-matrix differential equation.
Ex: Consider a system defined by y¨˙+4y¨+8y˙+6y=5u where u is the input and y is the output.
find the State Space Representation for the system.
x1=y⟹x˙1 =y˚=x2
x2=y˙⟹x˙2=y¨=x3
x3=y¨⟹x˙3 =y¨˙=−4y¨−8y˙−6y+5u
=−4x3−8x2−6x1+5u
the state space representation is
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡00−610−801−4⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡005⎦⎥⎥⎥⎤u and y=[100]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
Block Diagram
U(S)Y(S)=S3+4S2+8S+65
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡00−610−801−4⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡005⎦⎥⎥⎥⎤u and y=[100]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
Eigenvalues of an n x n Matrix A. The eigenvalues of an n x n matrix A are the roots of the characteristic equation
and can be obtained by solving ∣λI−A∣=0
Ex: find the poles and the eigen values of the system T⋅F⋅=S2+4S+610 then discuss system stability
the roots of the characteristic equation S2+4S+6=0 are S1,2=−2±1.2j
System is stable
⎣⎡x1˙x2˙⎦⎤=A⎣⎡0−61−4⎦⎤⎣⎡x1x2⎦⎤+⎣⎡010⎦⎤u
∣λI−A∣=∣∣∣∣∣∣λ6−1λ+4∣∣∣∣∣∣=0
λ(λ+4)−(−1×6)=0
λ2+4λ+6=0⟹λ1,2=−2±1.2j
Ex: for the system T⋅F⋅=S3+6S2+11S+65 find the SSR in diagonal form.
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=A⎣⎢⎢⎢⎡00−610−1101−6⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+B⎣⎢⎢⎢⎡005⎦⎥⎥⎥⎤u and y=C[100]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
∣λI−A∣=λ3+6λ2+11λ+6=0
λ1,2,3=−1,−2,−3
x˙=Ax+Bu⟹Pz˙=APz+Bu
z˙=P−1APz+P−1Bu
y=Cx⟹y=CPz
P=⎣⎢⎢⎢⎡1λ1λ121λ2λ221λ3λ32⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡1−111−241−39⎦⎥⎥⎥⎤ and P−1=⎣⎢⎢⎢⎡3−312.5−41.50.5−10.5⎦⎥⎥⎥⎤
P−1AP=⎣⎢⎢⎢⎡−1000−2000−3⎦⎥⎥⎥⎤ P−1B=⎣⎢⎢⎢⎡2.5−52.5⎦⎥⎥⎥⎤ CP=[111]
⎣⎢⎢⎢⎡z˙1z˙2z˙3⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡−1000−2000−3⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡z1z2z3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡2.5−52.5⎦⎥⎥⎥⎤u and y=[111]⎣⎢⎢⎢⎡z1z2z3⎦⎥⎥⎥⎤
Block Diagram
Ex: for the system T⋅F⋅=S3+3S2+3S+16 find the SSR in diagonal form.
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=A⎣⎢⎢⎢⎡00−110−301−3⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+B⎣⎢⎢⎢⎡006⎦⎥⎥⎥⎤u and y=C[100]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
∣λI−A∣=λ3+3λ2+3λ+1=0⟹λ1,2,3=−1
x˙=Ax+Bu⟹Pz˙=APz+Bu
z˙=P−1APz+P−1Bu
y=Cx⟹y=CPz
P=⎣⎢⎢⎢⎡1λ1λ12012λ1001⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡1−1101−2001⎦⎥⎥⎥⎤ and P−1=⎣⎢⎢⎢⎡111012001⎦⎥⎥⎥⎤
P−1AP=⎣⎢⎢⎢⎡−1001−1001−1⎦⎥⎥⎥⎤ P−1B=⎣⎢⎢⎢⎡006⎦⎥⎥⎥⎤ CP=[100]
⎣⎢⎢⎢⎡z˙1z˙2z˙3⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡−1001−1001−1⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡z1z2z3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡006⎦⎥⎥⎥⎤u and y=[100]⎣⎢⎢⎢⎡z1z2z3⎦⎥⎥⎥⎤
Block Diagram
Ex: for the system T⋅F⋅=S3+6S2+11S+66 find the SSR in partial fraction form.
UY=S3+6S2+11S+66=(S+1)(S+2)(S+3)6=S+13+S+2−6+S+33
Y=3x1S+1U−6x2S+2U+3x3S+3U
x1=S+1U⟹x˙1=−x1+u
x2=S+2U⟹x˙2=−2x2+u
x3=S+3U⟹x˙3=−3x3+u
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡−1000−2000−3⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡111⎦⎥⎥⎥⎤u and y=[3−63]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
Ex: for the system T⋅F⋅=(S+1)2(S+2)2S2+6S+5 find the SSR in partial fraction form.
UY=(S+1)2(S+2)2S2+6S+5=S+21+S+11+(S+1)21
Y=x1S+2U+x2S+1U+x3(S+1)2U
x1=S+2U⟹x˙1=−2x1+u
x2=S+1U⟹x˙2=−x2+u
x3=(S+1)2U=S+1x2⟹x˙3=x2−x3
⎣⎢⎢⎢⎡x1˙x2˙x3˙⎦⎥⎥⎥⎤=⎣⎢⎢⎢⎡−2000−1100−1⎦⎥⎥⎥⎤⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤+⎣⎢⎢⎢⎡110⎦⎥⎥⎥⎤u and y=[111]⎣⎢⎢⎢⎡x1x2x3⎦⎥⎥⎥⎤
Homogeneous
x˙=Axtake LaplaceSx(S)−x(0)=Ax(S)
Sx(S)−Ax(S)=x(0)
[SI−A]×x(S)=x(0)
x(S)=[SI−A]−1×x(0)
x(t)=L−1[SI−A]−1x(0)
Ex: Solve x˙=⎣⎡0−21−3⎦⎤⎣⎡x1x2⎦⎤ given that x(0)=⎣⎡1−1⎦⎤ then sketch x1(t)&x2(t)
[SI−A]=⎣⎡S2−1S+3⎦⎤
[SI−A]−1=S2+3S+21⎣⎡S+3−21S⎦⎤=(S+1)(S+2)1⎣⎡S+3−21S⎦⎤
[SI−A]−1=⎣⎡S+12+S+2−1S+1−2+S+22S+11+S+2−1S+1−1+S+22⎦⎤=⎣⎡2e−t−e−2t−2e−t+2e−2te−t−e−2t−e−t+2e−2t⎦⎤
x1(t)=(2e−t−e−2t)−(e−t−e−2t)=e−t and x2(t)=5(−2e−t+2e−2t)−(−e−t+2e−2t)=−e−t
Non-Homogeneous
Ex: Solve x˙=⎣⎡0−21−3⎦⎤⎣⎡x1x2⎦⎤+⎣⎡12⎦⎤u for unit step input given that x(0)=⎣⎡0−1⎦⎤ then sketch x1(t)&x2(t)
[SI−A]−1=(S+1)(S+2)1⎣⎡S+3−21S⎦⎤
x(0)+Bu(S)=⎣⎡0−1⎦⎤+⎣⎡S1S2⎦⎤=⎣⎡S1S−S+2⎦⎤
[SI−A]−1×[x(0)+Bu(S)]=(S+1)(S+2)1⎣⎡S+3−21S⎦⎤⎣⎡S1S−S+2⎦⎤
=(S+1)(S+2)1⎣⎡SS+3+S−S+2S−2+2−S⎦⎤=⎣⎡S5S−2+2S−S2⎦⎤=(S+1)(S+2)1⎣⎡S5S−2+2S−S2⎦⎤=⎣⎢⎡S(S+1)(S+2)5S(S+1)(S+2)−2+2S−S2⎦⎥⎤=⎣⎢⎡S2.5+S+1−5+S+22.5S−1+S+15+S+2−5⎦⎥⎤
x1(t)=2.5−5e−t+2.5e−2t and x2(t)=−1+5e−t−5e−2t
A system is said to be controllable at time t0 if it is possible by means of an unconstrained control vector to transfer the system from any initial state x(t0) to any other state in a finite interval of time.
A system is said to be observable at time t0 if, with the system in state x(t0), it is possible to determine this state from the observation of the output over a finite time interval.
Consider the system x˙=Ax+Bu and y=Cx
where x=state vector(n−vector)
u=control vector(r−vector)
y= output vector(m−vector)
A=n×n matrix
B=n×r matrix
C:m×n matrix
controllability condition ∣∣∣B⋮AB⋮⋯⋮An−1B∣∣∣=0 or rank =n
output controllability condition ∣∣∣CB⋮CAB⋮⋯⋮CAn−1B∣∣∣=0 or rank =m
observability condition ∣∣∣C∗⋮A∗C∗⋮⋯⋮(A∗)n−1C∗∣∣∣=0 or rank =n
Ex: A=n=3⎣⎢⎢⎢⎡200023001⎦⎥⎥⎥⎤,B=⎣⎢⎢⎢⎡010101⎦⎥⎥⎥⎤,C=m=2⎣⎡100100⎦⎤
controllability condition ∣∣∣B⋮AB⋮A2B∣∣∣=0 or rank =n
∣∣∣∣∣∣∣∣∣0101000232010494 01∣∣∣∣∣∣∣∣∣ rank =3 therefore, the system is completely controllable.
output controllability condition ∣∣∣CB⋮CAB⋮CA2B∣∣∣=0 or rank =m
∣∣∣∣∣∣011002200440∣∣∣∣∣∣ rank =2 therefore, the system is output controllable.
observability condition ∣∣∣C∗⋮A∗C∗⋮A∗2C∗∣∣∣=0 or rank =n
∣∣∣∣∣∣∣∣∣100010200020400040∣∣∣∣∣∣∣∣∣ rank =3 therefore, the system is not observable.