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Modern Control systemsLecture-3. Solution of State Equations
V. Sankaranarayanan
V. Sankaranarayanan Modern Control systems
Outline
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Outline
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Outline
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution to Homogeneous State Equations
Solution to Scalar D.E.s
Let us consider the scalar differential equation,x = ax, where
x(t) = b0 + b1t+ b2t2 + · · ·+ bktk + · · ·
On substituting x(t) in our scalar differential Equation, we get..
b1 + 2b2t+ 3b3t2 + · · ·+ kbktk−1 + · · · =
a(b0 + b1t+ b2t2 + · · ·+ bktk + · · · )
Equating the coefficients of equal powers of ’t’
b1 = ab0
b2 =1
2ab1 =
1
2a2b0
b3 =1
3ab2 =
1
2 ∗ 3a3b0
...
bk =1
k!akb0
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Continuation of Solution of scalar D.E.s...
The value of b0 can be determined by substituting t = 0 inx(t) = b0 + b1t+ b2t2 + · · ·+ bkt
k + · · ·x(0) = b0
Hence the solution x(t) can be written as,x(t) = (1 + at+ 1
2!a2t2 + 1
3!a3t3 + · · ·+ 1
k!aktk + · · · )b0
= (1 + at+ 12!a2t2 + 1
3!a3t3 + · · ·+ 1
k!aktk + · · · )x(0)
= eatx(0)
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Let us consider the vector differential equation,
x = Ax,
where, x ∈ Rn → n-vector
A ∈ Rn∗n → n ∗ n constant matrix
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Let,
x1(t) = a11x1(t) + a12x2(t) + · · ·+ a1nxn(t) (1)
x2(t) = a21x1(t) + a22x2(t) + · · ·+ a2nxn(t)
...
...
xn(t) = an1x1(t) + an2x2(t) + · · ·+ annxn(t)
This can be written in Matrix form as,
x1(t)...
xn(t)
=
a11 · · · an1
.... . .
...an1 · · · ann
x1(t)
...xn(t)
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector D.E.s
Let,
x1(t) = a0 + a1t+ a2t2 + · · ·+ ant
n (2)
x2(t) = b0 + b1t+ b2t2 + · · ·+ bnt
n (3)
x3(t) = c0 + c1t+ c2t2 + · · ·+ cnt
n (4)
...
Consider the equation,x1(t) = a0 + a1t+ a2t2 + · · ·+ antn
Differentiating with respect to t, we get,x1(t) = a1 + 2a2t+ 3a3t2 + · · ·+ nantn−1
Substituting this in equation (1) , we get,a1 + 2a2t+ 3a3t2 + · · ·+ nantn−1 =a11(a0 + a1t+ a2t2 + · · ·+ antn) + a12(b0 + b1t+ b2t2 + · · ·+ bntn)+a13(c0 + c1t+ c2t2 + · · ·+ cntn) + ....
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Equating coefficients of equal powers of ’t’,
a1 = a11a0 + a12b0 + a13c0 + ...
b1 = a21a0 + a22b0 + a23c0 + ...
c1 = a31a0 + a32b0 + a33c0 + ...
...
Similarly,a2 = a11a1 + a12b1 + a13c1 + ..
= a11(a11a0 +a12b0 +a13c0 + ...)+a12(a21a0 +a22b0 +a23c0 + ...)+ · · ·
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Substituting t = 0 in equations (2), (3), · · · we get
x1(0) = a0
x2(0) = b0
x3(0) = c0
...
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Summing up all the results obtained so far, we get
x1
x2
x3
...
=
1 · · · 0...
. . ....
0 · · · 1
a0
b0c0...
+
a11 · · · a1n
.... . .
...an1 · · · ann
a0
b0c0...
t+
a11 · · · a1n
.... . .
...an1 · · · ann
2a0
b0c0...
t2 + · · ·
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Solution of Vector-matrix D.E.s
Replacing a0, b0, · · ·withx1(0), x2(0), · · · we get,
x1
x2
x3
...
=
1 · · · 0...
. . ....
0 · · · 1
x1(0)x2(0)x3(0)
...
+
a11 · · · a1n
.... . .
...an1 · · · ann
x1(0)x2(0)x3(0)
...
t+
a11 · · · a1n
.... . .
...an1 · · · ann
2x1(0)x2(0)x3(0)
...
t2 + · · ·
If ,
x(t) =
x1
x2
x3
...
,x(0) =
x1(0)x2(0)x3(0)
...
A =
a11 · · · a1n
.... . .
...an1 · · · ann
I =
1 · · · 0...
. . ....
0 · · · 1
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Homogeneous State Equations
Continuation of Solution of Vector-matrix D.E.s..
the solution x(t) can be written as,
x(t) = (I + At+1
2!(A)2t2 +
1
3!(A)3t3 + · · ·+
1
k!(A)ktk + · · · )︸ ︷︷ ︸x(0)
The expression in the under brace on the R.H.S of the last equation is ann ∗ n matrix
It is similar to the infinite power series for a scalar exponential.It is calledmatrix exponential and can be written as:
eAt = I + At+ 12!
(A)2t2 + 13!
(A)3t3 + · · ·+ 1k!
(A)ktk + · · ·
Thus, the solution can be written as
x(t) = eAtx(0)
= φ(t)x(0)
where,φ(t) is called the ’State Transition Matrix’
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Non Homogeneous State equations
Scalar Case
Consider a scalar state equation,x = ax+ bux− ax = bu,
Multiplying this equation by e−at on both sides and integration between 0and t gives,
x(t) = eatx(0) +∫ t0 (ea(t−τ)u(τ)dτ)
Vector Case
Consider the non homogeneous state equation described byx = Ax + Bu
where, x ∈ Rn → n-vectoru ∈ Rm →m-vectorA∈ Rn∗n → n∗n-constant matrix,B∈ Rn∗m → n∗m-constant matrix,
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Solution of Scalar D.E.sSolution of Vector D.E.s
Solution of Non Homogeneous State Equations
Continuation of Vector Case...
The solution of x(t) can be written as
x(t) = eAtx(0) +∫ t0 (eA(t−τ)u(τ)dτ)
This equation can also be written asx(t) = φ(t)x(0) +
∫ t0 (φ(t− τ)u(τ)dτ)
where, φ(t)=eAt, is the State Transition Matrix’
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix ExponentialProperties of State Transition Matrix
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix ExponentialProperties of State Transition Matrix
State Transition Matrix- ’φ’
Properties of State Transition Matrix- ’φ’
For the time invariant system, x = Axφ(t) = eAt
The properties of the State Transition Matrix are:
φ(0) = I
φ−1(t) = φ(−t)φ(t1 + t2) = φ(t1)φ(t2)
[φ(t)]n = φ(nt)
φ(t2 − t1)φ(t1 − t0) = φ(t2 − t0)
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix ExponentialProperties of State Transition Matrix
State Transition Matrix- ’φ’
Example
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Methods to compute eAt
Matrix exponential can be computed by
Numerical MethodsAnalytic Methods
Numerical Methods
If matrix A is given with all elements in numerical values, MATLAB providesa simple way to compute eAT , where T is a constant.
Analytic Methods
Some of the Analytic methods to be discussed are given belowLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Laplace Transformation Approach
We know that,x(t) = Ax(t)
Applying Laplace Transformation on both sides, we get,sX(s)−X(0) = AX(s)⇒ (sI-A)X(s) = X(0)
Pre-multiplying with (sI-A)−1 on both sides and taking Inverse LaplaceTransform on both sides, we get
x(t) = L−1((sI-A)−1)x(0)
We know that,
x(t) = eAtx(0)
Comparing equations, (1) and (2) we can writeeAt = L−1((sI-A)−1))
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Example of Laplace Transformation approach
Consider the following matrix A =
[−1 0−3 −2
]
sI-A =
[s 00 s
]−[−1 −0−3 −2
]⇒ (sI-A)−1 =
[s+ 1 0
3 s+ 2
]−1
⇒ (sI-A)−1 =
[1s+1
0−3
(s+1)(s+2)1s+2
]
Applying Inverse Laplace Transformation on both sides we get,
eAt = L−1((sI-A)−1) =
[e−t 0
−3e−t + 3e−2t e−2t
]
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Diagonal Transformation
Matrix A is diagonalized using a diagonalizing matrix P
The resultant matrix is given by
eAt = PeΛtP−1 = P
eλ1t · · · 0
.... . .
...0 · · · eλnt
P−1
If matrix A can be transformed into Jordan Canonical form, then eAt can begiven by
eAt = SeJtS−1
where, S is a transformation matrix that transforms matrix A intoJordan canonical form J
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Example of Diagonal Transformation
Let, matrix A =
[0 1−2 −3
]characteristic equation of this matrix is given by λ2 − 3λ+ 2Eigenvalues of matrix A are λ1 = −1, λ2 = −2
The corresponding eigenvectors of the eigenvalues λ1 and λ2 are
[1−1
]and[
12
]respectively
The modal matrix formed by these eigenvectors is given by P =
[1 1−1 2
]the diagonal matrix Λ is obtained by the transformation P−1AP
Λ =
[−1 00 −2
]∴ eJt = eΛt =
[e−t 00 e−2t
]
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Continuation of Example of Diagonal Transformation
Having transformed Matrix A into Jordan Canonical form, Matrixexponential eAt can be obtained by the transformation PeJtP−1
eAt =
[1 1−1 2
] [e−t 00 e−2t
] [2 1−1 −1
]=
[2e−t − e−2t e−t − e−2t
−2e−t2e−2t −e−t + 2e−2t
]
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Outline
1 Solution of Differential EquationSolution of Scalar D.E.sSolution of Vector D.E.s
2 State Transition MatrixProperties of State Transition Matrix
3 Computational Methods of Matrix ExponentialLaplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Cayley-Hamilton Theorem Approach
For large systems, this method is far more convenient computationally ascompared to the other two methods discussed earlier.
Cayley-Hamilton theorem
Statement:Every square matrix A satisfies its own characteristic equation.
If,q(λ) = λn + a1λn−1 + · · ·+ an−1λ+ an = 0 is the characteristic equationof A, then
q(A) = An + a1An−1 + · · ·+ an−1A + an = 0
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Cayley-Hamilton theorem
Let λ1, λ2, · · · , λn be the eigenvalues of the matrix A
Consider the scalar polynomial f(λ) = k0 + k1λ+ k2λ2 + · · ·+ knλn + · · · ,where λ is the eigenvalue of the matrix.
The matrix polynomial f(A) = k0I + k1A + k2A2 + · · ·+ knAn + · · · an becomputed by considering the scalar polynomial f(λ)
Dividing f(λ) by q(λ) we getf(λ)q(λ)
= Q(λ) +R(λ)q(λ)
∴ f(λ) = Q(λ)qλ+R(λ)where, R(λ) = α0 + α1λ+ α2λ2 + · · ·+ αn−1λn−1 is the remainderpolynomial
For λ = λ1, λ2 · · · , λn (for eigenvalues) q(λ) = 0∴ f(λi) = R(λi); i = 1, 2, 3, · · ·
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Cayley-Hamilton theorem
The coefficients of the remainder polynomial α0, α1, · · · , αn−1 can beobtained by substituting λ = λ1, λ2, · · · , λn in the relation f(λi) = R(λi)
Replacing λ with matrix A we get,f(A) = R(A)
= α0I + α1A + · · ·+ αn−1An−1
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Cayley-Hamilton theorem
Procedure to compute eAt:
Step-1:
Find the eigenvalues of matrix A
Step-2:
Case-1: If all the eigenvalues are distinct, the coefficients α0, α1, · · · , αn−1
can be obtained by solving ′n′ simultaneous equations given by f(A) = R(A)Case-2: If A possess an eigenvalue λK of order ′m′ then,
Only one independent equation can be obtained by substituting λk in theequation f(A) = R(A)The remaining m− 1 linear equations can be obtained by differentiatingf(λ) = R(λ) on both sides
∴ djf(λ)
dλj λ=λk=djR(λ)
dλj λ=λk; j = 0, 1, 2 · · · ,m− 1
Step-3:
The required result is obtained by substituting the values of α0, α1, · · · , αn−1
in f(A) = α0I + α1A + · · ·+ αn−1An−1
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Example of Cayley-Hamilton theorem Approach
Consider the matrix, A=
[0 1−1 −2
]Step-1:
Eigenvalues of the matrix are λ1 = λ2 = −1
Step-2:
Since the eigenvalues equal and of order 2
∴ 2− 1 = 1 equation can be obtained by differentiating the coefficients off(A) = R(A)
Another equation is obtained by substituting λ = −1 directly in the equationf(A) = R(A)
V. Sankaranarayanan Modern Control systems
Solution of Differential EquationState Transition Matrix
Computational Methods of Matrix Exponential
Laplace Transformation ApproachDiagonal TransformationCayley-Hamilton Theorem Approach
Computational Methods of Matrix exponential-eAt
Example of Cayley-Hamilton theorem Approach
Since A is of second-order, The polynomial R(λ) will be of the form α0 + α1λ
The coefficients of α0, α1 can be obtained as follows:
f(λ) = eλt = α0 + α1λ
∴ f(−1) = e−t = α0 − α1 (4)
d
dλf(λ)
λ=−1=
d
dλeλt
λ=−1=
d
dλ(α0 + α1λ)
⇒ te−t = α1
Substituting the value of α1 in equation(4) we get α0 = (1 + t)e−t
Step-3:
The required result is f(A) = eAt = α0I + α1A
=
[(1 + t)e−t te−t
−te−t (1− t)e−t]
V. Sankaranarayanan Modern Control systems
