Outline
Modern Control systems
Lecture-5 State Space to Transfer Function
V. Sankaranarayanan
V. Sankaranarayanan Modern Control systems
Outline
Outline
1 State Space to Transfer Function
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
State Space to Transfer Function
Deriving Transfer Function from State Space
Previously we have seen different methods of obtaining State-Space from
Transfer function.
The process of converting Transfer Function to State-Space form is NOT
unique .
Deriving Transfer function model from a State-Space model is UNIQUE.
We know that,
ẋ = Ax + Bu
y = Cx + Du
Applying Laplace Transform with zero initial conditions we get,
sX(s) = AX(s) + BU (s)
Y (s) = CX(s) + D(s)
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
State Space to Transfer Function
Deriving Transfer Function from State Space
The state equation can also be written as
(sI − A)X(s) = BU (s)
Pre multiplying on both sides by (sI − A)−1 we get,
X(s) = (sI − A)−1 BU (s)
Substituting X(s) in the output equation we get,
Y (s) = C(sI − A)−1 BU (s) + DU (s)
h i
⇒ Y (s) = C(sI − A)−1 B + D U (s)
| {z }
Transfer Function Matrix T (s)
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
State Space to Transfer Function
Example on State Space to Transfer Function
Consider the State Space representation,
ẋ1 0 1 x1 0
= +
ẋ2 −15 −2 x2 1
x1
y = −66 −3 + [5]]u
x2
0 1 0
Here,A = ,B = , C = −66 −3 , D = [5]
−15 −2 1
We know that, Transfer Function Matrix
Y (s) h i
T (s) = = C(sI − A)−1 BU + D
U (s)
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
State Space to Transfer Function
Example on State Space to Transfer Function
s 0 0 1
(sI − A) = −
0 s −15 −2
s −1
=
15 s + 2
1 s+2 1
(sI − A)−1 =
s2 + 2s + 17 −15 s
!
h
−1
i 1 s+2 1 0
C(sI − A) B +D = −66 −3 + [5]
s2 + 2s + 17 −15 s 1
−66 − 3s
= +5
s2 + 2s + 17
2
5s + 7s + 9
=
s2 + 2s + 15
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
State Space to Transfer Function
Example on State Space to Transfer Function
−66 − 3s
= +5
s2 + 2s + 17
5s2 + 7s + 9
=
s2 + 2s + 15
Students are advised to compare this result with that of Example-2 of
Canonical Form-I in Lecture -4
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
Similarity Transformation
The state space equations of a system is given as
ẋ = Ax + Bu
y = Cx + Du
Choose any non-singular matrix ’T ’ such that
z =T x ⇒ x =T −1 z and ẋ =T −1 ż
Substituting in the system equations
ż = T AT −1 z + T Bu
y = CT −1 z + Du
It can be written as
ż = Az + Dz u
y = Cz z + Dz u
where,
Az = T AT −1 ; Bz = T B ; Cz = CT −1 ; Dz = D
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
Similarity Transformation
Example on Similarity transform
0 1
Consider the matrix A = with eigenvalues −2, −1
−2 −3
Let, Az be the transformed matrix
We know that, Az = T −1 AT , where T is a non-singular, n × n matrix.
The key to similarity transformation is finding the matrix T corresponding to
the desired canonical transformation.
b b12
Let, T = 11 and the desired transformation be the Diagonal
b21 b22
Transformation.
Therefore, the transformed matrix Az becomes
−1
−2 0 b11 b12 0 1 b11 b12
Az = =
0 −1 b21 b22 −2 −3 b21 b22
−1
0 1 b11 b12 −2 0 b11 b12
⇒ =
−2 −3 b21 b22 0 −1 b21 b22
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
Similarity Transformation
Example on Similarity transform
0 1 1 b11 b12 −2 0 b22 −b12
⇒ =
−2 −3 |T | b21 b22 0 −1 −b21 b11
!
1 b22 −b12
T −1 = |T |
, where |T | = b11 b22 − b12 b21
−b21 b11
0 1 1 −2b11 b22 + b12 b21 3b11 b12
=
−2 −3 b11 b22 − b12 b21 −b21 b22 2b12 b21 − b11 b22
We know that, if two matrices are equal then their corresponding elements
are equal.
Therefore, equating corresponding elements, we get,
−2b11 b22 + b12 b21 = 0
3b11 b12 = b11 b22 − b12 b21
−b21 b22 = −2(b11 b22 − b12 b21 )
2b12 b21 − b11 b22 = −3(b11 b22 − b12 b21 )
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
Similarity Transformation
Example on Similarity transform
On solving these equations we arrive upon
b21 = −2b11
b22 = −b12
If b11 = b12 = 1, we get b21 = −2, b22 = −1
1 1 −1 −1
Therefore, T = and T −1 =
−2 −1 2 1
V. Sankaranarayanan Modern Control systems
State Space to Transfer Function
Similarity Transformation
Example on Similarity transform
Verification:
We know that,
0 1
Az = T −1 T
−2 −3
−1 −1 0 1 1 1
Az =
2 1 −2 −3 −2 −1
−2 0
=
0 −1
Hence, the obtained ,matrix T is the desired matrix for transforming matrix
A into Diagonal Canonical form.
Students are advised to compare T with the modal matrix P formed by the
eigenvectors of A which is used to transform matrix A into a Diagonal matrix.
V. Sankaranarayanan Modern Control systems
