Exercise Set 7 - Control Systems (5ESD0) Oct 2021

1. Linear algebra:
(a) Consider the following differential equation

ẋ1 = −2x1 + 3x2 + u,

ẋ2 = x1 − 3x2 ,
ẋ3 = 2x1 − 5x3 + u;

and define x = [x1 x2 x3 ]T . Write this system in the form ẋ = Ax + Bu, give the
corresponding matrices A and B, and compute the inverse of A.
(b) For invertible matrices A and B, prove the following identities:
ˆ (AB)−1 = B −1 A−1 ,
ˆ (AB)T = B T AT ,
ˆ A(I + BA)−1 = (I + AB)−1 A.
Solution:

(a) We have the state-space representation as:

   
−2 3 0 1
ẋ = Ax + Bu =  1 −3 0  x + 0 u.
2 0 −5 1

We have the inverse of A as:

   
15 15 0 −1 −1 0
adj(A) −1 
A−1 = = 5 10 0 = −0.333 −0.667 0 .
det(A) 15
6 6 3 −0.4 −0.4 −0.2

(b) ˆ To prove (AB)−1 = B −1 A−1 , we have that

(AB)−1 AB = I ⇐⇒ (AB)−1 = B −1 A−1 since B −1 A−1 AB = I.

ˆ To prove (AB)T = B T AT , we have that

Pn
I. the (i, j) element of AB is aik bkj ,
T
Pk=1
n
II. the (i, j) element of (AB) is a b ,
Pnk=1 jk ki
III. the (i, j) element of B T AT is k=1 ki ajk .
b
Observe that II. and III. are equivalent; therefore, (AB)T = B T AT .

ˆ To prove A(I + BA)−1 = (I + AB)−1 A, we have

A(I + BA)−1 = (I + AB)−1 A ⇐⇒ (I + AB)A = A(I + BA)

⇐⇒ A + ABA = A + ABA.

Since the last statement is true the earlier statements hold as well.

2. Conversion between plant representations:

Consider the following plants represented by differential equations:

i. ÿ + 2ẏ + u = 0,
ii. ÿ + 2ẏ + u̇ + u = 0,

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

iii. ÿ + 2ẏ + ü + u̇ + u = 0.

Under the assumption that initial conditions equal to zeroes, do the following exercises—for
each case (i. - iii.):
(a) Derive a corresponding state-space model using the simulation diagram approach, then
compute its transfer function (from the obtained state-space model representation).
(b) Directly apply Laplace transform to the differential equation and obtain its transfer
function. Compare the transfer functions you have obtained from (2a) and (2b).
Solution:

i. (a) We first sketch the simulation diagram which is given in Figure 1.

Figure 1: Problem 2: Simulation diagram for case (i.)

We then observe that

ẋ1 = x2
ẋ2 = −2x2 − u
y = x1
 ⊤
With x = x1 x2 , we have that
(
ẋ = Ax + Bu
y = Cx + Du

with    
0 1 0    
A= , B= , C= 1 0 , and D = 0 .
0 −2 −1
We can compute the transfer function from the state-space representation using
−1 −1
G(s) = C (sI − A) B+D =
s(s + 2)
(b) Assume the zero initial conditions. We have

(s2 + 2s)y(s) + u(s) = 0.

Therefore, the corresponding transfer function is

y(s) −1
= .
u(s) s(s + 2)
This is the exact same transfer function as the one found in (a).

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

Figure 2: Problem 2: Simulation diagram for case (ii.)

ii. (a) Again, we first sketch the simulation diagram which is given in Figure 2.
We then observe that
ẋ1 = x2 − u
ẋ2 = −2ẋ1 − u = −2x2 + u
y = x1
 ⊤
With x = x1 x2 , we have that
(
ẋ = Ax + Bu
y = Cx + Du

with    
0 1 −1    
A= , B= , C= 1 0 , and D = 0 .
0 −2 1
We can compute the transfer function from the state-space representation using

−1 (s + 1)
G(s) = C (sI − A) B+D =−
s(s + 2)

(b) Assume the zero initial conditions. We have

(s2 + 2s)y(s) + (s + 1)u(s) = 0.

Therefore, the corresponding transfer function is

y(s) (s + 1)
=− .
u(s) s(s + 2)

iii. (a) Again, we first sketch the simulation diagram which is given in Figure 3.
We then observe that
ẋ1 = −2x1 + x2 + u
ẋ2 = −u
y = x1 − u
 ⊤
With x = x1 x2 , we have that
(
ẋ = Ax + Bu
y = Cx + Du

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

Figure 3: Problem 2: Simulation diagram for case (iii.)

with    
−2 1 1    
A= , B= , C= 1 0 , and D = −1 .
0 0 −1
We can compute the transfer function from the state-space representation using

−1 (s2 + s + 1)
G(s) = C (sI − A) B+D =−
s(s + 2)

(b) Assume the zero initial conditions, we have

(s2 + 2s)y(s) + (s2 + s + 1)u(s) = 0.

Therefore, the corresponding transfer function is

y(s) (s2 + s + 1)
=−
u(s) s(s + 2)

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

3. Transformation matrices and canonical forms:

For case ii. and iii. from problem 2, use the obtained transfer functions (from either (2a) or
(2b)) to do the following exercises:
(a) Derive the corresponding state-space model in the control canonical form.
(b) Derive the corresponding state-space model in the modal form.
(c) For each case, compare the state-space models obtained from (3a) and (3b). Find the
similarity transformation matrix T which brings the state-space representation in (3a)
to the one in (3b).
Solution:
Remember that if we have a transfer function
b1 sn−1 + · · · + bn
G(s) = + b0
sn + a1 sn−1 + · · · + an
We can have a state-space realization in control canonical form given by

ẋ = Ax + Bu
y = Cx + Du

where
 
−a1 −a2 ... ... −an  
1
 1 0 0  0
.. 
 
..  
B =  ...  ,
    
 0
A= 1 . . 
,
 
C = b1 b2 . . . bn , D = b0
 . ..
 
 .. 0

. 0 0 
0 ... 0 1 0 0

and if we have a transfer function

c1 cn
G(s) = + ··· + + c0
s + p1 s + pn
We can have a state-space realization in modal form given by

ẋ = Ax + Bu
y = Cx + Du

where
 
−p1 0 ... 0  
c1
 .. ..   c2 
 0 −p2 . .     
A= , B =  . , C= 1 1 ... 1 , D = c0
 
 . .. ..  .. 
 ..

. . 0 
0 ... 0 −pn cn

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

(Case ii.) We have the plant transfer function as

(s + 1) −s − 1
G(s) = − = 2
s(s + 2) s + 2s

(a) The control canonical form is given by

(
ẋc = Ac xc + Bc u
y = Cc xc + Dc u

with
   
−2 0 1    
Ac = , Bc = , Cc = −1 −1 , and Dc = 0 .
1 0 0

(b) We can perform a partial fraction expansion and have that:

(s + 1) c1 c2
G(s) = − = +
s(s + 2) s s+2

with c1 = −0.5 and c2 = −0.5.

Therefore we can derive the modal form as:

(
ẋm = Am xm + Bm u
y = Cm xm + Dm u

with
   
0 0 −0.5    
Am = , Bm = , Cm = 1 1 , and Dm = 0 .
0 −2 −0.5

(c) Both representations have different system matrices, except for the feed-through
term Dc = Dm = 0. To find the transformation matrix T , compute
   
Bc Ac Bc = T Bm Am Bm .

We then have
 
  −1 0 −2
T = Bc Ac B c B m Am B m = ,
−1 1

and we check that indeed

 
−1 0 0
Am = T Ac T = ,
0 −2
 
−0.5
Bm = T −1 Bc = ,
−0.5
 
Cm = Cc T = 1 1 ,
Dm = Dc = [0].

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

(Case iii.) We have the plant transfer function as

(s2 + s + 1) −s2 − s − 1 s−1

G(s) = − = = 2 −1
s(s + 2) s2 + 2s s + 2s

(a) The control canonical form is then given as

(
ẋc = Ac xc + Bc u
y = Cc xc + Dc u

with
   
−2 0 1    
Ac = , Bc = , Cc = 1 −1 , and Dc = −1 .
1 0 0

(b) We can perform a partial fraction and have that:

s−1 c1 c2
G(s) = −1= + − 1.
s2 + 2s s s+2
with c1 = −0.5 and c2 = 1.5.

Therefore we can derive the modal form as

(
ẋm = Am xm + Bm u
y = Cm xm + Dm u

with
   
0 0 −0.5    
Am = , Bm = , Cm = 1 1 , and Dm = −1 .
0 −2 1.5

(c) It can be observed that both representations have the same non-zero feed-through
term Dc = Dm = −1. To find the transformation matrix T , compute
   
Bc Ac Bc = T Bm Am Bm .

We then have
2
 
  −1 0 3
T = Bc Ac B c Bm Am Bm = ,
−1 − 13

and we check that indeed

 
0 0
Am = T −1 Ac T = ,
0 −2
 
−0.5
Bm = T −1 Bc = ,
1.5
 
Cm = Cc T = 1 1 ,
Dm = Dc = [−1].

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

4. Determination of poles/zeros from a state-space model

Consider the state-space model:
(
ẋ = Ax + Bu
y = Cx + Du

with
   
−2 0 2    
A= , B= , C = 0.5 −0.5 , and D = −1 .
1 0 0

(a) Determine the poles of this system.

(b) Determine the zeros of this system.
(c) Compute the transfer function and verify the poles and zeros found in questions (a)
and (b) are correct.

Solution:

(a) The poles of the system are any s ∈ C such that:

 
 s+2 0
det(sI − A) = 0 ⇐⇒ 
 −1 = 0,
s
⇐⇒ (s + 2)(s) = 0.

Therefore, we have the poles at s = 0, −2.

(b) The zeros are are any s ∈ C such that:
 
   s+2 0 −2
sI − A −B 
 −1
det =0 ⇐⇒ s 0  = 0,
C D 
 0.5 −0.5 −1
⇐⇒ −(s2 + s + 1) = 0
√
3
Therefore, we have the zeros at s = −0.5 ± 2 i.
(c) The transfer function is given by

−1 −(s2 + s + 1)
G(s) = C (Is − A) B+D = ,
s(s + 2)

which has poles and zeros equal to the obtained answers in questions (a) and (b).

5. Linearisation:
(a) The mathematical model of a stick-balancing problem is

θ̈ = sin (θ(t)) − f (t) cos (θ(t)) .

Derive a linearized state-space model around θ = θ̇ = f = 0, by setting x1 = θ, and

x2 = θ̇ = ẋ1 . Choose f (t) as the input.

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Exercise Set 7 - Control Systems (5ESD0) Oct 2021

(b) The mathematical model of a single-link robotic manipulator with a flexible joint is
given by
I θ̈1 (t) + mgl sin θ1 (t) + k(θ1 (t) − θ2 (t)) = 0
J θ̈2 (t) − k(θ1 (t) − θ2 (t)) = f (t),
where θ1 (t), θ2 (t) are the angular positions at time t, I, J are the moments of inertia,
m and l are the link mass and length, and k is the link spring constant.
i. Determine the state-space equations by considering x = [θ1 θ̇1 θ2 θ̇2 ].
ii. Linearize the system around the nominal operating point (xn , fn ).
Solution:

(a) By setting x1 = θ, x2 = θ̇, and u = f we have that ẋ = g(x, u), where

 
x2
g(x, u) =
sin x1 − u cos x1

We can then linearize around the point θ = θ̇ = f = 0 using a first order Taylor
approximation given by
g(x, u) ≈ g(x∗ , u∗ ) + ∇x g(x∗ , u∗ )(x − x∗ ) + ∇u g(x∗ , u∗ )(u − u∗ )
≈ Alin x̄ + Blin ū

where   
0 1 0
Alin = , Blin =
1 0 −1
 ⊤
(b) i. If x = θ1 θ̇1 θ2 θ̇2 then
˙ 
x2

θ1
θ¨1  − mgl sin(x1 )+k(x1 −x3 ) 
ẋ =  I
θ˙2  = 
  
x4 
¨ f +k(x1 −x3 )
θ2 J

and hence we have ẋ = g(x, f ), where

 
x2
− mgl sin x1 − k (x1 − x3 )
g (x, f ) =  I I 
 x4 
k 1
J (x1 − x3 ) + J f

ii. Linearizing around a nominal operating point then gives

g(x, f ) ≈ g(xn , fn ) + ∇x g(xn , fn )(x − xn ) + ∇u g(xn , fn )(f − fn )
≈ Alin x̄ + Blin ū
assuming our nominal operating point is chosen such that g(xn , fn ) = 0 and where
   
0 1 0 0 0
− k+mgl cos(x1,n ) 0 k
0 0
Alin =  I I  , Blin =  
0
 0 0 0 1
k k 1
J 0 −J 0 J

Assuming that the output variable is the link angular position, i.e., y = x1 , we
have    
C= 1 0 0 0 , D= 0