Automatic Control Laboratory
D-ITET
Prof. Dr. Roy Smith
Spring Term 2015
C ONTROL S YSTEMS II (R EGELSYSTEME II)
S OLUTIONS TO P ROBLEM S ET 7
April 15, 2015
Exercise 1
Internal stability of two degrees-of-freedom control configurations
r
Kr
+
H
_
G
y
K
Figure 1: General configuration
Consider the generic connection in Fig. 1, and for simplicity assume the plant G is SISO. Let
Kr =
Ng
Nr
Nh
Nk
, H=
, K=
, G=
.
Dr
Dh
Dk
Dg
Then the transfer function from r to y is
Dk Nh Ng
Nr
y(s)
= T (s) =
·
.
r(s)
Dr Dg Dh Dk + Nh Nk Ng
(1)
1. For perfect tracking, the configuration in Fig. 1.b (H = Ku , K = I) should not be used if Nh
contains RHP roots. A RHP root in Nh would either lead to a T (s) with a RHP zero, or to a
pole-zero cancellation with Dg , Dk or Dr , with consequent internal instability.
2. Similarly, the configuration in Fig. 1.c (H = I, K = Ky ) should not be used if Dk contains
RHP roots (= RHP poles of K), in particular if Dk has roots in s = 0. We would either obtain a
RHP zero in T (s), or a pole-zero cancellation with Dr , Ng or Nh , leading to internal instability.
3. The configuration in Fig 1.d may be used if both Nh and Dk do not contain RHP roots. When
Nk Nh and Dk Dh are designed for instance to place the closed-loop poles of the system, for
implementation one can include RHP roots in Nk and Dh . In Fig. 2, we have an equivalent
representation of Fig. 1.
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r+
_
H Kr
G
y
-1
Kr K
Figure 2: Moving Kr after the summing point
Exercise 2
Internal Model Control 1
1. Given the following identities the proof is straight-forward:
K(I + GK)−1 = Q
(I + GK)−1 = I − GK(I + GK)−1 = I − GQ
(I + KG)−1 = I − KG(I + KG)−1 = I − K(I + GK)−1 G = I − QG
G(I + KG)−1 = G(I − QG)
2. Note that following the definition of T and Q one can observe that
1
,
s+1
which yields that the unstable zero of the plant G must be contained as a pole in Q. This observation together with the result obtained in the first part implies that there is no stabilizing controller
1
so that T (s) = s+1
, since it leads to an unstable Q irrespective of the controller K.
T = (I + GK)−1 GK = G(I + KG)−1 K = GQ
Exercise 3
=⇒
GQ =
Internal Model Control 2
1. Starting from the definition of the complementary sensitivity function, plug in the expression for
the controller to obtain:
−1
−1 −1
−1
T = (I + GK)−1 GK = (I + GG−1
2 (I − G1 ) ) GG2 (I − G1 )
= (I + G1 (I − G1 )−1 )−1 G1 (I − G1 )−1
= (I + G1 (I − G1 )−1 )−1 (I − G1 )−1 G1
= ((I − G1 ) + G1 )−1 G1
= G1
2. Recall that Q = K(I + GK)−1 and plug in K(s) to obtain:
−1
−1
−1 −1
Q = G−1
2 (I − G1 ) (I + GG2 (I − G1 ) )
−1
−1 −1
= G−1
2 (I − G1 ) (I + G1 (I − G1 ) )
−1
= G−1
2 ((I − G1 ) + G1 )
= G−1
2
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3. In inverse based control, we select the (open–) loop transfer function, usually as L(s) = ωsc assuming the plant is stable and minimum phase. Conversely, in IMC we select the complementary
sensitivity T (s), i.e. the shape of the closed-loop, and use this choice to compute Q(s).
If the plant G(s) is stable, but non-minimum phase, it can be decomposed in a minimum phase
part and a non-minimum phase part G(s) = GM P (s) · GN M P (s). We need Q(s) to be stable
in order to obtain an internally stable system (Slide 8.10), thus we can only invert the minimum
phase part: Q(s) := GM P (s)−1 Tideal (s).
Exercise 4
Transfer functions
Using (1) M can be derived from (2) as:
−1
I
K
M =
−G I
I + KX −1 (−G) −KX −1
=
.
−X −1 (−G)
X −1
Since X = I+GK and noting (I+GK)−1 G = G(I+KG)−1 and I−KG(I+KG)−1 = (I+KG)−1 ,
I − K(I + GK)−1 G −K(I + GK)−1
M =
(I + GK)−1 G
(I + GK)−1
(I + KG)−1 −K(I + GK)−1
=
G(I + KG)−1
(I + GK)−1
This is the block transfer function matrix implied by (3) and (4).
Exercise 5
State space realizations
Note that the cascade G1 G2 is described by the equation
u1 = y2
Using this equation and the equations of the linear systems:
x˙ 1 (t) = A1 x1 (t) + B1 u1 (t)
y1 (t) = C1 x1 (t) + D1 u1 (t)
and
x˙ 2 (t) = A2 x2 (t) + B2 u2 (t)
y2 (t) = C2 x2 (t) + D2 u2 (t)
we get the desired results. Both expressions for the state space realizations of the cascade can be found
by reordering the state as x = [x1 , x2 ] or x = [x2 , x1 ]. We get e.g.
x˙ 1 (t) = A1 x1 (t) + B1 (C2 x2 (t) + D2 u2 (t))
y1 (t) = C1 x1 (t) + D1 (C2 x2 (t) + D2 u2 (t))
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or
x˙ 1 (t) = A1 x1 (t) + B1 C2 x2 (t) + B1 D2 u(t)
x˙ 2 (t) = A2 x2 (t) + B2 u(t)
y(t) = C1 x1 (t) + D1 C2 x2 (t) + D1 D2 u(t)
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