# Problem Set 3 Solutions 1

```Massachusetts Institute of Technology
Department of Electrical Engineering and Computer Science
6.245: MULTIVARIABLE CONTROL SYSTEMS
by A. Megretski
Problem Set 3 Solutions
1
For each of the functions of complex parameter√ s (where Re{s} &gt; 0,
log(s) = log |s| + j arg(s) with arg(s) ∈ (−π/2, π/2), and s = exp(0.5 log(s))) below, state whether they are proper extensions of functions from the
CT class H2 , or H∞ , or both, or none of the two. In other words, can
these functions serve as Fourier transforms of funite energy signals,
or as transfer functions of L2 gain stable LTI systems? Provide brief
reasoning.
√
(a) G1 (s) = exp( s);
Answer: not in H2 , not in H∞ .
Reasoning: recall that |ez | = eRe(z) , and
q
p
p
p
Re( σ + jω) = (σ + σ 2 + ω 2 /4)/2 ≥ |ω|/2.
√
Therefore G1 is not in H2 because Re( s)√&gt; 0 whenever Re(s) &gt; 0, which implies
|G1 (s)| &lt; 1. G1 is not in H∞ because Re( s) → ∞ as s is real and s → ∞.
√
(b) G2 (s) = exp(− s);
Answer: in both H2 and H∞ .
1
Version of October 21, 2011.
2
Reasoning: G2 is in H∞ because it is an analytical in C+ , and satisfies |G1 (s)| &lt; 1
there. G2 is in H2 because it is an analytical in C+ , and the integrals
Z ∞
Z ∞ √
1
1
2
|G2 (σ + jω)| dω ≤
e− |ω|/2 dω
2π −∞
2π −∞
are uniformly bounded for σ &gt; 0.
√
(c) G3 (s) = s/(s + 1);
Answer: not in H2 but in H∞ .
Reasoning: note that
|G3 (σ + jω)|2 =
1
1
|ω|
≤
≤
2
2
(1 + σ) + ω
2(1 + σ)
2
for σ &gt; 0, ω ∈ R. By the equality, |G3 (σ + jω)|2 is not integrable over ω ∈ R,
which implies G3 ∈ H2 . By the inequality, G3 ∈ H∞ because G3 is analytical and
uniformly ounded in C+ .
(d) G4 (s) = exp(−s3 );
Answer: not in H2 and not in H∞ .
Reasoning: since
Re[−(σ + jω)3 ] = 3σω 2 − σ 3
converges to +∞ as ω → ∞ for a given σ &gt; 0, the function ω → |G4 (σ + jω)|2 is
neither bounded nor integrable over ω ∈ R.
(e) G5 (s) = log(s)/(s + 1);
Answer: in H2 but not in H∞ .
Reasoning: since log(r) → −∞ for real positive r → 0, G5 is not bounded in the
right half plane. On the other hand, G5 is analytical in the right half plane (log(s)
is analytical as the well-defined inverse of exp(s) restricted to the strip Im(s) ∈
(−π/2, π/2), and ratio of analytical functions is analytical), hence it is sufficient to
show that L2 norms of functions ω → G5 (jω + σ) are uniformly bounded.
Indeed, by definition of the logarithm, |Im log(jω + σ)| is uniformly bounded by
π/2, hence
Im log(jω + σ) π
1
jω + σ + 1 ≤ 2 |jω + σ + 1| .
3
On the other hand, combining
|ω| ≤ |jω + σ| = |jω − 1 + 1 + σ| ≤ |jω + 1| + |1 + σ| =
√
1 + ω2 + 1 + σ
with the inequality
| log(a + b)| ≤ max{| log(a)|, | log(b)|} ≤ | log(a)| + | log(b)| ∀ a, b ∈ (0, ∞),
which based on the fact that the function a → | log(a)| decreases for a ∈ (0, 1), and
increases for a &gt; 1, and taking into account that
Re log(jω + σ) = log |jω + σ|,
yields
log |ω| ≤ Re log(jω + σ) ≤ log(1 + ω 2 ) + log(1 + σ),
hence
Re log(jω + σ) | log(|ω|)| + log(1 + ω 2 )
log(1 + σ)
+
.
jω + σ + 1 ≤
|jω + 1|
|jω + σ + 1|
Therefore
|G5 (jω + σ) ≤
| log(|ω|)| + log(1 + ω 2 ) π
1
+ log(1 + σ) +
.
|jω + 1|
2 |jω + σ + 1|
Since the first term on the right side is square integrable and does not depend on
σ, and, by the Parceval identity,
Z ∞
Z ∞
π
dω
e−2at dt =
= 2π
2
a
0
−∞ |jω + a|
for all a &gt; 0, the L2 norm of ω → G5 (jω + σ) is uniformly bounded.
(f) G6 (s) = Re{1/(s + 1)};
Answer: not in H2 and not in H∞ .
Reasoning: function s → G6 (s) is not analytical in the right half plane. for
example, (G(1 + δ) − G(1))/δ converges to −0.25 as δ → 0 along the real axis, but
converges to 0 as δ → 0 along the imaginary axis.
(g) G7 (s) = |s|2 .
Answer: not in H2 and not in H∞ .
Reasoning: function s → G7 (s) is not analytical in the right half plane. for
example, (G(1 + δ) − G(1))/δ converges to 2 as δ → 0 along the real axis, but
converges to 0 as δ → 0 along the imaginary axis.
4
Let S1 , S2 be two DT LTI models defined by transfer functions
1
G1 (z) = 2 + ,
z
G2 (z) =
1+z
0.5 + z
respectively. One expects a series interconnection of S1 and S2 to be
the DT LTI model defined by the transfer function G(z) = G1 (z)G2 (z).
It turns out that this is not completely true: out of the two series
interconnections S12 = S1 ◦S2 and S21 = S2 ◦S1 , only one is exactly a transfer
function model. The other can be represented as a sum of two systems,
one of which is LTI model with transfer function G, and the other is the
”autonomous” LTI model with an appropriately chosen transfer function
L = L(z): it has only boundary conditions and no input, and its output is
the same as that of the standard DT LTI system with transfer function
L, but with zero input. A block diagram for such representation is shown
on Figure 1. Models like this are usually called uncontrollable, as there is
no way to affect the output of the subsystem with transfer function L.
w
-
G(z)
-
L(z)
-g
6
y
-
0
Figure 1: Uncontrollable LTI model
Represent both systems S12 , S21 in the form shown on Figure 1, and find
the corresponding transfer functions L. Is L uniquely defined in each of
Answer: S1 ◦ S2 has the form Figure 1 with L(z) ≡ 0 (i.e. is equivalent to a transfer
matrix model with G = G1 G2 ), while S2 ◦S1 has the form Figure 1 with L(z) ≡ (0.5+z)−1
(i.e. is not equivalent to a transfer matrix model with G = G2 G1 ). The selection of L is
not unique in either case.
Reasoning: for G(z) = c + b/(z − a), where a ∈ (−1, 1) and b, c ∈ R, the set XG
(as defined in subsection 4.3.2 of lecture notes) consists of all functions x0 : Z → R
such that x0 (t) = 0 for t &lt; 0 and x0 (t) = at p for t ≥ 0, where p ∈ R is an arbitrary
5
constant, and, by convention, a0 = 1, i.e. for t ∈ Z+ we have 0t = δ(t), where δ(&middot;) is
the unit sample function. Indeed x0 must be approximable by functions y0 satisfying the
difference equation
y0 (t + 1) − ay0 (t) = bw0 (t),
where w0 (t) = 0 for t ≥ 0, which yields y0 (t + 1) = ay0 (t) for t ≥ 0, which justifies the
description of XG .
In particular, x1 ∈ XG1 means x1 (t) = p1 x10 for some p1 ∈ R, where x10 (t) = δ(t),
while x2 ∈ XG2 means x2 (t) = p2 x20 (t) for some p2 ∈ R, where x20 (t) = (−0.5)t , and, for
G( z) = G1 (z)G2 (z) = 2 + 2/z, condition x ∈ XG0 means x(t) = px10 (t) for some p0 ∈ R.
Since
LG1 x20 = 2x10 , LG2 x10 = 0.5x10 + 0.5x20 ,
we have
(S1 ◦ S2 )(w, (x1 , x2 )) = {LG1 G2 w + LG1 x2 + x1 }
= {LG0 w + (2x2 (0) + x1 (0))x10 },
while
(S2 ◦ S1 )(w, (x1 , x2 )) = {LG2 G1 w + LG2 x1 + x2 }
= {LG0 w + 0.5x1 (0)x10 + (0.5x1 (0) + x2 (0))x20 }.
m,k
m,k
We consider models M1 ∈ SDT
(X1 ) and M2 ∈ SDT
(X2 ) equivalent when there exist
functions φ : X1 → X2 and ψ : X2 → X1 such that S1 (w, x1 ) = S2 (w, φ(x1 )) and
S2 (w, x2 ) = S1 (w, ψ(x2 )) for all w ∈ `m , x1 ∈ X1 , x2 ∈ X2 . Acording to this definition,
S1 ◦ S2 is equivalent to the LTI model S defined by transfer function G, which is certified
by
φ(x1 , x2 ) = (2x2 (0) + x1 (0))x10 , ψ(x) = (0, x(0)x10 ).
By a similar argument, S1 ◦ S2 is equivalent to the system defined on Figure 1 with L ≡ 0.
In contrast, S2 ◦ S1 is not equivalent to S (for example, because the zero-input response
of S2 ◦ S1 can be equal to x20 , which is impossible for S). Instead, S2 ◦ S1 is equivalent to
the system defined on Figure 1 with L(z) ≡ 2 + 1/z.
This excercise is related to the handy Poisson identity. The integral
Z ∞
Re{G(jω)}
dω,
2
−∞ (ω + 1) + 2
6
where G is a function from the CT class H2 , equals cRe{G(s0 )}, where real
number c and complex number s0 (such that Re{s0 } &gt; 0) do not depend on
G. This special case of the Poisson identity can be easily derived from
the Parseval identity by finding the inverse Fourier transform of
h(ω) =
1
(ω + 1)2 + 2
explicitly, and by expressing inverse Fourier transform of Re{G(jω)} in
terms of the inverse Fourier transform of G. Find the constants c, s0 ,
and explain how √
you prove
√ the identity.
Answer: c = π/ 2, s0 = 2 − j.
Reasoning: the answer follows from the general Poisson identity
Z
1 ∞ aRe{G(jω)}
dω = Re{G(a + jb)},
(1)
π −∞ a2 + (ω − b)2
√
which is valid for all a &gt; 0 and b ∈ R and all functions G ∈ H2 , applied with a = 2 and
b = −1.
To prove (1), let g be the inverse Fourier transform of G. Then, since
(
g(−t), t &lt; 0,
g0 (t) =
g(t),
t ≥ 0,
is the inverse Fourier transform of
G0 (jω) = 2Re{G(jω)},
and
(
e−(a−jb)t , t ≥ 0,
f0 (t) =
e(a+jb)t ,
t &lt; 0,
is the inverse Fourier transform of
F0 (jω) =
a2
2a
,
+ (ω − b)2
the Parceval identity
1
2π
Z
∞
0
Z
∞
F0 (t) G0 (jω)dω =
−∞
−∞
f0 (t)0 g0 (t)dt
7
yields
1
π
Z
∞
−∞
∞
Z
1 0 (a−jb)t
e
g(t)dt +
e
g(−t)dt
2 −∞
0
Z
G(a + jb) 1 ∞ −(a+jb)t
=
+
e
g(t)dt
2
2 0
aRe{G(jω)}
1
dω
=
a2 + (ω − b)2
2
Z
−(a+jb)t
G(a + jb) + G(a + jb)
2
= Re{G(a + jb)}.
=
The following situation is quite common in LTI systems research: a theoretical statement is proven (or available in the literature) for the case
of discrete time systems, and one would like to get convinced that a
similar statement also holds in the CT case. Of course, the need to go
the other way (from a CT theorem to a similar DT statement) is just as
common.
A rather starightforward tool for such conversions is the ”bilinear
transform”. It is easy to see that if z is a complex number with |z| &gt; 1,
or z = ∞, and ω0 &gt; 0 is a positive real number then
s = ω0
z−1
z+1
(2)
is a complex number with Re{s} &gt; 0. Conversely, if s ∈ C and Re{s} &gt; 0
then
ω0 + s
(3)
z=
ω0 − s
(obtained by solving (2) with respect to z) is either a complex number with
|z| &gt; 1, or ∞ (when s = ω0 ). Transformations (2),(3) also work between
the unit circle and the imaginary axis, except that z = −1 is mapped to
∞ = j∞.
(a) Is it true that Gd = Gd (z) is a DT H∞ class function with
kGd k∞ = sup |Gd (z)| = γ
|z|&gt;1
8
if and only if
Gc (s) = Gd
ω0 + s
ω0 − s
(4)
is a CT H∞ class function with
kGc k∞ =
sup |Gc (s)| = γ ?
Re{s}&gt;0
Sketch a proof or give a counterexample.
Reasoning: consider the function f : D+ → C defined by
f (z) = ω0
z−1
,
z+1
where
D+ = {z ∈ C : |z| &gt; 1}.
The function is analytical, and maps D+ to a subset of
C+ = {s ∈ C : Re{(}s) &gt; 0}
(all points except s = ω0 ). Hence, if Gc : C+ → C is analytical and uniformly
bounded then Gd : D+ → C defined by Gd = Gc ◦ f , i.e. by
z−1
,
Gd (z) = Gc ω0
z+1
is also analytical and uniformly bounded.
Conversely, if Gd : D+ → C is analytical and uniformly bounded then it has Laurent
series representation
Gd (z) =
∞
X
gd (k)z −k where
k=0
hence Gc : C+ → C
Gc (s) =
∞
X
k=0
gd (k)
lim |gd (k)|1/k ≤ 1,
k→∞
ω0 − s
ω0 + s
k
is an analytical function. Since Gc (ω0 ) = gd (0) is finite, and for every s ∈ C+ ,
s 6= ω0 there exists z ∈ D+ such that Gc (s) = Gd (z), the fiunction Gc is uniformly
bounded.
9
(b) It is easy to see that transform (4) does not map every DT H2 class
function Gd to a CT H2 class function Gc . Give a counterexample.
Example: Gd (z) ≡ 1.
(c) For every ω0 &gt; 0 there exist positive real numbers a, b &gt; 0 such that
Gd = Gd (z) is a DT H2 class function with
1/2
Z π
1
2
kGd k2 = sup
|Gd (r exp(jΩ))| dΩ
=γ
(5)
r&gt;1 2π −π
if and only if
b
Gd
Gc (s) =
s+a
ω0 + s
ω0 − s
(6)
is a CT H∞ class function with
1/2
Z ∞
1
2
=γ .
|Gc (jω + σ)| dω
kGc k2 = sup
σ&gt;0 2π −∞
(7)
Find a, b as functions of ω0 , and explain why the statement is true.
√
√
Answer: b = 2ω0 (another possible choice is b = − 2ω0 ), a = ω0 .
Reasoning: whenever the integral converges absolutely, the substitution
ω = ω0
sin(Ω/2)
cos(Ω/2)
yields the identity
Z
∞
−∞
2ω0
g
2
ω + ω02
ω0 + s
ω0 − s
Z
π
dω =
g ejΩ dΩ,
−π
which suggests the choice for a and b in the answer.
In the case when Gd (z) is uniformly bounded on D+ , this identity proves that
kGd k2 = kGc k2 , as in that case finiteness of the integrals in (5), (7) is evident.
In general, we use the fact that functions from class H2 can be approximated arbitrarily well (in the L2 norm) by functions from H∞ . For example, if Gd ∈ H2
then
∞
∞
X
X
−k
Gd (z) =
gd (k)z
where
|gd (k)|2 = kGd k22 &lt; ∞,
k=0
k=0
10
hence kGd − Gd,m k2 → 0 as k → ∞, where
Gd,m (z) =
m
X
gd (k)z −k .
k=0
Let Gc,m : C+ → C be defined by
√
2ω0
Gd,m
Gc,m (s) =
s + ω0
ω0 + s
ω0 − s
.
Since Gd,m belong to the class H∞ , the inequality
Z ∞
1
|Gc (jω + σ)|2 dω ≤ kGd,m k22 ≤ kGd k22
2π −∞
is valid for all σ &gt; 0. Taking the limit as m → ∞ yields finiteness of the supremum
in (7).
```