Homework 1
24-677 Special Topics: Linear Control Systems Prof. D. Zhao
Due: Sept 4, 2019, 8:30 am. Submit within deadline.
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Exercise 1. Types of Systems
A system has an inputu(t) and an output y(t), which are related by the information pro-
vided below. Classify each system as a linear or non-linear and time invariant or time-varying.
1. y(t)=0forallt 2. y(t) = u3(t)
3. y(t) = u(3t)
4. y(t)=e−tu(t−T) 5. y(t)=u(t−1)
6. y(t)=t u(τ)dτ −∞
7. y(t)=t u2(τ)dτ −∞
0t≤0 8.y(t)= u(t) t>0
Solution:
1. y(t)=0forallt
y1(t) = 0 for u1(t) input y2(t) = 0 for u2(t) input y3(t) = α · 0 + β · 0 for input: αu1(t) + βu2(t)
Hence it is linear.
y(t + τ) = 0
output for u(t + τ ) = 0 Hence time invariant
2. y(t) = u3(t)
(a) Non-linear:
Consider with
u(t) = αu1(t) + βu2(t)
y1(t) = u31(t)
y2(t) = u31(t)
then the output y(t) corresponding to the input u(t) is
y(t) = u3(t)
= (αu1(t) + βu2(t))3
= α3u31(t) + β3u32(t) + 3α2βu21(t)u2(t) + 3αβ2u1(t)u2(t) ̸= αy1(t) + βy2(t)
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(b) Time-Invariant:
3. y(t) = u(3t) (a) Linear:
Consider with
Dτ {u(t)} = u3(t − τ) (1) Dτ {y(t)} = u3(t − τ) (2) = Dτ {u(t)} (3)
u(3t) = αu1(3t) + βu2(3t) y1(t) = u1(3t)
y2(t) = u1(3t)
then the output y(t) corresponding to the input u(3t) is
(b) Time-Varying:
4. y(t)=e−tu(t−T) (a) Linear:
Consider with
y(t) = u(3t)
= αu1(3t) + βu2(3t) = αy1(t) + βy2(t)
Dτ {u(3t)} = u(3t − τ ) (1) Dτ {y(t)} = u(3(t − τ )) (2) = u(3t − 3τ)) (3) ̸= Dτ {u(t)} (4)
u(t − T) = αu1(t − T) + βu2(t − T) y1(t) = e−tu1(t − T)
y2(t) = e−tu2(t − T) then the output y(t) corresponding to the input is
y(t) = e−tu(t − T)
= e−t (αu1(t − T) + βu2(t − T)) = αy1(t) + βy2(t)
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(b) Time Varying:
5. y(t)=u(t−1) (a) Linear:
Consider with
Dτ {e−tu(t)} = e−tu(t − τ − T ) (1) Dτ{y(t)}=e−(t−τ)u(t−τ −T) (2) ̸= Dτ {u(t)} (3)
u(t − 1) = αu1(t − 1) + βu2(t − 1) y1(t) = u1(t − 1)
y2(t) = u2(t − 1) then the output y(t) corresponding to the input is
(b) Time-Invariant:
6. y(t)=t u(τ)dτ −∞
(a) Linear: Consider
with
y(t) = u(t − 1)
= (αu1(t − 1) + βu2(t − 1)) = αy1(t) + βy2(t)
Dτ{u(t)}=u(t−τ−1) (1) Dτ{y(t)}=u(t−τ−1) (2) = Dτ {u(t)} (3)
u(t) = αu1(t) + βu2(t) t
then the output y(t) corresponding to the input is t
y1(t) = y2(t) =
−∞
t
−∞
u1(τ)dτ u2(τ)dτ
y(t) = =
= αy1(t) + βy2(t) 4
−∞
t −∞
u(τ)dτ
(αu1(t) + βu2(t)) dτ
(b) Time-Invariant:
7. y(t)=t −∞
u2(τ)dτ (a) Non-Linear:
8.y(t)=
t < 0
Consider with
y(t) = =
−∞
t −∞
0,
u(t) t≥0
DT {u(t)} = DT {y(t)} =
t −∞
then the output y(t) corresponding to the input is t
y1(t) = y2(t) =
−∞
t
−∞
u21(τ)dτ u2(τ)dτ
u(τ − T )dτ t−T
u(τ)dτ
(1) (2)
(3) (4)
u2(τ)
(αu1(τ) + βu2(τ))2 dτ
ttt = α2 u21(τ)dτ + β2 u2(τ)dτ + 2αβ
−∞ −∞ −∞
̸= αy1(t) + βy2(t) (b) Time-Invariant:
DT {u(t)} = DT {y(t)} =
u2(τ − T)dτ
(1) (2)
(3) (4)
−∞
τ′=τ+T t −→
u(τ′ −T)dτ′ = DT {u(t)}
u(t) = αu1(t) + βu2(t) t
t −∞
t−T −∞
τ′=τ+T t −→
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−∞
u2(τ)dτ
u2 (τ ′ − T ) dτ ′ = DT {u(t)}
−∞
u1(τ)u2(τ)dτ
(a) Linear: Consider
with
u(t) = αu1(t) + βu2(t)
(b) Time-Varying:
0, t < τ
Dτ{y(t)}= u(t−τ) t≥τ ̸= DT {u(t)}
0, y1(t) = u1(t)
0, y2(t) = u2(t)
t < 0 t ≥ 0
t < 0 t ≥ 0
then the output y(t) corresponding to the input is 0, t < 0
t < 0
t ≥ 0
= αy1(t) + βy2(t)
t < 0 t ≥ 0
y(t)= u(t) t≥0
0 ,
αu1(t) + βu2(t)
=
= αy1(t) + βy2(t)
0,
Dτ {u(t)} = u(t − τ )
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Exercise 2. Fields
Is this given set a field?
1. Is N∗ a field ? Where N∗ is set of all Natural numbers. 2. Is Q a field ? Where Q is set of all rational numbers. 3. Set of all polynomials
4. Set of rational functions with real coefficients.
Solution:
1. N∗ is a set of all natural numbers without 0. As the set does not contain 0, an
element’s additive inverse does not exists. Hence N∗ is not a Field.
2. Q is a set of all Rational numbers (p where p and q ∈ R). As the Q set satisfies all
q
the rules specified in slides M1-1 for a definition of a field. Hence Q is a Field.
3. The set of polynomials will not consist the multiplicative inverse of an element. Because, by definition of a polynomial, the only polynomial with a negative degree is 0. Hence is not a field
4. The set rational functions with real coefficients satisfies all the rules specified in slides M1-1 for a definition of a field. Hence is a field
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Exercise 3. Subspace
The set W of n × n matrices with real entries is known to be a linear vector space.
Determine which of the following set are subspaces of W : 1. The set of n × n skew-symmetric matrices?
2. The set of n × n diagonal matrices?
3. The set of n × n upper-diagonal matrices?
4. The set of n × n singular matrices?
Solution: For the given set to be a subspace, it should be a Vector space. Hence it should
satisfy all conditions A0 - A4 and SM0 - SM4 as provided in M1 slide.
1. The nxn skew symmetric matrix is in a subspace of W.
2. The nxn diagonal matrix is in a subspace of W .
3. The nxn upper-diagonal matrix is in a subspace of W.
100 000 100
4.0 0 0+0 1 0=0 1 0whichisnonsingular,
000 001 001
The given set is not closed under addition. It violates the condition A0. Hence not a subspace.
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Exercise 4. Linearization
Perform linearization on the given differential equation
y ̈+(1+y)y ̇−2y+0.5y3 =0
Solution: Let x = x1 x2T , also x1 = y and x2 = y ̇ gives the state variable model x ̇1 x2
x ̇ =2x−0.5x3−(1+x)x =f(x) 21112
Equilibrium points are solutions of f(x) = 0, so each must have x2 = 0 and 2x1 − 0.5x13 = 0, three solutions are
0 2 −2 xe1 = 0 xe2 = 0 xe3 = 0
and we have the Jacobian matrix 01
2− 3x12 −x2 −(1+x1) 2
such that at (0,0), (2,0) and (-2,0)
x ̇1 0 1 x ̇1 0 1 x ̇1 0 1
x ̇ =2−1x,x ̇ =−4−3x,x ̇ =−41x 222
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Exercise 5. State space representations
1. What is the dimension (or shape) of the matrix B and D in the given state space
representation:
y1
x ̇1 a11 a12 a13 x1 b11 b12 b13 b14
x ̇2 = a21 a22 a23 x2 + b21 b22 b23 b24 u(t)
x ̇3 a31 a32 a33 x3 b31 b32 b33 b34
x1
y2 T =[B] x2 +[D]u(t)
x3
State how may inputs, states and outputs are present in this system.
Solution:
The state vector is of shape 3 × 1.
Given the [bij] matrix is 3 × 4 the input vector must be 4 × 1 i.e. 4 inputs, for matrix multiplication to be defined.
In the 2nd equation, there are 2 outputs y1 and y2. Thus B should be a 2 × 3 matrix and D should be 2 × 4. There are 4 inputs, 3 states and 2 outputs in the system.
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Exercise 6. Linearization of system
Consider the below Inverted pendulum problem,
θ m, I
F
The system dynamics is illustrated as the below equations: 2l (m + M)x ̈ + mlθ ̈− mlθ ̇2sinθ = F
(I + ml2)θ ̈ + mlx ̈ cos θ − mgl sin θ = 0
M
Linearize the state equations about the equilibrium. Derive the A and B matrices by substituting the operating equilibrium point. Here m is mass of the rod, M mass of cart, l is length of the rod, I is moment of inertia of the rod, θ is the swing angle of the rod, g is gravitational acceleration, x is distance along cart, F is Force applied as in the diagram.
Solution:
The state equations are: (I + ml2)θ ̈ + mlx ̈ cos θ − mgl sin θ = 0
(m + M)x ̈ + mlθ ̈− mlθ ̇2sinθ = F Differentiate for linearizing and get:
dx
dt =x ̇
2 22 2 ̇2 dx ̇ = (ml +I)F−gm l sinθ+ml(ml +I)θ sinθ
dt (M +m)(ml2 +I )−m2 l2 cosθ
dθ =θ ̇ dt
dθ ̇ = glm(M+m)sinθ−mlcosθ(F+mlsinθθ ̇2) dt (M +m)(ml2 +I )−m2 l2 cosθ
x ̇ = 0 θ ̇ = 0
F = (M + m)g tan θ F=gm2l2 sinθ
⇒ θ = nπ,F = 0
The equilibrium is:
(ml2 +I )
Look at equilibria: from 1st and 3rd rows x ̇ = θ ̇ = 0
F = (M + m)gtanθ
⇒ at equilibrium F = gm2l2 sinθ , imply θ = nπ,F = 0
(ml2 +I )
Linearize about the unstable equilibrium θ ≈ 0
Compute Jacobian:
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The Jacobian is defined as -
∂f1 ∂f1 ∂f1 ∂f1 ∂ x ∂ x ̇ ∂ θ ∂ θ ̇
∂f2 ∂ θ ̇
∂f3
∂ θ ̇
∂f4 ∂f4 ∂f4 ∂f4 ∂ x ∂ x ̇ ∂ θ ∂ θ ̇
∂f1 =0,∂f1 =1,∂f1 =0,∂f1 =0 ∂x ∂x ̇ ∂θ ∂θ ̇
∂f2 =0,∂f2 =0 ∂x ∂x ̇
∂f2
∂ x A =
∂f2 ∂ x ̇
∂f2 ∂ θ
∂f3 ∂ x
∂f3 ∂ x ̇
∂f3 ∂ θ
Here,
∂f2 =((m+M)(I+ml2)−m2l2cosθ)(−gm2l2cosθ+(I+ml2)mlθ ̇2cosθ)
∂θ
((I + ml2)(m + M) − m2l2cosθ)2
−(F (I + ml2) − gm2l2sinθ + (I + ml2)mlθ ̇2sinθ)(m2l2sinθ)
((I + ml2)(m + M) − m2l2cosθ)2 ∂f2 = 2ml(ml3+I)θ ̇sinθ
∂θ ̇ (M + m)(ml2 + I) − m2l2cosθ ∂f3 =0,∂f3 =0,∂f3 =0,∂f3 =1
∂ x ∂ x ̇ ∂ θ ∂ θ ̇ ∂f4 =0,∂f4 =0
∂ x ∂ x ̇
∂f4 = ((m + M)(I + ml2) − m2l2cosθ)((m + M)mglcosθ + mlFsinθ − m2l2θ ̇2cos2θ + m2l2θ ̇2sin2θ)
∂θ ((I + ml2)(m + M) − m2l2cosθ)2 −((M + m)mglsinθ − mlF cosθ − m2l2θ ̇2sinθcosθ)(m2l2sinθ)
((I + ml2)(m + M) − m2l2cosθ)2
∂f4 = −2m2l2cosθsinθθ ̇
∂θ ̇ (M + m)(ml2 + I) − m2l2cosθ 12
Substitute equillibrium point
01 0 0
0
1
0 0
−m2l2g
(M +m)(ml2 +I )−m2 l2
⇒ A = 0 0
00 glm(M+m) 0
(M +m)(ml2 +I )−m2 l2 0
ml2+I
∂f (M+m)(ml2+I)−l2m2 B=∂F|x ̄0,u ̄0 = 0
−ml
(M +m)(ml2 +I )−m2 l2
0
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