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University of Pennsylvania ESE 500: Linear Systems Theory Homework II Due: on Wednesday 10/20 at 23:59 on Gradescope INSTRUCTIONS Read the following instructions carefully before beginning to work on...

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University of Pennsylvania
ESE 500: Linear Systems Theory
Homework II
Due: on Wednesday 10/20 at 23:59 on Gradescope
INSTRUCTIONS
Read the following instructions carefully before beginning to work on the homework.
• You must submit your solutions on Gradescope. It must be submitted as a single PDF
file, compiled in LATEX or written by hand and scanned into images included in you
pdf. It must be readable by our staff to be graded.
• Please start a new problem on a fresh page and mark on Gradescope all the pages co
e-
sponding to each problem. Failure to do so may result in your work not graded completely.
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E
ata (Last update 10/8)
1. Problem 8: The output should read y =
[
1 cos θ + (r2 − 1) sin θ
−r1 sin θ + (r2 − 1) cos θ
]
.
Problem 1 (Matrix differentiation - 13 points). Let A(t) be an n × n matrix, which depends
on t, i.e. its elements are functions of t:
[A(t)]ij = aij(t), for i, j = 1, . . . , n
The derivative Ȧ(t) of A(t) with respect to t, is also an n× n matrix and is defined by taking
element-wise derivatives:
[Ȧ(t)]ij = ȧij(t), for i, j = 1, . . . , n
a) (4 points) For n× n differentiable matrices A1(t), A2(t) prove:
d
dt
(
A1(t)A2(t)
)
= Ȧ1(t)A2(t) +A1(t)Ȧ2(t)
1
) (4 points) Using induction, prove that for n × n differentiable matrices A1(t), A2(t), ... ,
Ak(t), we have:
d
dt
(
A1(t)A2(t)...Ak(t)
)
= Ȧ1(t)A2(t)...Ak(t) +A1(t)Ȧ2(t)...Ak(t XXXXXXXXXXA1(t)A2(t)...Ȧk(t)
c) (5 points) The exponential of a n× n matrix A is defined as the following:
expA =
∞∑
i=0
Ai
i !
where A0 = I. Suppose A(t) and Ȧ(t) commute. Prove:
d
dt
expA(t) = Ȧ(t) expA(t)
Problem 2 (Functions of a matrix - 26 points). Let f, g be functions over matrices and A,B ∈
Rn×n. Suppose AB = BA.
a) (3 points) Prove f(A)g(B) = g(B)f(A).
) (3 points) Prove f(AT ) = f(A)T .
c) (3 points) Let A = QJQ−1 be any matrix decomposition. Prove f(A) = Qf(J)Q−1.
d) (4 points) eA+B = eAeB. Hint: if two functions satisfy the same differential equation, then the uniqueness
of solution of differential equations says they are equal.
e) (4 points) Prove det(eA) = etr(A). Note: you can use known facts about determinant and trace.
f) (4 points) Prove that BeAt = eAtB if and only if AB = BA.
g) (5 points) For the time invariant linear state equation ddtx(t) = Ax(t) show that given any
x0 there exists a constant α(x0) such that
det[x(t) Ax(t XXXXXXXXXXAn−1x(t)] = α(x0)e
tr(A)t
Problem 3 (State space representation - 8 points). Consider a system with input u(t) and
output y(t) which can be described using the following set of differential equations:
z̈1(t) = z1(t) + z2(t) + u̇(t)
ż2(t) = ż1(t) + z2(t) + u(t)
y(t) = ż1(t)
a) (4 points) Define the states of the system such that it can be represented as an 3-dimensional
LTI system, i.e., as the following:
ẋ(t) = Ax(t) +Bu(t)
y(t) = Cx(t) +Du(t)
where A,B,C,D are constant matrices.
2
) (4 points) Consider T defined below, as a new basis for the state space and let x̂(t) be the
epresentation of x(t) with respect to the basis T.
T =

 10
1
 ,
 01
1
 ,
 11
1

Compute Â, B̂, Ĉ, D̂ in the new representation of the system with respect to T :
˙̂x(t) = Âx̂(t) + B̂u(t)
y(t) = Ĉx̂(t) + D̂u(t)
Problem 4 (10 points). Using the definition of transition matrix, prove that:
a) (2 points) φA(t2, t1)φA(t1, t0) = φA(t2, t0), for all t0, t1, t2.
) (1 point) φA(t, t) = I, for all t.
c) (2 points) φ−1A (t, τ) = φA(τ, t), for all t, τ .
d) (5 points)
d

φA(t, τ) = −φA(t, τ)A(τ)
Problem 5 (8 points). Find the solution x(t) of the following time variant system
ẋ(t) =
[
(t+ 1)2 t+ 1
t+ 1 t2 − 1
]
x(t), x(0) = x0
Problem 6 (Periodic System - 13 points). Consider the system
ẋ(t) = A(t)x(t)
where A(t) is a periodic matrix with period T . That means A(t+ T ) = A(t) for all t ∈ R.
a) (2 points) First, consider the state transition matrix Φ(t1, t0) for the system. Define the
matrix Ψ(t, 0) = Φ(t+ T, 0). Show the Ψ satisfies:
Ψ̇(t, 0) = A(t)Ψ(t, 0)
Ψ(0, 0) = Φ(T, 0)
) (2 points) Show that Φ(t+ T, 0) = Φ(t, 0)Φ(T, 0).
c) (3 points) Since we know that Φ(T, 0) is invertible, there exists some complex n× n matrix
R such that Φ(T, 0) = eTR. Define
P (t)−1 := Φ(t, 0)e−tR
Show that P (t)−1 is periodic with period T . This implies that P (t) is periodic with period
T . Also show that P (T ) = I.
d) (4 points) Show that
Φ(t, t0) = P (t)
−1e(t−t0)RP (t0)
Hint: Note that Φ(t, t0) = Φ(t, 0)Φ(0, t0)
e) (2 points) Express the system using the coordinate frame z(t) = P (t)x(t). What do you
notice about this new system?
3
Problem 7 (Discretization of continuous time LTI systems - 12 points). The dymanics of an
aircraft consist of a set of nonlinear coupled differential equations. Under certain assumptions,
though, these can be decoupled and reformed in a linear system. Aircraft pitch is governed
y the longitudinal dynamics. Let’s assume that the aircraft is in steady-cruise at constant
altitude and velocity (thus, the thrust, drag, weight and lift forces balance each other in the
x- and y-directions) and that a change in pitch angle will not change the speed of the aircraft
under any circumstance (unrealistic but simplifies the problem). Under these assumptions, the
longitudinal equations of motion for the aircraft are:
ȧ = µΩσ[−(CL + CD)a+
1
µ− CL
q − (CW sin γθ + CL]
q̇ =
µΩ
2iyy
[[CM − η(CL + CD)]a+ [CM + σCM (1− µCL)]q + (ηCW sin γ)δ]
θ̇ = Ωq
where a is the angle of attack, θ is the pitch angle and q the rate of pitch angle, δ the el-
evator deflection angle, µ = ρSc̄4m , ρ the density of air, S the platform’s area of the wing, c̄
the average chord length, m the mass of the aircraft, Ω = 2Uc̄ , U the equili
ium flight speed,
CT , CD, CL, CW , CM the coefficients of thrust, drag,lift, weight and pitch moment, γ the flight
path angle, σ = 11+µCL , iyy the normalized moment of inertia and η = µσCM .
Using the parameters from the data from one of Boeing’s commercial aircraft, the above
system becomes:
ȧ = −0.313a+ 56.7q + 0.232δ
q̇ = −0.0139a− 0.426q + 0.0203δ
θ̇ = 56.7q
In this problem we will transform this simple continuous time model in discrete time one
upon which one may design an autopilot that controls the pitch of an aircraft.
a) (2 points) Rewrite the model in state-space space form:
ẋ(t) = Ax(t) +Bδ(t), y(t) = Cx(t)
Use x(t) := [a, θ, q]T . The elevator deflection angle δ is considered the input of the system
and the pitch angle θ of the aircraft is the output.
) (5 points) Let the sampling period be Ts = 0.01s. Compute the discrete-time system. Then,
check your derivations using the c2d function of Matlab.
Note: You may use MATLAB for simple numerical calculations (e.g., matrix inversion,
matrix multiplication), but you should show all the steps in your derivations.
4
c) (3 points) Using δ(t) = 0.2, simulate the continuous time system using ode45 of Matlab fo
t ∈ [0, 10]. Plot the output y(t) with zero initial conditions, i.e. x(0) = [0, 0, 0]T .
d) (2 points) Simulate the discrete time system, for δ(k) = 0.2, k = 0, 1, . . . , N − 1, where
N = 10/Ts. In the same plot as before, plot the discretized y(k). Is the response of the
discrete-time system close to the one of the continuous time? Repeat the simulation fo
Ts ∈ {0.1, 0.5, 1, 2, 5, 10} and plot the responses in the previous plot (don’t forget to label
what’s what in the plot). For all of the simulations assume zero initial conditions, i.e.
x(0) = [0, 0, 0]T . Comment on the graph.
Problem 8 (Local linearization around a trajectory: unicycle - 10 points). A single-wheel cart
(unicycle) moving on the plane with linear velocity v and angular velocity ω can be modeled
y the following system of nonlinear of ODEs:
̇1 = v cos θ
̇2 = v sin θ
θ̇ = ω,
(1)
where (r1, r2) denotes the Cartesian coordinates of the wheel and θ its orientation. Regard this
as a system with input u =
[
v
ω
]
and output y =
[
1 cos θ + (r2 − 1) sin θ
−r1 sin θ + (r2 − 1) cos θ
]
.
a) (0.5 points) Prove that the map that takes input u(t) into y(t) is nonlinear.
) (2 points) Construct a state-space model for this system with state
x =
x1x2
x3
 :=
 r1 cos θ + (r2 − 1) sin θ−r1 sin θ + (r2 − 1) cos θ
θ
 .
c) (0.5 points) Prove that xeq = 0, ueq = 0 is an equili
ium for the state-space system derived
in part 2. Is this the only equili
ium point?
d) (1.5 points) Compute a local linearization for the state-space system derived in part 2, around
the
Answered 15 days After Oct 16, 2021

Solution

Abhishek answered on Oct 21 2021
130 Votes
PROBLEM-1
a.    Use a principal lattice to address the IVP X′=(114−2)X,X(0)=(1−2). The eigenvalues of A=(114−2) are λ1 = 2 and λ2 = − 3 with relating eigenvectors v1=(11) and v2=(1−4), separately. A principal lattice is then given by Φ(t)=(e2te−3te2t−4e−3t). We work out Φ− 1(0) by seeing that Φ(0)=(111−4), so Φ−1(0)=1−4−1(−4−1−11)=(4/51/51/5−1/5). Henceforth, X(t)=Φ(t)Φ−1(0)X0=(e2te−3te2t−4e−3t)(4/51/51/5−1/5)(1−2)=(e2te−3te2t−4e−3t)(2/53/5)=(25e2t+35e−3t25e2t−125e−3t).
.    If A will be a steady lattice, Φ(t, t0) = eA(t−t0); thus Φ(t, s) = eA(t−s) and (64) decreases to (41). We concede the confirmation of the overall case, where A(t) relies upon t, until some other time in this part.
c.    Then, at that point, the evaluations [⋅] which were gotten in the past segment can be supplanted by ⋅+O(λ-1), where Corollary 8.3.2 has been utilized. The determinant of the trademark network M(λ) given by (8.1.14) is an outstanding aggregate in the feeling of Section A.2. Here we utilize the key network Y(⋅,λ) as determined in Corollary 8.3.2.
PROBLEM-2
a. a.    Now let V mean the a
angement of all a
angements of (LH) on J; let α1, α2 be scalars (i.e., α1, α2 ∈ F); and let ϕ1, ϕ2 be a
angements of (LH) (i.e., ϕ1, ϕ2 ∈ V). Then, at that point, it is handily confirmed that α1ϕ1 + α2ϕ2 will likewise be an answer of (LH) (i.e., α1ϕ1 + α2ϕ2 ∈ V). We have subsequently shown that V is a vector space. Presently on the off chance that we pick n straightly free vectors ξ1, … ,ξn in the n-dimensional x-space, then, at that point, there exist n a
angements ϕ1, … ,ϕn of (LH) with the end goal that ϕ1(τ) = ξ1, … ,ϕn(τ) = ξn. It is a simple make a difference to check that this a
angement of a
angements {ϕ1, … ,ϕn} is straightly free and that it traverses V. In this way, {ϕ1, … ,ϕn} is a premise of V and any a
angement ϕ can be communicated as a straight mix of the vectors ϕ1, … ,ϕn.
. b.    We expect that A(t) and g(t) are characterized and nonstop over R = (− ∞,∞) (i.e., every part aij(t) of A(t) and every part gk(t) of g(t) is characterized and consistent on R). As indicated in Section III, framework (LN) has for any τ, ∈ R and any ξ ∈ Rn an interesting a
angement fulfilling x(τ) = ξ. This a
angement exists on the whole genuine line R and is constant in (t, τ, ξ). Besides, in the event that An and g rely persistently upon boundaries λ ∈ Rl, the a
angement will likewise change consistently with λ. Without a doubt, in the event that we separate the capacity.
c. c.     In request to determine the above condition, we need to begin first with the meanings of the epipole and epipolar line. As displayed in Fig. 1, the epipole, indicated by e and e′ in two perspectives, separately, is the reason behind crossing point of the line joining the two camera places (called the standard) with the relating picture plane. Epipolar line is the convergence of the epipolar plane and the picture plane.
d. Thus, the a
angement of (LN) might be seen as comprising of a part due to the "compelling term" g(t) and one more part because of the underlying information ξ.
e. This kind of partition is in everyday conceivable just in straight frameworks of differential conditions. We call ϕp the specific a
angement and ϕh the homogeneous a
angement of (LN).Before continuing to straight frameworks with steady coefficients, we present adjoint conditions. Leave Φ alone a crucial lattice for the straight homogeneous framework (LH).
f.    We expect that A(t) and g(t) are characterized and nonstop over R = (− ∞,∞) (i.e., every part aij(t) of A(t) and every part gk(t) of g(t) is characterized and persistent on R). As verified in Section III, framework (LN) has for any τ, ∈ R and any ξ ∈ Rn a special a
angement fulfilling x(τ) = ξ. This a
angement exists on the whole genuine line R and is ceaseless in (t, τ, ξ). Besides, assuming An and g rely constantly upon boundaries λ ∈ Rl, the a
angement will likewise differ persistently with λ. To be sure, in the event that we separate the capacity (35)ϕ(t,τ,ξ)=φ(t,τ)ξ+∫τtφ(t,η)g(η)dη concerning t to get ϕ′(t, τ, ξ), and assuming we substitute ϕ and ϕ′ into (LN) (for x), it is a simple make a difference to check...
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