Continuous-time and Discrete-time systems, Schur conditions, Routh-Hurwitz

In summary, continuous-time systems involve continuous variables, while discrete-time systems involve discrete variables. The Schur conditions and Routh-Hurwitz criterion are both methods for determining the stability of a linear system, with the Schur conditions providing a necessary and sufficient condition and the Routh-Hurwitz criterion providing a sufficient condition. However, these methods cannot be applied to nonlinear systems, which require different techniques for stability analysis.
  • #1
EconStu
1
0
Hi guys,

I hope one of you can help me out with the following problem. I hope I posted the problems in the correct section. Problem:

Consider the following continuous-time and discrete-time systems with respect to
a common 2X2 matrix A:

x˙ = Ax (1) is a continuous- time system
xt+1 = xt + hAxt (2) is a discrete-time system

where 0 < h 1 and A is assumed to be asymptotically stable (AS). The intuition
is that for sufficiently small step sizes h, system (2) is a good approximation of
(1) and therefore AS, too. Your task is to find a positive number H such that
(2) is AS for all h < min{1,H}.
Hints:
• Write (2) in the form xt+1 = Bxt (where B depends on h, of course) and
use the Schur conditions.
• Denote the Routh-Hurwitz coefficients for the stability of (1) by a1, a2, and
the coefficients for the Schur conditions in (2) by b1, b2. Express the latter
in terms of a1, a2 and (of course) h, and use the information on the signs of
a1, a2 that you have from the stability assumption on A.
• To simplify matters, you are allowed to suppose that traceA + 2 > 0.

Here is what i have so far:

I applied Schur and Routh-Hurwitz conditions and expressed one in the form of the other as was advised in the hint.I got the following result:

My matrix A = [d e]
-------------[f g]

My matrix B = [1+dh eh]
------------[fh gh +1]

S1 (Schur Condition 1) dgh^2+ efh^2 >0 (B1)
S2 (Schur Condition 2) 4 + 2dh + 2gh + dgh^2 + efh^2 >0 (B2)
S3: -gh-dh-dgh^2+efh^2 (B3)

Routh-Hurwitz conditions:

RH1: -d-g < 0 (A1)
RH2: dg-ef > 0 (A2)

After expressing Shur Conditions in the form of Routh-Hurwritz Conditions, I got the following result:

-A2h^2 >0
4-2A1h-A2h^2 >0
A1h-A2h^2>0

I am not sure how to take it from there though.

I am a Master Student in Economics. My mathematics background is insufficient to solve this problem. I hope somebody can help.

Thanks in Advance
 
Mathematics news on Phys.org
  • #2


Hi there,

Thank you for posting your problem in the correct section. I am a scientist with a background in mathematics and I will do my best to help you with your problem.

Firstly, I would like to commend you for successfully applying the Schur and Routh-Hurwitz conditions to express one in the form of the other. This is a good start and shows that you have a good understanding of the problem.

To continue, let's take a closer look at the result you have obtained so far:

-A2h^2 >0
4-2A1h-A2h^2 >0
A1h-A2h^2>0

From the Schur conditions, we know that B1 and B2 must be greater than 0 for the discrete-time system (2) to be asymptotically stable. This means that both A2h^2 and 4-2A1h-A2h^2 must be positive. This leads to two inequalities:

-A2h^2 >0 (from B1)
4-2A1h-A2h^2 >0 (from B2)

We also know from the Routh-Hurwitz conditions that A1h-A2h^2 must be greater than 0 for the continuous-time system (1) to be asymptotically stable. This leads to the inequality:

A1h-A2h^2>0 (from A1)

Combining these three inequalities, we can solve for h:

-A2h^2 >0
4-2A1h-A2h^2 >0
A1h-A2h^2>0

Solving for h, we get:

h < min{A1/A2, (4-A2)/2A1}

Therefore, the positive number H that we are looking for is:

H = min{A1/A2, (4-A2)/2A1}

This means that for all h < min{1,H}, the discrete-time system (2) will be asymptotically stable.

I hope this helps you to solve your problem. If you have any further questions, please do not hesitate to ask. Good luck with your research!
Scientist
 

FAQ: Continuous-time and Discrete-time systems, Schur conditions, Routh-Hurwitz

What is the difference between continuous-time and discrete-time systems?

Continuous-time systems are those that are described using mathematical equations that involve continuous variables, such as time. Discrete-time systems, on the other hand, are described using mathematical equations that involve discrete variables, such as samples taken at specific time intervals. The main difference between these two types of systems is the way in which time is treated in the mathematical model.

What are the Schur conditions for stability?

The Schur conditions provide a set of necessary and sufficient conditions for determining the stability of a linear system. These conditions state that for a system to be stable, all of its eigenvalues (or poles) must have negative real parts. In other words, the system's transfer function must have all its poles in the left half of the complex plane.

What is the Routh-Hurwitz stability criterion?

The Routh-Hurwitz stability criterion is a method for determining the stability of a linear system by examining the coefficients of its characteristic polynomial. It states that for a system to be stable, all the coefficients in the first column of the Routh array (a table formed from the polynomial coefficients) must have the same sign. If any coefficient in the first column is zero, the next row of the array is used to determine the system's stability.

How do the Schur conditions and Routh-Hurwitz criterion relate to each other?

The Schur conditions and Routh-Hurwitz criterion are both methods for determining the stability of a linear system. The Schur conditions provide a necessary and sufficient condition for stability, while the Routh-Hurwitz criterion provides a sufficient condition. This means that if a system satisfies the Schur conditions, it will also satisfy the Routh-Hurwitz criterion. However, the Routh-Hurwitz criterion can still be used to determine stability for systems that do not meet the Schur conditions.

Can the Schur conditions and Routh-Hurwitz criterion be applied to nonlinear systems?

No, the Schur conditions and Routh-Hurwitz criterion are only applicable to linear systems. Nonlinear systems do not have characteristic polynomials or transfer functions, which are required for these methods to be used. Other techniques, such as Lyapunov stability analysis, must be used to determine the stability of nonlinear systems.

Similar threads

Replies
12
Views
1K
Replies
16
Views
3K
Replies
1
Views
1K
Replies
1
Views
1K
Replies
2
Views
1K
Back
Top