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Integral of controllability gramian

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I am having trouble finding a way to integrate the controllability gramian in Matlab. My system is unstable so I can't use the built in function. My system also has eigenvalues on the imaginary axis so I can't use a function I found online for unstable systems. So i'm trying to integrate the controllability gramian for some finite time interval on MATLAB, but it seems impossible. Here's what the integral looks like.
Where A is a 6x6 matrix and B is a 6x1 matrix. The matrix exponential in the equation is what's causing me the most trouble.
Any ideas?

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Accepted Answer

Star Strider
Star Strider on 1 Jun 2016
Edited: Star Strider on 1 Jun 2016
Try this with your matrix and vector:
A = rand(6); % Create Data
B = rand(6, 1); % Create Data
f = @(tau) expm(A*tau)*B*B'*expm(A'*tau); % Integrand
W = @(t) integral(f, 0, t, 'ArrayValued',1); % Controllability Gramian
Wt = W(1)
The integral function was introduced in R2012a. Before that, I believe the appropriate function is quadv.

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Star Strider
Star Strider on 1 Jun 2016
My pleasure!
The quadv function does not support the 'ArrayValued',1 name-value pair. Instead, it would interpret those as the tolerance and trace, with unpredictable results. It is best not to use them.
Instead, just use:
W = @(t) quadv(f, 0, t); % Controllability Gramian
Wt = W(1)
Meena Farag
Meena Farag on 2 Jun 2016
Awesome. This simple problem has been stressing me out for a while, lol. Thanks again!
Star Strider
Star Strider on 2 Jun 2016
As always, my pleasure!
I was surprised that your question hasn’t been asked before.

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More Answers (4)

Sheng Cheng
Sheng Cheng on 16 Feb 2017
Edited: Sheng Cheng on 20 Feb 2017
I have a different way for computing the controllability gramian matrix based on an early paper, 'Computing integrals involving the matrix exponential', by Charles Van Loan. The paper can be found here: https://www.cs.cornell.edu/cv/ResearchPDF/computing.integrals.involving.Matrix.Exp.pdf
I will skip the mathematical rigorous proof in the paper. In fact, all the result you need to compute the gramian matrix is written in the left column on the first page. Especially, equation (1.2) is the form we are looking for. (Please read the paper for the extremely simple structure of this integral (actually the controllability gramian is indeed an integral involving matrix exponential)).
The code is just in two lines
A = rand(6); % Create Data
B = rand(6, 1); % Create Data
temp = expm([-A B*B';zeros(6,6) A']); % Coming from the first equation below (1.4)
Wc = temp(7:12,7:12)'*temp(1:6,7:12); % Coming from the second equation below (1.4)
Here, you don't need to define a function and an integral like the one suggested by Star Strider. All you need is expm and then some very simple matrix operation.

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Star Strider
Star Strider on 16 Feb 2017
Interesting. Noted.
Thank you.
Star Strider
Star Strider on 20 Feb 2017
In all my control courses, I never encountered that. The paper you attached is in my collection.
+1 Vote!

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Ahmed Rashid
Ahmed Rashid on 31 May 2016
Why don't you check the rank of the controllability matrix?
C = rank([B AB A^2B ... A^(n-1)*B])
If C has full rank, then the system is controllable.

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Ahmed Rashid
Ahmed Rashid on 31 May 2016
Calculating the integral is equivalent to calculating the Lyapunov equation.
A*W+W*A^T+B*B^T=0
So you can find it by solving the above system of equations in Matlab.
Meena Farag
Meena Farag on 1 Jun 2016
It is only equivalent for stable systems, my system is unstable. I found that I have no choice but to compute the integral for some finite time.
Bryan Jevon
Bryan Jevon on 30 Oct 2018
May you help me with source or literature that provide information about how weak rank test by using controllability and observability matrix?

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Roger Stafford
Roger Stafford on 1 Jun 2016
In matlab there is a very important difference between e.^((A.’)*τ) and e^((A.’)*τ) (without the dot.) The first of these is an element-wise exponentiation and the second a matrix exponentiation. If you use exp((A.’)*τ), it will produce the element-wise version. I rather suspect you want the element-wise version.

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Meena Farag
Meena Farag on 1 Jun 2016
Yes, I am using the element-wise version (and expm, not exp). I am still not able to integrate. I still get the same error:
Error using .*
Matrix dimensions must agree.
Error in @(t)(expm(A.*t)*B1)*(expm(A.*t)*B1)'
Error in quad (line 76)
y = f(x, varargin{:});
Error in DOC (line 63)
quad(fn(:,1),0,1)
Star Strider
Star Strider on 1 Jun 2016
Actually, I looked this up in my control reference (and Wikipedia Controllability Gramian). It’s matrix exponentiation, expm.

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Rajani Metri
Rajani Metri on 1 Dec 2018
How to calculate Minimum control u*(t) required to state transfer from x1(t) to x2(t) and from it the states x1*(t) and x2(*)? also how to Plot them?
Thank you

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Star Strider
Star Strider on 1 Dec 2018
Post this as a new Question.
No one will respond to it here.

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