piecewise symbolic function within a symbolic matrix

Hallo, I would like to construct a matrix P whose elements depend on x (a vector that varies with time). Some elements of P are piecewise functions of x. The code is just to test whether P can be updated. Any help will be much appreciated.

sym x                  
P = sym(zeros(3,3))    %create empty symbolic matrix

P(1,1) = sym('1/(1+exp(-4.3*x(1)))')
P(2,1) = sym('0.2*x(2)')
P(3,1) = sym('0.4*sin(0.0192+x(3))')

P(1,2) = sym('1/(1+exp(11.8*x(2)))')
P(2,2) = .3435
P(3,2) = sym('.02 + (x(3)>=.04)*(.66*x(3))')    %piecewise

x = [.5, .5, .1]
eval(P)

But I get an error due to the piecewise function (without it, P easily updates):

Error using sym/eval (line 11)
Error: Unbalanced or unexpected parenthesis or bracket.

Outside of the sym function, .02 + (x(3)>=.04)*(.66*x(3)) works all right.

I have also tried:

sym('piecewise([x(3)<.04,.02],[x(3)>=.04,0.02+0.66*x(3)])')

but still no go...I wonder what I am doing wrong? Thanks for your time

Answer 1

[Star Strider]

You may not be doing anything actually ‘wrong’, but eval may not be your best choice.

Replace it with:

vpa(P)

to get it to work without errors.

[Andy]

I tried your suggestion:

...
x = [.5, .5, .1]
P1 = vpa(P)

But P1 is still a symbolic matrix (i.e. not a matrix of actual function values)

[Star Strider]

The Symbolic Math Toolbox may not be the best way to do it if you simply want a numeric answer. I copy-pasted the elements of your matrix to a numeric matrix and made an anonymous function out of it:

x = [.5, .5, .1]

P = @(x) [(1/(1+exp(-4.3*x(1))))  (1/(1+exp(11.8*x(2))));  (0.2*x(2))  .3435;  (0.4*sin(0.0192+x(3)))  (.02 + (x(3)>=.04)*(.66*x(3)))]

Px = P(x)

produces:

Px =

     895.6688e-003     2.7320e-003
     100.0000e-003   343.5000e-003
      47.5672e-003    86.0000e-003

Does that do what you want?

[Andy]

Thanks for this.

[Star Strider]

My pleasure!

The sincerest expression of appreciation here on MATLAB answers is to Accept the answer that most closely solves your problem.

[Walter Roberson]

Inside the Symbolic Toolbox, the result of comparisons is true (a logical value) or false (a logical value) or FAIL (which is its own object type) . The results are not numeric 0 or numeric 1, so

sym('.02 + (x(3)>=.04)*(.66*x(3))')

will not do what is desired.