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Asked by condor
on 9 Apr 2011

Hi.

I created this function (it is equal to the heaviside function):

function Y = fzine( X )

Y = zeros(size(X));

Y(X > 0) = 1;

eng = symengine;

if strcmp(eng.kind,'maple')

Y(X == 0) = nan;

else

Y(X == 0) = 0;

end

Now, why if

syms x

I get:

>> heaviside(x)

ans =

heaviside(x)

BUT

>> fzine(x) ??? Error using ==> sym.sym>notimplemented at 2621 Function 'gt' is not implemented for MuPAD symbolic objects.

Error in ==> sym.sym>sym.gt at 801 notimplemented('gt');

Error in ==> fzine at 7 Y(X > 0) = 1;

The two functions (fzine and heaviside) share the same code...?

Answer by condor
on 10 Apr 2011

Accepted answer

Yes...

and using heaviside is very fast, I wrote also this (very fast):

function Y = fzine( X ) %FZINE Summary of this function goes here % Restituisce 0 quando X<0 oppure X==0 % Restituisce 1 quando X>0 Y = heaviside(X); if Y==0.5 Y=0; end

Walter Roberson
on 10 Apr 2011

If your X is a symbolic object, then heaviside is defined to give you back a symbolic number, which you would have to use double() on in order to compare to 0.5

Andrew Newell
on 10 Apr 2011

The answers are out of sequence now, @Walter. This one is in reply to my suggestion that @condor do the calculation numerically.

Andrew Newell
on 10 Apr 2011

@condor, are we done? If so, it would really help sort out this mess if you accepted the most useful answer and voted for any others that you found useful.

Answer by Andrew Newell
on 9 Apr 2011

The symbolic toolbox cannot evaluate `X > 0` for a symbolic variable `X`. You can make a small change in your function so it can handle a symbolic variable:

function Y = fzine( X )

Y = zeros(size(X));

Y(double(X) > 0) = 1;

eng = symengine;

if strcmp(eng.kind,'maple')

Y(X == 0) = nan;

else

Y(X == 0) = 0;

end

Now if you try

fzine(sym(1))

you get

ans =

1

(you can also input a double variable). It still won't handle `fzine(x)`; but notice that

heaviside(x)

just returns the name of the function. It isn't trying to actually evaluate heaviside(x).

Show 9 older comments

condor
on 10 Apr 2011

Yes Walter....very very good! It seems working.

Just a question..why do I get 2 results:

>> fzine = @(x) subs('piecewise([x > 0,1],[Otherwise,0])','x',x);

>> syms MSP

>> costi=fzine(MSP-10)*2+fzine(MSP-50)*2.5+fzine(MSP-100)*3+15

costi =

piecewise([100 < MSP, 45/2], [not 10 < MSP, 15], [MSP in Dom::Interval(10, [50]), 17], [MSP in Dom::Interval(50, [100]), 39/2])

>> delta=1.3*costi+costi-MSP

delta =

piecewise([100 < MSP, 207/4 - MSP], [not 10 < MSP, 69/2 - MSP], [MSP in Dom::Interval(10, [50]), 391/10 - MSP], [MSP in Dom::Interval(50, [100]), 897/20 - MSP])

>> double(solve(delta))

ans =

39.1000

34.5000

Answer by Andrew Newell
on 10 Apr 2011

Here is a numerical version of what you wish to do. First, the function:

function y = heavy0(x) y = 0*x; y(x>0) = 1;

Now solve with it:

costi= @(x) heavy0(x-10)*2+heavy0(x-50)*2.5+heavy0(x-100)*3+15; delta = @(x) 1.3*costi(x)+costi(x)-x; fzero(delta,0)

ans =

39.1000

This also works for problems where the solution is the zero point of a heaviside function:

costi = @(x) heavy0(x-10)+x-10; fzero(costi,1)

ans =

10

Answer by Paulo Silva
on 9 Apr 2011

good question, I'm also clueless about why that happens

Answer by Walter Roberson
on 9 Apr 2011

Your fzine is probably not in the private directory of the symbolic functions, so probably you are not getting the correct methods invoked for all items.

Answer by condor
on 9 Apr 2011

So what you mean with private directory (where is it?)... where should I put the fzine?

Answer by condor
on 9 Apr 2011

ok I did:

>> which heaviside C:\Program Files\MATLAB\R2010b\toolbox\symbolic\symbolic\heaviside.m

so I cut and pasted the fzine in that folder; now:

>> syms x >> fzine(x) ??? Undefined function or method 'fzine' for input arguments of type 'sym'.

IT doesn't work, help please.

Answer by condor
on 9 Apr 2011

>> rehash toolbox >> syms x >> heaviside(x)

ans =

heaviside(x)

>> fzine(x) ??? Error using ==> sym.sym>notimplemented at 2621 Function 'gt' is not implemented for MuPAD symbolic objects.

Error in ==> sym.sym>sym.gt at 801 notimplemented('gt');

Error in ==> fzine at 7 Y(X > 0) = 1;

Still doesn't work...

Answer by condor
on 9 Apr 2011

Andrew you say:

"The symbolic toolbox cannot evaluate X > 0 for a symbolic variable X."

Why this doesn;t work with heaviside? It doens't evaluate the function but it accept a symbolic object as input so you can solve an equation that contains heaviside(x)...

??

condor
on 9 Apr 2011

An example:

>> syms x

>> delta=heaviside(x)+15-12*x

delta =

heaviside(x) - 12*x + 15

>> solve(delta)

ans =

4/3

>> delta=fzine(x)+15-12*x

??? Error using ==> sym.sym>notimplemented at 2621

Function 'gt' is not implemented for MuPAD symbolic objects.

Error in ==> sym.sym>sym.gt at 801

notimplemented('gt');

Error in ==> fzine at 8

Y(X > 0) = 1;

Answer by condor
on 10 Apr 2011

Here where the error gets generated:

function X = gt(A,B)

%GT Symbolic greater-than.

A = sym(A);

B = sym(B);

if isa(A.s,'maplesym')

X = A.s > B.s;

else

notimplemented('gt');

end

end

Show 1 older comment

Walter Roberson
on 10 Apr 2011

Odd... your path says 2010b, which is restricted to using MuPad and cannot use Maple. I wonder if the internals of the symbolic toolbox still say Maple in places ??

condor
on 10 Apr 2011

You are right Walter...and I don't know why! But heaviside (I don't know how) is able to bypass that check above...I mean that with heaviside

isa(A.s,'maplesym') returns 1 otherwise I would get the same error!

If I am able to discover how to bypass that I could replicatye it in my function fzine....

Answer by Andrew Newell
on 10 Apr 2011

Here is a different way you could approach this problem:

fzine = (x + abs(x))/2; delta = fzine+15+12*x; simplify(solve(delta))

ans =

-5/4

In fact, you could define your function this way:

function Y = fzine( X ) Y = (X + abs(X))/2; Y(X~=0) = Y/X;

This now behaves almost exactly like the Heaviside function. Even the derivative works:

>> fd = simplify(diff(fzine(x)))

fd =

-(abs(x) - x*sign(x))/(2*x^2)

>> subs(fd,-2)

ans =

0

>> subs(fd,2)

ans =

0

The only difference is for `x=0`:

>> subs(fd,0)

ans =

NaN

(the derivative of `heaviside` gives `Inf`).

**EDIT**: Another difference occurs if you try this:

y = fzine(x); subs(y,0)

ans =

NaN

**EDIT**: I think the problem with the original function is not just the greater than sign. It is the conditional commands, which make it impossible for the function to return an explicit symbolic expression. For more general cases, you could replace a lot of `if/then` conditions by expressions involving ... (drum roll) `heaviside`.

Show 3 older comments

Andrew Newell
on 10 Apr 2011

Sorry, I got a bit hasty there (I'm trying to keep up with your comments!). I added a line above.

Answer by condor
on 10 Apr 2011

I think the best solution up to know is the one that Walter suggested (it seems to be a little bit slow but that is ok for my use...):

fzine = @(x) subs('piecewise([x > 0,1],[Otherwise,0])','x',x);

however I still don't understand:

1) Why using the fzine defined as above and solving and equation as:

>> syms MSP >> costi=fzine(MSP-10)*2+fzine(MSP-50)*2.5+fzine(MSP-100)*3+15

costi =

piecewise([100 < MSP, 45/2], [not 10 < MSP, 15], [MSP in Dom::Interval(10, [50]), 17], [MSP in Dom::Interval(50, [100]), 39/2])

>> delta=1.3*costi+costi-MSP

delta =

piecewise([100 < MSP, 207/4 - MSP], [not 10 < MSP, 69/2 - MSP], [MSP in Dom::Interval(10, [50]), 391/10 - MSP], [MSP in Dom::Interval(50, [100]), 897/20 - MSP])

>> double(solve(delta))

ans =

39.1000 34.5000

returns 2 solutions?? The first is the correct one...the second is what we would have if in costi all the fzine are = zero so we would have: costi = 15 and costi*1.3+costi=34.5 BUT MSP cannot be 34.5 otherwise costi would not be 15.... !?!

2) I still don't understand why heaviside works with symbolic objects while another function that uses the same code does not...

Walter Roberson
on 10 Apr 2011

1) I don't know. I used the piecewise version in Maple and it gave only the first (391/10) solution.

2) I do not have the Symbolic Toolbox so I cannot check or test the heaviside source code myself.

Answer by Andrew Newell
on 10 Apr 2011

@Condor, I think you are mistaken that the function `heaviside` uses the same code as the other functions we have considered. `heaviside` is an interface to MuPAD code, which creates a MuPAD object that can be manipulated according to the rules in MuPAD. For example, operators like `diff` and `int` can act on `heaviside`, while they cannot on Walter's version of `fzine`. My function behaves a little more like a symbolic object, but I have described some shortcomings.

You can easily create a function that gives the correct *numerical* answer for a numerical input. But you want it to do more than that. You want it to create that MuPAD object even though we don't even know the properties of MuPAD objects.

It's an interesting exercise seeing where attempts to recreate `heaviside` fall short, but do we really need to go further with this? Why not just use `heaviside`?

Show 4 older comments

Andrew Newell
on 10 Apr 2011

The difference in value at x=0 only matters if your solution turns out to be 10, 50, or 100 in the example you've given. But you could check that these aren't solutions by substituting, e.g., subs(costi,10).

condor
on 10 Apr 2011

I cannot because the function that I am looking for will be used in a GUI with a long long code

Andrew Newell
on 10 Apr 2011

Then you would be better off treating this as a numerical problem in Matlab itself instead of MuPAD. Implementing a numerical version of heaviside is trivial, and you can use FZERO to find the solution.

Answer by Konstantinos
on 18 May 2013

Edited by Walter Roberson
on 18 May 2013

The way I found so you can do your job is for example:

I wanted to find the Z transform of the following:

syms n z x2 = n*heaviside(n) + (6-2*n)*heaviside(n-6) + (n-6)*heaviside(n-6); ztrans(x2, n, z)

and it would give me a wrong value because of the heaviside(n-6). So I fixed it by instead using:

x2 = n*heaviside(n) + (6-2*n)*heaviside(n-5.5) + (n-6)*heaviside(n-5.5);

Basically I guess using some double inbetween your discrete values will fix all the problems. I could have also changed heaviside(n) to heaviside(n+0.5) but there wasn't a point in this particular example. Hope this helps.

## 1 Comment

## Andrew Newell (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/5151#comment_10607

Note to readers of this question: To see how the accepted answer relates to the question, read How to change the heaviside function -- and how to use the new function with symbolic objects?, How to change the heaviside function -- and how to use the new function with symbolic objects?, and How to change the heaviside function -- and how to use the new function with symbolic objects? in that order.