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Kim
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Converting Mathcad to Matlab

Asked by Kim
on 18 Jul 2013
Latest activity Commented on by Kim
on 13 Nov 2013
I have a book with some good radar equations in it. However, everything is in Mathcad - I use Matlab. I've been converting everything I use from Mathcad to Matlab. I've run into one I'm not sure how to convert.
Mathcad has a function called root. The equation I'm interested in converting is: root(f1(u) - sqrt(0.5), u, 0.3, 0.7
As I understand it, 'root' returns the value of u to make the function f1 equal to zero. With 0.3 and 0.7 being specified, root finds u on this interval.
How do I accomplish the same thing using Matlab?
Thanks, Kim

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1 Answer

Eric
Answer by Eric
on 18 Jul 2013
Edited by Eric
on 18 Jul 2013
 Accepted Answer

Look at the documentation for Matlab's fzero() function. For example, to find the zero of cosine between 1 and 2:
fun = @cos; % function
x0 = [1 2]; % initial interval
x = fzero(fun,x0)
x is returned with a value of 1.570796326794897.
For your case you'll use something like
fun = @(x)f1(x) - sqrt(0.5);
and of course you'll define x0 to be [0.3 0.7].
Note that the documentation for Matlab R2013a has a typo. In my code above where I defined x0, the documentation actually defines this as x (i.e., they forgot the 0).
Good luck,
Eric

  3 Comments

Kim
on 6 Aug 2013
It has taken me forever to get back to this. When I tried your example it worked fine, when I tried mine it didn't. Got the following: >> x0 = [0.3 0.7]; >> fun = @(u)f1(u) - sqrt(0.5); >> x = fzero(fun,x0) Error using fzero (line 233) FZERO cannot continue because user supplied function_handle ==> @(u)f1(u)-sqrt(0.5) failed with the error below.
Undefined function 'f1' for input arguments of type 'double'.
The error is simple: you didn't define a function 'f1' anywhere. Did you create a file called f1.m with your function? What does this:
which f1
give you?
Kim
on 13 Nov 2013
Yep, that was the problem. In my problem f1(u) = sinc(pi*u). So I changed what I did to: fun = @(u) sinc(u) - sqrt(0.5); u0 = [0.3 0.7]; x = fzero(fun,u0)
Note - I had to learn that in mathcad sinc is straight sin(x)/x and for matlab it is sin(pi*x)/(pi*x) - so had to take care of pi.
But now it works.

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