ACCEPTED SOLUTION
Accepted Solutions
Mathcad won't be able to provide a symbolic solution because it cannot decide on the different branches of min, max or an if statement with a variable, symbolic argument
For a numeric solution you may use a simple solve block. In the example you provided it works OK var values of d up to 2. z(d)=1 for d>2, so z is not bijective outside the interval you have chosen and so the inverse can't works. rev(z) would yield 1 for z=1.
In your final application one problem could be that a single, constant guess value for d (in my sheet I had chosen d=1) does not work for the whole range of z-values provided. But we'll deal with this issue when we get to it.
Mathcad won't be able to provide a symbolic solution because it cannot decide on the different branches of min, max or an if statement with a variable, symbolic argument
For a numeric solution you may use a simple solve block. In the example you provided it works OK var values of d up to 2. z(d)=1 for d>2, so z is not bijective outside the interval you have chosen and so the inverse can't works. rev(z) would yield 1 for z=1.
In your final application one problem could be that a single, constant guess value for d (in my sheet I had chosen d=1) does not work for the whole range of z-values provided. But we'll deal with this issue when we get to it.
My apologies, by the way, I should have specified that I was working for d in [0,2[ (d is actually the sum of 2 duty cycles, each one between 0 and 1). It is such an "obvious" hypothesis for me, because I am so focused on the problem, that it became kind of implicit. So you are right, it wouldn't work out of this interval.
You are also right that, in this example, the min in the x function is not necesary. In the application, I do use x with d in the whole range of d, so I just mechanically put the min whith application in mind. Once again, hypothesis that are implicit considering the application, but are absolutely not obvious when asking for help.
No problem. If you stay with d in the range [0;2[ you would also not need the max() function either as 1/(2-d) would always be larger than 1 for d>1. But I already guessed that your original problem would be a bit more complex.
EDIT: Here is an alternative approach without the need of a guess value using the root() function:
Mine Solution:
When the function is an even function, Werner's method couldn't work, because root or find only return one root or can't find any root in the guess value range.
I guess you haven't looked at Adrien's original function because you have not access to Prime.
Here is a pdf printout so you can try your luck using the symbolics 
Keep in mind that we must use the min, max and if statements, as Adrien's original function is more complex and requires that.
Oh, sorry, Werner, the solve symbolic calculation really can't solve that function 
I haven't install Mathcad Prime, so I haven't read the original article. If use the numerical root or find, my method is the same as yours, and seems more troublesome...
zhu laojianke wrote:
Oh, sorry, Werner, the solve symbolic calculation really can't solve that function
Yes - the problem are the conditionals and also max and min. The symbolics can't evaluate them (symbolically woithout any concrete values) and so it fails.
