i Mathcad experts,
I have some questions about Mathcad function. I don’t have any useful idea to solve my issues.
I have one equation, and I can draw the curve of this equation.
And then I try to solve the peak value of curve (fig.1) by using the method in EX2.
Below are my questions.
1.Any method I can solve this equation to get the fn_Y2, please see EX1?
2.How do I use solve function to get the fn_Y2, if I have different inputs Q2 at the same time (please see EX3). I can do this when I only have one input, please see EX2.
3.If the issue 2 can be solved, and then I would try to input the fn_Y2 to get the peak value . Also I would like to draw the relationship of Q and peak value, please see the fig.2 (this fig is from TI’s datasheet), could you please give me some suggestions.
1 ACCEPTED SOLUTION
Accepted Solutions
Prime is not able to solve your equation symbolically.
But as you had shown yourself it can provide solution(s) if you provide numeric values for the other two variables
You may also use the "assume" modifier to limit the solution to the only positive, real one
But you are just looking for numeric results (depending on Q and L(ambda), so I would suggest that you use Mathcads numeric methods. Either a solve block with "find" or one of the two ways to use the "root" function.
Here is the latter, assuming the solution is within the range from 10^-10 to 10:
You can turn this calculation into a function of lambda and Q
So it can be used to get the result for any of these values
From here its not difficult to plot either the My function for various Q-values and their peaks
or the plot you seem to be looking for
Mathcad 15 sheet attached
I am not sure why do you want to differentiate with respect to fn_Y2, and not fn_Y.
Its too complicated to solve for fn_Y...also having these 2 as parameters
But you can do something like this:
as you already did in fact here:
Its based on a manual way to do this what you want as you did...
Here...someone will need to find a way on how to choose maximum value as we can see for example at Q4 index is 3 and at Q6 index is 5, so index is changing:
And then how to iterate Q parameter from 0.1, 0.2,...1, then for each Q value to find the maximum, and then to plot, these in a more automatical way...Maybe one could find how to do it...
Prime is not able to solve your equation symbolically.
But as you had shown yourself it can provide solution(s) if you provide numeric values for the other two variables
You may also use the "assume" modifier to limit the solution to the only positive, real one
But you are just looking for numeric results (depending on Q and L(ambda), so I would suggest that you use Mathcads numeric methods. Either a solve block with "find" or one of the two ways to use the "root" function.
Here is the latter, assuming the solution is within the range from 10^-10 to 10:
You can turn this calculation into a function of lambda and Q
So it can be used to get the result for any of these values
From here its not difficult to plot either the My function for various Q-values and their peaks
or the plot you seem to be looking for
Mathcad 15 sheet attached
Ok, I made this with waterfall plot also in Prime 10 following that first thread you had posted.
But I see a thing here: if I try to plot fn_Y from 0 the error with division by 0 arise. I am wondering how you managed to plot from 0 in Mathcad 15?
Prime 10:
Your plot in Mathcad 15:
Your f vector in Mathcad 15:
My f vector in Prime:
And we are both using the same :
A 2D plot will happily ignore failing function values and still plot the rest. The creation of a vector/matrix fails if just one of the calculation fails. So simply use
when you create the matrix, the plot will ignore the NaN's and work OK.
I this specific case you may also replace the NaN by a zero, because that's the correct limit
That said you may also use
@J_power wrote:
Thanks for your response.
May I know why you use Vector in M function and F function?
You are talking about the vectorization used when showing the five peak positions?
f.nY and M.Y are functions written to receive single scalar arguments. When you want to feed into those function a vector like Q.3 instead and expect that every single element of this vector should be used as function argument and all results the n collected in a vector, then this is exactly what vectorization is made for. Sometimes (depends how your function is defined) its not necessary to apply explicit vectorization because Mathcad would apply (implicit) vectorization automatically. But i won't rely on it and suggest to always use explicit vectorization in such cases.
Here two simply examples. Both function simply square the argument given. They do the very same when used with a simple scalar argument but return different results when called with a vector as argument.
Reason is that f uses multiplication and this is an operation defined for vectors, too (scalar product). Function g uses squaring and this is an operation NOT defined for vectors and instead of throwing an error Mathcad applies implicit vectorization. So using explicit vectorization is a good idea without having to think if its really necessary or if Mathcad would do it on its own.
BTW, the vectorization of M.Y in the plot was senseless and i shouldn't have used it because here I use your range (! not a vector) Q.2 to plot all five curves in one go (something not working in Prime anymore as Cornel also noticed). Vectorization has no effect if no vectors are involved (but it doesn't do any harm either).
I played around and added a 3D plot with the L (lambda), Q an max My values, showing the curve corresponding to L=6.3 in red
I also added a way to use the symbolics to get the necessary fnY-values, but I won't suggest using it:
@J_power wrote:
What's the function of this equation?
This expression does the very same (or something similar) as the function f.positive in my last reply (posting a worksheet with Lucs approach).
F is the result of calling the function "temp" which uses the symbolics to return a vector with six elements.
In case of your example just one of these six values is real and positive and this is the one to pick. I won't bet that its always the first or always the second one (even this may be the case here).
Furthermore sometimes because of round-off errors even the real solutions are returned with a very tiny imaginary part.
1) |Im(F)|<^10^.10 returns a 6-element vector with a 1 where F is a real value (Imaginary part 0 or very small) and a zero if F is non.real.
2) F is multiplied b the vector created in the first step, so all non-real values are set to zero
3) this new vector is sorted using the "sort" function. That way the largest (positive ?) value comes last in the vector returned
4) "reverse" changes the order of the vector elements so that that positive value is the first element
5) This first element is now chosen using the matrix index ORIGIN
And, no, this sure is not the most elegant or intuitive way to achieve this task 😉
After all I think its best to avoid the symbolics altogether and just use Mathcads numeric facilities.
The question is to find the peak values in:
an example of which is:
That means solving:
Which gives a huge result. You can just select the one result, that is real and positive:
And use it to calculate the peak value with:
Which you can plot over values of lambda and Q:
And of course you can pick the Peak value for any set of lambda and Q values:
Success!
Luc
@LucMeekes wrote:
Sure. I see from Werner's results that Mathcad 15 provides the positive as second element of the result array. So, define
positive := ORIGIN+1
and use that to index the f() function.
Success!
Luc
Seems to apply only to the symbolic solution using numeric values as in my sheet.
When using the pure symbolic approach, the solution yielding positive values is the first one as in your MC11.
BTW, it seems not to be necessary here, but to be on the safe side we could use a small utility function to be sure to get the positive value:
Lucs method using the symbolic "solve" with all the variables undefined only works because Mathcad is capable enough to find all solution for f and so we can later pick the one we are interested in. Depending on the equation provided is easily possible that the symbolic solve would fail (more likely in MC15 than in MC11).
Mathcad 15 sheet attached
