I know this is small stuff but I would like to use my Mathcad examples in a book someday and it would be nice if I didn't have to manually edit so much.
Is there and easy way to make this x/y?Īfter that I often see functions the look like this (-b+a)/3 instead of (a-b)/3. My next complaint is that often Mathcad produces transfer functions where the sign is wrong for both the numerator and the denominator.
#Polyroots mathcad 14 update
Then my pole/zero plots will update automatically too. I will try out the numer and denom functions with polyroots tomorrow to see if I can easily generate poles and zeros.
#Polyroots mathcad 14 generator
P.S.: Using your pseudounits as you did at the end of the new sheet you attached is dangerous as you loose most of the benefits the usage of units in Mathcad has.Wow! Mathcad sure has kept that a secret! I can't find numer or denom in the help at all but I tried it and it works! I can now generator coefficients for by difference equations automatically from the z domain transfer function. Of course you always can get one numerical solution with correct units if you use either root or a solve block Chosing the correct units might be tricky. So we have to divide the whole vector by length and multiply by 1/velocity. In your case the dimension of the coefficient of the linear element is length (the zero reults from L-L). If the dimensions of the rest are OK this will work. The rule is to make the second element (the coefficient of the linear element) show the reciproke of the unit you want to see for the result. If you want solutions with correct units, you will have to adapt the units in the vector accordingly (which does not feel natural on first sight). The obvious and most natural one is to strip all unitsĪnother approach is to make all elements the very same unitīoth ways above of course will give you a dimensionless solution. The unique Mathcad visual format and scratchpad interface integrate standard mathematical notation, text and graphs in a single worksheet - making Mathcad ideal for knowledge capture. Todays top engineers use Mathcad to perform, document and share calculation and design work. So you have a couple of choices to get polyroot to work. The Global Standard for Engineering Calculations.
Polyroot(v) will always solve so that the scalar product of v with the vector of powers of x is dimensionless " no matter which of the first three elelements has the wrong unit (and disregarding the setting of ORIGIN).īut if you use polyroot with a vector of more than three elements and the wrong unit is in the fourth or higher element it will give you the appropriate element count (no matter what ORIGIN is set to, the first element is always "Element 0"). It will always tell you that "Element 2 of this vector must have units. The error message itself is a confusing (not necessarily wrong. So you found yet another function in Prime which still will not work as expected when used with units So you will never get an error that element 0 or element 1 should have a different dimension. You can provide ANY units for those first two values, but the third (and all following for polynomial of higher order) have to be consistent. Mathcad can't tell the desired unit of the resulting variable unless you provide units for the first two elements of the vector. If you supply km (or m, it doesn't matter) you get the correct result.Įven the eror message makes sense (though it wasn't helpful in your case). The correct unit for the coefficient of the linear element is length. In your first attempt you supplied no unit at all and in your second the wrong unit. And as Prime has dynamic unit checking you also have to supply the correct unit for a zero value. Polyroots() is fully unit aware and works intuitiv as expected - you just have to take care and use the correct units all over. Forget (most) of what I had written above.
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