ACCEPTED SOLUTION
Accepted Solutions
@prop_design wrote:
is there a way to convert the odesolve output to a matrix, so that the max and min can be found. apparently the odesolve output is a function. i attached a worksheet. i want to convert t and x to matrices.
How about
Of course you could use any vector "tvec" as long as its elements are all in the range from 0 to tmax. You may simply replace tstep by any other value.
@prop_design wrote:
is there a way to convert the odesolve output to a matrix, so that the max and min can be found. apparently the odesolve output is a function. i attached a worksheet. i want to convert t and x to matrices.
How about
Of course you could use any vector "tvec" as long as its elements are all in the range from 0 to tmax. You may simply replace tstep by any other value.
hi luc,
i've been looking over the various files you mentioned. i'm afraid i can't figure out how to call the solver. i attached a MC15 file, but i'm not sure if you can open it. I think you are using MC11. i attached a picture as well.
ideally, i'd like to use three different forcing functions. cosine wave, sine wave, and square wave. i'd like to be able to change the damping ratio and other parameters as well. to operate in different regions. i'm not sure if your solver cares which region you are in or not. so stuff like beats, resonance, under damped, over damped, critically damped.
Note that LODEsolve is a SYMBOLIC ONLY function, it does not evaluate numerically. But of course, as with any symobilic expression, you can assign its result to a function and evaluate that numerically. You almost did that in your MCAD15 file: xsym(t):=LODEsolve(ODE,f(t),init,x,t).
The one thing you forgot was the symbolic evaluation, so it should be:
xsym(t):=LODEsolve(ODE,f(t),init,x,t) ->
(That is, press [CTL >] at the en of the expression.)
1. How did the file operate in Mathcad 15, that I attached specific to your problem (The file LM_20210429...)?
2. The LODEsolve() function should function with the first two of your proposed f(t), not with the third.
The reason for that is that the symbolic processor cannot handle the mod() function. You'll have to create a square wave using the heaviside step function.
As for functional properties of the ODE (whether or not damped, critical or whatever): The LODEsolve() function does not care. It executes a pure mathematical operation (an automation of what ttokoro showed) that, if the ODE meets the requirements for it, and a solution can be found, will produce the solution. And if you work with a pure symbolic expression you get a pure symbolic mathematical function.
I've elaborated your example here:
gives:
Note that the expression runs out to the right for a couple of pages. In Mathcad 15 it may look differently, due to another symbolic processor.
Now we can use the (result) function xsym:
and plot the results over time
Note that the Mathcad built-in odesolve function produces apparently the same result.
But it's not EXACTLY the same, if I subtract the ODEsolve result from the LODEsolve result:
With the very same xsym() function you can now experiment with different values of b, m, k and omega.f.
And you can easily calculate differentials:
The Mathcad file is attached.
Success!
Luc
The first response I get from the f3 function in your files is:
This is due to the fact that the numeric processor doesn't know x, and complains.
I had explained that LODEsolve is a symbolic-only function. You always have to evaluate it symbolically.
After I've added a symbolic evaluation I get this:
This is due to the fact that the number of cycles is unknown. While the symbolic processor in general can deal with unknown variables, the number of cycles is used by the pulse function in a summation operator. Well, in this case the symbolic processor would like to know how many summations there are. So we can help it by telling. I define nc before using LODEsolve in this definition, and I delete nc from the parameter list of xsym3, gives:
(This expression continues for several pages)
But it can be used. A bit further down you've plotted f3, I put xsym3 along:
.
Now I'll be the first to admit that working symbolically with square wave signals isn't ideal.
I've found no way to define a continuous square wave signal that is acceptable to the symbolic processor.
(With sines and cosines it's easy, they are already known symbolically).
You can define a continuous symmetric square wave as:
but when evaluated symbolically it becomes:
which leads to a Laplace transform of:
And that's not what we wanted.
A continuous asymmetric square wave can be defined with:
But:
So far the only square wave definition that works symbolically, and is Laplace transformable is one that is defined in terms of the Heaviside step functio:
But then you have to define every pulse individually, which can be automated using the summation operator:
but not out to infinity:
So square waves can be used symbolically with a limited number of cycles.
Here's a setup that works (at least in Mathcad 11):
(Again, the expressions above are huge).
Mathcad 11 file is attached.
Success!
Luc
P.S. Be advised that the symbolic process doesn't "know" about units. It doesn't "know" how to convert them etc. like the numeric processor does. At best, it treats them like unknown variables, and if you are careful you can have them cancel out if necessary.
If you want to work with symbolics and units, preferrably first use the symbolics to arrive at a function (with parameters for all your variables that need to carry units) that the numerical processor understands, then call that function with the united variables as parameters. So instead of:
(Note that you do not get t in seconds!)
do:
