Hi,
How to show all the values of ic1(t). If I put iC1(1ms) for example the calculation is done, but I want to have a vector of all the values of iC1(t). Is it possible?
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Here you are:
As you can see the evaluation fails for the first and the last value of t.
You are using numerical differentiation for a function derived with "odesolve" for arguments from 0s to t.end
The algorithm used by Prime for this seems to use the function values of arguments before and after the point in question. But in the result of a 'function' determined by "odesolve" there are no function values for arguments before the first initial value of t (0 s) or after the last value (t.end). So the numerical derivative can't be calculated for these end values.
The evaluation should be possible without this problem using the function i.C1(t), which, as I have shown, is determined together with the others in the solution block.
Here you are:
As you can see the evaluation fails for the first and the last value of t.
You are using numerical differentiation for a function derived with "odesolve" for arguments from 0s to t.end
The algorithm used by Prime for this seems to use the function values of arguments before and after the point in question. But in the result of a 'function' determined by "odesolve" there are no function values for arguments before the first initial value of t (0 s) or after the last value (t.end). So the numerical derivative can't be calculated for these end values.
The evaluation should be possible without this problem using the function i.C1(t), which, as I have shown, is determined together with the others in the solution block.
@AlfredFlaßhaar wrote:
I would like to complete this task in the classic way. To do this, I need the function n vE(t) so that I can use it in MC14. There is a sentimental reason for my request: I remember my student days when easy exercises of this kind were stressful.
vE(t) is a simple square wave signal
similar to the one posted here Why these 2 function implementation of square wave...
You also may use one of the alternatives I provided there (Tf = 1 ms, Vmax = 10 V) or if its more convenient for you, you may also use a definition using the Heaviside function:
But I guess that the periodic nature of the function vE will make deriving a closed form symbolic solution using the "classic way' to solve this 'easy exercise' somewhat difficult.
Good luck!
@AlfredFlaßhaar wrote:
Thank you very much! Yes, now it gets much more difficult. The inhomogeneity is a piecewise continuous periodic function. Let's see if a Fourier series for vE(t) helps?
You sure can get an approximation that way.
Best wishes for your efforts!
You are using boolean equals so you can't expect them to have any effect.
If your question is about the two boolean expressions you try to evaluate symbolically (why?) and which seems to take an infinite time - this is because you defined t as a range and the symbolics tries to evaluate your expressions using this implicit t-loop.
You can "undefine" t with respect to the symbolics by writing t:=t and then you get a result
But because you never defined the Wronski determinant (there is a difference between using := and the fat boolean = ) you just see W(t) in the results.
I also doubt that an approximating 'solution' using just a few summands of the Fourier series would be any better than the solution derived numerically by using odesolve. But then I may be proven wrong.
Here you are.
I made n a global variable so I can define it at the end.
Here the result for n=5. Feel free to set higher values if you dare 😉
The symbolic results (large sums) are so large that Mathcad refuses to display them, but nonetheless you can use them in further calculations, plots, etc.
I also changed vhom(t) to vhom1(t), because you never have defined a function vhom in your sheet.
Its up to @Cornel to decide if an approximation of that kind is helpful for him.
