I'm trying to make a template that can be used for circuit analysing using Laplace transform.
Here, we have a second order RC lowpass filter and I want to use an repetitive input signal.
In V2(s) equation, where we have the input voltage I would like to use the f(s) input function or even better g(s) repetitive input function.
In the file attached you can see that I have 3 plots: one with input voltage as a constant, where it works and the plot is ok, and another two using f(s) and g(s), where it returns this error "Replace complex values and NaNs by real numbers".
What can be improved in order to use the repetitive s domain function?
Mathcad using is Prime 4 and you cand find attached a .pdf too.
Try to evaluate V.2_f for, lets say, t=10^-5, numerically!
You run into an overflow error - the calculation of a subexpression yields a result greater than the numerical limit of 10^307 and throws an error.
If i rewrite your input signal, when I try to plot returns the following error: "This value must be a real number"
LE: I solved this step, I will go on and return with news.
It is not necessary to rewrite the whole worksheet. It is sufficient and very fast (a few seconds) to import the file into Prime. Then you have to do some adaptation.
If you've already done it, did you import the file in Prime correctly?
(Vsqw is dimensionless while V.sqw isn't.)
Hello again -MF-,
Sorry all for my slow reply.
Yes I imported as you shown but it does't have the same result. See picture attached
You have to check that the labels are the correct ones.
In any case, I attach the mcdx file.
In the graph there is a small anomaly (circled in red) to be corrected.
If C=2*10^-6 then tau is 10000 times faster than my result as already Werner_E wrote.
My results change the C to 2*10^-2, therefore, transient change in output signal is visible. However, Prime 5 of my devise only solve one or two input series of inverse Laplace transformation. It can also solve numerically three input pulse series. Therefore, I made the 3rd response as a series of rest 7 pulse responses.
Because the 1st to 2nd responses after the 3rd response is too small to affect the results in my graph scales.