ACCEPTED SOLUTION
Accepted Solutions
@SU_10542991 wrote:
Thank you for the response. I have checked Vo(0.3) and it is showing error. I understood that the limit operator (n-->infinity) is the issue. Is there any way to resolve this issue?
Actually the limit operator is also not helpful if you try to symbolically evaluate Vo(0.3) as the result of Primes symbolic is just kind of a building plan how to derive the invlaplace transformed if, yes if we could simplify that derivation limit.
It just means that Prime is not capable enough to give you the answer you are asking for.
I tried with the legacy symbolic engine which is available up to Prime 6 and it wasn't able to come up with a solution function either
(but it shows that a bit more clearly that its not capable enough):
So at the moment I see no way to derive the output voltage using invlaplace with Prime.
@SU_10542991
If it can help something or if further others colleagues come with other ideas starting from here::
Mathcad Prime 7 file attached.
Hmmm, the argument for sinh in your V.o2 is so large that for every value of t (other than t=0) we have an overflow (value greater than 10^308). Thats the reason the plot consists of just one point at the origin.
I am not sure if linearizing the exponential functions give a meaningful result, but adding a "simplify" will at least make the whole thing plot:
If we use series, 4 approximation (which seems to be a good approximation for e^-0.1*s and e^-0.2*s) then te result is:
Mathcad Prime 7 file attached.
Thank you for the response @Cornel. I tried using your methodology and it is working, the only issue was that as @Werner_E said in the below response that 10^308 error was showing. So, I used the simplify symbolic also, and then it is working. The same methodology I tried to take the inverse laplace transform of Vi(s), but the following is the error showing up: I don't understand why is this "s" term coming in the inverse laplace transform term. What does this delta(t) mean?
As @Werner_E wrote, it's probably about the fact that you don't have the same label for every s in your laplace equation. Some of the s in your equation have the label "Automatic", while others are likely labeled "Variable". But you must have the same label for all s of your equation (whichever label you want, you choose the label you want to use) to receive a correct answer. Always put a worksheet of Mathcad.
And I think the most likely correct answer to your problem is that in which you use series, 4 approximation because series, 4 approximation well approximates those exponential function, instead of series, 2 approximation.
@Werner_E @LucMeekes @ttokoro
I think the problem with the fact that Vi(t) is not displayed, meaning that error occurs: "This variable is undefined. Check that the label is set correctly", is I think because we have to define that delta(t). How can we define this delta(t)?
Mathcad Prime 7 file attached.
Thank you.
@Werner_E @ttokoro @LucMeekes @SU_10542991
Case 1: Using series, 2 approximation:
Case 2: Using series, 3 approximation:
This is because of:
Case 2: Using series, 4 approximation:
This is because of:
And in the output we also can see:
So, Vi(s), Vi(t) is an unstable input as it is.
Only if we approximate those e^-0.1*s and e^-0.2*s with series, 2 approximation (so, with a linear expression) - series, 3 and series, 4 approximation are nonlinear expression - then Vi(s), Vi(t) and Vo2(s), Vo2(t) become a stable input and stable output.
Mathcad Prime 7 file attached.
Hi @LucMeekes
I tried transforming the input signal Vi(s) alone to Vi(t). But it is showing error in plotting. I don't understand why while taking inverse laplace transform, "s" term is coming. And what does this delta(t) mean? Can you please correct me where I am wrong?
For example, we can see that in the case of series, 2 approximation, in Vo2(t) we receive a term that has staircase shape.
Maybe input signal is infinity number of triangle waves shown in green plot but that only shows 5 waves.
But cannot get Vout(t).
The Laplace transform of a repetitive waveform with a period of T is
where, V1 (s) is the Laplace transform of a one-period waveform.
Unfortunately, it cannot inverse transform to t domain waveform. Therefore, we must use finite number of Laplace transform of voltage waveform.
Left plot shows the transient current response of RL Circuit. (R=1, L=1)
Right plot shows RCRC circuit response of all is unit value.
In this case Mathcad can solve the answer. (In these plots, I use Maple2022 because it is faster than Mathcad.)
But using your R and C values, it shows nothing because C and R are so small and immediately charged to the applied voltage.
I think the time constant is so small and numerical calculation is failed.
I think the time constant is so small and numerical calculation is failed.
Output voltage of the system.
After about 3 microsecond, output signal is as same as input voltage of infinity triangle train.
Prime 8 sheet attached.
Hi @ttokoro Thank you for the response. I am unable to open this .mcdx file that you shared. It is saying that the file is not compatible with the current version. I am using Mathcad prime 7. I observed that the answers that you are getting by taking the invplaplce is not obtained when I am trying to do. This is what I am getting:
Is this difference because you are using a different version mathcad?
