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1-Visitor
March 9, 2020
Question

Error in plotting invlaplace

  • March 9, 2020
  • 3 replies
  • 2898 views

Hi,

I encountered an error when I tried to plot a graph including arguments (t and Δ(t)). The error says:This function has the form:f(f(Unitless)->Unitless)->Unitless, but must have the form:f(any1)->any2. I have attached the file. Can anyone help me out? Thanks a lot.

3 replies

23-Emerald I
March 9, 2020

Your function i(t,delta) is the problem--you don't have a variable delta, delta is required to be a function of time

Capture.JPG

Can you evaluate this function for any single set of parameters?  i(1,0.01)=

1-Visitor
March 9, 2020

Hi, 

 

Thank you for your reply. I was suspecting the error was caused by the lack of definition as well. But I was not too sure exactly the delta(t) function was representing since it is derived from the inverse Laplace. Any ideas?

23-Emerald IV
March 9, 2020

Are you sure your component values are OK? R1 and R4 are two extremely small resistors values...

Anyway, I threw the delta function out:

LucMeekes_0-1583758370603.png

Now the error is an overflow.

 

Success!
Luc

1-Visitor
March 9, 2020

Hi Luc,

 

Thank you so much for your reply. Before I posted this thread, I was suspecting that the error was caused by the lack of definition of the function. But I am not so sure what exactly the delta(t) function represents. May I ask why you substituted delta(t) to 0? Thanks again.

23-Emerald IV
March 9, 2020

Symbolically evaluate the delta function for a couple of parameter values, and observe the results. Make sure one of the values is 0...

LucMeekes_0-1583774921202.png

For more information: https://en.wikipedia.org/wiki/Dirac_delta_function

 

Success!
Luc

21-Topaz II
March 10, 2020

Hi,

I think there is an error in the analysis of the circuit. However the dimensions are not correct. I attach the file with the corrections, which I think needs to be done:

answer5.jpg

the Dirac impulse in the origin has been left out since the graph starts in an instant t> 0.

answer55.jpg

1-Visitor
March 10, 2020

Hi,

 

Thank you so much for your reply. I really appreciate it. Will have a look at the circuit.