Use of integrator in Simulink

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Tan Edwin
Tan Edwin el 15 de Feb. de 2011
Respondida: Vineet Kumar el 1 de Sept. de 2016
Im trying to find the area under the graph (called it graph A) from simulink graph A looks something like this: <a><[IMG]http://img171.imageshack.us/img171/5214/83269046.th.png[/IMG][/URL]</a>> (Sry dunno how to show image. Graph A looks sthing like a normally distributed curve)
but when i use an integrator from simulink, the graph i obtained showed a negative valuefor the initial part. There is no negative value for graph A. So is there something wrong with my method?
  4 comentarios
Kaustubha Govind
Kaustubha Govind el 15 de Feb. de 2011
What happens when you set continuous sample time for your S-function: [0 0]
James Allison
James Allison el 16 de Feb. de 2011
In the Diagnostics>Data Validity pane of configuration parameters, try setting 'Underspecified initialization detection' to 'Simplified'.

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Vineet Kumar
Vineet Kumar el 1 de Sept. de 2016
The Integrator block outputs the value of the integral of its input signal with respect to time.
The Integrator Limited block is identical to the Integrator block with the exception that the output of the block is limited based on the upper and lower saturation limits. See Limiting the Integral for details.
Simulink® treats the Integrator block as a dynamic system with one state.

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