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Photonics

Photonics

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how to launch a known power into FDTD simulation

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    • Brian West
      Subscriber

      This question was addressed on a previous thread, now closed.  But I find it hard to believe that the solution proposed there is the only way of doing this...so I'm going to re-open the discussion

      I want to launch a known power into an FDTD simulation (through a port, but the issue is also relevant to a mode source).

      Currently, you can choose the amplitude, which refers to the peak value of the calculated input mode.

      The previous thread suggests that you use sourcepower.  But unless I'm mistaken, this requires that the user sets up a "fake" FDTD of the input plane plus a negligible distance in the propagation direction, measures the sourcepower from that, then deletes the fake FDTD and creates the actual FDTD, setting amplitude to P/sqrt(sp), where P is the desire power and sp is the sourcepower from the previous step.

      Is this actually the best way of doing it?  It seems needlessly complicated, especially if you're scanning or optimizing over something that changes the input mode.

      Hopefully I'm just misinterpreting the requirements.

      Thanks,

      Brian

    • Amrita Pati
      Ansys Employee

      Hi Brian,

      If you are doing a linear simulation, then the input power of the source will not affect the simulation as all the results in FDTD are normalized to the source power. So, the results will be the same regardless of the amplitude. If you just want to know the power injected by the source in each simulation of a sweep/optimization you can use the sourcepower script command.

      But as you mention, if you do need to change the amplitude of the source to have a desired amount of power (which is not required for linear simulation), then you would have to calculate the power of the source using the sourcepower script command or by integrating the Poynting vector, calculate the amplitude required to have the desired power, and then set the source amplitude as the calculated value.

      Regards,
      Amrita

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