October 19, 2023 at 11:55 amSangmyeong YuSubscriber
The primitive plane wave has only one pulse, but I want to make multi pulse plane wave. I searched the FDTD manual and found this code. However, since this code is a Gaussian pulse, I want to creat a sine-shaped plane wave. I tried modifying the code, but I don't know why it doesn't work or how to make it work.
October 19, 2023 at 7:42 pmGuilin SunAnsys Employee
In your case you can comment out all the lines for saving the file. This file can directly assign the time signal to the source specified.
In addition, only in certain non-linear simulations you need sine wave. In linear cases you can still use the pulse but only set the monitor to record the center frequency with one point. This will be faster and accurate. However if you use sine wave you can get the time domain result accurately but not in the frequency domain, since the frequency domain result is obtained from Fourier transform, which requires the signal to decay to zero. Using a filter will modify the amplitude so the result amplitude is not accurate.
October 20, 2023 at 8:41 amSangmyeong YuSubscriber
Additionally, when using a plane wave of 600 nm to 1000 mn as a source, I am curious whether all wavelengths in that range are simulated or only the short wavelength corresponding to the intermediate value is simulated.
October 20, 2023 at 4:07 pmGuilin SunAnsys Employee
In such case you will use a pulse, which has broader spectrum than 600-1000nm. The algorithm Lumerical developes for the pulse is to guarantee the specified wavelength range such as 600-1000nm to have accurate Fourier transform (this is how the frequency domain result is obtaied) since they have higher amplitude in the source spectrum.
So to answer your question, all the spectrum in the pulse is simulated, but only the specified band in the monitor (the default is the same as the specified source bandwidth 600-1000nm) are recorded with specified frequency points. You can modify the monitor settings to specify range to record, but as mentioned above, the wavelengths outside the specified wavelegth range in the source may not be accurate.
Please write a new post if you have further questions since this post may not be monitored from today.
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