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Multiple frequency sources in comsol 4.1
Posted 2011/08/03 16:31 GMT-4 RF & Microwave Engineering, Modeling Tools & Definitions, Parameters, Variables, & Functions 7 Replies
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2D--RF-->EMW-->Frequency domain.
Goal: implement two sources (plane wave and a dipole) at different frequencies.
I am trying to excite a plane wave using a scattering boundary condition (orientation not important to the question; but in x direction). And a dipole source using a the provided feature of electric point dipole (moment in x direction)
I have tried ways to make them radiate at different frequencies but they both seem to pick the frequency value from the Study-->frequency domain; either the same value.
Implementing a different value of dipole:
Computed the problem once and studied the value of the weak form for the dipole (-test(Ex)*mu0_const*emw.iomega). The quantity related to frequency is Iomega and its units are rad/sec. I tried to define a new variable inside Model-->definitions-->Geometric entity level-->point (the dipole point source)
inside the tried to change the frequency using the variable freq.
Either freq = 5.3e14 or iomega=5.2e10 (any value). Comsol does not pick these values! and still generates the same frequency wave for each.
I understand that comsol converts all equations to their corresponding weak forms before it can solve by linear (or other) interpolation. But the only equation it displays for the dipole is the weak form which is crunched out as part of the solution. Changing a variable in it, or picking a variable from it and using it to manipulate frequency in a specific boundary or domain is a wrong approach?
Any alternate approaches?
Help!
Goal: implement two sources (plane wave and a dipole) at different frequencies.
I am trying to excite a plane wave using a scattering boundary condition (orientation not important to the question; but in x direction). And a dipole source using a the provided feature of electric point dipole (moment in x direction)
I have tried ways to make them radiate at different frequencies but they both seem to pick the frequency value from the Study-->frequency domain; either the same value.
Implementing a different value of dipole:
Computed the problem once and studied the value of the weak form for the dipole (-test(Ex)*mu0_const*emw.iomega). The quantity related to frequency is Iomega and its units are rad/sec. I tried to define a new variable inside Model-->definitions-->Geometric entity level-->point (the dipole point source)
inside the tried to change the frequency using the variable freq.
Either freq = 5.3e14 or iomega=5.2e10 (any value). Comsol does not pick these values! and still generates the same frequency wave for each.
I understand that comsol converts all equations to their corresponding weak forms before it can solve by linear (or other) interpolation. But the only equation it displays for the dipole is the weak form which is crunched out as part of the solution. Changing a variable in it, or picking a variable from it and using it to manipulate frequency in a specific boundary or domain is a wrong approach?
Any alternate approaches?
Help!
7 Replies Last Post 2014/10/06 0:23 GMT-4
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Posted:
1 decade ago
I think it is possible to use two frequencies only in transient problem.
In frequency domain is solution expected as E=Esolved*exp(i*omega*t) where omega is given by frequency.
In frequency domain is solution expected as E=Esolved*exp(i*omega*t) where omega is given by frequency.
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Posted:
1 decade ago
Hi Jakob,
Thanks for the prompt reply. Some more questions:
1-Why does Comsol not allow this in Frequency domain? is this a conceptual fault at my end? or limited functionality of the frequency domain?
2-"E=Esolved*exp(i*omega*t) where omega is given by frequency" Where did you get this equation from? and this is the solution to the dipole? by solution you mean the weak expression Comsol uses to simulate a dipole or the "master equation" of the dipole?
3-Have not noticed much difference in the dipole definition options in transient analysis relative to the frequency domain. Are you saying that what i tried to do in Frequency domain should work in the transient domain? why is it logically more correct in transient domain.
4-If its only about time: i tried adding time dependent steps in Studies. But i do not understand the logic of this implementation either. Not that it worked :)
More help please.
Thanks for the prompt reply. Some more questions:
1-Why does Comsol not allow this in Frequency domain? is this a conceptual fault at my end? or limited functionality of the frequency domain?
2-"E=Esolved*exp(i*omega*t) where omega is given by frequency" Where did you get this equation from? and this is the solution to the dipole? by solution you mean the weak expression Comsol uses to simulate a dipole or the "master equation" of the dipole?
3-Have not noticed much difference in the dipole definition options in transient analysis relative to the frequency domain. Are you saying that what i tried to do in Frequency domain should work in the transient domain? why is it logically more correct in transient domain.
4-If its only about time: i tried adding time dependent steps in Studies. But i do not understand the logic of this implementation either. Not that it worked :)
More help please.
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Posted:
1 decade ago
1- If you express solution E=Esolve(x,y,z)*exp(i*omega*t) and insert it into Maxwell equations all time derivatives can be then expressed as dE/dt=Esolve(x,y,z)*exp(i*omega*t)*i*omega=i*omega*E
So you eliminate time from equations and solve only for Esolve(x,y,z). But you can use only one frequency omega!
2- I do not know what you mean by dipole, it radiates to the different directions with different intensity, so I do not know how you mean to simulate it. If you mean by dipole sipmly time dependent potential on some conductor, define it as function f(x,y,z,t) and describe intensity on some surface.
4-if you have different sources with different omegas you must use time dependent solution
Read RF module manual
Look at examples
So you eliminate time from equations and solve only for Esolve(x,y,z). But you can use only one frequency omega!
2- I do not know what you mean by dipole, it radiates to the different directions with different intensity, so I do not know how you mean to simulate it. If you mean by dipole sipmly time dependent potential on some conductor, define it as function f(x,y,z,t) and describe intensity on some surface.
4-if you have different sources with different omegas you must use time dependent solution
Read RF module manual
Look at examples
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Posted:
1 decade ago
Hello,
I mentioned:
"And a dipole source using a the provided feature of electric point dipole (moment in x direction)" You can check comsol 4 for its e field/intensity patterns in 2 and 3d."
Incase you are implying that if we take out the time dependence from the equation, frequency domain, then the angular frequency is the only variable we have to implement wave progression?
My question was: can a plane wave generating boundary condition. (comsol feature) and an electric point dipole (comsol feature); two different sources of EMW, can have two different omegas in Freq domain? or you mean there can be only one omega for the entire model?
The equation you mentioned (the complex electric field solution) is something i cannot see in the defined electric point dipole feature of comsol 4.1
Perhaps you are refering to comsol 3.5?
I mentioned:
"And a dipole source using a the provided feature of electric point dipole (moment in x direction)" You can check comsol 4 for its e field/intensity patterns in 2 and 3d."
Incase you are implying that if we take out the time dependence from the equation, frequency domain, then the angular frequency is the only variable we have to implement wave progression?
My question was: can a plane wave generating boundary condition. (comsol feature) and an electric point dipole (comsol feature); two different sources of EMW, can have two different omegas in Freq domain? or you mean there can be only one omega for the entire model?
The equation you mentioned (the complex electric field solution) is something i cannot see in the defined electric point dipole feature of comsol 4.1
Perhaps you are refering to comsol 3.5?
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Posted:
1 decade ago
Hello,
I know the thread has more then 2 years, but since it still is unanswered and it was one of the first results google gave me, I decided to post here, sorry if I am breaking any forum rules.
If I got the question right, I think you could use conditional expressions and the variable "freq" to implement multiple sources at different frequencies.
If the RF module work the same as the acoustics module, you probably have a text box where you could insert the amplitude of your source. In this box you could simply use something like:
freq*(freq>=600)
With this your source amplitude would be linearly dependant to the frequency for frequencies of 600Hz or more, being zero for any frequency smaller then 600Hz.
Hope this help.
Cheers
I know the thread has more then 2 years, but since it still is unanswered and it was one of the first results google gave me, I decided to post here, sorry if I am breaking any forum rules.
If I got the question right, I think you could use conditional expressions and the variable "freq" to implement multiple sources at different frequencies.
If the RF module work the same as the acoustics module, you probably have a text box where you could insert the amplitude of your source. In this box you could simply use something like:
freq*(freq>=600)
With this your source amplitude would be linearly dependant to the frequency for frequencies of 600Hz or more, being zero for any frequency smaller then 600Hz.
Hope this help.
Cheers
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Posted:
1 decade ago
Hello,
I know the thread has more then 2 years, but since it still is unanswered and it was one of the first results google gave me, I decided to post here, sorry if I am breaking any forum rules.
If I got the question right, I think you could use conditional expressions and the variable "freq" to implement multiple sources at different frequencies.
If the RF module work the same as the acoustics module, you probably have a text box where you could insert the amplitude of your source. In this box you could simply use something like:
freq*(freq>=600)
With this your source amplitude would be linearly dependant to the frequency for frequencies of 600Hz or more, being zero for any frequency smaller then 600Hz.
Hope this help.
Cheers
Hi Gregorio,
Could you please elaborate on this? If I wanted to have a source at 800 Hz and another at 1500 Hz, how would I implement this?
Best,
Jerry
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Posted:
1 decade ago
Hi Gregorio,
Could you please elaborate on this? If I wanted to have a source at 800 Hz and another at 1500 Hz, how would I implement this?
Best,
Jerry
Hi Jerry,
I belive if you wanted to do something like two Dirac (with amplitude = 1) deltas, one in 800Hz and another in 1500Hz you could input the following in the Amplitude box:
1*(freq ==800)+1*(freq==1500)
Not sure if double or single equals, but with this, for every frequency appart from 800 and 1500 the values of the expressions inside the parentheses would be equal zero. But usually discontinuities like this are not solver friendly, maybe you could use a gaussian pulse, then your input would me something like (for gaussian):
1/sqrt(2*pi)*exp(-(freq-800).^2/(2*1^2))+1/sqrt(2*pi)*exp(-(freq-1500).^2/(2*1^2))
Basically you can input any mathematical equation in the amplitude box.
Hope this helps (and works out)
Greg