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Admittance for a piezoelectric transducer in Comsol 3.5a
Posted 30.11.2011, 13:22 GMT-5 Low-Frequency Electromagnetics, MEMS & Nanotechnology, Piezoelectric Devices Version 3.5a 1 Reply
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Hello,
I am performing a frequency response analysis on a 2D piezoelectric transducer in Comsol 3.5a, and I would like to plot the magnitude and phase of the admittance (relative to voltage). Following the "composite piezoelectric transducer" tutorial, the magnitude of the current is given by integrating -imag(nJ) over the boundary of interest; dividing that quantity by V gives the admittance. I am a bit confused about this definition - why is the imaginary part of current density integrated over the boundary instead of the real part?
It was posted elsewhere that the phase(A) = atan2(imag(A), real(A))*180/pi. Intuitively, to get the phase of current, I would think that "A" would be the integration of nJ over the boundary of interest. Is this correct? Also, what is this phase relative to?
In summary, my questions are:
1. Why is the magnitude of current = integrating the imaginary part of nJ?
2. How would I plot the phase of the admittance relative to the voltage?
Any help would be appreciated. Thanks.
I am performing a frequency response analysis on a 2D piezoelectric transducer in Comsol 3.5a, and I would like to plot the magnitude and phase of the admittance (relative to voltage). Following the "composite piezoelectric transducer" tutorial, the magnitude of the current is given by integrating -imag(nJ) over the boundary of interest; dividing that quantity by V gives the admittance. I am a bit confused about this definition - why is the imaginary part of current density integrated over the boundary instead of the real part?
It was posted elsewhere that the phase(A) = atan2(imag(A), real(A))*180/pi. Intuitively, to get the phase of current, I would think that "A" would be the integration of nJ over the boundary of interest. Is this correct? Also, what is this phase relative to?
In summary, my questions are:
1. Why is the magnitude of current = integrating the imaginary part of nJ?
2. How would I plot the phase of the admittance relative to the voltage?
Any help would be appreciated. Thanks.
1 Reply Last Post 07.12.2011, 19:53 GMT-5