- #1
tworitdash
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- TL;DR Summary
- I am trying to perform an integration for mutual admittance of two modes in a rectangular waveguide. The integral is in the spectral domain ([itex](k_x, k_y) [/itex]). However, it is very difficult to make it as a form of a single integral. Therefore, I tried doing the first one integral with respect to ([itex]k_y[/itex]) for each [itex]k_x[/itex]. However, I see a null value at the points where [itex]k_x = k_0[/itex]. I choose an integration path that doesn't contain the poles on the re axis.
The integral looks like
[tex] Y_{mut, mn} = -j^{m+n}nm \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{2 ab (k^2 - k_x^2) \sin^2(\frac{k_yb}{2}) \cos^2(\frac{k_xa}{2})}{\omega \mu k_z (\frac{k_yb}{2})^2 [(n\pi)^2 - (k_xa)^2][(m\pi)^2 - (k_xa)^2]} dk_x dk_y [/tex]Here,
[tex] k_z = -1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}[/tex]
. Therefore, this is in the bottom Riemann sheet. This root has been taken to make sure that [tex]k_z[/itex] remains purely imaginary for [itex] k_0^2 <(k_x^2+k_y^2) , Imag(k_z)<0 [/itex] and purely real when [itex] k_0^2<(k_x^2+k_y^2) , Real(k_z) >0 [/itex] . I tried only pulling the integral involving [itex]k_y[/itex].
[tex] D(k_x) = \int_{-\infty}^{\infty} \frac{ (k^2 - k_x^2) \sin^2(\frac{k_yb}{2})} {(-1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}) (\frac{k_yb}{2})^2} [/tex]
I couldn't get a closed-form integral for this one. If someone happens to know, please let me know. Coming to numerical integral, I performed a numerical integral for each [itex]k_x[/itex] point in the range of [−5k0,5k0] and performed the integral with respect to[itex]k_y[/itex]when [itex]k_y[/itex] varies as [−5k0,5k0]. And I have chosen the integration path as shown in below figure (Just imagine [itex]k_y[/itex] instead of [itex]k_x[/itex] ) here.
However, when in my problem, when [itex] (k_x,k_y)=(k0,0)[/itex] the function has a 0/0 form and this is on the integration path. Therefore, I get a discontinuity there. Basically, it can't avoid the pole when
[itex] k_x = k0[/itex]. I need to have all points continuos for this integration so that I can perform the other integral over [itex]k_x[/itex]. Am I doing something wrong?
[tex] Y_{mut, mn} = -j^{m+n}nm \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \frac{2 ab (k^2 - k_x^2) \sin^2(\frac{k_yb}{2}) \cos^2(\frac{k_xa}{2})}{\omega \mu k_z (\frac{k_yb}{2})^2 [(n\pi)^2 - (k_xa)^2][(m\pi)^2 - (k_xa)^2]} dk_x dk_y [/tex]Here,
[tex] k_z = -1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}[/tex]
. Therefore, this is in the bottom Riemann sheet. This root has been taken to make sure that [tex]k_z[/itex] remains purely imaginary for [itex] k_0^2 <(k_x^2+k_y^2) , Imag(k_z)<0 [/itex] and purely real when [itex] k_0^2<(k_x^2+k_y^2) , Real(k_z) >0 [/itex] . I tried only pulling the integral involving [itex]k_y[/itex].
[tex] D(k_x) = \int_{-\infty}^{\infty} \frac{ (k^2 - k_x^2) \sin^2(\frac{k_yb}{2})} {(-1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}) (\frac{k_yb}{2})^2} [/tex]
I couldn't get a closed-form integral for this one. If someone happens to know, please let me know. Coming to numerical integral, I performed a numerical integral for each [itex]k_x[/itex] point in the range of [−5k0,5k0] and performed the integral with respect to[itex]k_y[/itex]when [itex]k_y[/itex] varies as [−5k0,5k0]. And I have chosen the integration path as shown in below figure (Just imagine [itex]k_y[/itex] instead of [itex]k_x[/itex] ) here.
MATLAB code with the integration path on complex plane:
for i = 1:length(kx)
kxi = kx(i);
del = 0.01 .* k0;Y = @(ky) (k0^2 - kxi^2)./(-1j .* sqrt(-(k0.^2 - kxi.^2 - ky.^2))) .* (sinc(ky .* b./2/pi)).^2;
y(i) = integral(Y, -50.*k0-1j*del,50.*k0+1j*del);% 'Waypoints', [(-1-1j).*del, (1+1j).*del]);
end
However, when in my problem, when [itex] (k_x,k_y)=(k0,0)[/itex] the function has a 0/0 form and this is on the integration path. Therefore, I get a discontinuity there. Basically, it can't avoid the pole when
[itex] k_x = k0[/itex]. I need to have all points continuos for this integration so that I can perform the other integral over [itex]k_x[/itex]. Am I doing something wrong?
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