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
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_yHere,
k_z = -1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}
. Therefore, this is in the bottom Riemann sheet. This root has been taken to make sure that k_z[/itex] remains purely imaginary for k_0^2 &lt;(k_x^2+k_y^2) , Imag(k_z)&lt;0 and purely real when k_0^2&lt;(k_x^2+k_y^2) , Real(k_z) &gt;0 . I tried only pulling the integral involving k_y.<br /> <script class="js-extraPhrases" type="application/json"> { "lightbox_close": "Close", "lightbox_next": "Next", "lightbox_previous": "Previous", "lightbox_error": "The requested content cannot be loaded. Please try again later.", "lightbox_start_slideshow": "Start slideshow", "lightbox_stop_slideshow": "Stop slideshow", "lightbox_full_screen": "Full screen", "lightbox_thumbnails": "Thumbnails", "lightbox_download": "Download", "lightbox_share": "Share", "lightbox_zoom": "Zoom", "lightbox_new_window": "New window", "lightbox_toggle_sidebar": "Toggle sidebar" } </script> <div class="bbImageWrapper js-lbImage" title="Untitled Diagram (9).png" data-src="https://www.physicsforums.com/attachments/untitled-diagram-9-png.262876/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/untitled-diagram-9-png.262876/" data-url="" class="bbImage" data-zoom-target="1" style="width: 278px" alt="Untitled Diagram (9).png" title="Untitled Diagram (9).png" width="1000" height="787" loading="lazy" decoding="async" /> </div><br /> <div style="text-align: center"><br /> 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}​</div><br /> 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 k_x point in the range of [−5k0,5k0] and performed the integral with respect tok_ywhen k_y varies as [−5k0,5k0]. And I have chosen the integration path as shown in below figure (Just imagine k_y instead of k_x ) here.<br /> <br /> [CODE lang="matlab" title="MATLAB code with the integration path on complex plane"]for i = 1:length(kx)<br /> <br /> kxi = kx(i);<br /> <br /> del = 0.01 .* k0;Y = @(ky) (k0^2 - kxi^2)./(-1j .* sqrt(-(k0.^2 - kxi.^2 - ky.^2))) .* (sinc(ky .* b./2/pi)).^2;<br /> <br /> y(i) = integral(Y, -50.*k0-1j*del,50.*k0+1j*del);% 'Waypoints', [(-1-1j).*del, (1+1j).*del]);<br /> <br /> end[/CODE]<br /> <br /> However, when in my problem, when (k_x,k_y)=(k0,0) 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<br /> k_x = k0. I need to have all points continuos for this integration so that I can perform the other integral over k_x. Am I doing something wrong?<div class="bbImageWrapper js-lbImage" title="Screen Shot 2020-05-16 at 12.55.02 PM.png" data-src="https://www.physicsforums.com/attachments/screen-shot-2020-05-16-at-12-55-02-pm-png.262875/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/screen-shot-2020-05-16-at-12-55-02-pm-png.262875/" data-url="" class="bbImage" data-zoom-target="1" style="width: 318px" alt="Screen Shot 2020-05-16 at 12.55.02 PM.png" title="Screen Shot 2020-05-16 at 12.55.02 PM.png" width="922" height="574" loading="lazy" decoding="async" /> </div>
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_yHere,
k_z = -1j \sqrt{(-(k_0^2 - k_x^2 - k_y^2))}
. Therefore, this is in the bottom Riemann sheet. This root has been taken to make sure that k_z[/itex] remains purely imaginary for k_0^2 &lt;(k_x^2+k_y^2) , Imag(k_z)&lt;0 and purely real when k_0^2&lt;(k_x^2+k_y^2) , Real(k_z) &gt;0 . I tried only pulling the integral involving k_y.<br /> <script class="js-extraPhrases" type="application/json"> { "lightbox_close": "Close", "lightbox_next": "Next", "lightbox_previous": "Previous", "lightbox_error": "The requested content cannot be loaded. Please try again later.", "lightbox_start_slideshow": "Start slideshow", "lightbox_stop_slideshow": "Stop slideshow", "lightbox_full_screen": "Full screen", "lightbox_thumbnails": "Thumbnails", "lightbox_download": "Download", "lightbox_share": "Share", "lightbox_zoom": "Zoom", "lightbox_new_window": "New window", "lightbox_toggle_sidebar": "Toggle sidebar" } </script> <div class="bbImageWrapper js-lbImage" title="Untitled Diagram (9).png" data-src="https://www.physicsforums.com/attachments/untitled-diagram-9-png.262876/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/untitled-diagram-9-png.262876/" data-url="" class="bbImage" data-zoom-target="1" style="width: 278px" alt="Untitled Diagram (9).png" title="Untitled Diagram (9).png" width="1000" height="787" loading="lazy" decoding="async" /> </div><br /> <div style="text-align: center"><br /> 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}​</div><br /> 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 k_x point in the range of [−5k0,5k0] and performed the integral with respect tok_ywhen k_y varies as [−5k0,5k0]. And I have chosen the integration path as shown in below figure (Just imagine k_y instead of k_x ) here.<br /> <br /> [CODE lang="matlab" title="MATLAB code with the integration path on complex plane"]for i = 1:length(kx)<br /> <br /> kxi = kx(i);<br /> <br /> del = 0.01 .* k0;Y = @(ky) (k0^2 - kxi^2)./(-1j .* sqrt(-(k0.^2 - kxi.^2 - ky.^2))) .* (sinc(ky .* b./2/pi)).^2;<br /> <br /> y(i) = integral(Y, -50.*k0-1j*del,50.*k0+1j*del);% 'Waypoints', [(-1-1j).*del, (1+1j).*del]);<br /> <br /> end[/CODE]<br /> <br /> However, when in my problem, when (k_x,k_y)=(k0,0) 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<br /> k_x = k0. I need to have all points continuos for this integration so that I can perform the other integral over k_x. Am I doing something wrong?<div class="bbImageWrapper js-lbImage" title="Screen Shot 2020-05-16 at 12.55.02 PM.png" data-src="https://www.physicsforums.com/attachments/screen-shot-2020-05-16-at-12-55-02-pm-png.262875/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/screen-shot-2020-05-16-at-12-55-02-pm-png.262875/" data-url="" class="bbImage" data-zoom-target="1" style="width: 318px" alt="Screen Shot 2020-05-16 at 12.55.02 PM.png" title="Screen Shot 2020-05-16 at 12.55.02 PM.png" width="922" height="574" loading="lazy" decoding="async" /> </div>
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