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Homework answers / question archive / EEEE1043 Coursework Task: For an isotropic point source located in the specific position as indicated in Fig
EEEE1043 Coursework
Task:
For an isotropic point source located in the specific position as indicated in Fig. 1, develop a general program that calculates the electromagnetic field from point p to point q. Assume the operation frequency is at 2.4 GHz. Plot the electric field vs distance (from point p to point q). You will need to use the reflection coefficient equation to calculate the reflected rays.
Fig. 1. Geometry of the corridor.
Mathematically, the electric field of direct ray is written as follows:
1 ? j?r ? ??
Ei ? e ? ? r ?4??
where r is the distance along the direct ray path, λ is the wavelength, ? is the wave
2? ? j?r ? number (also denoted as k is some literature,?? ), e is the phase factor, and is
a parameter related to the antenna effect.
Sample solution
% Sample Program clear all;
f = 2.4*10^9; % frequency (given) c = 3*10^8; % speed of light beta = 2*pi*f/c; % wave number
% to solve for the direct ray
tx = 0; ty = 4; tz = 2; % tx [x,y,z] location rx = 10; ry = 3; rz = 2; % rx [x,y,z] location rxv = rx:.01:22; % from p to q d = sqrt(((rxv-tx).^2)+((ry-ty)^2)+((rz-tz)^2)); % distance Ed = (1./d).*exp(-j*beta*d); % direct ray (line of sight) Edf = 20*log10(abs(Ed)); % direct ray alone plot(rxv,Edf) xlabel('Line Segment "pq"/m'),ylabel('E-field/dB')
Below is the marking scheme: -
Sections |
Poor (0-1) |
Reasonable (2) |
Excellent (3) |
Demonstrate own efforts in coding without copying |
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Accuracy of simulation results |
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The number of rays considered, convergence of results |
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Structure of the submitted file; Quality of figures
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1 mark maximum |
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Please download the answer file using this link
https://drive.google.com/file/d/1E19GZdoTTXuMaQ_mKLV4GXuXHjCvSzRh/view?usp=sharing