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authorprashantsinalkar2017-10-10 12:27:19 +0530
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+// Example no 4.8
+// To determine a)the loss due to knife-edge diffraction b)the heigth of obstacle required to induce 6dB diffraction loss
+// Page no. 133
+
+clc;
+clear;
+
+// Given data
+f=900*10^6; // Operating frequency in Hz
+c=3*10^8; // Speed of ligth in m/s
+hr=25; // Heigth of receiver in m
+ht=50; // Heigth of transmitter in m
+h=100; // Heigth of obstruction in m
+d1=10*10^3; // Distance between transmitter and obstruction in m
+d2=2*10^3; // Distance between receiver and obstruction in m
+
+// a)To determine the loss due to knife-edge diffraction
+lambda=c/f; // Operating wavelength in m
+ht=ht-hr; // Hegth of transmitter after subtracting smallest heigth (hr)
+h=h-hr; // Heigth of obstruction after subtracting smallest heigth (hr)
+bet=atan((h-ht)/d1); // From geometry of environment in rad
+gamm=atan(h/d2); // From geometry of environment in rad
+alpha=bet+gamm; // From geometry of environment in rad
+v=alpha*sqrt((2*d1*d2)/(lambda*(d1+d2))); // Fresnel diffraction parameter
+
+// the loss due to knife-edge diffraction
+Gd=-20*log10(0.225/v); // Diffraction loss for v>2.4 in dB
+
+// Displaying the result in command window
+printf('\n The loss due to knife-edge diffraction = %0.1f dB',Gd);
+
+// b)To determine the heigth of obstacle required to induce 6dB diffraction loss
+Gd=6; // Diffraction loss in dB
+v=0; // Fresnel diffraction parameter from the plot of Knife-edge diffraction gain as a function of Fresnel diffraction parameter
+// v=0 is possible only if alpha=0. Therefore bet=-gamm
+// By considering this situation, the geometry of environment provides (h/d2)=(ht/(d1+d2))
+h=(ht*d2)/(d1+d2); // the heigth of obstacle required to induce 6dB diffraction loss
+
+// Displaying the result in command window
+printf('\n The heigth of obstacle required to induce 6dB diffraction loss = %0.2f m',h);
+