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+// Example no 4.9
+// To find a)the minimum mean square error b)the standard deviation about mean value c)received power at d=2 km d)the likelihood that the received signal level at 2 km e) the percentage of  area within 2 km 
+// Page no. 143
+
+clc;
+clear all;
+
+// Given data
+d0=100;                                        // First receiver distance in meter
+d1=200;                                        // Second receiver distance in meter
+d2=1000;                                       // Third receiver distance in meter
+d3=3000;                                       // Fourth receiver distance in meter
+p0=0;                                          // Receved power of first receiver in dBm
+p1=-20;                                        // Receved power of second receiver in dBm
+p2=-35;                                        // Receved power of third receiver in dBm
+p3=-70;                                        // Receved power of forth receiver in dBm
+
+// a)To find the minimum mean square error
+n=2887.8/654.306;                              // Loss exponent after differentiating and equating the squared error function with zero
+
+// Displaying the result in command window
+printf('\n Loss exponent = %0.0f',n);
+
+// b)To find the standard deviation about mean value
+P0=-10*n*log10(d0/100);                        // The estimate of p0 with path loss model
+P1=-10*n*log10(d1/100);                        // The estimate of p1 with path loss model
+P2=-10*n*log10(d2/100);                        // The estimate of p2 with path loss model
+P3=-10*n*log10(d3/100);                        // The estimate of p3 with path loss model
+J=(p0-P0)^2+(p1-P1)^2+(p2-P2)^2+(p3-P3)^2;     // Sum of squared error
+SD=sqrt(J/4);                                  // The standard deviation about mean value
+
+// Displaying the result in command window
+printf('\n The standard deviation about mean value = %0.2f dB',SD);
+// The decimal point is not given in the answer given in book.
+
+// c)To find received power at d=2 km
+d=2000;                                         // The distance of receiver
+P=-10*n*log10(d/100);                           // The estimate of p2 with path loss model
+
+// Displaying the result in command window
+printf('\n The received power (at d=2 km) = %0.2f dBm',P);
+// Answer is varying due to round off error
+
+// d)To find the likelihood that the received signal level at 2 km
+gam=-60;                                         // The received power at 2km will be greater than this power
+z=(gam-P)/SD;
+Pr=(1/2)*(1-erf(z/sqrt(2)));                     // The probability that received signal will be greater than -60dBm
+
+// Displaying the result in command window
+printf('\n The probability that received signal will be greater than -60dBm = %0.1f percent',Pr*100);
+// Answer is varying due to round off error
+
+// e)To find the percentage of  area within 2 km 
+A=92;                                            // From figure 4.18, area receives coverage above -60dBm
+
+// Displaying the result in command window
+printf('\n The percentage of  area within 2 km = %0.0f percent',A);
+
+