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+//ANALOG AND DIGITAL COMMUNICATION
+//BY Dr.SANJAY SHARMA
+//CHAPTER 6
+//NOISE
+clear all;
+clc;
+printf("EXAMPLE 6.4(PAGENO 283)");
+
+//given
+A_1 = 10//voltage gain for first stage
+A_2 = 25//volatage gain for second stage
+R_i1 = 600//input resistance for first stage in ohms
+R_eq1 = 1600//equivalent noise resistance for first stage
+R_01 = 27*10^3//Output resistance for first stage
+R_i2 = 81*10^3//input resistance for second stage
+R_eq2 = 10*10^3//Equivalent noise resistance for second stage
+R_02 = 1*10^6//putput resistance for second case
+
+//calculations
+R_1 = R_i1 + R_eq1
+R_2 = ((R_01*R_i2)/(R_01+R_i2)) + R_eq2
+R_3 = R_02
+R_eq = R_1 + (R_2/A_1^2) + R_3/(A_1^2 *A_2^2);
+
+//results
+printf("\n\nEquivalent input noise resistance = %.2f Ohms",R_eq);