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+//ANALOG AND DIGITAL COMMUNICATION
+//BY Dr.SANJAY SHARMA
+//CHAPTER 6
+//NOISE
+clear all;
+clc;
+printf("EXAMPLE 6.14(PAGENO 308)");
+
+//given
+A = 60//gain of noiseless amplifier
+V_n1 = 1*10^-3//output of the amplifier
+B = 20*10^3//initial bandwidth
+B1 = 5*10^3//change in bandwidth
+k = 1.38*10^-23//boltzman's constant
+T = 273 + 80//temperature in degree kelvin
+
+//calculaitons
+//since the bandwidth is reesuced to 1/4th of its value,therefore the noise voltage 
+//will be V_n proportional to sqrt(B)
+//Hence, the noise voltage at 5KHz will become half its value at 20KHz bandwidth i.e,
+V_n = .5*10^-3//noise voltage in volts
+V_no = V_n1/A;//noise ouput voltage 
+R = (V_no^2/(4*k * T * B ));//resistance at 80degree celcius
+
+//results
+printf("\n\ni.Meter reading in volts = %.10f V",V_n);
+printf("\n\nii.Resistance at 80 degree celcius = %.2f ohms",R);
+printf("\n\nNote: There is calculation mistake in textbook in the measurement of resistance they took constant in formula as 1 instead of 4");