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+//chapter16
+//example16.6
+//page347
+
+Av=100
+mv=0.1
+
+Avf=Av/(1+Av*mv)
+mv=(Av/Avf-1)/Av
+
+// fall in gain is 6dB so 20log(Av/Av1)=6
+// making Av1 as subject we get
+Av1=Av/exp(6*log(10)/20)
+Avf_new=Av1/(1+Av1*mv)
+change=100*(Avf-Avf_new)/Avf
+
+printf("percentage change in gain = %.3f percent \n",change)
+
+// the accurate answer is 8.297 percent but in book it is given as 8.36 percent