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//chapter 14
//example 14.7
//page 302
Ao_db=70 // db
Av_db=67 // db
Rout=1.5 // kilo ohm
// since 20*log(Ao)-20*log(Av)=Ao_db-Av_db we get
// 20*log(Ao/Av) = Ao_db-Av_db so
// Ao/Av = 10^((Ao_db-Av_db)/20)
// and also Ao/Av=1+Rout/Rl since Av/Ao=Rl/(Rl+Rout)
// so making Rl as subject we get
Rl=Rout/(10^((Ao_db-Av_db)/20)-1)
printf("minimum value of load resistance = %.3f kilo ohm \n",Rl)
// the accurate answer is 3.636 kilo ohm
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