4 - P In = 2 Vin R In 1.6-l I I iout ~-r Amplifier I v out...

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, 2 P, = Vin In R. In Amplifier ( 1.6) ,-- - - - - --l I I I i out I ~---r--'-" + I v out I I I I I I Load I '- --1 FIGURE 1.4 An illustration of the definition and use of the decibel (dB).
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The power delivered to the load is 2 P = V out out R L ( 1.7) The power gain of the amplifier is P G · Pout ·ower am -- Pin V;ut tc; --- vfn RL (1.8 ) The power gain expressed in decibels is defined as Power Gain dB - 10 loglo (P~ut) Pin (1.9 )
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G ' Vout Voltage am = - Vin ( 1.10) G ' lout Current am = -,- lin (1,11) In dB these are defined as Voltage Gain dB = 20 loglo (V~ut) v in (1.12) Current Gain dB - 20 loglo (i~ut) . 1 m (1.13) ( ) 2 , V out Power Gam = - Vin R L = tc; 0.14) In dB the power gain becomes, for Ri~ = R L , ( 1.1
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In dB the power gain becomes, for Ri~ = R L , ( 1.1 from which we obtain the defining relation for expressing the voltage gain dB as in (1.12), In summary, the ratio of two quantities in dB is given by dB = 10 log,o (;:) (power) (1.1 dB = 20 loglO C:) (voltage) ( 1.1' dB - 20 log,o C:) (current) ( 1.1:
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4 - P In = 2 Vin R In 1.6-l I I iout ~-r Amplifier I v out...

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