An amplifier has a gain of Av(mid)=100. At high cutoff frequency point, the gain will * rise to a value depends on the design of the amplifier rise to 141.44 rise to 200 reduce to a value depends on...


An amplifier has a gain of<br>Av(mid)=100. At high cutoff<br>frequency point, the gain will *<br>rise to a value depends on the design<br>of the amplifier<br>rise to 141.44<br>rise to 200<br>reduce to a value depends on the<br>design of the amplifier<br>reduce to 70.7<br>change (reduce or rise) to a value<br>O depends on the design of the<br>amplifiera<br>reduce to 0.707<br>be kept constant<br>reduce to 50<br>An amplifier has fL=100HZ due to<br>capacitor c=0.1 micro_Farad and<br>has a gain of Av(mid)=-10. Find<br>the gain Av at f=50HZ *<br>5 at 180 degree<br>10 at 180 degree<br>20 at 360 degree<br>(6_10) at (200_250) degree<br>20 at 180 degree<br>(1_5) at (200_250) degree<br>O (6_10) at (160_190) degree<br>(1_5) at (160_190)<br>5 at 90 degree<br>O O<br>O O O O<br>

Extracted text: An amplifier has a gain of Av(mid)=100. At high cutoff frequency point, the gain will * rise to a value depends on the design of the amplifier rise to 141.44 rise to 200 reduce to a value depends on the design of the amplifier reduce to 70.7 change (reduce or rise) to a value O depends on the design of the amplifiera reduce to 0.707 be kept constant reduce to 50 An amplifier has fL=100HZ due to capacitor c=0.1 micro_Farad and has a gain of Av(mid)=-10. Find the gain Av at f=50HZ * 5 at 180 degree 10 at 180 degree 20 at 360 degree (6_10) at (200_250) degree 20 at 180 degree (1_5) at (200_250) degree O (6_10) at (160_190) degree (1_5) at (160_190) 5 at 90 degree O O O O O O

Jun 10, 2022
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