Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_226

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1.071P Table the textbook Table in the textbook it is low pass filter for high frequencies the gain becomes From Table the textbook the maximum gain dB 40 dB value cut-off frequency which the gain is dB less than maximum gain Step 2 of 11 circuit gain y 201og Substitute From Table the maximum value 40 dB V. Substitute 20log equation 40 20 transfer function is 100 Substitute 1000 for in the equation 100 1000 100 100 100 100 1.0049 Step 5 of 11 Apply logarith sides 20log =20(1.9978) 39.9560 dB Therefore, the magnitude the frequency 1000 Hz dB Step 6 of 11 phase angle 1000 (0.1) the chase angle at the frequency 1000 Hz phase angle the frequency Hz (10) Therefore, the phase angle the frequency 10 -84.2894° frequency which gain equal 100 100 20log 100 100 =100 994 kHz which gain 999.94 kHz Step 9 of 11 the phase angle at the frequency 999.94 kHz tan (99.994) 4270° the angle at the frequency 999.94 kHz Step 10 of 11 values in table f(Hz) 0 40 0 100 40 0 1000 39.9560 dB -5.71059 37 -45° 20 0 Step frequency response Bode Diagram 40 35 dB) 20 10 Frequency (rad/sec) Figure Thus