Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_233

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Step of 8 1.078P Refer Figure P1.76 in the Refer Table in the textbook Determine the transfer function for the left side part of the circuit Therefore, the transfer function Step of 8 This transfer function it is a low pass filter. Thus, the cut-off frequency is, 2πx 15.91kHz Therefore, the corner frequency is 15.91kHz Step Use source transform technique to re-draw the right side circuit diagram. 100 nF + R3 Figure 1 Step Apply voltage division rule and determine the output negative sign because, the current in opposite direstion Gm Gm Gm Therefore, the transfer function is, Gm Step Determine the cut-off frequency = Therefore, the corner frequency Step of 8 Determine the over all transfer function = 1+ (666.667) Step 7 of 8 The corner frequencies are and =15.91kHz Determine the band width of the bode plot. B.W=15910-53.05 =15.857kHz Therefore, the band width 15.857kHz Step of 8 The circuit is combination of low pass and high pass filter, thus the circuit has a plot like band pass The constant magnitude is, The bode plot increases its magnitude +20 dB below the corner frequency and it maintains constant upto the corner frequency, =15.91kHz Draw the bode plot for the transfer function 56.478 dB +20 dB -20 dB Figure 2 the bode plot shown Figure 2.