Question

For each filter mentioned in the following cases, first simulate the circuit using Multisim. You can get a plot of the transfc) Active Band-pass Filter R2-10k C1-100nF R1-10K Vin Vout 10 fez 100 fc 0.01 fc fci fcz Frequency Vo/Vi 20logVo /Vi Note: Th

For each filter mentioned in the following cases, first simulate the circuit using Multisim. You can get a plot of the transfer function that is called the Bode plot. From the right toolbar, select "Bode Plotter". Change initial (I) and final (F frequencies to 1Hz and 200 KHz, respectively. Use a Voltage AC source as the input signal. You do not need to change any parameter from voltage AC source Connect "Bode Plotter" to input and output of your circuit accordingly Run the sim!ation and find the cutoff frequency. At the cutoff frequency, te amplitude drops by 3dB or to 70.79% of the maximum. You can find the cutoff frequency by setting the vertical I value to-3 dB Bode Plotter-XBP1 XOP Horiont Lin Log IN Controls Reverse Save V3 Measurement For each filter, follow these steps Magnitude Plot Build the filters using op-amp 741 Apply a 10-volt peak-peak sinusoidal waveform (set by observation on Channel 1 of the oscilloscope) at the input of the filter On Channel 2 of the oscilloscope, observe the output. Vary the frequency in order to find the cutoff frequency fo, which is the frequency at which the output voltage drops by 3 dB or to 70.79% of the maximum. Find IV)/ Vil as a ratio and in dB at 0.0L5, 0.1 105, and 100f. . Note: No/Vi in dB is calculated using 20log Vo /Vi Plot Vo/Vil in dB. Use a logarithmic scale for frequency
c) Active Band-pass Filter R2-10k C1-100nF R1-10K Vin Vout 10 fez 100 fc 0.01 fc fci fcz Frequency Vo/Vi 20logVo /Vi Note: The band-pass filter has two cutoff frequencies
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Answer #1

1nF XBP1 R2 IN OUT V3 -12V C1 R1 100nF V1 +10Vpk 1kHz 0° 741 V2 F12V

For fc1 and fc2

Design1 20 -35 -145 1G -200 10M 100k 1k 10 100m Cursor Bode Result 15.5006k ected Cursor: 2 -2.9666 156.9316 -3.0719 yl 2

for 0.01 fc1 and 10fc2

Design1 20 0 -145 -200 10M 100m 100k 1G 10 1k Cursor Bode Result ected Cursor: 2 1.5732 -40.1015 153.3455k 19.9136 y2 +0 .00

for 0.1fc1 and 100fc2

Design1 20 0 -35 90- -145 200 100k 10 1G 1m 100m 1k 10M Cursor Bode Result 15.4777 -20.2832 yl x2 y2 1.5087M 41.0295

0.01fc1 0.1fc1 fc1 fc2 10fc2 100fc2
Frequency 1.5693 15.693 156.9316 15.5K 155K 1.55M
Gain 100 10 0.707 0.707 10 100
Gain in dB -40 -20 -3 -3 -20 -40
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For each filter mentioned in the following cases, first simulate the circuit using Multisim. You can get a plot of the transfer function that is called the Bode plot. From the right toolbar, select &...
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