he resistor was measured to be 150.1 Ω and the capacitor to be 233.6 nF
A summary of the results is as follows:
Frequency = 1.01 kHz: Input = 1.51 V; Output = 1.44 V
Frequency = 10.06 kHz: Input = 1.37 V; Output = 0.62 V
Frequency = 102.8 kHz: Input = 1.31 V; Output = 90 mV
The resistor was measured to be 331.4 Ω and the capacitor to be 96.3 nF. (1) Calculate values of the transfer function, using the formula above, at the frequencies that were investigated and compare them to the ratios of the measured output and input voltages at each frequency.
A summary of the results is as follows:
Frequency = 1.01 kHz: Input = 1.50 V; Output = 1.44 V
Frequency = 10.06 kHz: Input = 1.20 V; Output = 0.50 V
Frequency = 102.8 kHz: Input = 1.13 V; Output = 70 mV
The resistor was measured to be 150.1 Ω and the capacitor to be 233.6 nF. (2) Are the ratios of the output voltage to input voltage for this filter similar to the ratios of the first low-pass filter? Does it make sense based on the transfer function how the ratios compare?

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