Given: The circuit shown above operates in steady state over a range of frequencies. V i is the input voltage and V o is the output voltageRequired:a. Determine the resistance value in ? for which the half-power frequency, f o , is 400 kHz b. Determine the gain in db, G db1 , for this filter at a frequency of 80 kHz, and the gain in db, G db2 , at a frequency of 1600 kHz.c. Is this a High-pass or a Low-pass filter
Solution
a) Half power frequency is also called cut off frequency, wc = 1/RC
OR 2*pi*fc = 1/RC
hence, R = 1/(2*pi*fc*C) = 15.9 Ohms
b) The gain can be found out by using the formula wRC/sqrt[1+(wRC)2], where w = 2*pi*f, put f =80kHz, C=25nF & R =15.9Ohms & the compute 20*log10(G).
Same formula to be applied to find out the gain at f = 1600 Hz.
This is a high pass filter, as the capacitor connected to the begining will allow only high frequencies to pass.
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Given- The circuit shown above operates in steady state over a range o.docx
1. Given: The circuit shown above operates in steady state over a range of frequencies. V i is the
input voltage and V o is the output voltageRequired:a. Determine the resistance value in ? for
which the half-power frequency, f o , is 400 kHz b. Determine the gain in db, G db1 , for this filter
at a frequency of 80 kHz, and the gain in db, G db2 , at a frequency of 1600 kHz.c. Is this a High-
pass or a Low-pass filter
Solution
a) Half power frequency is also called cut off frequency, wc = 1/RC
OR 2*pi*fc = 1/RC
hence, R = 1/(2*pi*fc*C) = 15.9 Ohms
b) The gain can be found out by using the formula wRC/sqrt[1+(wRC)2], where w = 2*pi*f, put
f =80kHz, C=25nF & R =15.9Ohms & the compute 20*log10(G).
Same formula to be applied to find out the gain at f = 1600 Hz.
This is a high pass filter, as the capacitor connected to the begining will allow only high
frequencies to pass.
![Given: The circuit shown above operates in steady state over a range of frequencies. V i is the
input voltage and V o is the output voltageRequired:a. Determine the resistance value in ? for
which the half-power frequency, f o , is 400 kHz b. Determine the gain in db, G db1 , for this filter
at a frequency of 80 kHz, and the gain in db, G db2 , at a frequency of 1600 kHz.c. Is this a High-
pass or a Low-pass filter
Solution
a) Half power frequency is also called cut off frequency, wc = 1/RC
OR 2*pi*fc = 1/RC
hence, R = 1/(2*pi*fc*C) = 15.9 Ohms
b) The gain can be found out by using the formula wRC/sqrt[1+(wRC)2], where w = 2*pi*f, put
f =80kHz, C=25nF & R =15.9Ohms & the compute 20*log10(G).
Same formula to be applied to find out the gain at f = 1600 Hz.
This is a high pass filter, as the capacitor connected to the begining will allow only high
frequencies to pass.](https://image.slidesharecdn.com/given-thecircuitshownaboveoperatesinsteadystateoverarangeo-230203002035-adc6a602/85/Given-The-circuit-shown-above-operates-in-steady-state-over-a-range-o-docx-1-320.jpg)
