Assume you are given an unknown first-order RC two port network but a

Assume you are given an unknown first-order RC two port network but are able to measure the magnitude and phase plots of its transfer function as shown in below: What common filter characteristic corresponds to these graphs? Draw the Bode plot for the unknown network and mark the break frequency and slope. What Is the transfer function of the RC circuit which leads to this response? Give the value for the break frequency and use a transfer function which is expressed in terms of this frequency. Draw the RC circuit diagram and label the input voltage source v1 and output v0. Assuming a 1 mu F capacitor is used in the circuit, what would be the correct value of the resistor to achieve this break frequency? Solution 1.   The graphs corresponds to a HIGH PASS FILTER, that allows signals of frequency greater than the break frequency. Break, frequency is the frequency at which phase difference is 45o or magnitude ot the transfer function is max/2       Form the curves given, the break frequency fB ˜ 40Hz ii) The circuit has a resistor R in sereis with the capacitor, the input signal is applied across them and the output is taken across the resistor. Let Vi be the input signal Let Vo be the output signal The impedance of the circuit Z = R - jXC The current in the circuit I = Vin / Z The output voltage is the voltage across the resistor Vo = IR Vo = VinR/Z Vo = Vin R / (R-jXC) Transfer function Vo / Vin = R / (R-jXC) Vo / Vin = 1 / (1-jXC /R) Vo / Vin = 1 / (1-j1/CR) Vo / Vin = 1 / (1-jB /) or Vo / Vin = 1 / (1-jfB /f)Where fB is the break frequency and is equal to fB = 1/2RC iii) The circuit has a resistor R in sereis with the capacitor, the input signal is applied across them and the output is taken across the resistor. iv) With C = 1F, and fB = 40Hz 40 = 1 / (2)(3.14)(1*10-6)R R = 3.98k Vo / Vin = 1 / (1-jXC /R) Vo / Vin = 1 / (1-j1/CR) Vo / Vin = 1 / (1-jB /) or Vo / Vin = 1 / (1-jfB /f) Where fB is the break frequency and is equal to fB = 1/2RC iii) The circuit has a resistor R in sereis with the capacitor, the input signal is applied across them and the output is taken across the resistor. iv) With C = 1F, and fB = 40Hz 40 = 1 / (2)(3.14)(1*10-6)R R = 3.98k Vo / Vin = 1 / (1-j1/CR) Vo / Vin = 1 / (1-jB /) or Vo / Vin = 1 / (1-jfB /f) Vo / Vin = 1 / (1-jfB /f) Where fB is the break frequency and is equal to fB = 1/2RC iii) The circuit has a resistor R in sereis with the capacitor, the input signal is applied across them and the output is taken across the resistor. iv) With C = 1F, and fB = 40Hz 40 = 1 / (2)(3.14)(1*10-6)R R = 3.98k