The goal of this project was to be able to control the relative strength of individual frequency bands from an input signal. Three signals were seperated from each other using filters. After each individual frequency band was adjusted and scaled by a desired magnitude using operational amplifiers and potentiometers, the signals were combined back together as one signal. The output signal's magnitude could also be adjusted at the end.
The input signal was seperated into 3 frequency bands:
The low pass filter works by only allowing signals with low frequencies to pass through. Our goal is to only allow frequencies below 320 Hz. One way I found helpful in understanding the low pass filter is to observe how the capacitor works at very low and very high frequencies. The I-V characteristics of a capacitor can be represented as a differential equation. I = C * dv/dt where I is current, C is the capacitance of the capacitor and dv/dt is the rate of change of the voltage.
When the signal Vin has very low frequencies close to 0, that means Vin is a DC constant signal. The rate of change or derivative of any constant is 0 and therefore I (the current) is 0. At this point, we can treat the capacitor as an open circuit since there is no current going through. There will be no current going through the resistor R1 which means there is no voltage drop across that resistor. Therefore Vout will be equal to Vin and the signal passes through.
When the signal Vin has a very high frequency, the capacitor can be treated as a short circuit. Based on Ohm's law V = IR, the voltage drop across a short circuit is 0. Therefore Vout is 0 and the signal is filtered out.
The cutoff frequency Fc can be calculated with the equation 1/2πRC where R is the resistance of the resistor and C the capacitance of the capacitor.
The high pass filter works by only allowing signals with high frequencies to pass through. Our goal is to only allow frequencies above 3200 Hz. It relies on the same concept behind the low pass filter. However, in order to filter out high frequencies instead of low ones, we will have to rearrange the positions of the resistor/capacitor and switch them around.
When the signal Vin has very low frequencies close to 0, we can treat the capacitor as an open circuit. Because there is no current flowing through the open circuit, there will be no current across the resistor R7. Therefore Vout will be 0 and the signal is filtered out.
When the signal Vin has a very high frequency, the capacitor can be treated as a short circuit. Because there is no voltage drop across the short circuit, Vin and Vout are both measured at the same nodes and thus will be the same. Therefore the signal passes through.
Again the cutoff frequency Fc can be calculated with the equation 1/2πRC where R is the resistance of the resistor and C the capacitance of the capacitor.
The band pass filter works by only allowing signals between certain frequencies to pass through (in this case between 320 Hz and 3200 Hz). In order to construct a band pass filter, we can combine a low pass filter and high pass filter. The purpose of the op amp is to serve as a buffer. Because the input terminals of the op amp have high impedance, this will ensure that no current is leaked while still allowing the signal to be received by the HPF from the LPF.
The bode plots for all three filters are shown on the picture to the most right. When the gain is 0 dB, that means our signal is going through completely. As Vout approaches 0, the gain (dB) decreases. If we observe the band pass filter's bode plot, the low pass filter's cutoff frequency FH corresponds to 3200 Hz and the high pass filter's cutoff frequency FL corresponds to 320 Hz.
Op amps amplify or scale the signal by a certain magnitude Av. This can be found by solving for the ratio between Vout and Vin. The main characteristics of the op amp is that the input impedance is extremely high and that the voltage at V+ and V- are the same. Because the input impedance is high, the current flowing in will have no choice but to flow through Rf. The circuits can be solved with KCL.
The design specifications of the audio equalizer (with a test condition of Vin being a 1 Vpp Sinewave) are as follows (V_amp is the final output signal):
The final calculated values for components are as follows:
Low Pass Filter