What Is A Low Pass Filter?

The RC Filter

An Operational Amplifier Filter

        The goal for this lesson is this.

• Given a requirement for a simple low-pass filter.
• Be able to meet that requirement using several different kinds of implementations including simple resistor-capacitor (RC) filters, and operational amplifier filters.

        A low-pass filter is a filter that passes low frequency signals and selectively attenuates higher frequency signals.  The simplest low-pass filter is an RC circuit.

In another lesson, we found the following information for this circuit.

• The circuit satisfies a differential equation:
• RC(dV_out(t)/dt) + V_out(t) = V_in(t)
• If the input voltage is a sinusoid, then the steady state sinusoidal output can be computed.
• If V_in(t) = A sin(wt)
• Then, V_out(t) = B sin(wt + f)
• Where:
• 
• f = - tan^-1(wRC)

        The circuit can also be analyzed using impedance and phasor concepts.  In that case, we know the following.

• The circuit is an AC voltage divider, with the output voltage taken across the capacitor.
• The impedance of the resistor is Z_R = R.
• The impedance of the capacitor is Z_C = 1/jwC
• The ratio of output to input is Z_C/(Z_R + Z_C)
• In phasor notation, the ratio of output voltage to input voltage is:
• V_out/V_in = (1/jwC)/[R + 1/jwC] = 1/[1 + jwRC]
• j is the square root of -1 (A unit imaginary number)
• So, when the input voltage is a sinusoid, then the steady state sinusoidal output can be computed.
• If V_in(t) = A sin(wt)
• Then, V_out(t) = B sin(wt + f)
• Where:
• B/A = 1/|1 + jwRC|
• f = - tan^-1(wRC)
• And, of course, this is exactly the same as the result above.

        Here is an interesting circuit.  It has an operational amplifier with some resistors and a capacitor.

By using KCL we can get a differential equation relating the input and the output voltages.  Let's write KCL at the inverting input node.  Here's a copy of the circuit with currents defined.

• Writing KCL, we get:
• I₁ + I_Cf+ I_Rf = 0
• (V_in/R₁) + (V_out/R_f) + (C_fdV_out/dt) = 0
• The KCL equation can be rearranged to get a differential equation:
• R_fC_fdV_out/dt + V_out= -V_inR_f/R₁
• This differential equation is remarkably like the differential equation for the low-pass filter.  The only difference is the factor -R_f/R₁ which multiplies the input voltage on the right hand side of the differential equation.  That leads to some pretty interesting conclusions.
• If the input voltage is a sinusoid, then the steady state sinusoidal output can be computed.
• If V_in(t) = A sin(wt)
• Then, V_out(t) = B sin(wt + f)
• Where:
• B/A = (-R_f/R₁)/|1 + jwRC|
• f = tan^-1(wRC)
• There is a remarkable similarity to what we found for the RC filter earlier.
• The primary difference is the factor of (-R_f/R₁) in the magnitude of the output.
• Also, be sure to note the minus sign, which could also have been accounted for by adding 180^o to the phase shift.

        This operational amplifier filter can also be analyzed using phasors.  In this situation, we view the impedances as combinations of impedances.  Here is a circuit diagram for the combined impedances.