RC Low-Pass Filter - RMS Attenuation

The circuit below is an RC low-pass filter with a voltage input consisting of sinusoids at several different frequencies. The filter cutoff frequency is f c f_c (see details below).

In steady-state, what is the ratio of the RMS output voltage to the RMS input voltage?

Note: This can be solved two different ways. Both methods yield the same result.

  1. The first way involves a superposition of three different AC steady states - one for each frequency. And then the RMS voltage is the square root of the sum of the squares of the individual voltage magnitudes at each frequency. V R M S = V 1 2 + V 2 2 + V 3 2 V_{RMS} = \sqrt{{|V|_1}^2 + {|V|_2}^2 + {|V|_3}^2}
  2. The second way involves a time-domain simulation to calculate the composite capacitor voltage, as well as the calculation of the following integral: V R M S = 1 T 0 T v 2 ( t ) d t V_{RMS} = \sqrt{\frac{1}{T} \int_0^T v^2(t) \, dt} . Note that here, T T must be an integer multiple of each of the constituent signal periods.

Details and Assumptions:

  • R = 1 R = 1
  • ( f 1 , f 2 , f 3 , f c ) = ( 60 , 120 , 180 , 90 ) (f_1,f_2,f_3,f_c) = (60,120,180,90) .
  • ( ω 1 , ω 2 , ω 3 , ω c ) = ( 2 π f 1 , 2 π f 2 , 2 π f 3 , 2 π f c ) (\omega_1, \omega_2, \omega_3, \omega_c) = (2 \pi f_1, 2 \pi f_2, 2 \pi f_3, 2 \pi f_c )
  • C = 1 R ω c \large{C = \frac{1}{R \, \omega_c}}


The answer is 0.53.

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1 solution

Karan Chatrath
Sep 5, 2019

General Outline of Solution

This circuit is analysed in steady-state by applying the principle of superposition.

Consider the voltage component:

V i 1 = cos ( ω 1 t ) V_{i1} = \cos(\omega_1t) The impedance of the circuit considering only this input is:

Z 1 = R + 1 j ω 1 C Z_1 = R + \frac{1}{j\omega_1C}

Here j = 1 j = \sqrt{-1} . The current is then:

I 1 = V i 1 , m a x Z 1 I_1 = \frac{V_{i1,max}}{Z_1}

The voltage drop across the capacitor is then:

V o 1 = I 1 ( 1 j ω 1 C ) V_{o1} = I_1\left( \frac{1}{j\omega_1C}\right)

This same process is repeated for the other two voltage input components which are:

V i 2 = 2 cos ( ω 2 t ) V_{i2} = 2\cos(\omega_2t) V i 3 = 3 cos ( ω 3 t ) V_{i3} = 3\cos(\omega_3t)

Finally, we obtain output voltages V o 1 V_{o1} , V o 2 V_{o2} , V o 3 V_{o3} , which are complex numbers.

The required answer is then:

A = V o 1 2 + V o 2 2 + V o 3 2 V i 1 , m a x 2 + V i 2 , m a x 2 + V i 3 , m a x 2 = 0.53 \boxed{A = \frac{\sqrt{\mid V_{o1}\mid^2 + \mid V_{o2}\mid^2 + \mid V_{o3}\mid^2}}{\sqrt{\mid V_{i1,max} \mid^2 + \mid V_{i2,max} \mid^2 + \mid V_{i3,max} \mid^2}}=0.53}

Solution verified by solving the system differential equation numerically and computing the required ratio by evaluating the required integrals, again numerically. A closed-form solution is possible but tedious.

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