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 (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.
Details and Assumptions:
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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 ) The impedance of the circuit considering only this input is:
Z 1 = R + j ω 1 C 1
Here j = − 1 . The current is then:
I 1 = Z 1 V i 1 , m a x
The voltage drop across the capacitor is then:
V o 1 = I 1 ( j ω 1 C 1 )
This same process is repeated for the other two voltage input components which are:
V i 2 = 2 cos ( ω 2 t ) V i 3 = 3 cos ( ω 3 t )
Finally, we obtain output voltages V o 1 , V o 2 , V o 3 , which are complex numbers.
The required answer is then:
A = ∣ V i 1 , m a x ∣ 2 + ∣ V i 2 , m a x ∣ 2 + ∣ V i 3 , m a x ∣ 2 ∣ V o 1 ∣ 2 + ∣ V o 2 ∣ 2 + ∣ V o 3 ∣ 2 = 0 . 5 3
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.