Question

How do you calculate the frequency response of a passive low-pass filter using transfer functions?

Answer 1

Calculating the frequency response of a passive low-pass filter using transfer functions involves representing the filter in the s-domain, where 's' is a complex variable representing frequency. The transfer function of the filter describes the relationship between the input and output signals in the s-domain. For a passive low-pass filter, the transfer function can be derived based on its circuit configuration.

The most common passive low-pass filter is the RC filter (Resistor-Capacitor). Let's go through the steps to calculate its frequency response using transfer functions:

Step 1: Write the circuit equation
Consider a simple first-order RC low-pass filter. The circuit consists of a resistor (R) and a capacitor (C) connected in series, with the input voltage (Vin) applied across the resistor and the output voltage (Vout) taken across the capacitor.

Step 2: Apply Kirchhoff's voltage law (KVL)
Apply KVL around the loop to obtain the equation:

Vin - Vout - VR - VC = 0

Step 3: Express the voltage across the components
The voltage across the resistor (VR) can be written as:

VR = R * I

The voltage across the capacitor (VC) can be written as:

VC = (1/C) * ∫I dt

Where I is the current flowing through the resistor and capacitor.

Step 4: Find the relationship between Vin and Vout
Substitute the expressions for VR and VC into the KVL equation:

Vin - Vout - R * I - (1/C) * ∫I dt = 0

Step 5: Take the Laplace transform
To move from the time domain to the s-domain, take the Laplace transform of the equation:

V(s) = Vin(s) - Vout(s) - R * I(s) - (1/C) * (s * I(s))

Where V(s) is the Laplace transform of Vout(t), Vin(s) is the Laplace transform of Vin(t), and I(s) is the Laplace transform of the current I(t).

Step 6: Apply Ohm's law to find I(s)
The relationship between the current and the voltages is given by Ohm's law:

I(s) = Vout(s) / R

Step 7: Substitute I(s) into the equation
Replace I(s) with Vout(s) / R in the equation:

V(s) = Vin(s) - Vout(s) - R * (Vout(s) / R) - (1/C) * (s * (Vout(s) / R))

Step 8: Solve for the transfer function
Now, isolate Vout(s) on one side of the equation:

Vout(s) * (1 + R * sC) = Vin(s)

Finally, the transfer function H(s) is given by:

H(s) = Vout(s) / Vin(s) = 1 / (1 + R * sC)

Step 9: Frequency response
The frequency response of the filter is obtained by evaluating the transfer function H(s) at s = jω, where ω is the angular frequency in radians per second. For a low-pass filter, we are interested in the magnitude response:

|H(jω)| = |1 / (1 + jωRC)|

The magnitude response |H(jω)| gives the relationship between the output voltage and the input voltage as a function of frequency. You can then plot this magnitude response on a logarithmic scale to visualize how the filter attenuates high-frequency components.

Keep in mind that this example assumes an ideal RC filter without any parasitic elements or deviations. In practice, real passive filters may have additional components or characteristics to consider in the analysis.