Question

Consider a low-pass filter having transfer function, H(s) = 6 / s²+9s+18. Find its 3-dB cut odd frequency, ω, which, by definition, satisfies the criterion, |H(jω)|² = |H(j0)|² / 2

Answer 1

To find the 3-dB cutoff frequency of a low-pass filter with the given transfer function, solve the equation |H(jω)|² = |H(j0)|² / 2 for ω.

Step-by-step explanation:

A low-pass filter is a type of electronic circuit that allows low-frequency signals to pass through while attenuating high-frequency signals. The cutoff frequency of a low-pass filter is the frequency at which the filter attenuates the signal by 3 decibels (dB) or half its power. In this case, we have the transfer function H(s) = 6 / (s² + 9s + 18), and we want to find the cutoff frequency ω that satisfies |H(jω)|² = |H(j0)|² / 2. To find ω, we need to solve the equation |H(jω)|² = |H(j0)|² / 2 for ω. Let's proceed with the calculations:

1. Replace s with jω in the transfer function H(s) = 6 / (s² + 9s + 18).
2. Square the magnitude of H(jω) and equate it to half the magnitude of H(j0) squared, which simplifies to 36 / (ω⁴ + 81ω² + 324).
3. Solve for ω by finding the roots of the quadratic equation ω⁴ + 81ω² + 324 - 72 = 0.
4. Use a calculator or numerical methods to solve the equation and find the values of ω.

The solution will give you the cutoff frequencies of the low-pass filter. Make sure to check if the obtained values are within the range of frequencies that the filter is designed to pass.