Question

A particular band-pass function has a network function as $H(s) = \frac{3s}{s^2+4s+3}$ then its quality factor Q is defined by}

• A. $\sqrt{3}/4$
• B. $2/\sqrt{3}$
• C. $\sqrt{3}/2$
• D. $4/\sqrt{3}$

Hint:

For any quadratic denominator $s^2 + as + b$, the resonant frequency is always $\omega_0 = \sqrt{b}$ and the quality factor is $Q = \frac{\sqrt{b}}{a}$.
This is a very useful shortcut for competitive exams.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This problem asks for the quality factor $Q$ of a second-order band-pass filter given by its transfer function $H(s)$.

Step 2: Key Formula or Approach:
The standard transfer function of a second-order band-pass filter is:
\[ H(s) = \frac{K \cdot s}{s^2 + \left(\frac{\omega_0}{Q}\right)s + \omega_0^2} \]
where:
- $\omega_0$ is the undamped natural frequency (resonant frequency).
- $Q$ is the quality factor.

Step 3: Detailed Explanation:

• We are given the transfer function:
\[ H(s) = \frac{3s}{s^2 + 4s + 3} \]

• By comparing the denominator coefficients with the standard form:
\[ \omega_0^2 = 3 \implies \omega_0 = \sqrt{3}\text{ rad/s} \]

• Comparing the coefficient of $s$:
\[ \frac{\omega_0}{Q} = 4 \]

• Rearranging to solve for $Q$:
\[ Q = \frac{\omega_0}{4} \]

• Substituting the value of $\omega_0$:
\[ Q = \frac{\sqrt{3}}{4} \]

Step 4: Final Answer:
The quality factor $Q$ is $\sqrt{3}/4$, which corresponds to Option (A).