RLC Circuit Dynamic Filter Transfer (Electrical Engineering) Solution

Problem Overview

We differentiate the bandpass filter gain transfer function:

H(\omega) = \frac{10\omega}{\omega^2 + 25}

and solve for the peak frequency where

\frac{dH}{d\omega} = 0

Step-by-Step Mathematical Solution

Step 1: Component Dictionary

Numerator ( $u$ ): $u = 10\omega \implies u' = 10$
Denominator ( $v$ ): $v = \omega^2 + 25 \implies v' = 2\omega$

Step 2: Apply the Quotient Rule

\frac{dH}{d\omega} = \frac{u'v - uv'}{v^2} = \frac{(10)(\omega^2 + 25) - (10\omega)(2\omega)}{(\omega^2 + 25)^2}

Step 3: Expand and Simplify the Numerator

\frac{dH}{d\omega} = \frac{10\omega^2 + 250 - 20\omega^2}{(\omega^2 + 25)^2} = \frac{250 - 10\omega^2}{(\omega^2 + 25)^2} = \frac{10(25 - \omega^2)}{(\omega^2 + 25)^2}

Step 4: Find Peak Frequency ( $\frac{dH}{d\omega} = 0$ )

Set the numerator equal to zero:

10(25 - \omega^2) = 0 \implies \omega^2 = 25 \implies \omega = 5\text{ krad/s}

(since frequency

\omega > 0

Engineering Conclusion

Executing the Quotient Rule proves that the maximum signal gain occurs exactly at the resonant frequency $\omega = 5\text{ krad/s}$ (where $H(5) = \frac{50}{50} = 1.0$ ). At frequencies below $5\text{ krad/s}$ , the derivative is positive (amplifying incoming low bands), while above $5\text{ krad/s}$ , the derivative is negative (attenuating high-frequency noise).

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