Question

For the given Bode magnitude plot of the transfer function, the value of R is \(\underline{\hspace{2cm}}\) Ω. (Round to 2 decimals). 
 

Hint:

Use Bode peak magnitude & resonance frequency to recover unknown R in RLC circuits.

Answer 1

Correct Answer: 0.09

The resonant peak at 26 dB occurs around \[ \omega = 2000\ \text{rad/s} \] For an RLC circuit: \[ \omega_0 = \frac{1}{\sqrt{LC}} \] Given: \[ L = 1\text{ mH}, C = 250\mu F \] \[ \omega_0 = \frac{1}{\sqrt{0.001 \times 250\times10^{-6}}} = 2000\ \text{rad/s} \] Peak magnitude (in dB): \[ M = 20 \log \left(\frac{1}{R} \sqrt{\frac{L}{C}}\right) \] Convert dB: \[ 26 = 20 \log\left(\frac{1}{R}\sqrt{\frac{0.001}{250\times10^{-6}}}\right) \] \[ \sqrt{\frac{0.001}{250\times10^{-6}}} = 2 \] Thus, \[ 26 = 20 \log\left(\frac{2}{R}\right) \] \[ \frac{2}{R} = 10^{1.3} = 20 \] \[ R = \frac{2}{20} = 0.1\ \Omega \] \[ \boxed{R = 0.10\ \Omega} \]