Question

A Bode plot of a transfer function G(s) is shown in the figure. The gain 'A' in dB at frequency f = 100Hz is:

• A. $-20\text{ dB/dec}$
• B. $+60\text{ dB/dec}$
• C. $+40\text{ dB/dec}$
• D. $+40\text{ dB/dec}$

Hint:

A slope of $-20\text{ dB/dec}$ means that for every 10-fold increase in frequency, the gain decreases by exactly $20\text{ dB}$.
Since $100\text{ Hz}$ is 10 times $10\text{ Hz}$, the gain drops from $80\text{ dB}$ to $60\text{ dB}$ instantly.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The problem involves reading a Bode magnitude plot. We are given the gain at $f = 10\text{ Hz}$ ($80\text{ dB}$) and a constant roll-off slope of $-20\text{ dB/dec}$.
We need to find the gain 'A' in dB at $f = 100\text{ Hz}$.

Step 2: Key Formula or Approach:
The relationship between gains at two different frequencies on a Bode plot with a constant slope is given by:
\[ \text{Gain}(f_2) = \text{Gain}(f_1) + \text{Slope} \times \log_{10}\left( \frac{f_2}{f_1} \right) \]

Step 3: Detailed Explanation:

• Given parameters from the graph:
Initial frequency, $f_1 = 10\text{ Hz}$.
Initial gain, $\text{Gain}(f_1) = 80\text{ dB}$.
Final frequency, $f_2 = 100\text{ Hz}$.
Slope of the line, $\text{Slope} = -20\text{ dB/dec}$.

• Let us calculate the number of decades between $10\text{ Hz}$ and $100\text{ Hz}$:
\[ \text{Decades} = \log_{10}\left( \frac{100}{10} \right) = \log_{10}(10) = 1\text{ decade} \]

• Apply the formula:
\[ \text{Gain}(100\text{ Hz}) = 80\text{ dB} + (-20\text{ dB/dec} \times 1\text{ decade}) \]
\[ \text{Gain}(100\text{ Hz}) = 80 - 20 = 60\text{ dB} \]

• Note: The units in the options are written as "dB/dec" due to a typographical error in the exam paper. The value $+60$ corresponds directly to the calculated gain of $60\text{ dB}$.

Step 4: Final Answer:
The gain at $100\text{ Hz}$ is $60\text{ dB}$ (listed as $+60\text{ dB/dec}$), which corresponds to Option (B).