Question

The positive RF peaks of an AM voltage rise to maximum value of 12V and drop to a minimum value of 4V. Assuming single tone modulation, the modulation index is

• A. 3
• B. 2
• C. \( \frac{1}{2} \)
• D. \( \frac{1}{3} \)

Hint:

For AM, \(V_{max} = V_c(1+m)\) and \(V_{min} = V_c(1-m)\).
Modulation index \(m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}\).
Also, \(V_c = \frac{V_{max} + V_{min}}{2}\) and \(V_m = \frac{V_{max} - V_{min}}{2}\), then \(m = V_m/V_c\).

Answer 1

The Correct Option is C

For an Amplitude Modulated (AM) wave, let \(V_{max}\) be the maximum peak amplitude of the modulated wave and \(V_{min}\) be the minimum peak amplitude of the modulated wave.

Given:

• \(V_{max} = 12\text{V}\) 
• \(V_{min} = 4\text{V}\)

The modulation index (\(m\) or \(\mu\)) can be calculated using the formula:

\[ m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} \]

Substitute the given values:

\[ m = \frac{12\text{V} - 4\text{V}}{12\text{V} + 4\text{V}} = \frac{8\text{V}}{16\text{V}} = \frac{8}{16} = \frac{1}{2} \]

The modulation index is \(\frac{1}{2}\) or 0.5.

Alternatively, let \(V_c\) be the carrier amplitude and \(V_m\) be the message signal amplitude.

\(V_{max} = V_c + V_m = V_c(1+m)\)

\(V_{min} = V_c - V_m = V_c(1-m)\)

Adding these: \(V_{max} + V_{min} = 2V_c\)

12 + 4 = 2V_c \(\Rightarrow\) 16 = 2V_c \(\Rightarrow\) V_c = \(8\text{V}\)

Subtracting these: \(V_{max} - V_{min} = 2V_m\)

12 - 4 = 2V_m \(\Rightarrow\) 8 = 2V_m \(\Rightarrow\) V_m = \(4\text{V}\)

Modulation index \(m = \frac{V_m}{V_c} = \frac{4\text{V}}{8\text{V}} = \frac{1}{2}\).

Final Answer:

\(\frac{1}{2}\)