Question

For an amplitude modulated wave, if the maximum amplitude is \(400%\) more than its minimum amplitude, then the modulation index is

• A. \(\frac{3}{4}\)
• B. \(1\)
• C. \(\frac{2}{3}\)
• D. \(\frac{1}{2}\) \bigskip

Hint:

For AM waves: \[ m=\frac{A_{\max}-A_{\min}} {A_{\max}+A_{\min}} \] If \(A_{\max}=5A_{\min}\), immediately remember that \[ m=\frac{5-1}{5+1}=\frac{2}{3}. \]

Answer 1

The Correct Option is C

Concept: For an amplitude modulated (AM) wave, \[ m=\frac{A_{\max}-A_{\min}} {A_{\max}+A_{\min}} \] where \[ m=\text{modulation index} \] \[ A_{\max}=\text{maximum amplitude} \] \[ A_{\min}=\text{minimum amplitude} \] This relation is frequently used in communication systems.

Step 1: Interpret the statement given in the question.
The maximum amplitude is \(400%\) more than the minimum amplitude. This means \[ A_{\max}=A_{\min}+400%\,A_{\min} \] \[ A_{\max}=A_{\min}+4A_{\min} \] \[ A_{\max}=5A_{\min} \] Let \[ A_{\min}=A \] Then, \[ A_{\max}=5A \]

Step 2: Substitute into the modulation index formula.
\[ m=\frac{A_{\max}-A_{\min}} {A_{\max}+A_{\min}} \] Substituting, \[ m=\frac{5A-A}{5A+A} \] \[ m=\frac{4A}{6A} \] \[ m=\frac{2}{3} \]

Step 3: Write the final result.
Therefore, the modulation index is \[ \boxed{\frac{2}{3}} \] Hence, the correct option is \[ \boxed{\text{(C)}} \]