Question

For an amplitude modulated (AM) wave, the minimum amplitude is 4.0 V and the modulation index is 0.4. The maximum amplitude is:

• A. 8.55 V
• B. 12.25 V
• C. 6.50 V
• D. 9.33 V

Hint:

In AM waves, the modulation index \(m\) relates max and min amplitudes: \[ m = \frac{A_\text{max} - A_\text{min}}{A_\text{max} + A_\text{min}} \] and carrier amplitude \(A_c = (A_\text{max} + A_\text{min})/2\). Use these formulas for any AM amplitude problem.

Answer 1

The Correct Option is D

Step 1: AM envelope relations.
For amplitude modulation, A_max = A_c(1 + m) and A_min = A_c(1 - m), where A_c is the unmodulated carrier amplitude and m the modulation index.

Step 2: Isolating carrier amplitude.
From A_min = A_c(1 - m), we get A_c = A_min/(1 - m).

Step 3: Numerical substitution.
A_c = 4.0/(1 - 0.4) = 4.0/0.6 ≈ 6.667 V.

Step 4: Computing maximum amplitude.
A_max = 6.667 × (1 + 0.4) = 6.667 × 1.4 ≈ 9.33 V.

Step 5: Conceptual note.
The modulation index scales the carrier upward to produce the peak envelope value.

Step 6: Cross-check.
A_min = 6.667 × 0.6 = 4.0 V, confirming consistency.

Step 7: Conclusion.
Hence, the maximum AM wave amplitude is roughly 9.33 V.