The modulation index is 0.75. Then the useful power available in the side bands nearly

Options

A. `11%`
B. `22%`
C. `33%`
D. `44%`

Answer

To solve the problem of finding the useful power available in the sidebands when the modulation index is given as 0.75, we can follow these steps:

### Step 1: Understand the Modulation Index
The modulation index (m) is given as:
\[ m = 0.75 \]

### Step 2: Write the Formula for Total Power
The total power (P_total) in an amplitude modulated signal is given by:
\[ P_{total} = P_c \left(1 + \frac{m^2}{2}\right) \]
where \( P_c \) is the carrier power.

### Step 3: Write the Formula for Sideband Power
The power available in the sidebands (P_side) can be expressed as:
\[ P_{side} = P_c \left(\frac{m^2}{2}\right) \]

### Step 4: Calculate the Ratio of Sideband Power to Total Power
To find the useful power available in the sidebands as a fraction of the total power, we can write:
\[ \text{Useful Power Ratio} = \frac{P_{side}}{P_{total}} \]

Substituting the formulas from Steps 2 and 3, we have:
\[ \text{Useful Power Ratio} = \frac{P_c \left(\frac{m^2}{2}\right)}{P_c \left(1 + \frac{m^2}{2}\right)} \]

### Step 5: Simplify the Expression
Since \( P_c \) appears in both the numerator and the denominator, we can cancel it out:
\[ \text{Useful Power Ratio} = \frac{\frac{m^2}{2}}{1 + \frac{m^2}{2}} \]

### Step 6: Substitute the Value of m
Now, substituting \( m = 0.75 \):
\[ \text{Useful Power Ratio} = \frac{\frac{(0.75)^2}{2}}{1 + \frac{(0.75)^2}{2}} \]

Calculating \( (0.75)^2 \):
\[ (0.75)^2 = 0.5625 \]
So we have:
\[ \text{Useful Power Ratio} = \frac{\frac{0.5625}{2}}{1 + \frac{0.5625}{2}} = \frac{0.28125}{1 + 0.28125} = \frac{0.28125}{1.28125} \]

### Step 7: Calculate the Final Value
Now, calculating the ratio:
\[ \text{Useful Power Ratio} \approx 0.2195 \]

### Step 8: Convert to Percentage
To express this as a percentage, multiply by 100:
\[ \text{Useful Power Percentage} = 0.2195 \times 100 \approx 21.95\% \]

### Step 9: Round the Result
Rounding this to the nearest whole number gives approximately:
\[ \text{Useful Power Percentage} \approx 22\% \]

### Conclusion
The useful power available in the sidebands is approximately 22%.

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