Consider the following amplitude modulated (AM) signal, where `f_m lt B`
`x_(AM) (t) = 10 ( 1+ 0.5 sin 2 pi f_m t) cos 2 pi f_c t `
The average side-band power for the AM signal given above is

Options

A. 25
B. 12.5
C. 6.25
D. 3.125

Explanation

To find the average side-band power for the given amplitude modulated (AM) signal, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the AM Signal Parameters**:
The given AM signal is:
\[ x_{AM}(t) = 10 \left(1 + 0.5 \sin(2 \pi f_m t)\right) \cos(2 \pi f_c t) \]
From this equation, we can identify:
- \( P_C = 10 \) (the carrier amplitude)
- \( M_A = 0.5 \) (the modulation index)

2. **Average Power of the Carrier**:
The average power of the carrier signal \( P_C \) is given by:
\[ P_C = \frac{(A_C)^2}{2} = \frac{(10)^2}{2} = \frac{100}{2} = 50 \text{ watts} \]

3. **Average Side-band Power Formula**:
The average side-band power \( P_{SB} \) for an AM signal is given by the formula:
\[ P_{SB} = \frac{M_A^2 P_C}{4} \]

4. **Substituting Values into the Formula**:
Now we can substitute the values of \( M_A \) and \( P_C \) into the formula:
\[ P_{SB} = \frac{(0.5)^2 \cdot 50}{4} \]
\[ P_{SB} = \frac{0.25 \cdot 50}{4} = \frac{12.5}{4} = 3.125 \text{ watts} \]

5. **Conclusion**:
Therefore, the average side-band power for the given AM signal is:
\[ P_{SB} = 3.125 \text{ watts} \]

### Final Answer:
The average side-band power for the AM signal is **3.125 watts**.

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