A message signal of frequency 10 kHz and peak voltage 10 volt is used to modulate a carrier of frequency 1 MHz and peak voltage 20 volt. Determine (a) modulation index (b) the side bands produced.

Options

A. 1000 KHz, 990KHz
B. 1010KHz,990KHz
C. 990KHz,1000KHz
D. 1010KHz,1000KHz

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the Modulation Index
The modulation index (m) is defined as the ratio of the peak voltage of the message signal (Vm) to the peak voltage of the carrier signal (Vc). The formula for the modulation index is given by:

\[ m = \frac{V_m}{V_c} \]

Where:
- \( V_m = 10 \, \text{V} \) (peak voltage of the message signal)
- \( V_c = 20 \, \text{V} \) (peak voltage of the carrier signal)

Substituting the values:

\[ m = \frac{10 \, \text{V}}{20 \, \text{V}} = 0.5 \]

### Step 2: Determine the Sidebands
The sidebands produced in amplitude modulation are calculated using the carrier frequency (fc) and the message frequency (fm). The lower sideband (LSB) and upper sideband (USB) can be calculated as follows:

- Lower Sideband (LSB):
\[ LSB = f_c - f_m \]
- Upper Sideband (USB):
\[ USB = f_c + f_m \]

Given:
- \( f_c = 1 \, \text{MHz} = 1000 \, \text{kHz} \)
- \( f_m = 10 \, \text{kHz} \)

Calculating the lower sideband:

\[ LSB = 1000 \, \text{kHz} - 10 \, \text{kHz} = 990 \, \text{kHz} \]

Calculating the upper sideband:

\[ USB = 1000 \, \text{kHz} + 10 \, \text{kHz} = 1010 \, \text{kHz} \]

### Final Results
(a) The modulation index is \( 0.5 \).

(b) The sidebands produced are:
- Lower Sideband: \( 990 \, \text{kHz} \)
- Upper Sideband: \( 1010 \, \text{kHz} \)