A signal of frequency 20 kHz and peak voltage of 5 Volt is used to modulate a carrier wave of frequency 1.2 MHz and peak voltage 25 Volts. Choose the correct statement.

In amplitude modulation, the extent of modulation is expressed through the modulation index, denoted by $$m$$. The definition is

$$m \;=\;\frac{V_m}{V_c},$$

where $$V_m$$ is the peak (maximum) voltage of the modulating or message signal, and $$V_c$$ is the peak voltage of the unmodulated carrier wave.

We have been given

$$V_m = 5\ \text{V}, \qquad V_c = 25\ \text{V}.$$

Substituting these values into the formula, we obtain

$$m \;=\;\frac{5\ \text{V}}{25\ \text{V}} = \frac{5}{25} = 0.2.$$

Now, for a carrier wave of frequency $$f_c$$ that is amplitude-modulated by a signal of frequency $$f_s$$, two additional frequencies—called sideband frequencies—appear in the spectrum. Their values are found from the relations

$$f_{\text{USB}} = f_c + f_s, \qquad f_{\text{LSB}} = f_c - f_s,$$

where $$f_{\text{USB}}$$ is the frequency of the upper sideband and $$f_{\text{LSB}}$$ is the frequency of the lower sideband.

The numerical data supplied are

$$f_c = 1.2\ \text{MHz} = 1200\ \text{kHz}, \qquad f_s = 20\ \text{kHz}.$$

Carrying out the additions and subtractions step by step:

$$f_{\text{USB}} = 1200\ \text{kHz} + 20\ \text{kHz} = 1220\ \text{kHz},$$

$$f_{\text{LSB}} = 1200\ \text{kHz} - 20\ \text{kHz} = 1180\ \text{kHz}.$$

So, the complete set of results is

$$m = 0.2,\qquad f_{\text{USB}} = 1220\ \text{kHz},\qquad f_{\text{LSB}} = 1180\ \text{kHz}.$$

Examining the options, we see that these values coincide exactly with Option C.

Hence, the correct answer is Option C.