Question

The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5 V amplitude are $\frac{a}{10}$V and $\frac{b}{10}$V respectively. Then the value of $\frac{a}{b}$ is _________.

Hint:

In a simple AM wave, the two sidebands ($f_c+f_m$ and $f_c-f_m$) are always symmetrical and have equal amplitudes. The value of this amplitude is $\frac{m_a A_c}{2}$. Therefore, the ratio of their amplitudes will always be 1.

Answer 1

Correct Answer: 1

In standard Amplitude Modulation (AM), the amplitudes of the upper sideband (USB) and the lower sideband (LSB) are equal.
The amplitude of each sideband is given by the formula:
$A_{sideband} = \frac{m_a A_c}{2}$
where $A_c$ is the amplitude (peak voltage) of the carrier wave and $m_a$ is the modulation index.
First, we calculate the modulation index, $m_a$:
$m_a = \frac{A_m}{A_c}$
where $A_m$ is the amplitude of the modulating wave.
Given:
Carrier amplitude, $A_c = 15$ V.
Modulating wave amplitude, $A_m = 5$ V.
$m_a = \frac{5 \text{ V}}{15 \text{ V}} = \frac{1}{3}$.
Now, we calculate the amplitude of the sidebands:
$A_{sideband} = \frac{(1/3) \times 15 \text{ V}}{2} = \frac{5 \text{ V}}{2} = 2.5$ V.
The amplitude of the upper sideband is 2.5 V, and the amplitude of the lower sideband is also 2.5 V.
We are given that the amplitudes are $\frac{a}{10}$ V and $\frac{b}{10}$ V.
Amplitude of upper sideband: $\frac{a}{10} = 2.5 \implies a = 25$.
Amplitude of lower sideband: $\frac{b}{10} = 2.5 \implies b = 25$.
The value of the ratio $\frac{a}{b}$ is:
$\frac{a}{b} = \frac{25}{25} = 1$.