The amplitude of upper and lower side bands of AM 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 _________.

We start by translating the words of the question into the symbols used in amplitude-modulation theory. The unmodulated carrier has a peak (maximum) voltage $$V_c = 15\ \text{V}$$ and an angular frequency $$\omega_c = 2\pi f_c$$ with $$f_c = 11.21\ \text{MHz}$$. The modulating (message) signal is a sinusoid whose peak voltage is $$V_m = 5\ \text{V}$$ and whose frequency is $$f_m = 7.7\ \text{kHz}$$.

In ordinary (double-side-band) amplitude modulation the time-domain equation for the modulated voltage is written first. The standard formula is

$$v(t)=V_c\bigl[1+m\sin(\omega_m t)\bigr]\sin(\omega_c t),$$

where the dimensionless quantity $$m$$ is called the modulation index or modulation depth. Its definition is

$$m=\frac{V_m}{V_c}.$$

Substituting the given numerical values, we have

$$m=\frac{V_m}{V_c}=\frac{5\text{ V}}{15\text{ V}}=\frac13\;.$$

Next we want to isolate the upper and lower sideband terms. To do this we expand the product in the time-domain expression. First we note the trigonometric identity

$$\sin A\,\sin B=\frac12\bigl[\cos(A-B)-\cos(A+B)\bigr].$$

Applying this identity to $$\bigl[m\sin(\omega_m t)\bigr]\sin(\omega_c t)$$ gives

$$m\sin(\omega_m t)\sin(\omega_c t)=\frac{m}{2}\Bigl[\cos\bigl((\omega_c-\omega_m)t\bigr)-\cos\bigl((\omega_c+\omega_m)t\bigr)\Bigr].$$

Hence the complete expansion of the modulated signal becomes

$$ \begin{aligned} v(t) &=V_c\sin(\omega_c t)+V_c\Bigl[\tfrac{m}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)-\tfrac{m}{2}\cos\bigl((\omega_c+\omega_m)t\bigr)\Bigr] \\ &=V_c\sin(\omega_c t)+\frac{mV_c}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)-\frac{mV_c}{2}\cos\bigl((\omega_c+\omega_m)t\bigr). \end{aligned} $$

We see three distinct sinusoidal terms:

• $$V_c\sin(\omega_c t)$$ is the carrier itself, still having peak amplitude $$V_c = 15\ \text{V}$$.
• $$\dfrac{mV_c}{2}\cos\bigl((\omega_c-\omega_m)t\bigr)$$ is the lower sideband (LSB) at frequency $$f_c-f_m$$.
• $$-\dfrac{mV_c}{2}\cos\bigl((\omega_c+\omega_m)t\bigr)$$ is the upper sideband (USB) at frequency $$f_c+f_m$$. The minus sign only changes phase, not amplitude, so its peak amplitude is the same magnitude.

The important point is that each sideband has a peak voltage of

$$V_{\text{SB}}=\frac{mV_c}{2}.$$

Substituting $$m=\dfrac13$$ and $$V_c=15\ \text{V},$$ we obtain

$$ \begin{aligned} V_{\text{SB}} &=\frac{1}{2}\,\bigl(\tfrac13\bigr)\,(15\ \text{V}) \\ &=\frac{1}{2}\times5\ \text{V} \\ &=2.5\ \text{V}. \end{aligned} $$

The question states that the upper sideband amplitude is expressed as $$\dfrac{a}{10}\ \text{V}$$ and the lower sideband amplitude as $$\dfrac{b}{10}\ \text{V}$$. Since both sidebands are equal in a normal AM wave, we identify

$$\frac{a}{10}=2.5 \quad\Longrightarrow\quad a=25,$$

and also

$$\frac{b}{10}=2.5 \quad\Longrightarrow\quad b=25.$$

The ratio sought is therefore

$$\frac{a}{b}=\frac{25}{25}=1.$$

So, the answer is $$1$$.