The modulation index \( m \) is given by the ratio of the peak voltage of the modulating signal \( V_m \) to the peak voltage of the carrier wave \( V_c \): \[ m = \frac{V_m}{V_c} \] Given that \( m = 0.75 \) and \( V_c = 16 \, {V} \), we can solve for \( V_m \): \[ 0.75 = \frac{V_m}{16} \] \[ V_m = 0.75 \times 16 = 12 \, {V} \] Thus, the peak voltage of the modulating signal is 12 V.
Final Answer: 12 V.
Match List-I with List-II:
| List-I (Modulation Schemes) | List-II (Wave Expressions) |
|---|---|
| (A) Amplitude Modulation | (I) \( x(t) = A\cos(\omega_c t + k m(t)) \) |
| (B) Phase Modulation | (II) \( x(t) = A\cos(\omega_c t + k \int m(t)dt) \) |
| (C) Frequency Modulation | (III) \( x(t) = A + m(t)\cos(\omega_c t) \) |
| (D) DSB-SC Modulation | (IV) \( x(t) = m(t)\cos(\omega_c t) \) |
Choose the correct answer:

If the roots of $\sqrt{\frac{1 - y}{y}} + \sqrt{\frac{y}{1 - y}} = \frac{5}{2}$ are $\alpha$ and $\beta$ ($\beta > \alpha$) and the equation $(\alpha + \beta)x^4 - 25\alpha \beta x^2 + (\gamma + \beta - \alpha) = 0$ has real roots, then a possible value of $y$ is: