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.