A message signal of peak voltage 20V is used to modulate a carrier wave of amplitude 40V in amplitude modulation. The ratio of maximum amplitude and minimum amplitude in amplitude modulated signal is

3 : 1

The correct answer is Option (3) → 3 : 1

The modulation index (μ) is -

$μ=\frac{V_m}{V_c}=\frac{20}{40}=0.5$

Maximum Amplitude, $V_{max}=V_c(1+μ)$

$=40×(1+0.5)=60V$

Maximum Amplitude, $V_{min}=V_c(1-μ)$

$=40×(1-0.5)=20V$

$\frac{V_{max}}{V_{min}}=\frac{60}{20}=\frac{3}{1}$