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
Answer & explanation
Correct answer: option 3
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, $V_{min}=V_c(1-μ)$
$=40×(1-0.5)=20V$
$\frac{V_{max}}{V_{min}}=\frac{60}{20}=\frac{3}{1}$