Question

An amplitude modulated wave is as shown in the figure.
find out:
(a) modulation index
(b) peak carrier voltage
(c) peak information voltage

Answer 1

Maximum voltage, $V_{max}=\frac{100}{2}=50V$

Minimum voltage, $V_{min}=\frac{20}{2}=10V$

Percentage modulation, $\mu =\frac{V_{max}-V_{min}}{V_{max}+V_{min}}\times 100$

$=\frac{50-10}{50+10}\times 100=\frac{40}{60}\times 100=66.67\%$

Peak carrier voltage, $V_c=\frac{V_{max}+V_{min}}{2}=\frac{50+10}{2}=30V$

Peak value of information voltage, $V_m=\mu V_c=\frac{66.67}{100}\times 30=20V$