Question

Assuming that an oscilloscope displays a V_max of 5.22V and V_min of 1.33V. Calculate the percentage of modulation

Answer 1

$$m=\dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}$$

$$m=\dfrac{5.22V-1.33V}{5.22V+1.33V}\approx0.5939$$

The percentage of modulation is $59.39\%.$