Answer to Question #244709 in Other for ivan

Question

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

Answer 1

Given $V_{max}=5.22V$ and $V_{min}=1.33V,$ modulation(m) is given by,

$m=((V_{max}-V_{min})/(V_{max}+V_{min}))$

$m=(5.22-1.33)/(5.22+1.33)$

$m=3.89/6.55=0.5939(4\displaystyle\space decimal\displaystyle\space places)$

Therefore, percentage of modulation is given as,

$percentage\displaystyle\space of\displaystyle\space modulation=m*100=0.5939*100=59.39\%$