Q.The maximum and minimum amplitudes of a sinusoidal modulated wave are 4V and 1V. Determine the percentage modulation.
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Concept understanding — AM Modulation Index
The modulation index (or modulation factor, or depth of modulation) tells us how strongly the message controls the carrier amplitude. For AM it is the ratio of peak modulating voltage to peak carrier voltage:
ma=VCVm
and it is often quoted as a percentage, M=(Vm/VC)×100. For distortion-free modulation Vm must be less than VC, so ma lies between 0 and 1, with a maximum permitted value of 1.0 (100% modulation). If Vm>VC then ma>1, giving over-modulation: the negative peaks are clipped, the envelope no longer matches the message, and distortion known as splattering appears.
When an AM wave is displayed on an oscilloscope, the index is read directly from the largest and smallest envelope amplitudes:
Vmax=VC(1+ma),Vmin=VC(1−ma)
which rearrange to the standard working formula
ma=Vmax+VminVmax−Vmin
The same relation holds with antenna currents, ma=(Imax−Imin)/(Imax+Imin).
When several sinusoidal signals of different frequencies modulate the carrier at once, their indices combine as the root of the sum of squares, mt=m12+m22+⋯+mn2, and this total should still not exceed unity. Because it fixes both the quality and the strength of the transmitted signal, the modulation index is one of the most important quantities in the whole study of AM.
When only the envelope's maximum and minimum amplitudes are known, the modulation index is found from their difference over their sum.
[!ANSWER]
ma=Vmax+VminVmax−Vmin=4+14−1=0.6 i.e. 60 %
[!TLDR]
Using the envelope extremes, ma=(4−1)/(4+1)=0.6, or 60 % modulation.
The amplitude-modulated envelope varies between a maximum Vmax=Vc+Vm and a minimum Vmin=Vc−Vm. Eliminating Vc and Vm gives the modulation index directly from these extremes:
ma=Vmax+VminVmax−Vmin
Substituting Vmax=4V and Vmin=1V:
ma=4+14−1=53=0.6
Expressed as a percentage, the percentage modulation is 0.6×100=60%.