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Solved Examples · Example 7

Q.The output of an amplitude modulator is found to have maximum and minimum amplitudes of 12V and 4V respectively. Calculate

(a) modulation index
(b) carrier amplitude and
(c) signal amplitude.
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[!TLDR]

From the envelope peaks, ma=(12−4)/(12+4)=0.5m_a = (12-4)/(12+4) = 0.5, VC=(12+4)/2=8V_C = (12+4)/2 = 8 V and Vm=maVC=4V_m = m_a V_C = 4 V.

The modulation index of an AM wave can be measured straight from an oscilloscope display of the envelope, using the maximum and minimum envelope amplitudes. From Vmax=VC+maVCV_{max} = V_C + m_a V_C and Vmin=VC−maVCV_{min} = V_C - m_a V_C, the standard results are

ma=Vmax−VminVmax+Vmin,VC=Vmax+Vmin2,Vm=maVCm_a = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}, \qquad V_C = \frac{V_{max} + V_{min}}{2}, \qquad V_m = m_a V_C

Given Vmax=12V_{max} = 12 V and Vmin=4V_{min} = 4 V: …

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