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Question Bank (3 marks) · Q3

Q.Derive an expression for modulation index in terms V_max and V_min.

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[!TLDR]

From the peaks of the AM envelope, ma=(Vmax−Vmin)/(Vmax+Vmin)m_a = (V_{max}-V_{min})/(V_{max}+V_{min}).

The amplitude of an AM wave varies as A=Ec(1+macos⁡ωmt)A = E_c(1+m_a\cos\omega_m t), where ma=Em/Ecm_a=E_m/E_c. The envelope therefore swings between a maximum and a minimum value.

When cos⁡ωmt=+1\cos\omega_m t = +1, the envelope reaches its maximum:

Vmax=Ec+Em.V_{max} = E_c + E_m.

When cos⁡ωmt=−1\cos\omega_m t = -1, the envelope reaches its minimum:

Vmin=Ec−Em.V_{min} = E_c - E_m.

Adding the two equations:

Vmax+Vmin=2Ec  ⇒  Ec=Vmax+Vmin2.V_{max}+V_{min} = 2E_c \;\Rightarrow\; E_c = \frac{V_{max}+V_{min}}{2}.

Subtracting the second from the first:

Vmax−Vmin=2Em  ⇒  Em=Vmax−Vmin2.V_{max}-V_{min} = 2E_m \;\Rightarrow\; E_m = \frac{V_{max}-V_{min}}{2}.

The modulation index is the ratio of the modulating amplitude to the carrier amplitude:

ma=EmEc=(Vmax−Vmin)/2(Vmax+Vmin)/2=Vmax−VminVmax+Vmin.m_a = \frac{E_m}{E_c} = \frac{(V_{max}-V_{min})/2}{(V_{max}+V_{min})/2} = \frac{V_{max}-V_{min}}{V_{max}+V_{min}}.

This is convenient because VmaxV_{max} and VminV_{min} can be read directly from the AM waveform displayed on an oscilloscope.

[!ANSWER]

ma=Vmax−VminVmax+Vminm_a = \dfrac{V_{max}-V_{min}}{V_{max}+V_{min}}.

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