Question

What is the modulation index if an audio signal of amplitude one-half of the carrier amplitude is used in AM?
A. 1
B. 0
C. 0.5
D. >1

Answer 1

Hint: The modulation index of AM is given by the ratio of the maximum information signal to the amplitude of the carrier. Hence, the modulation index, $m=\dfrac{{{E}_{0m}}}{{{E}_{0c}}}$, where ${{E}_{0m}}$ is the amplitude of the of the information and ${{E}_{0c}}$ is the amplitude of the carrier wave.

Step by step solution:
Let’s start by understanding what amplitude modulation (AM) and how it works. Amplitude modulation is the simplest of the three possible kinds of modulation used. The other two kinds of modulation are frequency modulation (FM) and phase modulation (PM).

The amplitude modulation process consists of a transmitter, which produces the modulated wave. The two initial waves that are used to produce the AM wave are the information signal $({{E}_{0m}}(t))$ and the carrier signal $({{E}_{c}})$, given by ${{E}_{c}}(t)={{E}_{0c}}\sin ({{\omega }_{c}}t+\phi )$. The modulated wave produced by the transmitter is the AM wave. The waveform of the AM wave is, ${{E}_{AM}}(t)={{E}_{0AM}}\sin ({{\omega }_{c}}t+\phi )$, where the amplitude of the AM wave is the sum of the amplitudes of information signal and carrier signal. Hence, ${{E}_{0AM}}={{E}_{0m}}(t)+{{E}_{0c}}$.

As per the question, it is given that the amplitude of the information signal is one-half of the carrier amplitude. Hence, ${{E}_{0m}}(t)=(0.5){{E}_{0c}}$.

Using this information and substituting in this value into the modulation index (m), we find the value of the modulation index (m) for the AM wave is, $m=\dfrac{{{E}_{0m}}}{{{E}_{0c}}}\Rightarrow m=\dfrac{(0.5){{E}_{0c}}}{{{E}_{0c}}}\Rightarrow m=0.5$.

Note: The physical significance of the modulation index in this problem where we found the modulation index to be 0.5 is that, m=0.5 fraction, needs to be changed into percentage, which would become 50%. Hence, this means that the carrier amplitude varies by 50% above and below its original value of ${{E}_{0m}}(t)$.

An alternate method to solve this problem is using the concept of maximum and minimum amplitude of the AM wave. The maximum amplitude of the AM wave is, \[{{E}_{0AM}}(\max )={{E}_{0c}}+{{E}_{0m}}(t)={{E}_{0c}}+(0.5){{E}_{0c}}=1.5{{E}_{0c}}\] and the minimum amplitude of the AM wave is, \[{{E}_{0AM}}(\min )={{E}_{0c}}-{{E}_{0m}}(t)={{E}_{0c}}-(0.5){{E}_{0c}}=0.5{{E}_{0c}}\]. Hence, the modulation index is, $m=\dfrac{{{E}_{0AM}}(\max )-{{E}_{0AM}}(\min )}{{{E}_{0AM}}(\max )+{{E}_{0AM}}(\min )}=\dfrac{1.5{{E}_{0c}}-0.5{{E}_{0c}}}{1.5{{E}_{0c}}+0.5{{E}_{0c}}}=\dfrac{{{E}_{0c}}}{2{{E}_{0c}}}=0.5$.